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CBSE Term II, , 2022, , Physics, Class XII, Complete Theory Covering NCERT, Cased Based Questions, Short/Long Answer Questions, 3 Practice Papers with Explanations, , CLICK HERE FOR MORE, , Author, Manish Dangwal, , ARIHANT PRAKASHAN (School Division Series)

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ARIHANT PRAKASHAN (School Division Series), , © Publisher, No part of this publication may be re-produced, stored in a retrieval system or by any means,, electronic, mechanical, photocopying, recording, scanning, web or otherwise without the written, permission of the publisher. Arihant has obtained all the information in this book from the sources, believed to be reliable and true. However, Arihant or its editors or authors or illustrators don’t take any, responsibility for the absolute accuracy of any information published and the damage or loss suffered, thereupon., , All disputes subject to Meerut (UP) jurisdiction only., , Administrative & Production Offices, Regd. Office, ‘Ramchhaya’ 4577/15, Agarwal Road, Darya Ganj, New Delhi -110002, Tele: 011- 47630600, 43518550, , Head Office, Kalindi, TP Nagar, Meerut (UP) - 250002, Tel: 0121-7156203, 7156204, , Sales & Support Offices, Agra, Ahmedabad, Bengaluru, Bareilly, Chennai, Delhi, Guwahati,, Hyderabad, Jaipur, Jhansi, Kolkata, Lucknow, Nagpur & Pune., , ISBN : 978-93-25796-89-8, PO No : TXT-XX-XXXXXXX-X-XX, Published by Arihant Publications (India) Ltd., For further information about the books published by Arihant, log on to, www.arihantbooks.com or e-mail at [email protected], Follow us on, , CBSE Term II, , 2022

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Contents, CHAPTER, Electromagnetic Waves, , -, , CHAPTER, Ray Optics and Optical Instruments, , -, , CHAPTER, Wave Optics, , -, , CHAPTER, Dual Nature of Radiation and Matter, , -, , CHAPTER, Atoms, , -, , CHAPTER, Nuclei, , -, , CHAPTER, Semiconductor Electronics: Materials, Devices, and Simple Circuits, , -, , Practice Papers, , -, , -, , Watch Free Learning Videos, Subscribe arihant, , Channel, , þ Video Solutions of CBSE Sample Papers, þ Chapterwise Important MCQs, þ CBSE Updates

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Syllabus, CBSE Term II Class 12, Time: Hours, , Max Marks:, , Units, Unit V, , Periods, , Marks, , Electromagnetic Waves, Chapter : Electromagnetic Waves, , Unit VI, , Optics, Chapter : Ray Optics and Optical Instruments, Chapter, : Wave Optics, , Unit VII, , Dual Nature of Radiation and Matter, Chapter, , Unit VIII, , Atoms and Nuclei, Chapter, Chapter, , Unit IX, , : Dual Nature of Radiation and Matter, , : Atoms, : Nuclei, , Electronic Devices, Chapter, : Semiconductor Electronics: Materials,, Devices and Simple Circuits, Total, , UNIT-V, , Electromagnetic waves, , Chapter-, , Electromagnetic waves, Electromagnetic waves, their characteristics, their Transverse nature qualitative ideas, only . Electromagnetic spectrum radio waves, microwaves, infrared, visible, ultraviolet,, X-rays, gamma rays including elementary facts about their uses., , UNIT-VI, , Optics, , Chapter-, , Ray Optics and Optical Instruments, Ray Optics: Refraction of light, total internal reflection and its applications, optical, fibers, refraction at spherical surfaces, lenses, thin lens formula, lensmaker s formula,, magnification, power of a lens, combination of thin lenses in contact, refraction of light, through a prism., Optical instruments: Microscopes and astronomical telescopes reflecting and, refracting and their magnifying powers., , CBSE Term II, , 2022, , Periods, , Periods

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Chapter-, , Wave Optics, Wave optics: Wave front and Huygen s principle, reflection and refraction of plane wave, at a plane surface using wave fronts. Proof of laws of reflection and refraction using, Huygen s principle. Interference, Young s double slit experiment and expression for, fringe width, coherent sources and sustained interference of light, diffraction due to a, single slit, width of central maximum., , UNIT-VII, , Dual Nature of Radiation and Matter, , Chapter-, , Dual Nature of Radiation and Matter, Dual nature of radiation, Photoelectric effect, Hertz and Lenard s observations;, Einstein s photoelectric equation-particle nature of light., Experimental study of photoelectric effect, Matter waves-wave nature of particles,, de-Broglie relation., , UNIT-VIII Atoms and Nuclei, , Periods, , Periods, , Chapter-, , Atoms, Alpha-particle scattering experiment; Rutherford s model of atom; Bohr model, energy, levels, hydrogen spectrum., , Chapter-, , Nuclei, Composition and size of nucleus, Nuclear force, Mass-energy relation, mass defect,, nuclear fission, nuclear fusion., , UNIT-IX, , Electronic Devices, , Chapter-, , Semiconductor Electronics: Materials, Devices and Simple Circuits Energy bands in, conductors, semiconductors and insulators qualitative ideas only , Semiconductor, diode - I-V characteristics in forward and reverse bias, diode as a rectifier;, Special purpose p-n junction diodes: LED, photodiode, solar cell., , CBSE Term II, , 2022, , Periods

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1, , CBSE Term II Physics XII, , CHAPTER 01, , Electromagnetic, Waves, In this Chapter..., l Maxwell's Equations, l Electromagnetic Waves, l Electromagnetic Spectrum, , Maxwell’s Equations, These are the basic equations of electricity and magnetism., These equations give complete description of all, electromagnetic interactions. There are four Maxwell’s, equations (for free space), which are given below, q, (i) ò E × dS =, (Gauss’s law of electrostatics), e0, (ii), , ò B × dS = 0, , (iii), , ò E× dl = -, , (iv), , ò, , (Gauss’s law of magnetostatics), , dfB, dt, (Faraday’s law of electromagnetic induction), dfE ö, æ, (Ampere-Maxwell law), B × dl = m0 ç Ic + e0, ÷, dt ø, è, , These charges produce an oscillating electric field in space,, which produces an oscillating magnetic field, which in turn is, a source of oscillating electric fields and so on., The oscillating electric and magnetic fields regenerate each, other as a continuous wave which propagates through space., The frequency of EM wave is equal to the frequency of, oscillation of charge., 1, i.e., n=, 2p LC, , Nature of Electromagnetic Waves, In an electromagnetic wave, electric and magnetic fields are, perpendicular to each other and to the direction of wave, propagation. A plane electromagnetic wave propagating, along the z-direction is shown below, X, , Maxwell on the basis of his equations predicted the existence, of electromagnetic waves., , Direction of wave propagation, , O, , Electromagnetic Waves, These waves are produced due to the change in electric field, E and magnetic field B sinusoidally and propagating through, space such that, the two fields are perpendicular to each, other and perpendicular to the direction of wave, propagation., , Source of Electromagnetic Waves, Maxwell found that, the accelerated or oscillating charge, radiate electromagnetic waves., , E, B, , B, Z, E, , Y, , A plane EM wave travelling along Z-axis, , The electric field E x is along x-direction and dotted curve, shows magnetic field B which is along y-direction. Both E, and B vary sinusoidally and become maximum at same, position and time. As E and B are mutually perpendicular to, each other, so they are transverse in nature.

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2, , CBSE Term II Physics XII, , The EM wave propagating in the positive z-direction is represented as, E = E x = E 0 sin ( kx - wt ), B = By = B0 sin( ky - wt ), where, k is propagation vector or wave vector = 2p/ l and wis angular frequency = 2pn., , Important Characteristics of Electromagnetic Waves, Some features of EM waves are given below, (i) EM waves do not require any material medium for propagation., (ii) These waves travel in free space with the speed of light (3 ´ 10 8 ms -1 ), given by c = 1 / m 0 e0 , which shows that light, waves are electromagnetic in nature., (iii) Speed of electromagnetic wave in a medium is v = 1 / me, where e and m are the permittivity and magnetic, permeability of a material medium, respectively. This means, the speed of EM wave in a medium depends on electric, and magnetic properties of a medium., (iv) The direction of variations of electric and magnetic fields are perpendicular to each other and also perpendicular to the, direction of wave propagation., Thus, electromagnetic waves are transverse in nature., (v) In free space, the magnitudes of electric and magnetic fields in electromagnetic waves are related by E 0 / B0 = c., (vi) The energy in electromagnetic waves is divided, on an average, equally between electric and magnetic fields., Ue = Um, where, U e = energy of electric field and U m = energy of magnetic field., 1, B2, (vii) The energy density (energy per unit volume) in an electric field E in vacuum is e 0 E 2 and that in magnetic field B is, ., 2m 0, 2, (viii) Electromagnetic waves, being uncharged, are not deflected by electric and magnetic fields., (ix) An electromagnetic wave carries energy and momentum. An electromagnetic wave also exerts pressure called, radiation pressure. If wave is incident on a completely absorbing surface, then momentum delivered is given by, U, p=, c, , Electromagnetic Spectrum, The orderly arrangement of EM waves in increasing or decreasing order of wavelength l and frequency n is called, electromagnetic spectrum. The range varies from 10 -12 m to 104 m, i.e. from g-rays to radio waves., Electromagnetic wave spectrum is shown below, Frequency (Hz), , 1023, 1022, 1021, 1020, 1019, 1018, 1017, 1016, 1015, 1014, 1013, 1012, 1011, 1010, 109, 108, 107, 106, 105, 104, 103, 102, 101, , Wavelength (m), , Gamma rays, , X-rays, Ultraviolet, Visible, Infrared, Microwaves, Short radio waves, Television and FM radio, AM radio, , Long radio waves, , 10–14, 10–13, 10–12, 10–11, 10–10, 10–9, 10–8, 10–7, 10–6, 10–5, 10–4, 10–3, 10–2, 10–1, 1, 101, 102, 103, 104, 105, 106, 107, , 400 nm, Violet, 450 nm, Blue, 500 nm, Green, 550 nm, Yellow, 600 nm, Orange, 650 nm, Red, 700 nm, , Electromagnetic spectrum with common names for various parts of it

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3, , CBSE Term II Physics XII, , Various Electromagnetic Waves of Electromagnetic Spectrum with their Features and Uses, Name of, Wave, , Frequency, Range, , Radio waves 500 kHz to, 1000 MHz, , Wavelength, Range, > 0.1 m, , Production, , Detection, , Rapid acceleration, and decelerations of, electrons in aerials., , Receiver’s aerials, , Uses, l, , l, , l, , Microwaves 1 GHz to, 300 GHz, , 0.1 m to, 1 mm, , Klystron valve or, magnetron valve., , Point contact diodes, , l, , l, , l, , Infrared, waves, (heat waves), , 3 ´ 1011 Hz to, 4 ´ 1014 Hz, , 14, , Light rays, 4 ´ 10 Hz to, (visible rays) 7 ´ 1014 Hz, , Ultraviolet, rays, , 1014 Hz to, 1016 Hz, , 1 mm to, 700 nm, , Vibration of atoms, and molecules., , Thermopiles, bolometer and, infrared, photographic film, , 700 nm to, 400 nm, , Electrons in atoms, Eye,, emit light, when they photocells and, move from a higher, photographic film, energy level to a lower, energy level., , 400 nm, to 1 nm, , Inner shell electrons Photocells and, in atoms moving from photographic film, higher energy level to, a lower energy level., , l, , l, , l, , l, , l, , l, , l, , l, , l, , X-rays, , 3 ´ 1016 Hz to, 3 ´ 1021 Hz, , 1 nm to, 10-3 nm, , X-ray tubes or inner, Photographic film,, shell electrons,, Geiger tubes and, bombarding metals by ionisation chamber, high energy electrons., , l, , l, , l, , Gamma ( g ), rays, , 3 ´ 1018 Hz, < 10-3 nm, to 5 ´ 1022 Hz, , Radioactive decay of, the nucleus., , Photographic film, and ionisation, chamber, , l, , l, , l, , These are used in AM (Amplitude, Modulation) from 530 kHz to 1710 kHz and, ground wave propagation., These are used in TV waves ranging from, 54 MHz to 890 MHz., These are used in FM (Frequency, Modulation) ranging from 88 MHz to, 108 MHz., These are used in RADAR systems for, aircraft navigation., These are used in microwave oven for, cooking purpose., These are used in study of atomic and, molecular structures., These are used in physical therapy., These are used in satellites for army, purpose., These are used in weather forecasting., These are used by the optical organs of, humans and animals for three primary, purposes given below, (i) To see things, avoid bumping into, them and escape danger., (ii) To look for food., (iii) To find other living things with which, to copulate, so as to prolong the, species., These are used in burglar alarm., These are used in checking mineral sample., These are used to study molecular, structure., To kill germs in water purifiers., Used in LASER eye surgery., These are used in medicine to detect the, fracture, diseased organs, stones in the, body, etc., These are used in engineering to detect, fault, cracks in bridges and testing of welds., These are used at metro-stations to detect, metals or explosive material., These are used to produce nuclear, reactions., These are used in radio therapy for the, treatment of tumour and cancer., These are used in food industry to kill, pathogenic micro-organism.

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4, , CBSE Term II Physics XII, , Solved Examples, Example 1. An electromagnetic wave of frequency, 40 MHz travels in free space in the x-direction., (i) Determine the wavelength of the wave., (ii) At some point and at some instant, the electric, field has its maximum value of 750 NC -1 and is, along the Y-axis. Calculate the magnitude and, direction of the magnetic field at this position, and time., 8, Sol. (i) Wavelength of the wave, l =, , c 3.0 ´ 10, =, = 7.5 m, n 40 ´ 106, , (ii) Given, maximum value of electric field, E0 = 750 NC -1, E, 750, \ Magnetic field, B0 = 0 =, = 2.5 ´ 10– 6 T, c, 3 ´ 108, Since, E and B are mutually perpendicular and they both, are perpendicular to the propagation of wave. Thus, we, concluded that, magnetic field is in negative z-direction., , Example 2. Find the amplitude of electric and, magnetic fields in a parallel beam of light of, intensity 4.0 Wm -2 ., Sol. The intensity of plane electromagnetic wave, I =, \Amplitude of electric field, E0 =, =, , 1, e0 E02 c, 2, , 2I, e0 c, 2´4, 8.86 ´ 10–12 ´ 3 ´ 108, , Example 4. About 5% of the power of a 100 W light, bulb is converted to visible radiation. What is the, average intensity of visible radiation at a distance of, (a) 1m from the bulb and, (b) 10 m?, Assume that, the radiation is emitted isotropically, and neglect reflection., Sol. Total power = 100 W, Visible radiation power = 5% of total power, 5, =, ´ 100 = 5 W, 100, (a) At a distance of 1m, the energy distributed in the form of, sphere., Area of sphere = 4p (radius) 2, Power, 5, Intensity of visible radiation =, =, Area, 4 ´ 3.14 ´ (1) 2, = 0.4 W/m 2, (b) Intensity of visible radiation at a distance of 10 m, 5, =, 4 ´ 3.14 (10) 2, = 4 ´ 10- 3 W/m 2, , Example 5. Evaluate the amplitude of electric and, magnetic fields produced by the radiation coming, from a 20pW bulb at distance of 2m. Assume that, the, efficiency of the bulb 20% and it is a point source., Sol. Consider the situation shown below, X, , -1, , = 54.87 NC, Further, amplitude of magnetic field,, E, 54.87, B0 = 0 =, T, c 3.0 ´ 108, = 1.83 ´ 10-7 T, , Example 3. Light with an energy flux of 18 Wcm -2 falls, on a non-reflecting surface at normal incidence. If, the surface has an area of 20 cm 2 , then find the, average force exerted on the surface during a span, of 30 min., , r, S, P, , Intensity at distance from a point source (bulb), I =, Efficiency, h =, Þ, , Sol. Total energy falling on the surface,, U = Energy flux ´ area ´ time, = (18 Wcm -2 )(30 ´ 60 s ) (20 cm 2 ), = 6.48 ´ 105 J, Therefore, the total momentum delivered is,, U 6.48 ´ 105, p= =, kg- ms -1, c, 3.0 ´ 108, = 2.16 ´ 10–3 kg- ms -1, The average force exerted on the surface,, p 2.16 ´ 10–3, (Q F × t = change in momentum), F= =, t, 30 ´ 60, = 1.2 ´ 10–6 N, , P, 4pr 2, , Output P, =, Input, P¢, , æ 20 ö, P = hP ¢ = ç, ÷ ( 20p ) = 4pW, è 100 ø, P, 4p, 1 W, I=, =, =, 4pr 2 4p( 2) 2 4 m 2, , Also, intensity of EM wave is given by, 1, 1 1, I = e0 E02 c Þ = e0 E02 ´ 3 ´ 108, 2, 4 2, 10-8, 10-8, 104, =, =, 6e0, 6 ´ 8.85 ´ 10-12 6 ´ 8.85, 100, = 13.73 V/m, \Amplitude of electric field, E0 =, 6 ´ 8.85, Þ, , E02 =, , \Amplitude of magnetic field, B0 =, , E0 13.73, =, = 457, . ´10-8 T, c 3 ´ 108

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5, , CBSE Term II Physics XII, , Chapter, Practice, PART 1, Objective Questions, l, , Multiple Choice Questions, 1. Which of the following statement is correct in, regards to the Maxwell’s equations?, (a) The most important prediction to emerge from Maxwell’s, equations is the existence of magnetic waves., (b) Maxwell’s equations involves only electric and magnetic, fields., (c) The total current has the same value of current i for all, surfaces., (d) We can rephrase Faraday’s law of electromagnetic, induction that there is an induced emf equal to the rate, of change of magnetic flux., , 2. The electric field of an electromagnetic wave, travelling through vacuum is given by the equation, E = E 0 sin ( kx - wt ). The quantity that is, independent of wavelength is, k, w, (c) w, , (a), , (b) kw, (d) k, , 3. The electric and the magnetic fields, associated, with an electromagnetic wave, propagating along, the + Y-axis, can be represented by, (a) ( E = E0 k$ , B = B0 i$ ), (b) ( E = E $j, B = B i$ ), 0, , harmonic electromagnetic wave in vacuum is, B 0 = 510 nT. What is the amplitude of the electric, field part of the wave?, (a) 130 NC -1, , (b) 153 NC -1, , -1, , (d) 190 NC -1, , (c) 170 NC, , 6. An electromagnetic wave travelling along Z-axis is, given as E = E 0 cos( kz - wt ). Choose the incorrect, statement from the following., (a) The associated magnetic field is given as, 1, 1, B = k$ ´ E = ( k$ ´ E ), c, w, (b) The electromagnetic field can be written in terms of the, associated magnetic field as E = c ( B ´ k$ ), (c) k$ ´E = 0, k$ ´ B = 0, (d) None of the above, , 7. A plane electromagnetic wave of frequency, 25 GHz is propagating in vacuum along the, z-direction. At a particular point in space and time,, the magnetic field is given by B = 5 ´ 10 - 8 $j T. The, corresponding electric field E is (Take, speed of, light, c = 3 ´ 10 8 ms - 1 ), (a) - 1 .66 ´ 10- 16 $i Vm -1, (b) 1 .66 ´ 10- 16 $i V m -1, (c) - 15 $i Vm - 1, (d) 15 $i Vm -1, , 0, , (c) ( E = E0 $j, B = B0, (d) ( E = E0 i$, B = B0, , 8. Suppose that, the amplitude of electric field of an, , k$ ), $j ), , 4. Light wave is travelling along y-direction. If the, corresponding E vector at any time is along the, X-axis, the direction of B vector at that time is along, Y, , O, , X, , Z, , (a) Y -axis, (c) + Z-axis, , 5. The amplitude of the magnetic field part of a, , (b) X-axis, (d) -Z-axis, , electromagnetic wave is E 0 = 120 NC -1 and its, frequency is n = 50.0 MHz. The expressions for E, will be (if wave travels along X-axis), , (a) [(120 NC -1 ) sin {(1.05 rad m -1 ) x, - (3.14 ´ 108 rad s -1 ) t}] $i, (b) [(120 NC -1 ) sin {(1.05 rad m -1 ) x, - (3.14 ´ 108 rad s -1 ) t}] k$, (c) [(120 NC -1 ) sin {(1.05 rad m -1 ) x, - (3.14 ´ 108 rad s -1 ) t}] $j, (d) [(120 NC -1 ) cos {(1.05 rad m -1 ) x, - (3.14 ´ 108 rad s -1 ) t}] $j

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6, , CBSE Term II Physics XII, , 9. The electric field part of an electromagnetic wave, , 15. An electromagnetic wave given as, , in a medium is represented by E x = 0;, éæ, N, rad ö æ, - 2 rad ö ù, E y = 2.5 cos êç 2p ´10 6, ÷ t - ç p ´10, ÷x ;, C, m ø è, s ø úû, ëè, E z = 0., The wave is, , E = E 0 $i cos ( kz - wt) is incident normally on a, perfectly reflecting infinite wall at z = a. Assuming, that, the material of the wall is optically inactive,, the reflected wave will be given as, (a) E r = E0 $i( kz - wt ), (c) E r = - E0 $i cos( kz + wt ), , (a) moving along y-direction with frequency 106 Hz and, wavelength 200 m, (b) moving along x-direction with frequency 106 Hz and, wavelength 100 m, (c) moving along x-direction with frequency 106 Hz and, wavelength 200 m, (d) moving along – x-direction with frequency 106 Hz and, wavelength 200 m, , 1 $ $, ( j + k), 2, 1, (d), ( 2i$ + $j), 5, , (a) Electromagnetic waves cannot be deflected by any field., 1, , thus it, (b) The velocity of light in a medium is v =, me, depends on both the electric and magnetic properties of, the medium., (c) The constant velocity of electromagnetic waves in, vacuum is used to define a standard of time., (d) The direction of propagation vector k describes the, direction of propagation of the wave., , 13. An electromagnetic wave of frequency n = 3.0 MHz, passes from vacuum into a dielectric medium with, permittivity e = 4.0, then, (a) wavelength is doubled and the frequency remains, unchanged, (b) wavelength is halved and frequency becomes half, (c) wavelength is halved and frequency remains unchanged, (d) wavelength and frequency both remain unchanged, , 14. The speed of electromagnetic wave in vacuum, depends upon the source of radiation, (a) increases as we move from g-rays to radio waves, (b) decreases as we move from g-rays to radio waves, (c) is same for all of them, (d) None of the above, , (b) 0111, ., ´ 10-8 Nm -2, , (c) 0.083 ´ 10-8 Nm -2, , (a) 0.3 ´ 10-17 kg-ms -1, , (b) 1 . 0 ´ 10-17 kg-ms -1, , (c) 3.0 ´ 10-17 kg-ms -1, , (d) 9.0 ´ 10-17 kg-ms -1, , 18. One requires 11 eV of energy to dissociate a carbon, monoxide molecule into carbon and oxygen atoms., The minimum frequency of the appropriate, electromagnetic radiation to achieve the, dissociation lies in, [NCERT Exemplar], , speed of 2 ´ 10 8 ms -1 . The relative magnetic, permeability of the medium is 1. The relative, electrical permittivity is, , 12. Which of the following statement is incorrect?, , (b) 0.332 ´ 10-8 Nm -2, , completely by a small object initially at rest. Power, of the pulse is 30 mW and the speed of light is, 3 ´ 10 8 ms -1 . The final momenum of the object is, , (b), , (b) 1, (d) 2.25, , (a) 0.166 ´10-8 Nm -2, , 17. A pulse of light of duration 100 ns is absorbed, , 11. Electromagnetic waves travel in a medium with a, , (a) 1.25, (c) 1.8, , -2, , are striking a, metal plate. The pressure on the plate is, , electric field and magnetic field are represented by, k$ and 2$i - 2$j, respectively. What is the unit vector, along direction of propagation of the wave?, 1 $ $, ( i + j), 2, 1 $, (c), ( i + 2$j), 5, , 0, , r, , 16. Radiations of intensity 0.5 Wm, , 10. In a plane electromagnetic wave, the directions of, , (a), , (b) E r = E0 $i cos( kz + wt ), (d) E = E $i sin( kz - wt ), , (a) visible region, (c) ultraviolet region, l, , (b) infrared region, (d) microwave region, , Assertion-Reasoning MCQs, Direction (Q. Nos. 19-23) Each of these questions, contains two statements, Assertion (A) and Reason (R)., Each of these questions also has four alternative, choices, any one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given, below., (a) Both A and R are true and R is the correct, explanation of A, (b) Both A and R are true, but R is not the correct, explanation of A, (c) A is true, but R is false, (d) A is false, but R is true, , 19. Assertion When the sun shines on our hand, we, feel the energy being absorbed from the, electromagnetic waves (our hands get warm)., Reason Electromagnetic waves transfer, momentum to our hand but because c is very large,, the amount of momentum transferred is extremely, small and we do not feel the pressure., , 20. Assertion Like light radiations, thermal radiations, are also an electromagnetic radiations., Reason The thermal radiations require no medium, for propagation.

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7, , CBSE Term II Physics XII, , 21. Assertion An electromagnetic wave is a, self-sustaining oscillating wave., Reason An oscillating charge produces an electric, field in space, which produces an oscillating, magnetic field, which in turn, is a source of electric, field and so on., , 22. Assertion The frequency of the electromagnetic, wave is naturally equal to the frequency of, oscillation of the charge., Reason The energy associated with the propagating, wave comes at the expense of the energy of the, source., , (iii) A plane electromagnetic wave of frequency 25 MHz, travels in free space along the x-direction. At a, particular point in space and time, E = 6.3 $j V / m., What is B at this point?, (a) 2.1 ´ 10-8 k$ T, (c) 3.5 ´ 106 k$ T, , 1, (iv) The correct dimension of e 0 E 2 (e 0 is the, 2, permittivity of free space and E is electric field), is, , frequency wave are dangerous to human being., Reason Ultraviolet radiations are absorbed by the, atmosphere., , 24. Oscillating Charge, An oscillating charge is an example of accelerating, charge. It produces an oscillating electric field in, space, which produces an oscillating magnetic, field, which in turn produces an oscillating electric, field and so on. The oscillating electric and, magnetic fields regenerate each other as a wave, which propagates through space., X, Direction of wave propagation, O, , E, B, , B, Z, , E, , Y, , (i) Total energy density of electromagnetic waves in, vacuum is given by the relation, (a), , 1 E2, B2, ×, +, 2 e0, 2m 0, , (b), , 1, 1, e0 E 2 + m 0 B 2, 2, 2, , (c), , E2 + B2, c, , (d), , 1, B2, e0 E 2 +, 2, 2m 0, , (ii) The magnetic field of plane electromagnetic wave, is given by, B y = 2 ´ 10 -7 sin (0.5 ´10 3 x + 1.5 ´ 1011 t )., This electromagnetic wave is, (a) a visible light, (b) an infrared wave, (c) a microwave, (d) a radio wave, , (b) [ML-1T -2 ], , (c) [ML2 T -2 ], , (d) [MLT -1 ], , (a) will have frequency of 107 Hz, (b) will have frequency of 2 ´ 107 Hz, (c) will have wavelength of 05, . m, (d) falls in the region of radio waves, , Case Based MCQs, Direction Read the following passage and answer the, questions that follows, , (a) [ML2 T -1 ], , (v) A charged particle oscillates about its mean, equilibrium position with a frequency of 10 9 Hz., The electromagnetic waves produced, , 23. Assertion Ultraviolet radiations of higher, , l, , (b) 2.1 ´ 108 k$ T, (d) 3.0 ´ 105 k$ T, , PART 2, Subjective Questions, l, , Short Answer (SA) Type Questions, 1. How are electromagnetic waves produced by, oscillating charges? What is the source of the energy, associated with the EM waves?, [All India 2020], , 2. (i) An electromagnetic wave is travelling in a, medium with a velocity v = v $i. Draw a sketch, showing the propagation of the electromagnetic, wave indicating the direction of the oscillating, electric and magnetic fields., (ii) How are the magnitudes of the electric and, magnetic fields related to velocity of the, electromagnetic wave? [Delhi 2013, All India 2008 C], Or Depict the fields diagram of an electromagnetic, wave propagating along positive X-axis with its, electric field along Y-axis., [Delhi 2020], , 3. A plane electromagnetic wave travels in vacuum, along z-direction. What can you say about the, directions of its electric and magnetic field vectors?, If the frequency of the wave is 30 MHz, what is its, wavelength?, , 4. The electric field of an electromagnetic wave is, given by E = (50 NC -1 )sin w( t - x / c). Find the, energy contained in a cylinder of cross-section, 10 cm 2 and length 50 cm along the X-axis.

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8, , CBSE Term II Physics XII, , 5. An electromagnetic wave is travelling in vacuum, , 14. Name the electromagnetic waves with their, , 8, , with a speed of 3 ´ 10 m/s. Find its velocity in a, medium having relative electric and magnetic, permeabilities 2 and 1, respectively., [Delhi 2019], , frequency range, produced in, (a) some radioactive decay,, (b) sparks during electric welding and, (c) TV remote., [All India 2020], , 6. Even though an electric field E exerts a force qE on, a charged particle yet electric field of an, electromagnetic wave does not contribute to the, radiation pressure (but transfers energy). Explain., [NCERT Exemplar], , 7. Show that the radiation pressure exerted by an EM, I, wave of intensity I on a surface kept in vacuum is ., c, [NCERT Exemplar], , 8. (i) Why are infrared waves often called heat waves?, Explain., (ii) What do you understand by the statement,, “electromagnetic waves transport, momentum”?, [CBSE 2018], , 9. Identify the electromagnetic wave, whose, wavelengths vary as, (i) 10 -12 m < l < 10 -8 m and, (ii) 10 -3 m < l < 10 -1 m., Write one use for each., , [All India 2017], , 10. (i) Arrange the following electromagnetic waves in, the descending order of their wavelengths., (a) Microwaves, (b) Infrared rays, (c) Ultraviolet radiation, (d) g-rays, (ii) Write one use each of any two of them., [Delhi 2013 C], , 11. Name the constituent radiation of electromagnetic, spectrum which is used for, (i) aircraft navigation and, (ii) studying the crystal structure., Write the frequency range for each., , [Delhi 2011C], , 12. Use the formula l m T = 0.29 cm-K to obtain the, characteristic temperature ranges for different, parts of the electromagnetic spectrum. What do the, numbers that you obtain tell you?, , 13. Answer the following questions., (i) Find the energy stored in a 90 cm length of a, laser beam operating at 6 mW., (ii) Find the amplitude of electric field in a parallel, beam of light of intensity 17.7 W/ m 2 ., , l, , Long Answer (LA) Type Questions, 15. Suppose that, the electric field amplitude of an, electromagnetic wave is E 0 = 120 N/C and that its, frequency is n = 50.0 MHz., (i) Determine, B 0 , w, k and l., (ii) Find expressions for E and B., , 16. Suppose that, the electric field part of an, electromagnetic wave in vacuum is, E = [3.1 cos {1.8 y + (5.4 ´ 10 6 t )}] $i, (i) What is the direction of propagation?, (ii) What is the wavelength l?, (iii) What is the frequency n?, (iv) What is the amplitude of the magnetic field part, of the wave?, (v) Write an expression for the magnetic field part of, the wave., [NCERT], , 17. In a plane electromagnetic wave, the electric field, oscillates sinusoidally at a frequency of, 2.0 × 10 10 Hz and amplitude 48 V/m., (i) What is the wavelength of the wave?, (ii) What is the amplitude of the oscillating magnetic, field?, (iii) Show that, the average energy density of the E, field equals the average energy density of the B, field. (Take, c = 3 × 10 8 m/s), , 18. (i) Which segment of electromagnetic waves has, highest frequency? How are these waves, produced? Give one use of these waves., (ii) Which EM waves lie near the high frequency, end of visible part of EM spectrum? Give its one, use. In what way, this component of light has, harmful effects on humans?, [Foreign 2016], , 19. Answer the following questions., (i) Show, by giving a simple example, how EM waves, carry energy and momentum., (ii) How are microwaves produced? Why is it, necessary in microwaves ovens to select the, frequency of microwaves to match the resonant, frequency of water molecules?, (iii) Write two important uses of infrared waves., [Delhi 2014 C]

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9, , CBSE Term II Physics XII, , 20. Answer the following questions., (i) Name the EM waves which are used for the, treatment of certain forms of cancer. Write, their frequency range., (ii) Thin ozone layer on top of stratosphere is, crucial for human survival. Why?, (iii) Why is the amount of the momentum, transferred by the EM waves incident on the, surface so small?, [Delhi 2014], , 21. State clearly how a microwave oven works to, heat up a food item containing water molecules., Why are microwaves found useful for the raw, systems in aircraft navigation?, [Foreign 2011], , 22. Name the parts of the electromagnetic spectrum, which is, (i) suitable for RADAR systems in aircraft, navigations., (ii) used to treat muscular strain., (iii) used as a diagnostic tool in medicine., Write in brief, how these waves can be, produced?, [All India 2015], , 23. Given below are some famous numbers, associated with electromagnetic radiations in, different contexts in Physics. State the part of the, electromagnetic spectrum to which each belongs., (i) 21 cm (wavelength emitted by atomic hydrogen, in interstellar space)., (ii) 1057 MHz ( frequency of radiation arising from, two close energy levels in hydrogen, known as, Lamb shift )., , (iii) 2.7 K (temperature associated with the isotropic, radiation filling all space thought to be a relic of, the big-bang origin of the universe)., (iv) 5890 Å-5896 Å (double lines of sodium)., (v) 14.4 keV (energy of a particular transition in 57 Fe, nucleus associated with a famous high resolution, spectroscopic method (Mössbauer spectroscopy)., [NCERT], l, , Case Based Questions, Direction Read the following passage and answer the, questions that follows, , 24. X-ray, X-ray is a type of radiation known as electromagnetic, waves. It helps in creating pictures of the inside of, human body. These images shows the different parts, of the body in various shades of black and white. It is, due to the difference in amount of absorption by, various tissues in the body., As calcium in bones absorbs most of the X-rays, so, bones look white in colour. Fat and other soft tissues, absorbs less and depicts grey colour., (i) To which part of the electromagnetic spectrum, does a wave of frequency 2 ´ 1018 Hz belong?, (ii) What is the range of wavelength for X-rays?, (iii) How are the X-rays produced?, (iv) What are the techniques by which X-rays can be, detected?, (v) Mention any two uses of X-rays.

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Chapter Test, Multiple Choice Questions, , Short Answer Type Questions, , 1. If e0 and m 0 are the electric permittivity and magnetic, permeability of free space and e and m are the, corresponding quantities in the medium, the index of, refraction of the medium in terms of above parameter is, (a), , æe m ö, (c) çç 0 0 ÷÷, è em ø, , 1/ 2, , (ii) the direction of its electric and magnetic field, vectors., , (i) ratio of the magnitudes and, , 2. The ratio of contributions made by the electric field and, magnetic field components to the intensity of an EM, wave is, (b) c 2 : 1, , (c) 1 : 1, , (d) c : 1, , 3. An electromagnetic wave going through vacuum is, described by E = E0 sin (kx - wt ) and B = B0 sin ( kx - wt ) ., Which of the following equations is true?, (a) E 0k = B0w, , (b) E 0w = B0k, , (c) E 0B0 = wk, , (d) None of these, , 4. An electromagnetic wave travelling in the x-direction is, described by the electric field, xö, æVö, æ, E y = 300 ç ÷ sin wçt - ÷, m, cø, è ø, è, , (a) 9. 4 ´ 10, , (b) 4.8 ´ 10, , N, , (c) 4.8 ´ 10 -7 N, , 5 ´ 10 -5 cm., , (i) What is the frequency (in Hz) and period (in s) in, vacuum?, (Ans. 6 ´ 10 14 and 0.16 ´ 10 - 14 ), (ii) What is the wavelength in glass, if refractive index, of glass is 1.5?, (Ans. 3.3 ´ 10 - 3 m), , 9. Name the electromagnetic radiation to which waves, of wavelength in the range of 10 -2 m belong. Give, one use of this part of electromagnetic spectrum., , 10. Find the wavelength of electromagnetic wave of, , frequency 5 ´ 10 10 Hz in free space. Give its two, applications., (Ans. 6 ´ 10 - 12 m), description., , -17, , N, , (d) 9. 4 ´ 10 -17 N, , 5. A plane electromagnetic wave, has frequency of, , 2.0 ´ 10 10 Hz and its energy density is 1.02 ´ 10 -8 J / m 3 in, vacuum. The amplitude of the magnetic field of the wave, 1, N - m2, is close to (Take,, and speed of light, = 9 ´ 10 9, 4 pe0, C2, = 3 ´ 10 8 ms -1 ), (a), (b), (c), (d), , 8. Green light of mercury has a wavelength, , 11. How are X-rays different from g-rays? Give a detailed, , An electron is constrained to move along the y-direction, with a speed of 2 ´ 10 7 ms -1 . The maximum electric force, on the electron is, -7, , 7. When a plane electromagnetic wave travels in, vacuum along y-direction. Write the, , æe m ö, (d) çç 0 0 ÷÷, è em ø, , (a) c : 1, , 16 MHz band. What is the corresponding wavelength, band?, (Ans. 54.5 m, 18.75 m), , 1/ 2, , æ em ö, ÷, (b) çç, ÷, è e0 m 0 ø, , em, e0 m 0, , 6. A radio can tune into any station from 5.5 MHz to, , Long Answer Type Questions, , 12. Answer the following questions., (i) Which part of electromagnetic spectrum is, absorbed from sunlight by ozone layer?, (ii) Welders wear special glass goggles while working., Explain, why., (iii) Why are infrared waves often called as heat, waves? Give their one application., , 13. (i) Identify the part of the electromagnetic spectrum, used in (a) radar and (b) eye surgery. Write their, frequency range., (ii) Prove that the average energy density of the, oscillating electric field is equal to that of the, oscillating magnetic field., , 190 nT, 160 nT, 180 nT, 150 nT, , Answers, Multiple Choice Questions, 1. (b), , 2. (c), , 3. (a), , 4. (b), , 5. (b), , For Detailed Solutions, Scan the code

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11, , CBSE Term II Physics XII, , EXPLANATIONS, PART 1, , As,, , 1. (c) We can rephrase Faraday’s law of electromagnetic, induction by saying that a magnetic field, changing with, time, give rise to an electric field., Thus, the statement given in option (c) is correct, rest are, incorrect., 2. (a) Given, E = E0 sin ( kx - wt ), Comparing this equation with standard equation, we get, k, Wavelength, l =, w, 1, 1, or, l=, =, nl c, k, Therefore, is independent of wavelength., w, 3. (a) In electromagnetic waves, electric vector, magnetic, vector and velocity of wave are perpendicular to one, another., Hence, as v = v $j, 0, , E = E0 k$ ü, $ $ $, Þ, ýÞ k ´ i = j, B = B0 i$ þ, 4. (d) The given wave is an electromagnetic wave. Electric and, magnetic components oscillate at right angles to each other, and to the direction of propagation, i.e. wave is travelling, along E ´ B., Y, , X, E´B, , \, , E0, c, 1 $, 1, B = ( k ´E ) = ( k$ ´ E ), c, w, , B0 =, , The associated electric field can be written in terms of, magnetic field as, E = c ( B ´ k$ ), Angle between k$ and E is 90° and between k$ and B is 90°., Therefore, k$ × E = k E cos 90° = 0 and k$ × B = k B cos 90°= 0 ., 7. (d) Given, B = 5 ´ 10-8 $j T and v = 3 ´ 108 k$, Y, B, , X, E, Z, , v, , Using E = B ´ v, we have, E = ( 5 ´ 10-8 $j) ´ ( 3 ´ 108 k$ ) = 15 $i Vm -1, 8. (c) Given, E0 = 120 NC-1, n = 50.0 MHz =50 ´ 106 Hz, \ Angular frequency, w = 2pn, = (2 ´ 3.14 rad) (50 ´ 106 Hz), = 3.14 ´ 108 rads -1, w 3.14 ´ 108 rads -1, Wave constant, k = =, = 1.05 rad m -1, c, 3 ´ 108 ms -1, According to the condition given in the question, the wave, is propagating along X-axis, this means E should be along, Y-axis and B should be along Z-axis., Clearly, E = E0 sin ( kx - wt ) $j, , Z, , = [120 NC -1 sin { (1.05 rad m -1 ) x, , Hence, B is along the –Z-axis at that time., 5. (b) Given, magnetic field part of harmonic electromagnetic, wave, B0 = 510 nT, E, Speed of light in vacuum, c = 0, B0, where, E0 is the electric part of the wave., E0, 3 ´ 108 =, Þ, 510 ´ 10- 9, or, , E0 = 153 NC -1, , Thus, the amplitude of the electric field part of wave is, 153 NC -1., 6. (c) Suppose an electromagnetic wave is travelling along, negative z-direction. Its electric field is given by, E = E0 cos( kz - wt ), which is perpendicular to Z-axis. It acts along negative, y-direction., The associated magnetic field B in electromagnetic wave is, along X-axis, i.e. along k$ ´ E., , - (3.14 ´ 108 rad s -1 ) t}] $j, 9. (c) Comparing the given equation,, N, rad ö, éæ, æ, -2 rad ö ù, cos ê ç 2p ´ 106, ÷ t - ç p ´ 10, ÷x, C, mø, s ø úû, è, ëè, With the standard equation,, Ey = E0 cos ( wt - kx ), we get, w = 2pn = 2p ´ 106, Ey = 2.5, , \, , n = 106 Hz, , Moreover, we know that,, 2p, = k = p ´ 10-2 m -1, l, Þ, l = 200m, As direction of field E of electromagnetic wave is in, y-direction, so the wave is moving along positive x-direction, with frequency 106 Hz and wavelength 200 m., 10. (a) Direction of propagation of an electromagnetic, wave is given by E ´ B., \A unit vector in the direction of propagation.

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12, , CBSE Term II Physics XII, , =, , E ´ B k$ ´ ( 2$i - 2$j), =, |E ´B|, |E ´ B |, æQ E ´ B = k$ ´ ( 2$i - 2$j), ö, ç, ÷, $, $, ç, $, $, = 2( k ´ i ) - 2( k ´ j) ÷, ç, ÷, = 2( $j) - 2( - $i ), ç, ÷, ç, ÷, $, $, = 2 j + 2i, ç, ÷, ç, ÷, 2, 2, |, E, ´, B, |, =, +, =, Q, 2, 2, 2, 2, è, ø, , 2$j + 2$i, =, 2 2, $i + $j, =, 2, 11. (d) Given, v = 2 ´ 108 ms -1 and m r = 1, , The speed of electromagnetic waves in a medium is given by, 1, …(i), v=, me, where, m and e are absolute permeability and absolute, permittivity of the medium, respectively., Now,, m = m 0 m r and e = e0 er, 1, 1, 1, \ Eq. (i) becomes, v =, =, ´, m 0m r e0 er, m 0 e0, m r er, Þ, , v=, , c, m r er, , æ, çQ c = 1, ç, m 0 e0, è, , ö, ÷, ÷, ø, , On squaring both sides, we get, c2, (3 ´ 108 ) 2, er = 2, =, = 2.25, v m r (2 ´ 108 ) 2 ´ 1, 12. (c) The statement in option (c) is incorrect and it can be, corrected as,, The constancy of the velocity of electromagnetic waves in, vacuum is used to define a standard of length. The meter is, defined as the distance travelled by light in vacuum in a time, ( 1 / c ) second., 13. (c) In vacuum, e0 = 1,, In medium, e = 4, So, refractive index,, n = e/ e0 = 4 / 1 = 2, l l, =, n 2, c c, and wave velocity, v = =, n 2, , Wavelength, l ¢ =, , cö, æ, çQ n = ÷, vø, è, Hence, it is clear that wavelength and velocity will become, half but frequency remains unchanged when the wave is, passing through any medium., 1, 14. (c) Speed of electromagnetic waves in vacuum =, m 0 e0, Therefore, speed of EM wave in vacuum is same for all of, them., 15. (b) When a wave is reflected from denser medium, then the, type of wave does not change but only its phase changes by, 180° or p rad., , Thus, for the reflected wave z$ = - z$ , $i = - $i and additional, phase of p in the incident wave., Given,, E = E0 i$ cos( kz - wt ), The reflected electromagnetic wave is given by, E r = E0 ( - i$ ) cos[ k( - z ) - wt + p ], = - E i$ cos[ -( kz + wt ) + p ], 0, , = E0 $i cos[ -( kz + wt ) = E0 i$ cos( kz + wt )], 16. (a) Intensity or power per unit area of the radiations,, I, I = pc Þ p =, c, 0.5, =, = 0.166 ´ 10-8 Nm -2, 3 ´ 108, Energy, 17. (b) As we know, momentum, p =, c, Power ´ Time, =, c, Given, P = 30mW = 30 ´ 10-3 W, t = 100ns = 100 ´ 10-9 s, c = 3 ´ 108 ms -1, 30 ´ 10-3 ´ 100 ´ 10-9, Þ, p=, 3 ´ 108, = 1.0 ´ 10-17 kg-ms -1, 18. (c) Given, energy required to dissociate a carbon monoxide, molecule into carbon and oxygen atoms E = 11 eV, We know that, E = hn, where h = 6.62 ´ 10-34 J-s, and, Þ, , n = frequency Þ 11 eV = hn, 11 ´ 1.6 ´ 10-19, J, n=, h, 11 ´ 1 . 6 ´ 10-19, J, =, 6.62 ´ 10-34, = 2.65 ´ 1015 Hz, , This frequency radiation belongs to ultraviolet region., 19. (a) When the sun shines on our hand, we feel the energy, being absorbed from the electromagnetic waves (our hands, get warm)., hn, 1, Since, momentum, p =, (Q hn = constant), Þ pµ, c, c, Electromagnetic waves also transfer momentum to our hand, but because c is very large, the amount of momentum, transferred is extremely small and we do not feel the, pressure., Therefore, both A and R are true and R is the correct, explanation of A., 20. (b) Light radiations and thermal radiations both belong to, electromagnetic spectrum. Light radiation belongs to visible, region while thermal radiation belongs to infrared region of, EM spectrum., Also, EM radiations require no medium for propagation., Therefore, both A and R are true but R is not the correct, explanation of A., 21. (a) An oscillating charge produces an electric field in space,, which produces an oscillating magnetic field, which in turn,, is a source of electric field.

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13, , CBSE Term II Physics XII, , Thus, EM wave is a self-sustaining oscillating wave., Therefore, both A and R are true and R is the correct, explanation of A., 22. (b) The oscillating electric and magnetic fields, thus, regenerate each other, as the wave propagates through the, space and the frequency of the electromagnetic wave, naturally equals the frequency of oscillation of the charge., The energy associated with the propagating wave comes at, the expense of the energy of the source, the accelerated, charge., Therefore, both A and R are true but R is not the correct, explanation of A., 23. (b) The wavelength of these waves ranges between 4000Å to, 100 Å, i.e. smaller wavelength and higher frequency. They, are absorbed by atmosphere and convert oxygen into ozone., They cause skin diseases and they are harmful to eye and, cause permanent blindness., Therefore, both A and R are true but R is not the correct, explanation of A., 24. (i) (d) The energy in EM waves is divided equally between, the electric and magnetic fields., The total energy per unit volume, U = U e + U m, 1, 1 B2, = e0 E2 +, 2, 2m0, (ii) (c) We have, By = 2 ´ 10-7 sin ( 0.5 ´ 103 x + 1 .5 ´ 1011 t ), Comparing with the standard equation, we get, By = B0 sin ( kx + wt ), Þ, k = 05, . ´ 103, 2p, Þ, l=, = 0.01256, 05, . ´ 103, The wavelength range of microwaves is 10-3 to 0.3. The, wavelength of this wave lies between 10-3 to 0.3, so the, equation represents a microwave., (iii) (a) According to Maxwell equation, the magnitude of the, electric and magnetic fields in an electromagnetic wave, are related as, E, 6.3$j Vm-1, B= =, c 3 ´ 108 $i ms-1, $, = 21, . ´ 10-8 kT, 1, (iv) (b) The quantity enE2 represents energy per unit, 2, volume., Thus, it has dimensions of, Energy [ML2 T -2 ], =, Volume, [L3 ], = [ML-1T -2 ], (v) (d) Given, frequency by which the charged particles, oscillates about its mean equilibrium position = 109 Hz., So, frequency of electromagnetic waves produced by the, charged particle, n = 109 Hz., c 3 ´ 108, =, = 0.3 m, n, 109, The range of radiowaves is 10-1 to 104 m, so frequency of, 109 Hz falls in the region of radiowaves., Wavelength, l =, , PART 2, 1. An oscillating charge is an example of accelerating charge. It, produces an oscillating electric field, which produces an, oscillating magnetic field, which in turn produces an, oscillating electric field and so on. The oscillating electric, and magnetic fields regenerate each other as a wave which, propagates through space., Electric and magnetic fieds are the source of energy, associated with EM waves., 2. (i) Given that, velocity v = v $i, i.e. the wave is propagating, along X-axis , so electric field E is along Y-axis and, magnetic field B is along Z-axis. The propagation of, electromagnetic wave is shown in the figure, Y E, E, , B, , E, , B, , O, B, , Z, , B, , E, , B, , v = v $i, X, , E, , Direction of propagation, , (ii) Speed of electromagnetic wave can be given as, E, E, c= 0 =, B0 B, where, E0 and B0 are peak values of E and B or, instantaneous values of E and B., 3. As we know that, the direction of electromagnetic wave is, perpendicular to both electric and magnetic fields. Here,, electromagnetic wave is travelling in z-direction, then, electric and magnetic fields are in xy-direction and are, perpendicular to each other., Frequency of waves, n = 30 MHz = 30 ´ 106 Hz, Speed, c = 3 ´ 108 m/s, Using the formula, c = nl, Wavelength of electromagnetic waves,, c, 3 ´ 108, 300, l= =, =, = 10 m, n 30 ´ 106, 30, Thus, the wavelength of electromagnetic waves is 10 m., 4. The average value of energy density (energy / volume) is, given by, 1, U av = e0 E02, 2, Total volume of the cylinder, V = A × l, Total energy contained in the cylinder,, æ1, ö, U = ( U av )( V ) = ç e0 E02 ÷( Al ), è2, ø, Substituting the values, we have, 1, U = ´ ( 8.86 ´ 10-12 )( 50) 2 ( 10 ´ 10-4 )( 50 ´ 10-2 )J, 2, = 5.5 ´ 10-12 J, 5. Given, velocity of electromagnetic wave in vacuum,, c = 3 ´ 108m/s, Relative electric permeability, er = 2, and magnetic permeability, m r = 1

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14, , CBSE Term II Physics XII, , Since, velocity of electromagnetic wave in a medium can be, calculated by, 1, 1, v=, =, e0 erm 0m r, e0m 0 ´ m r er, 1, where,, =c, e0m 0, c, …(i), Þ, v=, m r er, Therefore,, , 3 ´ 108, v=, 2´1, , 3, Þ v=, ´ 108 m/s, 2, , 6. Electric field of an electromagnetic wave is an oscillating, field which causes force on the charged particle. This, electric force averaged over an integral number of cycles is, zero, because its direction changes with every half cycle. So,, electric field is not responsible for radiation pressure., Force F, 7. Pressure =, =, Area, A, Force is the rate of change of momentum., dp, i.e., F=, dt, Energy in time dt,, U, U = dp × c Þ dp =, c, 1 U, \Pressure = ×, A c × dt, I, U ö, æ, =, ÷, çQ intensity, I =, A × dt ø, c, è, 8. (i) Infrared waves have frequencies lower than those of, visible light, they can vibrate not only the electrons, but, also the entire atoms or molecules in the structure of the, surface., This vibration increases the internal energy and hence, the temperature of the structure, which is why infrared, waves are often called heat waves., (ii) Electromagnetic wave transports linear momentum as it, travels through space. If an electromagnetic wave, transfer a total energy U to a surface in time t, then total, linear momentum delivered to the surface is given as, U, p=, c, where, c is the speed of electromagnetic wave., 9. (i) 10-12 m-10-8 m = 01, . Å - 100Å ® X-ray, It is used in crystallography., (ii) 10-3 m - 10-1 m = 01, . cm - 10 cm ® Microwaves, It is used in microwave over for cooking purpose., 10. (i) The decreasing order of wavelengths of electromagnetic, waves is, Microwaves > Infrared > Ultraviolet radiation > g-rays, (ii) Microwaves They are used in RADAR devices., g - rays It is used in radio therapy., 11. (i) Microwaves are used for aircraft navigation, their, frequency range is 109 Hz to 1012 Hz., (ii) X - rays are used to study crystal structure, their, frequency range is 1016 Hz to 1020 Hz., , 12. Given, l mT = 0.29 cm-K, 0.29, m, lm =, T ´ 100, Let we take, l m = 10- 6 m, Required absolute temperature, T =, Let we take,, , 0.29, = 2900 K, 100 ´ 10- 6, , l m = 5 ´ 10- 5 m, , Required absolute temperature,, 0.29, T=, = 6000 K, 100 ´ 5 ´ 10- 5, We can find the temperature for other parts of the, electromagnetic spectrum. These number tell us about the, temperature ranges for particular part of EM waves., 13. (i) The time taken by wave to move a distance 90 cm,, 90 ´ 10-2, t=, = 3 ´ 10-9 s, 3 ´ 108, Energy contained in 90 cm length,, U = Pt, = 6 ´ 10-3 ´ 3 ´ 10-9, = 18 ´ 10-12 J, 1, (ii) Intensity of light, I = e0 E02 c, 2, 1, Þ, 17.7 = ( 8.85 ´ 10-12 ) E02 ´ 3 ´ 108, 2, 4, 2, Þ, E0 = ´ 104, 3, 2, Þ, E0 =, ´ 102 V/m, 3, Therefore, the amplitude of electric field in parallel beam,, 2, E0 =, ´ 102 V/m, 3, 14. (a) Gamma rays-3 ´ 1018 Hz to 5 ´ 1022 Hz, (b) Ultraviolet rays-1014 Hz to 1016 Hz, (c) Radio waves-54 MHz to 890 MHz, 15. Given, amplitude of an electromagnetic wave, E0 = 120 N/C, Frequency of wave, n = 50 MHz = 50 ´ 106 Hz, E, (i) Speed of light in vacuum, c = 0, B0, E0, 120, B0 =, =, = 40 ´ 10- 8, c, 3 ´ 108, or, , B0 = 400 ´ 10- 9 T = 400 nT, , Angular frequency of wave,, w = 2pn = 2 ´ 3.14 ´ 50 ´ 106, w = 3.14 ´ 108 rad/s, Wave number of electromagnetic waves,, w 3.14 ´ 108, k= =, = 1.05 rad/m, c, 3 ´ 108, Wavelength of electromagnetic wave,, c, 3 ´ 108, l= =, = 6.00 m, n 50 ´ 106

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15, , CBSE Term II Physics XII, , (ii) Expression of electric field,, E = E0 sin ( kx - wt ), E = 120 sin (1.05 x - 3.14 ´ 108 t ), , Speed of electromagnetic waves, c =, Putting in Eq. (ii), we get, 1, 1, uE = e0 B02 ×, 4, m 0 e0, , Expression of magnetic field,, B = B0 sin ( kx - wt ), B = 4 ´ 10- 7 sin (1.05 x - 3.14 ´ 108 t ), , uE =, , 16. (i) The given equation signifies that, the electromagnetic, wave is moving along Y-axis and also in negative, direction, so it moves in - $j -direction., (ii) The electric part of electromagnetic wave in vacuum,, E = [3.1 cos {1.8 y + (5.4 ´ 106 t )}] $i, Comparing with standard equation,, E = E0 cos ( ky + wt ), we get, Angular frequency, w = 5.4 ´ 106 rad/s, Wave number, k = 1.8 rad/m, The amplitude of the electric field part of the wave,, E0 = 3.1 N/C, 2p 2p, l=, =, = 3.491 m, k 1.8, Þ, l = 3.5 m, (iii) Angular frequency, w = 2pn, n=, , w 5.4 ´ 106 ´ 7, =, 2p, 2 ´ 22, = 0.86 ´ 106 Hz, , (iv) As, c =, , E0, B0, , Amplitude of magnetic field,, E, 3.1, B0 = 0 =, c, 3 ´ 108, = 1.03 ´ 10– 8 T, (v) Expression for the magnetic field part of wave,, B = B cos ( ky + wt ) k$, 0, , B = 1.03 ´ 10– 8 cos ( 1.8 y + 5.4 ´106 t ) k$, 17. Given, frequency of oscillation = 2 ´ 1010 Hz,, Speed of wave, c = 3 ´ 108 m/s, and electric field amplitude, E0 = 48 V/m, c, 3 ´ 108, (i) Wavelength of waves, l = =, = 1.5 ´ 10- 2 m, f 2 ´ 1010, E, (ii) Using the formula, c = 0, B0, The amplitude of the oscillating magnetic field,, E, 48, B0 = 0 =, = 1.6 ´ 10- 7 T, c, 3 ´ 108, (iii) The average energy density of electric field,, 1, uE = e0 E02, 4, E0, We know that,, =c, B0, Putting in Eq. (i), we get, 1, \, uE = e0 . c 2 B02, 4, , 1, m 0 e0, , …(i), , …(ii), , 1 B02, ×, 4 m0, , We may express the average energy density in EM waves, 1, B2, u = e0 E02 = 0, 2, 2m 0, Thus, the average energy density of the E field equals the, average energy density of B field., 18. (i) Gamma rays has the highest frequency in the, electromagnetic waves. These rays are of the nuclear, origin and are produced in the disintegration of, radioactive atomic nuclei and in the decay of certain, sub-atomic particles. They are used in the treatment of, cancer and tumours., (ii) Ultraviolet rays lie near the high frequency end of, visible part of EM spectrum. These rays are used to, preserve food stuff. The harmful effect from exposure, to ultraviolet (UV) radiation can be life threatening and, include premature aging of the skin, suppression of the, immune systems, damage to the eyes and skin cancer., 19. (i) Consider a plane perpendicular to the direction of, propagation of the wave. An electric charge, on the, plane will be set in motion by the electric and, magnetic fields of EM wave, incident on this plane., This is only possible, if EM wave constitutes, momentum and energy. Thus, this illustrates that EM, waves carry energy and momentum., (ii) Microwaves are produced by special vacuum tube like the, klystron, magnetron and Gunn diode. The frequency of, microwaves is selected to match the resonant frequency, of water molecules, so that energy is transformed, efficiently to increase the kinetic energy of the, molecules. Thus, facilitating the food to cook properly., (iii) Uses of infrared rays, (a) In knowing the molecular structure and therapy to, heal muscular pain., (b) In remote control of TV, VCR, etc., 20. (i) g-rays are used for the treatment of certain forms of, cancer. Its frequency range is, 3 ´ 1019 Hz to 5 ´ 1022 Hz., (ii) The thin ozone layer on top of stratosphere absorbs, most of the harmful ultraviolet rays coming from the, sun towards the earth. They include UVA, UVB and, UVC radiations, which can destroy the life system on, the earth., Hence, this layer is crucial for human survival., (iii) An electromagnetic wave transports linear momentum as, it travels through space. If an electromagnetic wave, transfers a total energy U to a totally absorbing surface, in time t, then total linear momentum delivered to the at, surface,, U, hn, p=, Þ p=, c, c

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16, This means, the momentum range of EM waves is, 10-19 to 10- 41. Thus, the amount of momentum, transferred by the EM waves incident on the surface, is very small., 21. In microwave oven, the frequency of the microwaves is, selected to match the resonant frequency of water, molecules. This leads to the vibrations of these molecules., As these vibrations increase with time, the temperature, increases leading to production of heat and this is the heat, which is responsible for the cooking of food in the oven., As, microwaves are short wavelength radio waves, with, frequency of order of GHz. Due to short wavelength, they, have high penetrating power with respect to atmosphere, and less diffraction in the atmospheric layers. So, these, waves are suitable for the radar systems used in aircraft, navigation., 22. (i) Microwaves are suitable for RADAR systems that are, used in aircraft navigation. These rays are produced by, special vacuum tubes, namely klystrons, magnetrons, and gunn diodes., (ii) Infrared rays are used to treat muscular strain. These, rays are produced by hot bodies and molecules., (iii) X-rays are used as a diagnostic tool in medicine., These rays are produced, when high energy electrons, are stopped suddenly on a metal of high atomic, number.., 23. (i) This wavelength (21 cm) corresponds to the radio, waves., (ii) This frequency (1057 MHz) also corresponds to the, radio waves (short wavelength)., (iii) As, T = 2.7 K, , CBSE Term II Physics XII, , Using the formula, l m T = b = 0.29 cm-K, 0.29, cm = 0.11 cm, lm =, 2.7, This wavelength corresponds to the microwaves region, of the electromagnetic spectrum., (iv) This wavelength lies in the visible region of the, electromagnetic spectrum., (v) Energy, E =14.4 keV = 14.4 ´ 103 ´ 1.6 ´ 10-19 J, Frequency of wave,, E 14.4 ´ 1.6 ´ 10-16, n= =, h, 6.6 ´ 10-34, 18, = 3.5 ´ 10 Hz, This frequency lies in the X-ray region of the electromagnetic, spectrum., 24. (i) A wave of frequency 2 ´ 1018 belong to X-rays of, electromagnetic spectrum., (ii) The range of wavelength for X-rays is around 1 nm to, 10-3 nm., (iii) X-rays are produced due to change in speed of fast, moving electrons, when they collide and interact with, the target anode., (iv) X-rays can be detected with the help of Geiger-Muller, tube or GM counter and also with ionisation chamber., (v) Two applications or uses of X-rays, (a) X-rays are used in medical diagnosis and to cure, malignant growths., (b) These rays are used in detecting faults, cracks, etc., in, metal products.

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17, , CBSE Term II Physics XII, , CHAPTER 02, , Ray Optics and, Optical Instruments, In this Chapter..., l, , Ray Optics, , l, , Lens, , l, , Refraction, , l, , Prism, , l, , Total Internal Reflection (TIR), , l, , Optical Instruments, , l, , Refraction at a Spherical Surface, , Ray Optics, , sin i 1, = m 2 or m 21, sin r, , i.e., , A light wave can be considered to travel from one point to, another, along a straight line joining them is called a ray of, light. A bundle of such rays constitutes a beam of light and, the branch of study of light is called optics., , where, m 21 is constant, called refractive index of, second medium with respect to first medium., This is called also as Snell’s law of refraction., , Refraction, , Refractive Index, , It is the phenomenon of bending of ray of light, when they, pass from one transparent medium to another depending on, their optical densities., , The refractive index or index of refraction m of a material is, the ratio of the speed of light ( c) in vacuum to the speed of, light in the medium (v)., Mathematically, refractive index is given by the relation, Speed of light in the vacuum c, m=, =, Speed of light in the material v, , Incident ray, , Normal, Reflected, ray, i, , i, , Reflecting, surface r, , Refracted ray, , Laws of Refraction, There are two laws of refraction which are given below, (i) The incident ray, the refracted ray and the normal to, the refracting surface at the point of incidence, all lie, in the same plane., (ii) The ratio of the sine of angle of incidence to the sine, of angle of refraction is constant., , Following are few important points related to refractive, index, (i) If n 21 > 1, r < i, then the refracted ray bends towards, the normal. In such a case, medium 2 is said to be, optically denser than medium 1., (ii) If n 21 < 1, r > i, the refracted ray bends away from the, normal. This is the case, when incident ray in a denser, medium refracts into a rarer medium., Rarer, i, , 90°, , Denser, i, , 90°, r, , r, Denser, , Rarer

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18, , CBSE Term II Physics XII, , Principle of Reversibility of Light, When a light ray, after suffering any number of reflections, and refractions, has its final path reversed, it travels back, along its entire initial path. This is called principle of, reversibility of light. In the figure given below, OA is an, incident ray in medium 1 and AB is the refracted ray in, medium 2., O, Medium 1, i, A, r, B, , m2 =, , 1, 2, , m1, Thus, the refractive index of medium 2 relative to medium 1, is equal to the reciprocal of the refractive index of medium 1, relative to medium 2., , Refraction of Light Through a Rectangular, Glass Slab, Let ABCD be a rectangular glass slab. A ray of light is, incident along MN on the face AB of the rectangular slab at, Ði1 . It is refracted along NK with Ðr1 ., The refracted ray NK falls on face CD with Ði2 and emerges, out along KL with Ðr2 ., M, ma, A, , i1, B, , N, r1 d, mg, , D, , ma, , i2, , N¢, , K r, 2, , t, , L¢, , C, L, , Again, applying Snell’s law at K,, m g ´ sin i2 = m a ´ sin r2, m a sin i2 g, Þ, =, = ma, m g sin r2, , sin r2 a, = mg, sin i2, , ...(iii), , Multiplying Eqs. (ii) and (iii), we get, sin i2 sin r2 g, ´, = m a ´ am g, sin r2 sin i2, 1 = g m a ´ am g ,, 1, a, mg = g, ma, ...(iv), , As,, (alternate angles), i2 = r1, \, sin i2 = sin r1, From Eq. (iv), we get, sin r2 = sin i1 or r2 = i1, Hence, the emergent ray KL is parallel to the incident ray, MN as shown in the figure. We observe that the incident ray, MN is displaced laterally, on suffering two refractions, through a glass slab., , Expression for Lateral Displacement, Now, from K, draw KL¢ ^ MN produced., \ Lateral displacement of the ray on passing through the, parallel slab = KL¢., Let ÐKNL ¢ = d = deviation on first refraction., KL¢, In D NKL¢,, sin d =, NK, ...(v), \, KL¢ = NK sin d, NN ¢, In D NN ¢K,, cos r1 =, NK, NN ¢, t, \, NK =, =, cos r1 cos r1, where, t = NN ¢ = thickness of glass slab., t, From Eq. (v), we get KL¢ =, sin d, cos r1, , P, , Applying Snell’s law at N,, m a ´ sin i1 = m g ´ sin r1, sin i1 m g a, or, =, = mg, sin r1 m a, , ma, , =, , From Eqs. (i) and (iii), we get, sin i1 sin r2, =, sin r1 sin i2, , Medium 2, , 1, , mg, , ....(i), , .... (ii), , According to the principle of reversibility of light, when final, path of a light ray after suffering a number of reflections and, refractions is reversed, then the ray retraces its entire path., Now, imagine a plane mirror P held normal to KL so that on, reflection from mirror, path KL is reversed. The ray would, retrace its entire path. For the reversed ray, the application, of Snell’s law at K gives, m a ´ sin r2 = m g ´ sin i2, , KL¢ =, , t sin( i1 - r1 ), cos r1, , This is the required expression for lateral displacement (or, shift), which is obviously proportional to thickness (t) of glass, slab. Further, lateral displacement (or shift) will increase with, increasing angle of incidence ( i1 )., , Apparent Depth and Normal Shift, When an object is in denser medium and observer is in rarer, medium, then object appears to be at lesser depth than its, actual depth., Real and apparent depth are related as follows, Real depth ( h ), n 21 =, Apparent depth ( h ¢ )

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19, , CBSE Term II Physics XII, , Effect of Atmospheric Refraction at, Sunrise and Sunset, The refraction of light through the atmosphere is responsible, for many natural phenomena. e.g., (i) Due to refraction, the sun is visible a little before, the actual sunrise and until a little after the actual, sunset., (ii) The apparent flattening (oval shape) of the sun at, sunset and sunrise is also due to the same, phenomenon., , Critical Angle, The angle of incidence is in denser medium for which the, angle of refraction in rarer medium becomes 90° is called, critical angle., m2, Denser, , ic, , 90 Rarer, m1, , 1, sin ic, where, ic is critical angle., 1, , Þ, , m2 =, , Optical Fibres These fibres are fabricated with high quality, composite glass/quartz fibres. Each fibre consists of a core, and cladding such that refractive index of core is higher than, that of the cladding. When a signal in the form of light is, directed at one end of the fibre at a suitable angle, it, undergoes repeated total internal reflection along the length, of the fibre and finally, comes out from other end., Thus, these are extensively used for transmitting audio and, video signals through long distances., Prisms Prisms are designed to bend ray by 90° and 180° or to, invert image without changing its size by the use of total, internal reflection., , Refraction at a Spherical Surface, If an object is placed in a medium of refractive index n1 at a, distance u from the pole of a spherical surface of radius of, curvature R and after refraction, its image is formed in a, medium of refractive index n 2 at a distance v, then, n 2 n1 n 2 - n1, =, v, u, R, This equation holds for any curved spherical surface., , A ray of light travelling from denser medium to rarer medium, is incident at the interface of two media at an angle greater, than the critical angle for the two media, then the ray is, totally reflected back to denser medium and this, phenomenon is called total internal reflection., , r¢, , r, , i, Denser, medium, (water), A, , N, , O, , C, M, u, , I, , R, v, , Refraction at a spherical surface, , Cartesian Sign Convention for Spherical Surfaces, The principal axis of the spherical surface is taken as X-axis, and the optical centre as origin, here the principal axis is, the diameter extended., The direction of the incident light is taken as the positive, direction of X-axis and opposite to it is taken as negative., The upward direction is taken as positive and the, downward direction as negative., l, , B, , O1, , i, r, , Total Internal Reflection (TIR), , Rarer, medium, (air), , n2, , N, , n1, , O2, , O3, , i¢ N, , ic, , Water-air, D O4 interface, i > ic, N, , Totally, reflected ray, , Partially, C reflected rays, , There are some applications of total internal reflection which, are given below, Mirage It is the phenomenon, in which an inverted image of, distant tall objects cause an optical illusion of water. This, type of mirage is especially common in hot deserts., Looming The optical illusion of an object floating in air is, called superior mirage. It is also known as looming., This occurs in very cold regions due to total internal, reflection., Diamond The critical angle for diamond-air interface is very, small, therefore once light enters a diamond, it is very likely, to undergo total internal reflection inside it. Due to this,, diamond shines brilliantly., , l, , l, , Lens, A lens is a transparent medium bounded by two surfaces of, which one or both surfaces are spherical., Lenses are of two types as given below, (i) Convex or Converging Lens A lens which is thicker, at the centre and thinner at its ends is called convex, lens. Convex lenses are of three types as shown below, , (a) Double convex, lens, , (b) Plano-convex, lens, , (c) Concavo-convex, lens

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20, , CBSE Term II Physics XII, , (ii) Concave or Diverging Lens A lens which is thinner, at the centre and thicker at its ends is called a, concave lens., , (a) Double concave, lens, , (b) Plano-concave, lens, , (c) Convexo-concave, lens, , Converging and Diverging Action of Lenses, Convex lens or converging lens are the lens which converges, all the light rays, coming parallel to its principal axis., Concave lens or diverging lens diverges all the light rays, coming parallel to its principal axis., , Converging lens, , P, , P, , C, O, , O, S, , S, , Optical, centre, , (ii) Centre of Curvature The centres of the two, imaginary spheres of which the lens is a part, are, called centres of curvature of the lens., (iii) Radii of Curvature The radii of the two imaginary, spheres of which the lens is a part are called radii of, curvature of the lens., (iv) Principal Axis The imaginary line joining the two, centres of curvature is called principal axis of lens., (v) Principal Focus Lens has two principal foci, (a) First Principal Focus It is a point on the, principal axis of lens, the rays starting from this, point in convex lens or rays directed to this point, in concave lens become parallel to principal axis, after refraction., , O, , F1, , O, , f2, , f2, , Both the foci of convex lens are real, while that of, concave lens are virtual., (vi) Aperture The effective diameter of the circular, outline of a spherical lens is called its aperture., (vii) Refractive Axis It is an imaginary axis at the optical, centre perpendicular to the principal axis which, represents the lens., , (a) Real path of ray, , (b) Path of ray as shown with, reference to refractive axis, , Image Formation in Lenses, Using Ray Diagrams, We can represent image formation in lenses using ray, diagrams. For drawing ray diagrams in lenses like spherical, mirrors, we consider any two of the following rays, (i) Rays which are parallel to the principal axis after, refraction, will pass through principal focus in case of, convex lens and will appear to be coming from, principal focus in case of concave lens., F1, , F1, , F2, , F2, , (ii) Rays passing through or directed to the focus will, emerge parallel to the principal axis., F2, , F1, O, , 2F1, , F1 O, , F1, f1, , 2F2, , F2, , (iii) Rays directed to optical centre will emerge out, undeviated., , F2, F1, , f1, , O, , F2, , F2, , Diverging lens, , Some Definitions Related to Lens, Important terminologies related to lenses are given below, (i) Optical Centre The optical centre is a point lying on, the principal axis of the lens, directed to which, incident rays pass without any deviation in the path., , Optical, centre, , (b) Second Principal Focus It is a point on the, principal axis at which the rays coming parallel to, the principal axis converge (convex lens) or, passing through it appear to diverge (concave, lens) at this point after refraction from the lens., , O, , F2, , F1, , O

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21, , CBSE Term II Physics XII, , If f 1 = f 2 , then f is infinite, i.e. combination will behave, like plane glass sheet., If the lenses are placed d distance apart, then, 1, 1, 1, d, =, +, f, f1 f 2 f1 f 2, , Refraction by a Lens : Lens Maker’s Formula, n2 ö, æ 1, æ, 1, 1 ö, = ( 1 n 2 - 1) ç, ÷, ÷, çQ 1 n 2 =, f, n1 ø, è R1 R 2 ø, è, This is known as lens Maker's formula., If the lens is thin, then lens Maker’s formula is given as, 1 1 1, - =, v u f, This is the thin lens formula., where, f is focal length of lens, v is image distance and u is, object distance., , Linear Magnification, Produced by a Lens (m), , Prism, A prism is a portion of a transparent medium bounded by two, plane faces inclined to each other at a suitable angle., , Refraction of Light Through a Prism, The figure below shows the passage of light through a, triangular prism ABC., A, , It is defined as the ratio of the height of the image to height, of the object., h¢ v, Linear magnification, m =, =, h u, For erect (and virtual) image, m is positive and for an, inverted (and real) image, m is negative., , i, , If several thin lenses of focal lengths f 1 , f 2 , f 3 , ¼ are in, contact, then the effective focal length and power of their, combination is given by, 1, 1, 1, 1, =, +, +, +.. ., f, f1, f2, f3, and, , P = P1 + P2 + P3 +.. ., , Magnification by Combination of Lenses, Combination of lenses helps to obtain diverging or, converging lens of desired magnification. It also enhances, sharpness of the image. Thus, the net magnification of such a, combination (m) is given as, m = m1 ´ m 2 ´ m 3 ´ …, l, , l, , If combination of lenses consists of one convex lens ( f 1 ), and one concave lens ( - f 2 ), then, f f, f = 1 2, f 2 - f1, If f 1 > f 2 , then f is negative, i.e. combination will behave, like concave lens, when focal length of convex lens is larger., If f 1 < f 2 , then f is positive, i.e. combination will behave, like convex lens, when focal length of convex lens is smaller., , d, , Q, , e, r2, , r1, , R, S, , P, , O, C, , B, , The angles of incidence and refraction at first face AB are i, and r1 . The angle of incidence at the second face AC is r2 and, the angle of emergence is e., The angle between the emergent ray RSand incident ray PQ, is called angle of deviation (d)., d = ( i + e) - ( r1 + r2 ), Q, r1 + r2 = A, \, , d = ( i + e) - A, , If m is the refractive index of material of the prism, then, d = (m - 1) A, This is the angle through which a ray deviates on passing, through a thin prism of small refracting angle A., , Prism Formula, If the angle of incidence is increased gradually, then the, angle of deviation first decreases, attains a minimum value, ( d m ) and then again starts increasing., Angle of deviation, , Combination of Thin Lenses in Contact, , T, , N, , Power of a Lens, It is the ability to converge or diverge the rays of light, incident on it. The SI unit of power of lens is dioptre (D)., The power of a lens is measured as the reciprocal of its focal, length (in metre)., 1, P=, f (in m), , K, , A, , d, dm, , i =e, , i, Angle of incidence, , e, , When angle of deviation is minimum, the prism is said to be, placed in the minimum deviation position.There is only one, angle of incidence for which the angle of deviation is, minimum.

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22, , CBSE Term II Physics XII, , When d = d m (prism in minimum deviation position), e=i, and, r2 = r1, Q, r1 + r2 = A, Þ, r+r=A, A, or, r=, 2, Also, we have, A+ d = i+ e, Putting d = d m and e = i in Eq. (ii), we get, A + dm = i + i, æ A + dm ö, Þ, i= ç, ÷, è 2 ø, From Snell’s law, m =, , Case I When the image is formed at the near point, A¢, , …(i), , a, FB, , B¢, , b, C, , D, , ...(ii), , sin i, sin r, , æ A + dm ö, sin ç, ÷, è 2 ø, m=, \, A, sin, 2, This relation is called a prism formula., For thin prisms (i.e. A is very small), the value of d m is also, very small., æ A + dm ö, sin ç, ÷, è 2 ø A + dm, So,, m=, », A, 2, sin, 2, A/ 2, Þ, , A, , A¢¢, , Magnifying power, m =, , D, f, (Q v = - D, because image is formed at near point), When the eye is placed behind the lens at a distance a, then, D-a, m=1+, f, Case II When the image is formed at infinity, D, m=, f, m=1+, , Compound Microscope, It consists of two convex lenses coaxially separated by some, distance. The lens nearer to the object is called the objective., The lens through which the final image is viewed is called, the eyepiece., Eyepiece, , Objective lens, , B, C, , A Fo, uo, , Fo, , A¢¢, Q, a, , Using the reflecting and refracting properties of mirrors,, lenses and prisms, many optical instruments have been, designed like microscopes and telescopes. Our eye is a, natural optical device., , The Eye, The eye lens is a convex lens whose focal length can be, modified by the ciliary muscles. This property of eye is called, accommodation. The image is formed on a film of nerve, fibres called retina., Near point is the closest distance for which the lens can form, image and its value is 25 cm for a normal eye. The far point of, a normal eye is infinity., , Simple Microscope, It is an optical instrument which forms large image of close, and minute objects. It is a converging lens of small focal, length. When an object is at a distance less than the focal, length of the lens, the image obtained is virtual, erect and, magnified., , b C¢, , Fe A¢, , fo, B¢, , d m = (m - 1) A, , Optical Instruments, , b, a, , ue, B¢¢, vo, fe, D, , Angular magnification or magnifying power of a compound, microscope is defined as the ratio of the angle b subtended by, the final image at the eye to the angle a subtended by the, object seen directly, when both are placed at least distance of, distinct vision., b, \ Angular magnification, m =, a, tan b, For small angles, m =, tan a, The magnification produced by the compound microscope is, the product of the magnification produced by the eyepiece, and objective., m = me ´ mo, where, m e and m o are the magnifying powers of the eyepiece, and objective, respectively., Case I When the final image is formed at near point, Linear magnification is given by, D, me = 1 +, fe

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23, , CBSE Term II Physics XII, , where, f e is focal length of the eyepiece., v öæ, æ, Dö, m = ç1 - o ÷ ç1 + ÷, fo ø è, fe ø, è, where, v o , f o and f e are image distance, focal length, of objective lens and focal length of eyepiece,, respectively., Case II When the final image is at infinity, If u o is the distance of the object from the objective, and v o is the distance of the image from the objective,, v, then the magnifying power of the objective, m o = o, uo, When the final image is at infinity, then angular, magnification is given by, D, me =, fe, The total magnification, when image is at infinity is, given by, æv, Dö, m = mo ´ m e = ç o ´ ÷, fe ø, è uo, If the object is very close to the principal focus of the, objective and the image formed by the objective is, very close to the eyepiece, then, -L D, m=, ×, fo fe, where, L = length of the tube of microscope., , Case II When final image is formed at near point, ue, , fo, Parallel rays from, object at infinity, a, , Fe Fo b, B¢, C2, A¢, fe, , B¢¢, C1, , A¢¢, , Eye, , D, , b, a, f, m= o, - ue, , Angular magnification, m =, , m=-, , fo æ, fe ö, ç1 + ÷, fe è, Dø, , Reflecting (Cassegrain) Telescope, It consists of concave mirror of large aperture and large focal, length (objective). A convex mirror is placed between the, concave mirror and its focus. A small convex lens works as, eyepiece., Objective, mirror, , Secondary, mirror, , Eyepiece, , Astronomical (Refracting) Telescope, An optical instrument which is used for observing distinct, images of heavenly bodies like stars, planets, etc., when the, final image is formed at infinity., Case I When the final image is formed at infinity, , One popular configuration of mirror and eyepiece is called, the Newtonian reflecting type telescope, named after its, designer Newton., , Objective lens, , a, C1, , a, , Eyepiece, , fe, Eyepiece, , fo, , Parall, from oel rays, infinity bject at, , B, , Fo, Fe, b, I, , C2, , Eye, , ge, , a, im, al inity, n, i, f, F in, at, , Angular magnification is given by, b, m=, a, Since, b and a are very small., I, I, and tan b =, tan a =, fo, - fe, where, I is the image formed by the objective, f o and, f e are the focal lengths of objective and eyepiece,, respectively., f, m=- o, fe, , M1= Parabolic mirror, , Rays arrive, parallel from, very distant, object, M2= Plane mirror, , The plane mirror reflects the beam and a real image is, formed infront of eyepiece. The eyepiece acts as a magnifier, and the final magnified image of the distant object can be, observed by the eye., Advantages of Reflecting Telescope, over Refracting Telescope, For astronomical telescope, the mirror affords several, advantages over the objective lens. A mirror is easier to, produce with a larger diameter, so that it can intercept rays, crossing a larger area and direct them to the eyepiece., The mirror can be made parabolic to reduce spherical, aberration. Aberration is further reduced because passage, through one layer of glass (the objective lens) is eliminated.

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24, , CBSE Term II Physics XII, , Solved Examples, Example 1. (i) Find the speed of light of wavelength, , l = 780 nm (in air) in a medium of refractive index, m = 1 . 55., (ii) What is the wavelength of this light in the given, medium?, , Sol. (i) Speed of light, v =, , c, 3.0 ´ 10, =, m, 1 .55, , 8, , = 1.94 ´ 108 m/s, (ii) Wavelength of light in the given medium,, l, 780, l medium = air =, = 503 nm, m, 1 .55, , (m = 1 .5 ) of edge 9.0 cm. By what amount will the, printed letters appear to be shifted When viewed from, the top?, The shift in the position of the printed letters,, æ, 1ö, 1 ö, æ, Dd = ç1 - ÷ d = ç1 ÷ ´ 9.0 cm = 3.0 cm, è, mø, 1 .5 ø, è, , Example 3. Figs. (a) and (b) show refraction of an incident, ray in air at 60° with the normal to a glass-air and, water-air interface, respectively. Predict the angle of, refraction in glass, when the angle of incidence in, water is 45° with the normal to a water-glass interface, [Fig. (c)]., , Air, , Glass, Water, , Water, 60°, , 47°, (a), , (b), , 45°, (c), , Sol. From Fig. (a),, i = angle of incidence = 60°, r = angle of refraction = 35°., \ Refractive index of glass with respect to water,, 0.8660, sin i sin 60°, =, = 1 .51, =, am g =, ., sin r sin 35°, 05736, From Fig. (b), here, i = 60° and r = 47°, \Refractive index of water with respect to air,, sin i sin 60°, 0.8660, =, =, = 1 .18, am w =, sin r sin 47°, 0.7314, \ We have,, or, , am g, , = am w ´ wm g, 1 .51, am g, =, = 1 .28, wm g =, 1 .18, am w, , m = 1.0, , m = 4/3, , O, P, , C, 30 cm, , Sol. Given, m1 = 1, m 2 = 4 / 3, u = - 10 cm and R = 30 cm, m, m, m - m1, Using the relation, 2 - 1 = 2, v, u, R, Substituting the values, we get, 4/ 3, 1, 4/ 3 - 1, 4, 1, 1, Þ, =, Þ, +, =, v, -10, 30, 3v 10 90, 4, 1, 1, -8, Þ, =, =, Þ, 3v 90 10 90, v = - 15 cm, The image is formed 15 cm left of spherical surface and, is virtual., , Example 5. Double-convex lenses are to be, , a, , 60°, , The point C is centre of curvature of the, spherical surface., , 10 cm, , Sol. The thickness of the cube, t = 9.0 cm., , 35°, , sin 45° 1 / 2, =, 1.28, 1 .28, 0.707, =, = 0552, . 5 = sin 33 .54¢, 1 .28, \ Angle of refraction, r = 33 .54°, sin r =, , or, , Example 4. Locate the image of the point object O., , Example 2. A printed page is kept pressed by a glass cube, , Glass, Air, , From Fig. (c), i = angle of incidence = 45°., \ Using the relation,, sin i, sin 45°, or 1 .28 =, wm g =, sin r, sin r, , manufactured from a glass of refractive index, 1.55, with both faces of the same radius of, curvature. What is the radius of curvature, required, if the focal length is to be 20 cm?, Sol. Using the relation,, æ 1, 1, 1 ö, ÷ , we get, = ( m - 1 ) çç, ÷, f, R, R, è 1, 2ø, 1, é 1 æ 1 öù, = ( 1 .55 - 1 ) ê - ç, ÷ú, 20, ë R è -R ø û, 1ù, 2, é1, = 055, . ê + ú = 055, . ´, R, R, R, ë, û, or, R = 1.10 ´ 20 = 22 cm, , Example 6. The distance between two point, sources of light is 24 cm. Find out where would, you place a converging lens of focal length 9 cm,, so that the images of both the sources are, formed at the same point.

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25, , CBSE Term II Physics XII, f = 9 cm, , Example 9. One face of a prism with a refractive angle, , S1, , of 30° is coated with silver. A ray incident on, another face at an angle of 45° is refracted and, reflected from the silver coated face and retraces its, path. What is the refractive index of the prism?, , S2, , x, , 24-x, , 1 1 1, Sol. Using the relation, = =, v u f, 1, 1, 1, 1 1 1, For S1 , =, Þ, \, = v1 - x 9, v1 9 x, 1, 1, 1, For S2 ,, –, =, v2 - ( 24 - x ) 9, 1 1, 1, \, = v2 9 24 - x, , Sol., 90 °, 45°, , …(i), , …(ii), , Since, sign convention for S1 and S 2 is just opposite. Hence,, 1, 1, v1 = - v2 or, =v1, v2, 1 1, 1, 1, \, – =, 9 x 24 - x 9, Solving this equation we get, x = 6 cm. Therefore, the lens, should be kept at a distance of 6 cm from either of the object., , Example 7. An object of size 3.0 cm is placed at 14 cm, from concave lens of focal length 21 cm. Describe, the image produced by the lens. What happens, if, the object is moved further away from the lens?, Sol. Using the relation,, , 1 1 1, = - , we get, f v u, 1 1 1, 1, 1, = + =, +, v f, u -21 ( -14), , \, v = - 8.4 cm, \ The image is virtual, erect and located at 8.4 cm from the, lens on the same side as the object., Also, we know that,, I v, m= =, O u, v, -8.4, \, I = ´ O=, ´ 3 = 1.8 cm, u, -14, i.e. The image is of diminished size., If the object is moved away from the lens, the virtual image, moves towards the focus of the lens (but never beyond, focus)., , Example 8. A converging lens of focal length 5.0 cm is, placed in contact with a diverging lens of focal, length 10.0 cm. Find the combined focal length of, the system., Sol. Given,, , f 1 = + 5.0 cm and f 2 = - 10.0 cm, , Therefore, the combined focal length F is given by, 1, 1, 1, 1, 1, 1, =, –, =+, =, +, 10.0, F f1, f 2 5.0 10.0, \, F = + 10.0 cm, i.e. The combination behaves as a converging lens of focal, length 10.0 cm., , Given,, A = 30°, i1 = 45° and r2 = 0, Since,, r1 + r2 = A, \, r1 = A = 30°, Now, refractive index of the prism,, sin i1 sin 45° 1 / 2, m=, =, =, = 2, sin r1 sin 30°, 1/ 2, , Example 10. A compound microscope has a, magnifying power of 100, when the image is formed, at infinity. The objective has a focal length of 0.5 cm, and the tube length is 6.5 cm. Find the focal length, of the eyepiece., Sol. When the final image is at infinity,, ue = f e = tube length - vo, \, f e = 6.5 – vo, v D, Since,, M¥ = o ×, uo f e, v 25, vo, or, \, 100 = o ×, =-4, uo f e, uo f e, For the objective,, 1, 1, 1, 1, =, =, vo uo f o 0.5, , Þ, , 1, 1, =2, vo uo, , …(i), , …(ii), , …(iii), , We have three unknowns vo, uo and f e solving Eqs. (i), (ii), and (iii), we get, f e = 2 cm, , Example 11. A small telescope has an objective lens of, focal length 140 cm and an eyepiece of focal length, 5.0 cm. What is the magnifying power of the, telescope for viewing distance objects, when, (i) the telescope is in normal adjustment (i.e. when, the final image is at infinity) and, (ii) the final image is formed at the least distance of, distinct vision 25 cm?, Sol. (i) When the telescope is in normal adjustment, the, magnifying power is given by, f, 140, M= o =+, = 28, | f e|, 5, (ii) When the final image is formed at the least distance of, distinct vision, then M is given by, f æ, f ö 140 æ, 5.0 ö, M = o ç1 + e ÷ =, ç1 +, ÷ = 33.6, | f e| è, Dø, 5 è, 25 ø

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26, , CBSE Term II Physics XII, , Chapter, Practice, PART 1, Objective Questions, l, , 6. For the refraction shown below, the correct relation, is, i, , n1, , n2, , r, , Multiple Choice Questions, , P, , O, u, , C, , 1. A ray of light strikes an air-glass interface at an, angle of incidence ( i = 60° ) and gets refracted at an, angle of refraction r. On increasing the angle of, incidence ( i > 60° ), the angle of refraction r, (a) decreases, (c) is equal to 60°, , (b) remains same, (d) increases, , 2. The refractive indices of water and glass with respect, to air are 4/3 and 5/3, respectively. The refractive, index of glass with respect to water will be, (a) 1 / 3, (c) 5 / 4, , (b) 4 / 3, (d) 20 / 9, , 3. A ray of light is incident at the glass-water interface, at an angle (as shown below) and it emerges finally, parallel to the surface of water, then the value of n g, would be, Air, , I, , R, v, , (a), , n 2 n1 n 2 - n1, =, v, u, R, , (b), , n1 n 2 n 2 - n1, =, v, u, R, , (c), , n1 n 2 n1 - n 2, =, v, u, R, , (d), , n 2 n1 n1 - n 2, =, v, u, R, , 7. First and second focal lengths of spherical surface, of refractive index n are f1 and f 2 , respectively. The, relation between them, is, (a) f 2 = f1, (b) f 2 = - f1, (c) f 2 = nf1, (d) f 2 = - nf1, , 8. Which of the following is true for rays coming, from infinity?, m1, , n w = 4/ 2, Water, Glass, , m2, , i, , (a) ( 4 / 3) sin i, , (b) 1 /sin i, , (c) 4 / 3, , (d) 1, , 4. If the critical angle for light going from medium A, , to B is q. Then, find the speed of light in medium B,, if speed of light is v in medium A., (a) v( 1 - cos q) (b), , v, cos q, , (c), , v, sin q, , (d) v( 1 - sin q), , 5. The phenomena involved in the reflection of, radiowaves by ionosphere is similar to, [NCERT Exemplar], (a) reflection of light by a plane mirror, (b) total internal reflection of light in air during a mirage, (c) dispersion of light by water molecules during the, formation of a rainbow, (d) scattering of light by the particles of air, , (a) Two images are formed at two different points, (b) Continuous image is formed between focal points of, upper and lower lens, (c) One image is formed by the lens, (d) None of the above, , 9. The radius of curvature of the curved surface of a, plano-convex lens is 20 cm. If the refractive index, of the material of the lens be 1.5, it will, [NCERT Exemplar], (a) act as a convex lens only for the objects that lie on its, curved side, (b) act as a concave lens for the objects that lie on its curved, side, (c) act as a convex lens irrespective of the side on which the, object lies, (d) act as a concave lens irrespective of side on which the, object lies

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27, , CBSE Term II Physics XII, , 10. A double convex lens whose refractive index is 1.33, , 16. A prism has refractive angle 60°. When a light ray, , has both radii of curvature of magnitude 10 cm. If, an object is placed at a distance of 5 cm from this, lens, the position of the image formed (in cm) is, , is incident at 50°, then minimum deviation is, obtained. What is the value of minimum deviation?, (a) 40°, , (a) 7.46 same side of the object, (b) 7.46 opposite side of the object, (c) 14.45 same side of the object, (d) 14.45 opposite side of the object, , (a) object size, (c) focal length of the lens, , eyepiece respectively, of the telescope. The angular, magnification of the given telescope is equal to, , R, 2( n1 - n 2 ), 2R, (d), n 2 - n1, , two identical plano-concave parts. The power of, each part will be, P, 2, P, (d), 2, , (a) 2P, , (b), , (c) P, , 14. A light ray incident normally on one of the face of a, triangular prism follow the path as shown below,, then the angle of refraction r2 at the second face is, 30º, , (b) 90°, (d) 45°, , 15. White light is incident on one of the refracting, surfaces of a prism of angle 5°. If the refractive, indices for red and blue colours are 1.641 and 1.659, respectively, the angular separation between these, two colours when they emerge out of the prism is, (a) 0.9°, (c) 1.8°, , F1, F2, , (b), , F2, F1, , (c), , FF, 1 2, F1 + F2, , (d), , F1 + F2, FF, 1 2, , magnification of magnitude 5 for distant objects., The separation between the objective and the, eyepiece is 36 cm and the final image is formed at, infinity. The focal length f o of the objective and the, focal length f e of the eyepiece are, (a) 45 cm and - 9 cm, (c) 50 cm and 10 cm, l, , (b) - 7.2 cm and 5 cm, (d) 30 cm and 6 cm, , Assertion-Reasoning MCQs, Direction (Q. Nos. 20-25) Each of these questions, contains two statements Assertion (A) and Reason (R)., Each of these questions also has four alternative, choices, any one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given, below., (a) Both A and R are true and R is the correct, explanation of A., (b) Both A and R are true, but R is not the correct, explanation of A., (c) A is true, but R is false., (d) A is false and R is also false., , 20. Assertion Refractive index of glass with respect to, , r2, , (a) 60°, (c) 30°, , (a), , 19. An astronomical telescope has an angular, , 12. Which of the following statement is correct?, , 13. A biconcave lens of power P vertically splits into, , (b) aperture of the lens, (d) power of the lens, , 18. F1 and F2 are focal lengths of objective and, , (b), , (a) Power of a lens is a measure of the convergence or, divergence, which a lens introduces in the light falling on it., (b) Power is defined as the cosine of angle by which it, converges or diverges a beam of light falling at unit, distance from the optical centre., (c) Cutting of lenses helps to obtain diverging or converging, lenses of desired magnification., (d) Total magnification m of the combination is the sum of, magnification ( m1 , m 2 , m 3 .....) of individual lenses., , (d) 60°, , simple microscope, one should increase the, , lens. Their plane surfaces are parallel to each other. If, lenses are made of different materials of refractive, indices n1 and n 2 and R is the radius of curvature of, the curved surface of the lenses, then the focal length, of the combination is, R, 2( n1 + n 2 ), R, (c), n1 - n 2, , (c) 50°, , 17. In order to increase the angular magnification of a, , 11. A plano-convex lens fits exactly into a plano-concave, , (a), , (b) 45°, , (b) 0.09°, (d) 1.2°, , air is different for red light and violet light., Reason Refractive index of a pair of media does, not depends on the wavelength of light used., , 21. Assertion The refractive index of diamond is 6, and that of liquid is 3. If the light travels from, diamond to the liquid, it will be totally reflected, when the angle of incidence is 30°., Reason n = sin i c , where n is the refractive index of, diamond with respect to liquid and i c is the critical, angle.

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28, , CBSE Term II Physics XII, , (ii) A ray of light will undergo total internal reflection, inside the optical fibre, if it, , 22. Assertion When monochromatic light is incident, on a surface separating two media, the reflected, and refracted light both have the same frequency, as the incident frequency., Reason Speed of light and wavelength of light, does not changes in refraction and hence the ratio, n = c/ l is not a constant., , 23. Assertion Propagation of light through an optical, fibre is due to total internal reflection taking place, at the core-clade interface., Reason Refractive index of the material of the, core of the optical fibre is greater than that of, cladding., , (a) goes from rarer medium to denser medium, (b) incident at an angle less than the critical angle, (c) strikes the interface normally, (d) incident at an angle greater than the critical angle, , (iii) If in core, incidence angle is equal to critical angle,, then refraction angle will be, (a) 0°, (c) 90°, , (iv) In an optical fibre (shown), correct relation of, refractive indices of core and cladding is, n2, , 24. Assertion The property of convergent lens of, , Cladding, , (a) n1 = n 2, (c) n1 < n 2, , primary rainbow., , (a) 3 ´ 108 m/s, (c) 6 ´108 m/s, , Reason Secondary rainbow is formed by a four step, process and hence, the intensity of light is reduced at, the second reflection inside the rain drop., , Direction Read the following passage and answer the, questions that follows, , 26. Optical Fibre, , Light, ray, , Core, Cladding, , These fibres are fabricated in such a way that, light, reflected at one side of the inner surface strikes the, other at an angle larger than critical angle. Even, if, fibre is bent, light can easily travel along the length., (i) Which of the following is based on the, phenomenon of total internal reflection of light?, (a) Sparkling of diamond, (b) Optical fibre communication, (c) Instrument used by doctors for endoscopy, (d) All of the above, , (b)1.5 ´108 m/s, (d) 4.5 ´108 m/s, , PART 2, Subjective Questions, l, , An optical fibre is a thin tube of transparent, material that allows light to pass through, without, being refracted into the air or another external, medium. It makes use of total internal reflection., , (b) n1 > n 2, (d) n1 + n 2 = 2, , (v) If the value of critical angle is 30° for total internal, reflection from given optical fibre, then speed of, light in that fibre, , 25. Assertion Secondary rainbow is fainter than, , Case Based MCQs, , n1, , Core, , converging rays remain same in all media., Reason Property of lens, whether the rays are, diverging or converging does not depends on the, surrounding medium., , l, , (b) 45°, (d) 180°, , Short Answer (SA) Type Questions, 1. When monochromatic light travels from a rarer to a, denser medium, explain the following, giving, reasons., (i) Is the frequency of reflected and refracted light, same as the frequency of incident light?, (ii) Does the decrease in speed imply a reduction in, the energy carried by light wave?, [Delhi 2013], , 2. Mention any two situations in which Snell’s law of, refraction fails., , 3. A ray of light is incident at an angle of 45° on one, face of a rectangular glass slab of thickness 10 cm, and refractive index 1.5. Calculate the lateral shift, produced., , 4. Why does the sun rising in the sky appear oval in, shape?, , 5. Show analytically from the lens equation that when, the object is at the principal focus, the image is, formed at infinity.

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29, , CBSE Term II Physics XII, , 6. A student measures the focal length of a convex, lens by putting an object pin at a distance u from, the lens and measuring the distance v of the, image pin. What will be the graph drawn between, u and v ?, , From which face will the ray emerge? Justify your, answer., [All India 2016], A, , P, Q, , 7. A magician during a show makes a glass lens, n = 1.47 disappear in a trough of liquid. What is the, refractive index of the liquid? Could the liquid be, water?, , 8. A tank is filled with water to a height of 12.5 cm., The apparent depth of a needle lying at the bottom, of the tank is measured by a microscope to be, 9.4 cm. What is the refractive index of water? If, water is replaced by a liquid of refractive index 1.63, upto the same height, by what distance would the, microscope have to be moved to focus on the needle, again?, [NCERT], , 9. What should be the position of the object relative, to the biconvex lens, so that this lens behaves like a, magnifying glass?, , 10. How does the magnification of a magnifying glass, differ from its magnifying power?, , B, , 60°, , C, , 16. An equilateral glass prism has a refractive index 1.6, in air. Calculate the angle of minimum deviation of, the prism, when kept in a medium of refractive, 4 2, ., index, [Delhi 2019], 5, , 17. Is it possible to increase the range of a telescope by, increasing the diameter of the objective lens?, , 18. Explain two advantages of a reflecting telescope, over a refracting telescope., , 19. A small telescope has an objective lens of focal, length 144 cm and an eyepiece of focal length 6 cm., What is the magnifying power of the telescope?, What is the separation between the objective and, the eyepiece?, [NCERT], , 20. The objective of an astronomical telescope has a, , 11. Calculate the radius of curvature of an, equi-concave lens of refractive index 1.5, when it is, kept in a medium of refractive index 1.4, to have a, power of -5D ?, [Delhi 2019], , 12. An equi-convex lens of focal length f is cut into, two equal halves in thickness. What is the focal, length of each half ?, , 13. What is the focal length of a convex lens of focal, length 30 cm in contact with a concave lens of focal, length 20 cm? Is the system a converging or a, diverging lens? Ignore thickness of the lenses., [NCERT], , 14. The figure shows a ray of light falling normally on, the face AB of an equilateral glass prism having, refractive index 3/2, placed in water of refractive, index 4/3. Will this ray suffer total internal, reflection on striking the face AC? Justify your, answer., [CBSE 2018], , diameter of 150 mm and a focal length of 4 m. The, eyepiece has a focal length of 25 mm. Calculate the, magnifying power of telescope (l = 6000 Å for, yellow colour)., [Delhi 2011C], , 21. Define power of a lens. Write its units. Deduce the, 1 1, 1, for two thin lenses kept, = +, f f1 f 2, in contact co-axially., [Foreign 2012], relation, , 22. A symmetric biconvex lens of radius of curvature R, and made of glass of refractive index 1.5, is placed, on a layer of liquid placed on the top of a plane, mirror as shown in the figure below., An optical needle with its tip on the principal axis, of the lens is moved along the axis until its real and, inverted image coincides with the needle itself., The distance of the needle from the lens is, measured to be x., , A, , B, , C, , 15. A ray PQ incident normally on the refracting face, BA is refracted in the prism BAC made of material, of refractive index 1.5. Complete the path of ray, through the prism., , On removing the liquid layer and repeating the, experiment, the distance is found to be y. Obtain, the expression for the refractive index of the liquid, in terms of x and y., [CBSE 2018]

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30, , CBSE Term II Physics XII, , view the moon, find the diameter of the image of the, moon formed by the objective lens. The diameter of, the moon is 3.48 ´ 10 6 m and the radius of lunar, [Delhi 2019], orbit is 38, . ´ 10 8 m., , 23. State the conditions of total internal reflection., Refractive indices of the given prism material for, red, blue and green colours are 1.39, 1.48 and 1.42,, respectively. Trace the path of rays through the, prism., [All India 2019], A, , l, , Long Answer (LA) Type Questions, 28. (i) A point object O is kept in a medium of refractive, , G, B, R, 45º, , B, , C, , `, , 24. A ray of light incident on the face AB of an isosceles, triangular prism makes an angle of incidence i and, deviates by angle b as shown in the figure. Show, that in the position of minimum deviation Ðb = Ða., Also find out the condition, when the refracted ray, [All India 2019], QR suffer total internal reflection., , index n1 infront of a convex spherical surface of, radius of curvature R which separates the second, medium of refractive index n 2 from the first one,, as shown in the figure., Draw the ray diagram showing the image, formation and deduce the relationship between, the object distance and the image distance in, terms of n1 , n 2 and R., , O, , n1, , n2, , u, , R, , C, , A, , i, , Q b, , (ii) When the image formed above acts as a virtual, object for a concave spherical surface separating, the medium n 2 from n1 ( n 2 > n1 ), draw this ray, diagram and write the similar [similar to (i)], relation. Hence, obtain the expression for the, lens Maker’s formula., [All India 2015], , R, , P, , a, , a, C, , B, , 25. (i) A ray of light incident of face AB of an equilateral, glass prism, shows minimum deviation of 30°., Calculate the speed of light through the prism., A, , B, , C, , (ii) Find the angle of incidence at face AB, so that, the emergent ray grazes along the face AC., [Delhi 2017], , 26. An optical instrument uses an objective lens of, power 100 D and an eyepiece of power 40 D. The, final image is formed at infinity when the tube, length of the instrument is kept at 20 cm., (i) Identify the optical instrument., (ii) Calculate the angular magnification produced by, the instrument., [Delhi 2020], , 27. Draw a labelled ray diagram of an astronomical, telescope in the near point adjustment position., A giant refracting telescope at an observatory has an, objective lens of focal length 15 m and an eyepiece, of focal length 1.0 cm. If this telescope is used to, , 29. (i) Define the term focal length of a mirror. With, the help of a ray diagram, obtain the relation, between its focal length and radius of curvature., (ii) Calculate the angle of emergence ( e) of the ray of, light incident normally on the face AC of a glass, prism ABC of refractive index 3. How will the, angle of emergence change qualitatively, if the, ray of light emerges from the prism into a liquid, of refractive index 1.3 instead of air? [Delhi 2020], , 30. (i) Under what conditions is the phenomenon of, total internal reflection of light observed? Obtain, the relation between the critical angle of, incidence and the refractive index of the, medium., (ii) Three lenses of focal lengths +10 cm, -10 cm and, +30 cm are arranged co-axially as in the figure, given below. Find the position of the final image, formed by the combination., [All India 2019], +10 cm –10 cm, , +30 cm, , O, 30 cm, 5 cm, , 10 cm, ]

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31, , CBSE Term II Physics XII, , The focal lengths of the objective and eyepiece of a, microscope are 1.25 cm and 5 cm, respectively., Find the position of the object relative to the, objective in order to obtain an angular, magnification of 30 in normal adjustment., , 31. (i) Two thin lenses are placed co-axially in contact., Obtain the expression for the focal length of this, combination in terms of the focal lengths of the, two lenses., (ii) A converging lens of refractive index 1.5 has a, power of 10 D. When it is completely immersed, in a liquid, it behaves as a diverging lens of focal, length 50 cm. Find the refractive index of the, liquid., [All India 2020], , [Delhi 2012], , 36. (i) Draw a labelled ray diagram showing the image, formation of a distant object by refracting, telescope., Deduce the expression for its magnifying power, when the final image is formed at infinity., (ii) The sum of focal lengths of the two lenses of a, refracting telescope is 105 cm. The focal length, of one lens is 20 times that of the other., Determine the total magnification of the, telescope when the final image is formed at, infinity., [All India 2014], , 32. (i) A ray PQ of light is incident on the face AB of a, glass prism ABC (as shown in the figure) and, emerges out of the face AC. Trace the path of the, ray. Show that,, i + e= A +d, A, , i, , Q, l, , P, B, , C, , 34. Refraction of Light, , where, d and e denote the angle of deviation and, angle of emergence, respectively., Plot a graph showing the variation of the angle of, deviation as a function of angle of incidence., State the condition under which Ðd is minimum., (ii) Find out the relation between the refractive, index m of the glass prism and ÐA for the case,, when the angle of prism A is equal to the angle, of minimum deviation d m . Hence, obtain the, value of the refractive index for angle of prism, [Delhi 2015], A = 60°., , 33. Define magnifying power of a telescope. Write its, expression. A small telescope has an objective lens, of focal length 150 cm and an eyepiece of focal, length 5 cm. If this telescope is used to view a, 100 m high tower 3 km away, find the height of the, final image, when it is formed 25 cm away from the, eyepiece., [Delhi 2012], , 34. Draw a ray diagram to show the working of a, compound microscope. Deduce an expression for, the total magnification, when the final image is, formed at the near point., In a compound microscope, an object is placed at a, distance of 1.5 cm from the objective of focal length, 1.25 cm. If the eyepiece has a focal length of 5 cm, and the final image is formed at the near point., Estimate the magnifying power of the microscope., [Delhi 2010], , 35. How is the working of a telescope different from, that of a microscope?, , Case Based Questions, Refraction involves change in the path of light due, to change in the medium., Incident ray, , Normal, Reflected, ray, i, , i, , Reflecting, surface r, , Refracted ray, , When a beam of light encounters another, transparent medium, a part of light gets reflected, back into the first medium, while the rest enters, the other. The direction of propagation of an, obliquely incident ray of light, that enters the other, medium, changes at the interface of two media., This phenomenon is called refraction of light., (i) For the same value of angle of incidence, the, angles of refraction in three media A, B and C are, 15°, 25° and 35°, respectively. In which medium,, would the velocity of light be minimum?, [All India 2012], , (ii) Why does a crack in a glass window pane appear, silvery?, (iii) The refractive index of diamond is much higher, than that of glass. How does a diamond cutter, make use of this fact?, [All India 2011], (iv) What is the apparent position of an object below, a rectangular block of glass 6 cm thick, if a layer, of water 4 cm thick is on the top of the glass?, [Delhi 2015 C], (Take, n ga = 15, . and n wa = 1.33)

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Chapter Test, Multiple Choice Questions, , 8. A beam of light converges at a point P. Now, a lens is, , 1. Two convex and concave lens are in contact and, having focal lengths 12 cm and 18 cm, respectively., Focal length of joint lens will be, (a) 50 cm, , (b) 45 cm, , (c) 36 cm, , (d) 18 cm, , placed in the path of the convergent beam 12 cm from, P. At what point does the beam converge, if the lens, is (i) a convex lens of focal length 20 cm and, (ii) a concave lens of focal length 16 cm?, [Ans. (i) 7.5 cm and (ii) 48 cm], , 2. Two lenses are kept in contact with powers + 2 D and, - 4 D. The focal length of this combination will be, (a) + 50 cm, (c) - 25 cm, , of the surface of water through which light from the, bulb can emerge out? Refractive index of water is 1.33., (Consider the bulb to be a point source) (Ans. 2.58 m 2 ), , 9. A ray of light PQ enters an isosceles right angled prism, ABC of refractive index 1.5 as shown in figure., A, , (b) - 50 cm, Q, , (d) + 25 cm, , 90º, , 3. A thin lens of glass (m = 1 . 5) of focal length ± 10 cm is, immersed in water (m = 1 . 33 ). The new focal length is, (a) 20 cm, , 45º, B, , (b) 40 cm, , (c) 48 cm, , (d) 12 cm, , 4. A plot of angle of deviation D versus angle of incidence, , Angle of deviation, , i for a triangular prism is shown below., The angle of incidence for which the light ray travels, parallel to the base is, , (i) Trace the path of the ray through the prism., (ii) What will be the effect on the path of the ray, if the, refractive index of the prism is 1.4?, , 10. Which two of the following lenses L1, L2 and L3 will you, select as objective and eyepiece for constructing best, possible (i) telescope and (ii) microscope? Give reason, to support your answer., , 50°, , 40°, 0, , C, P, , 30° 45° 60°, Angle of incidence, , (a) 30°, (b) 60°, , Lens, , Power (P), , Aperture (A), , L1, , 6D, , 1 cm, , L2, , 3D, , 8 cm, , L3, , 10D, , 1 cm, , Long Answer Type Questions, , (c) 45°, , 11. (i) When a convex lens of focal length 30 cm is in, , (d) Data insufficient, , 5. An equilateral prism is in condition of minimum, deviation. If incidence angle is 4/5 times of prism, angle, then minimum deviation angle is, (a) 72°, , (b) 60 °, , (c) 48 °, , (d) 36 °, , Short Answer Type Questions, , 6. You are given two converging lenses of focal lengths, 1.25 cm and 5 cm to design to compound microscope., If it is desired to have a magnification of 30, find out, the separation between the objective and the eyepiece., , contact with a concave lens of focal length 20 cm,, find out if the system is converging or diverging., (ii) Obtain the expression for the angle of incidence of, a ray of light which is incident on the face of a, prism of refracting angle A, so that it suffers total, internal reflection at the other face. (Given, the, refractive index of the glass of the prism is m)., , 12. An angular magnification (magnifying power) of 30 is, desired using an objective of focal length 1.25 cm and, an eyepiece of focal length 5 cm. How will you, set up the compound microscope?, (Ans. 11.67 cm), , (Ans. 11.67 cm), , 7. A small bulb is placed at the bottom of a tank, containing water to a depth of 80 cm. What is the area, , Answers, Multiple Choice Questions, 1. (c), , 2. (b), , 3. (b), , 4. (c), , 5. (d), , For Detailed Solutions, Scan the code

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33, , CBSE Term II Physics XII, , EXPLANATIONS, PART 1, 1. (d) From Snell’s law of refraction,, sin i, a, …(i), mg =, = constant, sin r, Since, angle of incidence increase, so the angle of refraction, æ sin i ö, has to increase. Hence, the ratio ç, ÷ is a constant, è sin r ø, according to Eq. (i)., 4, 5, 2. (c) Given, a n w = , a n g =, 3, 3, \ a n w ´ wn g = a n g, 5/ 3 5 3 5, a ng, =, = ´ =, w ng =, n, 4/ 3 3 4 4, a w, 3. (b) The given ray of light is as shown below, 90°, , Air, r, , r, , Water, Glass, , i, , For glass-water interface, applying Snell’s law,, sin i n w, =, sin r n g, n w sin r, …(i), sin i, For water-air interface, angle of incidence in water = r, sin r, n, 1, Again,, = a =, sin 90° n w n w, 1, …(ii), Þ, sin r =, nw, From Eqs. (i) and (ii), we get, æ1 ö, ( n w ) ´ çç ÷÷, è nw ø Þ n = 1, ng =, g, sin i, sin i, n2, 4. (c) We know that, sin q =, n1, c, As, refractive index of a medium, n = , where c and v are, v, the speed of light in vacuum and medium, respectively., v, Þ, sin q =, v¢, Here, v is speed in medium A and v¢ is speed in medium B., v, Þ, v¢ =, sin q, Þ, , ng =, , 5. (b) The phenomenon involved in the reflection of, radiowaves by ionosphere is similar to total internal, reflection of light in air during a mirage, i.e. angle of, incidence is greater than critical angle., , 6. (a) The refraction formula for curved surface,, n 2 n1 n 2 - n1, =, v, u, R, 7. (b) When medium is equal on both sides of lens, then the, numerical value of both focal lengths is equal, hence, f 2 = - f1 ., 8. (a) Since, lens is made of two layers of different refractive, indices, for a given wavelength of light it will have two, different focal lengths or will have two images at two, 1, different points as, µ ( m - 1 ) (from lens Maker’s formula)., f, 9. (c) Given, R = 20 cm, and m = 1.5, on substituting the values, R, 20, in f =, =, = 40 cm, of converging nature as, m - 1 1.5 - 1, f >0. Therefore, lens act as a convex lens irrespective of the, side on which the object lies., 10. (a) Using lens Maker’s formula,, 1, = ( n - 1), f, , æ1, 1 ö, çç ÷÷, è R1 R 2 ø, , …(i), , Given, R1 = 10 cm, R 2 = - 10 cm, u = - 5 cm and n = 1.33, Substituting the given values in Eq. (i), we get, 1, 1ö, æ1, = ( 1.33 - 1 ) ç + ÷, f, è 10 10 ø, 1, 2 0.33, = 0.33 ´ =, Þ f = 15.15 cm, f, 10, 5, 1 1 1, Now, from thin lens formula, = f v u, uf, - 5 ´ 15.15 - 75.75, or image distance, v =, =, =, u + f - 5 + 15.15, 1015, ., = - 7.46 cm, Since, v is negative, hence image will be formed on the same, side of the object., 11. (c) Focal length of the combination of two given lenses,, 1 1, 1, …(i), = +, f, f1, f2, where, f1 and f 2 are the focal lengths of plano-convex lens, and plano-concave lens, respectively., As for plano-convex lens, R1 = R and R 2 = ¥, 1, æ 1 1 ö n -1, So,, = ( n1 - 1 ) ç - ÷ = 1, f1, R, èR ¥ø, Similarly, for plano-concave lens,, R1 = - R and R 2 = ¥, 1, 1 ö -( n2 - 1 ), æ 1, So,, = ( n2 - 1 ) ç, - ÷=, f2, R, ¥, R, è, ø, 1, 1, Putting these values of and in Eq. (i), we get, f1, f2, 1 ( n1 - 1 ) ( n 2 - 1 ) ( n1 - 1 - n 2 + 1 ) n1 - n 2, =, =, =, f, R, R, R, R, R, f =, n1 - n 2

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34, , CBSE Term II Physics XII, , 12. (a) The statement given in option (a) is correct but rest are, incorrect and these can be corrected as,, Power is defined as the tangent of angle by which it, converges or diverges a beam of light falling at unit distance, from the optical centre., Combination of lenses helps to obtain diverging or, converging lenses of desired magnification. It also enhances, sharpness of the image., Also, total magnification m of the combination of lenses is a, product of magnification ( m1 , m 2 , m 3 ,.....) of individual, lenses, i.e. m = m1 m 2 m 3 … ., 13. (b) If a symmetrical biconcave lens of focal length f (say) is, vertically splitted into two identical plano-concave parts (as, shown below), then focal length of each part will be 2 f ., f, , 2f, , 2f, , 17. (d) For least distance of distinct vision, the angular, magnification of simple microscope,, D, m =1 +, f, Þ, , æ, 1ö, çç\ Power, P = ÷÷, fø, è, , m = 1 + DP, , and for normal adjustment, m =, , D, f, , Þ, m = DP Þ m µ P, 18. (a) Given, f o = F1 and f e = F2, We know that, angular magnification for telescope,, f, F, F, | m| = o = 1 = 1, fe, F2, F2, | fo |, =5, | fe |, and length of the telescope,, L = | f o | + | f e | = 36, , 19. (d) For telescope,| m | =, , 1, focal length, 1, So, power of the biconcave lens, P =, f, , As we know, power of a lens =, , … (i), , Similarly, power of each part of plano-concave lens,, 1 1, [using Eq. (i)], P¢ =, = P, 2f 2, 14. (c) The given prism can be shown as below, A, 30º, i=0º, r1=0º, , B, , r2, , C, , Given, A = 30°, i = 0°, Þ, r1 = 0°, As, the angle of prism, A = r1 + r2, Þ 30° = 0 + r2 Þ r2 = 30°, 15. (b) Given, angle of prism = 5 °; n r = 1.641 , n b = 1.659, For this prism, the deviation, D = ( n - 1 ) A, So, for blue colour light, Db = ( n b - 1 ) A, Similarly, for red colour light, Dr = ( n r - 1 ) A, \Angle between the emergent blue and red rays, = Db - Dr = ( n b - n r ) A, = (1.659 - 1.641) ´ 5, = 0.018 ´ 5 °, = 0.09°, 16. (a) Given, incidence angle, i = 50°, Refractive angle, A = 60°, Minimum deviation, d = 2i - A = 2 ´ 50° - 60° = 40°, , …(i), , …(ii), , From Eqs. (i) and (ii), we get, Þ, f e = 6 cm, and, f o = 30 cm, 20. (c) Refractive index of any pair of media is inversely, proportional to wavelength of light., As,, , lv < lr, , Þ, nv > nr, where, l v and l r are the wavelengths of violet and red light,, respectively and n r and n v are refractive index of violet and, red light, respectively., Hence, refractive index of glass with respect to air is different, for red light and violet light., Therefore, A is true but R is false., 21. (d) Refractive index of diamond w.r.t. liquid,, 1, n, l, nd =, = d, sin ic n l, 6, 1, =, 3 sin ic, 1, Þ, sin ic =, = sin 45 °, 2, \, ic = 45 °, This means that, when the ray of light while travelling from, diamond to liquid is incident at angle of 45°, then it will be, totally reflected due total internal reflection., Therefore, A is false and R is also false., 22. (c) Reflection and refraction arise through interaction of, incident light with constituents of matter. When, monochromatic light is incident on a surface separating two, media, then the frequency of scattered light equals the, frequency of incident light., The speed of light and wavelength of light both changes in, c, refraction, then the frequency of light n = remains, l, constant., Therefore, A is true but R is false., Þ

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35, , CBSE Term II Physics XII, , 23. (b) In optical fibre communication, propagation of signal, through optical fibre takes place, which is based on the, phenomenon of total internal reflection at core-clade, interface., The refractive index of the material of the cladding is less, than that of core, hence light striking at core-cladding, interface gets totally internally reflected., Therefore, both A and R are true but R is not the correct, explanation of A., 24. (d) When a convergent lens is placed inside a transparent, medium having refractive index greater than that of material, of lens, it behaves as a divergent lens., Hence, property of a lens, whether the rays are converging, or diverging depends on the surrounding medium., Therefore, A is false and R is also false., 25. (a) Secondary rainbow is a result of four-step process, which, are as follows, (i) Refraction at the first surface of raindrop., (ii) Two total internal reflection from the second surface of, raindrop., (iii) Again refraction from the first surface of raindrop from, where the light finally emerges out., Thus, the intensity of light is reduced at the second, reflection and hence the secondary rainbow is fainter than, the primary rainbow., Therefore, both A and R are true and R is the correct, explanation of A., 26. (i) (d) Total internal reflection is the basis for following, phenomenon, (a) Sparkling of diamond., (b) Optical fibre communication., (c) Instrument used by doctors for endoscopy., (ii) (d) Total internal reflection (TIR) is the phenomenon that, involves the reflection of all the incident light off the, boundary. TIR only takes place, when both of the, following two conditions are met., The light is in the more denser medium and approaching, the less denser medium., The angle of incidence is greater than the so-called, critical angle., (iii) (c) If incidence angle, i = critical angle C, then refraction, angle, r = 90°., (iv) (b) In optical fibres, core is surrounded by cladding,, where the refractive index of the material of the core is, higher than that of cladding to bound the light rays inside, the core. i.e. n1 > n 2 ., v, (v) (b) From Snell’s law, sin C = 1 n 2 = 1, v2, where, C = critical angle = 30°, and v1 and v2 are speed of light in medium and vacuum,, respectively., We know that, v2 = 3 ´ 108 m/s, v1, \, sin 30° =, 3 ´ 108, 1, Þ, v1 = 3 ´ 108 ´, 2, Þ, v1 = 1 .5 ´ 108 m/s, , PART 2, 1. (i) The frequency of reflected and refracted light remains, same as that of incident light because frequency only, depends on the source of light., (ii) Since, the frequency remains same, hence there is no, reduction in energy., 2. Snell’s law of refraction fails in two situations, (i) When TIR (Total Internal Reflection) takes place at angle, greater than the critical angle., (ii) When light is incident normally on a surface, as i = 0,, r = 0., 3. Given, i1 = 45 °, t = 10 cm = 01, . m and m =1 .5, sin i1, By Snell’s law, m =, sin r1, sin i1 sin 45 °, Þ sin r1 =, =, m, 1 .5, æ\ sin 45 ° = 1 / 2 ö, 0.707, ç, ÷, Þ, sin r1 =, ç here, 2 = 1.414÷, 1 .5, è, ø, Þ, sin r1 = 0.4713, Þ, r1 = sin -1 (0.4713), Þ, r1 = 2812, . °, t sin ( i1 - r1 ) 0.1 sin ( 45 ° - 2812, . °), Lateral shift =, =, cos r1, cos 2812, . °, 0.1 sin 16.88° 01, . ´ 0.2904, =, =, cos 2812, . °, 0.8819, = 0.033 m, 4. It is due to the refraction of sunlight as it travels through, different layers of the earth’s atmosphere. Refraction of light, by these layers can make the sun appear flattened or, distorted. The rays of light from the upper part and lower, part of the periphery of the sun bend unequally on travelling, through earth’s atmosphere, making the sun appears oval in, shape., 5. Given, u = - f, 1 1 1, \From lens formula, - =, v u f, 1 1 1, 1, + =, Þ, =0, Þ, v, f, f, v, 1, Þ, v = = infinity, 0, 1 1 1, 6. As we know that, - =, v u f, \Graph between u and v is as shown below, v (in cm), , O, , u (in cm), , 7. If m1 = m 2 , then f = ¥, Hence, the lens in the liquid acts like a plane sheet, when, refractive index of the lens and the surrounding medium is, the same. Therefore, m1 = m 2 = 1.47.

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36, , CBSE Term II Physics XII, , Hence, the liquid medium is not water, refractive index for, water = 1.33., 8. Case I When tank is filled with the water., Given, the apparent depth = 9.4 cm, Height of water, t = 12.5 cm, So, real depth = 12 .5 cm, Refractive index of water,, Real depth, 12 .5, mw =, =, = 1.33, Apparent depth, 9.4, Case II When tank is filled with the liquid., Refractive index of liquid, m l = 1.63, Real depth, Again,, ml =, Apparent depth, 12.5, 1.63 =, Þ, Apparent depth, 12.5, Apparent depth =, = 7.67 cm, 1.63, \ The microscope is shifted by 9.4 – 7.67 = 1.73 cm., 9. Whenever object is placed within the focus of the biconvex, lens, we will obtain enlarged image, hence the biconvex lens, behaves like a magnifying lens., 10. The magnification of a magnifying glass depends upon,, where it is placed between the user’s eye and the object, being viewed and the total distance between them, while, the magnifying power is equivalent to angular magnification., 11. Given, m1 = 1 .4, m 2 = 1.5, P = - 5 D, Using lens Maker’s formula,, 1 æ m - m1 ö æ 1, 1 ö, ÷ç ÷, P = = çç 2, F è m1 ÷ø çè R1 R 2 ÷ø, æ 1.5 - 1.4 ö æ 1 1 ö, -5 = ç, ÷ ç- - ÷, è 1.4 ø è R R ø, (for equi-concave lens, R1 = - R and R 2 = R), 0.1 æ 2 ö, -5 =, ç- ÷, 1.4 è R ø, 1 2 1, Þ, R=, ´ =, = 0.0286 m = 2.86 cm, 14 5 35, 12. Focal length can be given as,, æ1, 1, 1 ö, = ( m - 1 ) çç - ÷÷, f, è R1 R 2 ø, where, m is the refractive index of the lens medium and, R1 & R 2 are radii of curvature., Focal length = f, , 1, 2 (m - 1 ), =, f¢, R, \, f¢ = 2f, Hence, focal length of each half becomes twice of the, original value., 13. Given, focal length of convex lens, f1 = 30 cm, Þ, , Focal length of concave lens, f 2 = - 20 cm, Using the formula of combination of lenses,, 1 1, 1, 1, 1, = +, =, f, f1, f 2 30 20, 2-3, 1, =, =60, 60, Þ, f = - 60 cm, Since, the focal length of combination is negative in nature., So, the combination behaves like a diverging lens, i.e. as a, concave lens., 14. Given, refractive index of water, m w = 4 / 3, 3, Refractive index of glass prism, m g =, 2, A, 60°, 30°, i=60°, B, , C, , For total internal reflection occurrence the incident angle, must be greater than critical angle., \ Let us calculate critical angle C., 1, As we know that, sinC =, m, refractive index of glass ( a m g ), where, m =, refractive index of water ( a m w ), 1, 1, 1, \ sin C =, =, =, æ a m g ö æ 3/ 2 ö 9/ 8, çç, ÷÷ çç, ÷÷, è a m w ø è 4/ 3 ø, 8, = 0.88, 9, C = 61 . 6°, , or sin C =, Þ, , (as, sin 60° = 3 / 2 = 0.86), , As the critical angle, i.e. 61 . 6° is greater than the angle of, incidence, i.e. 60°, hence TIR will not occurs., 15. Given, refractive index of the material of the prism, m = 1 .5, A, P, , 30°, Q, 60°, 30°, , Equi-convex lens have the same radius of curvature, i.e., R1 = - R 2, 1, é 1 æ 1 öù, \, = (m - 1 ) ê - ç - ÷ ú, f¢, ë R è R øû, , m =1.5, B, , 60°, , C

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37, , CBSE Term II Physics XII, , \Critical angle for the material,, 1, 1, sin C = =, = 2/ 3, m 1 .5, Þ, , æ 2ö, C = sin -1 ç ÷ ~, - 42°., è 3ø, , From the ray diagram, it is clear that angle of incidence, i = 30° < C., Therefore, the ray incident at the face AC will not suffer, total internal reflection and merges out through this face., 16. Given, A = 60° (for equilateral prism), m1 =, , 4 2, , m 2 = 1.6, 5, , The refractive index is given by, æ A + Dö, sin ç, ÷, m2, è 2 ø, =, m1, æ Aö, sin ç ÷, è 2ø, where, D = angle of minimum deviation., æ 60° + D ö, sin ç, ÷, 1.6 ´ 5, 2 ø, è, =, æ 60° ö, 4 2, sin ç, ÷, è 2 ø, æ 60° + D ö, 2 ´ sin 30° = sin ç, ÷, 2 ø, è, 1, æ 60° + D ö, Þ, = sin ç, ÷, 2 ø, 2, è, æ 60° + D ö, sin 45 ° = sin ç, Þ, ÷, 2 ø, è, 60° + D, Þ, 45 ° =, 2, D = 90° - 60° = 30°, 17. By increasing the diameter of the objective lens, we can, increase the range of the telescope because as the diameter, of lens increases, the area covered by the lens also increases., i.e. Lens is able to focus on a large area thereby helping us, to view the object better., 18. Advantages of reflecting telescope over refracting telescope, are as follows, (i) In reflecting telescope, image formed is free from, chromatic aberration defect. So, it is sharper than image, formed by a refracting type telescope., (ii) A mirror is easier to produce with a large diameter, so, that it can intercept rays crossing a large area and direct, them to the eyepiece., 19. Given, focal length of objective lens, f o = 144 cm, Focal length of eyepiece, f e = 6 cm, Magnifying power of the telescope in normal adjustment, (i.e. when the final image is formed at ¥),, f, 144, m = - o == - 24, fe, 6, \ Separation between lenses,, L = f o + f e = 144 + 6 = 150 cm, , 20. The diameter of objective of the telescope = 150 ´ 10-3 m, f o = 4 m, f e = 25 ´ 10-3 m, and, D = 25 mm = 0.25 m, f æ, Dö, ÷, Magnifying power, m = - o çç1 +, fe è, f e ÷ø, 4, 0.25 ö, æ, ç1 +, ÷, 25 ´ 10-3 è, 25 ´ 10-3 ø, = - 1760, 1.22l 1.22 ´ 6 ´ 10-7, Now,, dq =, =, D, 0.25, = 2.9 ´10- 6 rad, 21. The power of a lens is equal to the reciprocal of its focal, length, when it is measured in metre. Power of a lens,, P = 1 / f (in m) and its SI unit is dioptre (D)., =-, , A B, , P, , O, , I, , I1, , v, u, , v1, , Consider two lenses A and B of focal lengths, f1 and f 2 placed, in contact with each other. An object is placed at a point O, beyond the focus of the first lens A., The first lens produces an image (real image) at I1 , which, serves as a virtual object for the second lens B producing the, final image at I., Since, the lenses are thin, we assume the optical centres P of, the lenses to be coincident. For the image formed by the first, lens A, we obtain, 1 1 1, …(i), - =, v1 u f1, For the image formed by the second lens B, we get, 1 1, 1, - =, v v1, f2, , …(ii), , Adding Eqs. (i) and (ii), we obtain, 1 1 1, 1, - = +, v u f1, f2, , …(iii), , If the two lenses system is regarded as equivalent to a single, lens of focal length f , then we have, 1 1 1, …(iv), - =, v u f, From Eqs. (iii) and (iv), we get, 1, 1, 1, +, =, f1, f2, f, 22. First measurement gives the focal length ( feq = x ) of, combination of the convex lens and the plano-convex liquid, lens. Second measurement gives the focal length ( f1 = y) of, the convex lens., Focal length ( f 2 ) of plano-convex lens is given by, 1, 1, 1 1 1, =, - = f2, feq, f1 x y

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38, , CBSE Term II Physics XII, , Þ, , f2 =, , xy, y-x, , …(i), , For equi-convex glass lens using lens Maker’s formula, we get, æ1, 1, 1 ö, ÷÷, = ( n g - 1 ) çç f1, è R1 R 2 ø, 1, æ 2ö, = (1.5 - 1 ) ç ÷ (as, R1 = R and R 2 = - R), y, èRø, 1 1 2, = ´ Þ R=y, y 2 R, , Þ, , P¢, , P, Q, , Q¢, , Now, we apply lens Maker’s formula for plano-convex lens., Here, R1 = R, R 2 = ¥ and let n l = refractive index of liquid., 1, 1ö, æ1, = ( nl - 1 ) ç - ÷, f2, èR ¥ø, 1, æ1ö, Þ, = ( nl - 1 ) ç ÷, f2, èRø, R, y, Þ, nl = 1 +, =1 +, f2, æ xy ö, çç, ÷÷, è y - xø, y-x y, =1 +, =, x, x, 23. There are two conditions for total internal reflection as, follows, (i) Light must travel from denser to rarer medium., (ii) Angle i > ic, Given, m red = 1.39, m blue = 1.48 and m green = 1.42, A, G, i=45º, i=45º, R, 45º, 45º, , Q, , sin iC =, , C, , 1, m, , æ 1 ö, ( iC ) red = sin -1 ç, ÷ = 46°, è 1.39 ø, æ 1 ö, ( iC ) green = sin -1 ç, ÷ = 44.8°, è 1.42 ø, æ 1 ö, ( iC ) blue = sin -1 ç, ÷ = 43°, è 1.48 ø, Q Angle of incidence at face AC is 45° which is more than, the critical angle for blue and green colours, therefore blue, and green colours will undergo total internal reflection but, red colour will refract to other medium., \, , r1 = 90° - b and r2 = b - 30°, For minimum deviation, r1 = r2, Þ, 90° - b = b - 30°, Þ, 2 b = 120° or b = 60° = a, 1, For total internal reflection,, £m, sin iC, 1, £m, sin 30°, , (Q r2 = iC = 30° ), , Þ, m2 ³ 2, 25. (i) Given, angle of minimum deviation, dm = 30°, \ Angle of prism, A = 60°, By prism formula, reflected index, d + A, 30° + 60°, sin m, sin, sin 45 °, 2, 2, m =, =, =, sin A / 2, sin 30°, sin 30°, 1, =, ´2 = 2, 2, speed of light in vacuum (c), Also, m =, speed of light in prism (v), Þ, , v = c / m = ( 3 ´ 108 / 2 ) m/s, , Hence, speed of light through prism is ( 3 ´ 108 / 2 ) m / s., (ii) As the emergent ray grazes along the face AC, e = 90°. At, the interface AC, using Snell’s law,, sin i, =m, sin e, sin i/sin e = 2, Þ, sin i = 2 sin e = 2 ´ sin 90°, i = sin -1 ( 2 ), 26. Given, power of objective lens, Po = 100 D, Power of eyepiece, Pe = 40 D, As we know, power of a lens, 1, 100, =, =, focal length of lens (in m) f (in cm), Focal length of objective lens,, 1 100, fo =, =, = 1 cm, Po 100, , B, , B, , 24. Given, a = 60° (for isosceles triangle), , Similarly, focal length of eyepiece,, 1 100, fe =, =, = 2.5 cm, Pe, 40, (i) Since, the focal length of eyepiece is more than the focal, length of objective. So, the optical instrument is, compound microscope., (ii) Since, the final image is formed at infinity, so the angular, magnification is given as, L D, m=fo fe, where, L is the tube length of the instrument = 20 cm, (given) and D is the least distance of distinct vision, = 25 cm., Substituting the values in the above equation, we get, - 20 ´ 25, m=, = - 200, 1 ´ 25, .

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39, , CBSE Term II Physics XII, , For DNOC, Ði is the exterior angle., , 27. The ray diagram of an astronomical telescope in the near, point adjustment position is as shown below, fo, Parallel rays from, object at infinity, a, , \, , Ði = ÐNOM + ÐNCM, MN MN, For small angles, i =, +, OM, NC, , fe, fe f o b, , B¢¢, , B¢, , C1, , r = ÐNCM - ÐNIM, MN MN, r=, NC, NI, , Similarly,, C2, , A¢, , Eye, , Þ, , ...(i), , ...(ii), , By Snell's law, we get, A¢¢, , n1 sin i = n 2 sin r, , D, , Given, diameter of the moon, do = 3.48 ´ 10 m, Radius of lunar orbit, r = 3.8 ´ 108 m, Focal length of objective lens, f o = 15 m, The diameter of the image of moon formed by the objective, lens is given by, d, dI = o ´ f o, r, 3.48 ´ 106, =, ´15, 3.8 ´ 108, = 01373, ., m, = 13.74 cm, 28. (i) Let a spherical surface separate a rarer medium of, refractive index n1 from the second medium of refractive, index n 2 . Let C be the centre of curvature and R = MC be, the radius of the surface., Consider a point object O lying on the principal axis of the, surface. Let a ray starting from O incident normally on the, surface along OM and pass straight. Let another ray of light, incident on NM along ON and refract along NI. From M ,, draw MN perpendicular to OI., 6, , For small angles, n1 i = n 2 r, Putting the values of iand r from Eqs. (i) and (ii), we get, æ MN MN ö, æ MN MN ö, n1 ç, +, ÷ = n2 ç, ÷, MI ø, è OM MC ø, è MC, Þ, , n1, n, n - n1, + 2 = 2, OM MI, MC, , Applying new cartesian sign conventions, we get, OM = - u, MI = + v and MC = + R, Substituting this in Eq. (iii), we get, n 2 n 2 n 2 - n1, ...(iv), =, v, u, R, (ii) Now, the image I¢ acts as a virtual object for the second, surface that will form a real at I. As, refraction takes place, from denser to rarer medium,, r¢, n2, , n1, , c¢, R¢, , M¢ I¢, , A, , v, Denser n2, , i, r, , O, , P M, , \, C, , - n 2 n1 n 2 - n1, + ¢ =, v, v, R1, , ...(v), , I, , On adding Eqs. (iv) and (v), we get, 1 1, 1 ö, æ1, - = ( n 2 - n1 ) ç - ÷, v¢ u, è R R¢ø, , R, B, u, , I, , v¢, , N, , Rarer n1, , ...(iii), , v, , The above figure shows the geometry of the formation of, image I of an object O and the principal axis of a spherical, surface with centre of curvature C and radius of curvature R., Here, we have to make following assumptions, (a) the aperture of the surface is small as compared to the, other distance involved., (b) NM will be taken as nearly equal to the length of the, perpendicular from the point N on the principal axis., MN, tan ÐNOM =, OM, MN, tan ÐNCM =, MC, MN, tan ÐNIM =, MI, , ö æ1, 1 1 æ n2, 1 ö, - = ç - 1 ÷÷ ç - ÷, ¢ø, v u çè n1, R, R, è, ø, 1, 1 ö, æ1, = ( n 21 - 1 ) ç - ¢ ÷, f, èR R ø, æ, n 1 1 1ö, ççQ n 21 = 2 , = - ÷÷, n1 f v u ø, è, 29. (i) Focal length The distance of the principal focus from, the pole of the mirror is called the focal length of the, mirror., Relation between focal length and radius of curvature, of mirror, Consider a ray parallel to the principal axis striking the, mirror at point M, then CM will be perpendicular to the, mirror at point M.

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40, , CBSE Term II Physics XII, , Let q be the angle of incidence and MD be perpendicular, to the principal axis., M, , q, q, q, , 2q, , C, , P, , D, , F, , (a), q, , q, , 30. (i) Following are the criteria for total internal reflection, (a) Light must pass from a optically denser to a optically, rarer medium., (b) Angle of incidence in denser medium is must be greater, than critical angle for two media., From Snell’s law, m 2 sin iC = m1 sin 90°, m2, 1, =, m1 sin iC, 1, (where, iC is the critical angle), 1m 2 =, sin iC, , M, , iC, , 2q, P, , D, , Denser (m2), , q, C, , F, , 90°, , (b), , Then, ÐMCP = q and ÐMFP = 2q, MD, MD, Now, tan q =, and tan 2q =, …(i), CD, FD, For small q (condition true for paraxial rays),, tan q » q and tan 2q » 2q, Therefore, from Eq.(i), we get, MD, MD, CD, or FD =, …(ii), =2, FD, CD, 2, Again, for small q, we can observe that the point D is very, close to the point P. Therefore, FD = f and CD = R., From Eq.(ii), we have, R, f =, 2, (ii) Given, refractive index of the prism ABC, m = 3, A, 30°, 60°, e, , 30°, , 60°, B, , C, , Applying Snell’s law at interface AB., sin i 1, 1, = =, sin e m, 3, Þ, , 3 sin 30° = sin e, 1, Þ, 3 ´ = sin e or e = 60°, 2, Now, if the ray of light emerges from prism in a liquid of, refractive index 1.3, then, , (ii), , 1 1 1, - =, v u f, , O, , Þ, or, , 3, = 1.3sin e, 2, sin e = 0.666, e = sin -1 ( 0.666) = 41.76°, , +10 cm –10 cm +30 cm, , Final, , 30 cm 5 cm, , 10 cm, , 30 cm, , Here, for first lens,, u = - 30 cm and f = + 10 cm, 1 1 1 1, 1, = + =, v1, f, u 10 30, 1, 2, =, v1 30, Þ, v1 = 15 cm, For second lens, u = + 10 cm and f = - 10 cm, 1, 1 1 1, 1, = + =, =¥, v2, f, u 10 10, Thus, for last lens the object is at infinity, hence the image, formed at the focus of the lens, which is at a distance, of 30 cm., 1, 31. (i) Power of first lens, P1 =, f1, 1, Power of second lens, P2 =, f2, Net power of lens combination,, Pnet = P1 + P2, , 3 sin 30° = 1.3sin e, Þ, , Rarer (m1), , Þ, , 1, 1, 1, = +, f net f1, f2, , Þ, , f net =, , f1 f 2, f1 + f 2

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41, , CBSE Term II Physics XII, , i - d graph is shown in the figure,, Deviation angle (d), , (ii) Given, m g = 1 .5 and P = 10 D, As we know, focal length of the lens, 1, =, power (in dioptre), 1, Þ, fa =, = 01, . m = 10 cm, 10, Using the lens Maker’s formula,, Focal length of the lens when it is in air,, æ1, 1, 1 ö, ÷÷, = ( m g - 1 ) çç fa, è R1 R 2 ø, , … (i), , and focal length of the lens when it is immersed in liquid,, öæ1, 1 æm g, 1 ö, ÷÷, … (ii), =ç, - 1 ÷÷ çç f l çè m l, R, R, øè 1, 2ø, Dividing Eq. (i) by Eq. (ii), we get, (m g - 1 ), fl, =, fa æ m g, ö, çç, - 1 ÷÷, m, è l, ø, Substituting the given values, we get, - 50 ( 1.5 - 1 ), Þ, =, (Q f l = - 50 cm), 10, æ 1.5, ö, çç, - 1 ÷÷, èml, ø, 1.5, - 05, ., -1 =, = - 01, ., ml, 5, 1.5 5, Þ, ml =, = = 1.67, 0.9 3, \Refractive index of liquid is 1.67., 32. (i) Let PQ and RS are incident and emergent rays and, incident ray get deviated by d by the prism., i.e., ÐTMS = d, Let d1 and d2 are deviation produced at refractions taking, place at AB and AC, respectively., A, , T, D, , M, , d, , Q, i, , e, S, , N, , P, , B, , \, , R, r2, , r1, , dm, Angle of incidence (i), , C, , d = d1 + d2 = ( i - r1 ) + ( e - r2 ), …(i), = ( i + e ) - ( r1 + r2 ), But in DFNR,, ÐQNR + ÐRQN + ÐQRN = 180°, or, …(ii), ÐQNR = 180° - ( r1 + r2 ), In, ~QARN, ÐAQN and ÐARN are right angles., So,, …(iii), ÐQNR = 180° - A, where, A is angle of prism., From Eqs. (ii) and (iii), we have, …(iv), A = r1 + r2, From Eqs. (i) and (iv), we have, …(v), d = ( i + e) - A, or, i+ e = A + d, , The conditions for the angle of minimum deviation are, given as below, (a) Angle of incidence i and angle of emergence e are, equal., i.e., Ði = Ðe, (b) In equilateral prism, the refracted ray is parallel to, base of prism., (c) The incident and emergent rays are bent on same, angle from refracting surfaces of the prism., i.e., Ðr1 = Ðr2, For minimum deviation position,, Putting r = r1 = r2 and i = e in Eq. (iv), A, …(vi), 2r = A Þ r =, 2, From Eq. (i), dm = 2 i - A, A + dm, …(vii), i=, 2, \Refractive index of material of prism,, sin i, m =, sin r, From Eqs. (vi) and (vii), we get, æ A + dm ö, sin ç, ÷, è 2 ø, Þ, m =, sin ( A/ 2), (ii) As per the question,, Angle of prism, A = Angle of minimum deviation, dm, i.e., …(viii), ÐA = Ðdm, Substituting the value of Ðdm from Eq. (viii) we get, æ A + Aö, sin ç, ÷, è 2 ø, m =, Þ, sin ( A / 2 ), sin A, Þ, m=, sin ( A / 2 ), 2sin ( A / 2 ) × cos ( A / 2 ), Þ, m =, sin ( A / 2 ), Þ, m = 2 cos( A / 2 ), This is the required relation between refractive index of, the glass prism and angle of prism., Since, ÐA = 60°, (given), æ 60° ö, Þ, m = 2 cos ç, ÷, è 2 ø, Þ, m = 2 cos 30°, 3, Þ, m =2´, 2, Þ, m = 3, Þ, , m = 1 .732

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42, , CBSE Term II Physics XII, , 33. Magnifying power of a telescope is the ratio of the angle b, subtended at the eye by the image to the angle a subtended, at the eye by the object., Objective lens, , fo, , Parall, from oel rays, infinity bject at, , a, , fo, fe, b, , B, , a, , C1, , fe, Eyepiece, , I, , C2, , Eye, , ge, , a, im, al nity, Fin infi, at, , The objective lens forms real and inverted magnified image, A ¢ B ¢ of object AB in such a way that AB¢ fall some, where, between pole and focus of eye lens., So, A ¢ B ¢ acts as an object for eyepiece and its virtual magnified, image A ¢¢ B ¢¢ formed by the lens., The magnifying power of a compound microscope is defined, as the ratio of the visual angle subtended by final image at, eye ( b ) and the visual angle subtended by object at naked, eye, when both are at the least distance of distinct vision from, the eye., Visual angle with instrument ( b ), m Þ, Visual angle when object is placed at least, , The final image is magnified and inverted, , i.e., m = b/ a = f o / f e, For telescope, Focal length of objective lens, f o = 150 cm, Focal length of eyepiece, f e = 5 cm, , Þ, , æ B¢ A¢ ö D, =ç, ÷ ´ = mo me, è BA ø ue, , When final image forms at D = 25 cm,, \ Magnification, M = - f o / f e ( 1 + f e / D), 150 æ, 5 ö, 150 6, ´, ç1 +, ÷=5 è, 25 ø, 5, 5, M = - 36, Let height of final image is h cm., \, tan b = h / 25, b = visual angle formed by final image at eye, a = visual angle subtended by object at objective, 100 m, 1, tan a =, =, 3000 m 30, tan b, ( h / 25 ), But,, M=, Þ - 36 =, tan a, ( 1 / 30 ), h, 6h, Þ, -36 =, ´ 30 =, 25, 5, 36 ´ 5, h== - 30 cm, 6, Negative sign indicates inverted image., 34. A compound microscope consists of two convex lenses, co-axially separated by some distance. The lens nearer to, the object is called the objective. The lens through which, the final image is viewed is called the eyepiece. The focal, length of objective lens is smaller than eyepiece., =-, , Eyepiece, E, , Objective lens, A, , B, , Fo, , C1, , uo, , B¢¢, a, , Fo, , B¢, b, , Fe, , distance of distinct vision ( a ), b, tan b B ¢ A ¢ / ue, m= =, =, a tan a, BA / D, , C2, , Fo, , m = m o m e , where m o and m e are magnification produced by, objective and eyepiece, respectively., B¢ A¢, v, Now,, mo =, = o, BA, - uo, D, D, (by lens formula), me =, =1 +, ue, fe, æv ö æ, Dö, m = - çç o ÷÷ çç1 + ÷÷, fe ø, è uo ø è, This is the required expression., , \, , Also,, , uo = + 1.5 cm, , Þ, , f o = 1.25 cm, f e = 5 cm, ve = - D = - 25 cm, For objective lens,, 1, 1, 1, =, f o vo uo, , Þ, Þ, , 1, 1, 1, =, +, 1 .25 vo 1 . 5, 1, 1, 1, =, vo 1 .25 1 .5, =, , 1 .5 - 1 .25, 1 .5 ´ 1 .25, , =, , 0 .25, 1, =, 1 .5 ´ 1 .25 750, ., , vo = 7.5 cm, 7.5 æ, 25 ö, M=, ç1 +, ÷ = - 5 ´ 6 = - 30, 2.5 è, 5 ø, 35. Differences between telescope and microscope are as given, below, , A¢, u, , A¢¢, , D, , Compound microscope, final image at D, , Characteristics, , Telescope, , Microscope, , Position of object, , At infinity, , Near objective at a distance, lying between f o and 2 f o, , Position of image, , Focal plane, of objective, , Beyond 2 f o , where f o is the, focal length of objective.

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43, , CBSE Term II Physics XII, , When the final image is formed at infinity, angular, b, magnification is given by M = ., a, However, b and a are very small., \, b » tan b or a » tan a, tan b, Þ, M=, tan a, , For microscope, f o = 1.25 cm and f e = 5 cm, When final image forms at infinity, then magnification, produced by eyepiece is given by, L D, L, 25, M=- ×, Þ - 30 = ´, fo fe, 1 .25 5, 30 ´ 1 .25, Þ L = 7.50 cm, 5, For objective lens, vo = L = 7.5 cm, and, f o = 1.25 cm, Applying lens formula,, 1, 1, 1, =, f o vo uo, 1, 1, 1, Þ, =, 1 .25, 75, ., uo, 1, 1, 1, =, uo, 7 .5, 1 .25, 1 .25 - 75, ., 6.25, =, =7 .5 ´ 1 .25, 7 .5 ´ 1 .25, L=, , Þ, , uo = -, , I is the image formed by the objective, f o and f e are the, focal length of objective and eyepiece, respectively., I, tan a =, fo, I, or, tan b =, - fe, , 37. (i), , æ, Dö, çç1 + ÷÷, fe ø, è, , Paralle, l, from o rays, b, infinity ject at, , Fo Fe, B¢ b, , a, , e, ag, im, al inity, n, i, f, F in, at, L, , (ii), , (iii), , (iv), , fo, fe, , C1, , M = - fo / fe, , f o = 20 ´ 5 = 100 cm, , f o 100, =, = 20, fe, 5, sin i c, From Snell’s law, m =, =, sin r v, Þ v µ sin r, for given value of i., Smaller the angle of refraction, smaller the velocity of, light in medium., Velocity of light is minimum in medium A, as the angle of, refraction is minimum, i.e. 15°., Whenever rays of light travels through glass, they strike, the glass-air interface at an angle greater than critical, angle of glass. They are totally reflected, hence crack, appears silvery., The refractive index of diamond is much higher than that, of glass. Due to high refractive index, the critical angle, for diamond-air interface is low. The diamond is cut, suitably, so that the light entering the diamond from any, face suffers multiple total internal reflections at the, various surfaces. This gives sparkling effect to the, diamonds., real depth / thickness of object, Here, m =, apparent depth, M=, , . öæ, 25 ö, æ 75, = - ç ÷ ç1 +, ÷ = -5 ´6, 5 ø, è 1 .5 ø è, m = - 30, 36. It consists of an objective lens of a large focal length ( f o ) and, large aperture, also an eyepiece of small aperture and focal, length., (i) Magnification when final image is formed at infinity,, f, Magnification, m = - o and length of telescope,, fe, L = | f o| + | f e|, , a, , or, , Þ, , The object must be at a distance of 1.5 cm from objective, lens., \ Magnifying power,, , Objective lens, , M=, , (ii) Given, f o + f o = 105, f o = 20 f e, 105, fe =, =5, 21, , 7.5 ´ 1 .25, = - 1.5 cm, 6 .25, , æv ö, m = - çç o ÷÷, è uo ø, , - I / fe, I / fo, , \, , Eyepiece, , C2, , Eye, , Now, due to refraction at two different boundaries, the, apparent depth of object, thickness of glass thickness of water, =, +, m glass, m water, 6, 4, =, +, 1.5 1.3, = 4 + 3 = 7 cm

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44, , CBSE Term II Physics XII, , CHAPTER 03, , Wave Optics, In this Chapter..., l, , Wavefront, , l Superposition Principle, , l, , Huygens' Principle, , l Young's Double Slit Experiment, , l, , Doppler's Effect in Light, , l Diffraction of Light, , Wavefront, It is the locus of points (wavelets) having the same phase, (a surface of constant phase) of oscillations. A wavelet is the, point of disturbance due to propagation of light., Depending on the shape of source of light, wavefronts can, be of three types, which are explained as given below, (i) Spherical wavefront When the source of light is a, point source, the wavefront is a sphere with centre, as the source., Ray, , portion of spherical or cylindrical wavefront appears to, be plane. Such a wavefront is called a plane wavefront., , Light, rays, , Plane, wavefront, Plane wavefront, , Huygens’ Principle, Wavefront, , S, , Spherical wavefront, , (ii) Cylindrical wavefront When the source of light is, linear, all the points equidistant from the source, lie on a cylinder. Therefore, the wavefront is, cylindrical in shape., , According to Huygens’ principle,, (i) Each point on the given wavefront (called primary, wavefront) is the source of a secondary disturbance, (called secondary wavelets) and the wavelets emanating, from these points spread out in all directions with the, speed of the wave., (ii) A surface touching these secondary wavelets,, tangentially in the forward direction at any instant gives, the new wavefront at that instant. This is called, secondary wavefront., G1, , F1, , G1, , F1, , Wavefront, S, , D1, A, , O, D2, , Cylindrical wavefront, , (iii) Plane wavefront When the point source or linear, source of light is at very large distance, a small, , F2, G2, (a), , A¢, , A1, , A2, , B1, , B2, , C1, , C2, , D1, , D2, , F2 G2, t=0 t=t, (b)

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45, , CBSE Term II Physics XII, , In Fig.(a), F1 F2 is the section of the given spherical wavefront, and G1 G 2 is the new wavefront in the forward direction., In Fig.(b), F1 F2 is the section of the given plane wavefront, and G1 G 2 is the new wavefront in the forward direction., , Laws of Refraction (Snell’s Law) at a Plane Surface, Let 1, 2, 3 be the incident rays and 1 ¢, 2¢, 3 ¢ be the, corresponding refracted rays., N, , Refraction and Reflection of Plane Waves, Using Huygens’ Principle, , 2, , X, Medium 2, , 2, , B, , r, , A, , r, G, D Refracted, wavefront 2¢, , Rarer medium, v 1, m 1, Y, Denser medium, v2, m2, 3¢, , Laws of refraction by Huygens’ principle, 2¢, , G, r, i, , C, , 1¢, , D, , E, , F, , r, , 1¢, , i, X, , i, , N¢, Incident, wavefront, , 1, , E, , i, Medium 1, , Laws of Reflection at a Plane Surface, Let 1, 2, 3 be the incident rays and 1 ¢, 2¢, 3 ¢ be the, corresponding reflected rays., 3, , Incident, wavefront B, , 1, , Huygens’ principle can be used to explain the phenomena of, reflection and refraction of light on the basis of wave theory of, light., , N, , N¢, 3, , r, , A, C, F, Laws of reflection by Huygens’ principle, , 3¢, Reflected, wavefront, Y, , If c is the speed of light, t is the time taken by light to go from, B to C or A to D or E to G through F, then, EF FG, …(i), t=, +, c, c, EF, In DAEF, sin i =, AF, FG, In DFGC, sin r =, FC, AF sin i FC sin r, or, t=, +, c, c, AC sin r + AF (sin i - sin r ), Þ, t=, c, (Q FC = AC - AF), For rays of light from different parts on the incident, wavefront, the values of AF are different. But light from, different points of the incident wavefront should take the, same time to reach the corresponding points on the reflected, wavefront., So, t should not depend upon AF. This is possible only, if, sin i - sin r = 0, i.e., sin i = sin r, or, …(ii), Ði = Ðr, which is the first law of reflection., Further, the incident wavefront AB, the reflecting surface XY, and the reflected wavefront CD are all perpendicular to the, plane of the paper., Therefore, incident ray, normal to the mirror XY and, reflected ray all lie in the plane of the paper. This proves the, second law of reflection., , If v 1 , v 2 are the speeds of light in the two media and t is the, time taken by light to go from B to C or A to D or E to G, EF FG, through F, then, t=, +, v1, v2, EF, In DAFE,, sin i =, AF, FG, In DFGC,, sin r =, FC, AF sin i FC sin r, …(iii), Þ, t=, +, v1, v2, æ sin i sin r ö, AC sin r, Þ, t=, + AF çç, ÷, v2, v 2 ÷ø, è v1, For rays of light from different parts on the incident, wavefront, the values of AF are different. But light from, different points of the incident wavefront should take the, same time to reach the corresponding points on the refracted, wavefront. So, t should not depend upon AF. This is possible, only, if, sin i sin r, sin i v 1, …(iv), =0 Þ, =, v1, v2, sin r v 2, Now, if c represents the speed of light in vacuum, then, c, c, and m 2 =, are known as the refractive indices of, m1 =, v1, v2, medium 1 and medium 2, respectively., In terms of refractive indices, Eq. (iv) can be written as, sin i, m 1 sin i = m 2 sin r Þ m =, sin r, This is known as Snell’s law of refraction., Further, if l1 and l 2 denote the wavelengths of light in, medium 1 and medium 2, respectively and if the distance BC, is equal to l1 , then the distance AD will be equal to l 2 , thus, l1 BC v 1, v, v, or 1 = 2, =, =, l 2 AD v 2, l1 l 2, Þ, , n1 = n 2, , æQ v = n ö, ç, ÷, è l, ø

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46, , CBSE Term II Physics XII, , Hence, the frequency does not change on refraction., Thus, frequency n being a characteristic of the source, remains, the same as light travels from one medium to another., Also, wavelength is directly proportional to the (phase) speed, and inversely proportional to refractive index., l, l cl c, \, l¢ = , m = =, =, m, l¢ v l v, , where, a and b are the respective amplitudes of the two, waves and fis the constant phase angle by which second, wave leads the first wave., , Doppler’s Effect in Light, , and tan q =, , According to this effect, whenever there is a relative motion, between a source of light and observer, the apparent, frequency of light received by observer is different from the, true frequency of light emitted from the source of light., Astronomers call the increase in wavelength due to Doppler, effect as red shift, since a wavelength in the middle of the, visible region of spectrum moves towards the red end of the, spectrum. When waves are received from a source moving, towards the observer, there is an apparent decrease in, wavelength, this is referred to as blue shift., The fractional change in frequency is given by, v, Dn, = - radial, n, c, where, v radial is the component of the source velocity along, the line joining the observer to the source relative to the, observer., v radial is considered positive, when the source moves away, from the observer. The above formula is valid only, when the, speed of the source is small compared to that of light., , Superposition Principle, According to this principle, at a particular point in the, medium, the resultant displacement (y) produced by a, number of waves is the vector sum of the displacements, produced by each of the waves ( y1 , y2 , ... )., i.e., y = y1 + y 2 + y 3 + y4 + ..., This principle was stated first for mechanical waves but is, equally applicable to the electromagnetic (light) waves., , Applying superposition principle, the resultant displacement, of wave is, y = A sin( wt + q), b, , where, A = a 2 + b 2 + 2ab cos f or f, b sin f, a + b cos f, , A, f, , q, , a, , Intensity is directly, Resultant of amplitudes a and b, proportional to the square of, the amplitude of the wave. i.e. I µ a 2 ., In general, resultant intensity,, I R = I1 + I 2 + 2 I1 I 2 cos f, Þ, For constructive interference, cos f = maximum = +1, \ Phase difference, f = 0, 2p, 4 p, ..., i.e., f = 2np, where n = 1, 2 , ..., Path difference,, , Dx = nl, I max µ ( a + b) 2, , For destructive interference, cos f = minimum = -1, \ Phase difference, f = p, 3 p, 5 p, ..., f = ( 2n - 1 ) p, where n = 1, 2, ..., l, Path difference, Dx = ( 2n - 1 ), 2, I min µ ( a - b) 2, Comparison of Intensities of Maxima and Minima, I max ( r + 1 ) 2, =, I min ( r - 1 ) 2, where, r =, , a, (ratio of amplitudes)., b, , Interference of Light Waves, , Coherent and Incoherent Sources, , The phenomenon of formation of maximum intensity at some, points and minimum intensity at some other points by two, identical light waves travelling in same direction is called the, interference of light., At the points, where the resultant intensity of light is, maximum, interference is said to be constructive. At the, points, where the resultant intensity of light is minimum,, interference is said to be destructive., , Light sources are of two types, i.e. coherent and incoherent, light sources. The sources of light, which emit light waves of, same wavelength, same frequency and are in same phase or, having constant phase difference are known as coherent, sources. Two such sources of light, which do not emit light, waves with constant phase difference are called incoherent, sources., , Theory of Interference of Waves, Let the waves from two sources of light be represented as, y1 = a sin wt, and, y 2 = b sin ( wt + f), , Young’s Double Slit Experiment, Suppose S1 and S2 are two fine slits, a small distance d apart., They are illuminated by a strong source Sof monochromatic, light of wavelength l and MN is a screen at a distance D from, the slits.

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47, , CBSE Term II Physics XII, , Consider a point P at a distance y from O, the centre of, screen., The path difference between two waves arriving at point P is, equal to S2 P - S1 P., M, , Fringe Width, , P, S1, , y, d, , S, , O, , S2, D, N, 2, 2, é, d ù é, d ù, Now, (S2 P ) 2 - (S1 P ) 2 = êD 2 + æç y + ö÷ ú - êD 2 + æç y - ö÷ ú, 2 ø úû êë, 2 ø úû, è, è, êë, = 2yd, 2yd, Thus,, S2 P - S1 P =, S2 P + S1 P, , But, \, , The separation between two consecutive dark fringes,, Dl, Dl Dl, b¢ = ( 2n - 1 ), - {2( n - 1 ) - 1}, =, 2d, 2d, d, , S2 P + S1 P » 2D, dy, S2 P - S1 P », D, , For constructive interference (Bright fringes), dy, Path difference =, = nl, where n = 0, 1 , 2 , 3, ..., D, nDl, \, y=, (Q n = 0, 1, 2, 3, ... ), d, Hence, for n = 0, y 0 = 0 at O for central bright fringe, Dl, for n =1, y1 =, for 1st bright fringe, d, 2Dl, for 2nd bright fringe, for n = 2, y 2 =, d, nDl, for n = n, y n =, for nth bright fringe, d, The separation between two consecutive bright, nDl ( n - 1 )Dl Dl, fringes,, b=, =, d, d, d, For destructive interference (Dark fringes), dy, l, Path difference =, = ( 2n - 1 ), D, 2, Dl, or, y = ( 2n - 1 ), , where n = 1, 2, 3, ..., 2d, Dl, Hence, for n = 1,, for 1st dark fringe, y1¢ =, 2d, 3 Dl, for 2nd dark fringe, for n = 2,, y ¢2 =, 2d, Dl, for n = n, y ¢n = ( 2n - 1 ), for nth dark fringe, 2d, , The distance between two consecutive bright or dark fringes, is called fringe width., Dl, i.e., b or w =, d, If YDSE apparatus is immersed in a liquid of refractive index, m, then wavelength of light and hence fringe width decreases, m times., , Intensity of the Fringes, For a bright fringe, f = 2np, \, I R = I max = I1 + I 2 + 2 I1 I 2 = 4I, For a dark fringe, f = ( 2n - 1 )p, \, , I R = I min = I1 + I 2 - 2 I1 I 2 = 0, , Distribution of Intensity, If w1 and w 2 are widths of two slits from which intensities of, light I1 and I 2 emanate, then, I 1 a 2 w1, =, =, I2 b2 w2, where, a and b are the respective amplitudes of two waves., Fringe Shift, If refracting slab of thickness t is placed infront of one of the, two slits of YDSE, then fringe pattern gets shifted by n, fringes and is given by, (m - 1 ) t = nl, Fringe shift due to insertion of two slabs having thicknesses, t 1 & t 2 and refractive indices m 1 & m 2 ., (m 2 - m 1 ) t = nl, , Diffraction of Light, The phenomenon of bending of light around the sharp, corners and the spreading of light within the geometrical, shadow of the opaque obstacles is called diffraction of light., The dimensions of the aperture or the obstacle are, comparable to the wavelength of light., Diffracted, wave, , Incident, wave, a, l, Screen, a>l

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48, , CBSE Term II Physics XII, , Diffraction of Light at a Single Slit, A parallel beam of light with a plane wavefront is made to fall, on a single slit LN. As width of the slit LN = a is of the order, of wavelength of light, therefore diffraction occurs when, beam of light passes through the slit., The wavelets from the single wavefront reach the centre C on, the screen in same phase. Hence, interfere constructively to, give central maximum (bright fringe)., The diffraction pattern obtained on the screen consists of a, central bright band, having alternate dark and weak bright, bands of decreasing intensity on both sides., P, , Beam of, light from source a, , L, M1, M, M2, N, , y, , q, q, Q, , q, , C, , M2, , D, , Geometry of single slit diffraction, , Consider a point P on the screen at which wavelets travelling, in a direction, make an angle q with MC. The wavelets from, points L and N will have a path difference equal to NQ., From the right angled D LNQ, we have, NQ = LN sin q, or, NQ = a sin q, To establish the condition for secondary minima, the slit is, divided into 2, 4, 6,… equal parts such that corresponding, wavelets from successive regions interfere with path, difference of l/ 2 or for nth secondary minima, the slit can be, divided into 2n equal parts., Hence, for nth secondary minima,, a, l, Path difference = sin q =, 2, 2, nl, or, sin q n = , where n = 1, 2, 3, ..., a, To establish the condition for secondary maxima, the slit is, divided into 3, 5, 7, … equal parts such that corresponding, wavelets from alternate regions interfere with path difference, of l/ 2.., For nth secondary maxima, the slit can be divided into, ( 2n + 1 ) equal parts., Hence, for nth secondary maxima,, l, a sin q n = ( 2n + 1 ), 2, l, or, sin q n = ( 2n + 1 ), 2a, , ( n = 1, 2, 3, ... ), , Hence, the diffraction pattern can be graphically shown as, below, Y, , Intensity(I), I0, , XN, , X, C, –2l, –l, l, 2l, Path difference ( d sin q), , –3l, , 3l, , The point C corresponds to the position of central maxima, and the position -3 l, -2l, -l, l, 2l,3l… are secondary, minima. The above conditions for diffraction maxima and, minima are exactly reverse of mathematical conditions for, interference maxima and minima., Width of Central Maximum, It is the distance between first secondary minimum on either, side of the central bright fringe C., 2Dl, Width of central maximum = 2y =, a, As, the slit width a increases, width of central maximum, decreases., 2l, Angular width of central maxima, 2q =, \, a, , Fresnel’s Distance, When a slit or hole of size a is illuminated by a parallel beam,, l, it is diffracted with an angle q » ., a, , q = l/a, , a, , zl, a, , z, , Diffraction of a parallel beam, , In travelling a distance z, size of beam is z l / a., zl, a2, So, taking, ³ a or z ³, a, l, Now, distance z F is called Fresnel’s distance ( z F = a 2 / l)., , Difference between, Interference and Diffraction, (i) The interference pattern has a number of equally, spaced bright and dark bands whereas the diffraction, pattern has a central bright maximum, which is twice as, wide as the other maxima. The intensity falls as we go to, successive maxima away from the centre on either side., (ii) We calculate the interference pattern by superposing, two waves originating from the two narrow slits. The, diffraction pattern is a superposition of a continuous, family of waves originating from each point on a, single slit.

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49, , CBSE Term II Physics XII, , Solved Examples, Example 1 A plane wavefront incident on a reflecting, surface at an angle of 30° with horizontal. Find the, angle of reflected wavefront with horizontal., Sol. Consider a plane wavefront OA incident at 30° with the, horizontal, ON is normal to the reflecting surface, Ði is angle, of incidence and Ðr is angle of reflection. According to law, of reflection, Ði = Ðr., N, , A, , B, , 30°, O, , Now, from figure, we have, Ði + 30° = 90°, Þ, Ði = 60° = Ðr, Angle with horizontal = 90° - Ðr = 30°, , Example 2 A plane wavefront is incident from air, ( m = 1) at an angle of 37° with a horizontal boundary, of a refractive medium from air of refractive index, 3, m = . Find the angle of refracted wavefront with, 2, the horizontal boundary., Sol. It has been given that, incident wavefront makes 37° with, horizontal. Hence, incident ray makes 37° with normal as, the ray is perpendicular to the wavefront., N, , Incident, wavefront, , Incident ray, 37°, , Intensity, IR = I1 + I2 + 2 I1I2 cos f, (i) At f = 0,, Intensity, IR = I + 4I + 2 I × 4I cos 0° = 9I = Imax, p, (ii) At f = ,, 2, Intensity, IR = I + 4I + 2 I × 4I cos( p/ 2) = 5 I, , (iv) Ratio of maximum and minimum intensities,, Imax 9I, =, = 9:1, Imin, I, , Example 4 In a YDSE, green light of wavelength, 500 nm is used. Where will be the second bright, fringe be formed for a set-up in which separation, between slits is 4 mm and the screen is placed 1m, from the slits?, Sol. Position of second bright fringe, y2 =, =, , r, , Refracted, wavefront, Refracted, ray, , sin 37° 3, =, sin r, 2, 2 3 2, Þ sin r = ´ =, 3 5 5, 2ö, æ, Þ r = sin -1 ç ÷ , which is same as angle of refractive, 5, è ø, wavefront with horizontal., Now, by Snell’s law,, , Example 3 Two sources of intensity I and 4I are used, in an interference experiment. Find the intensity at, a point, where the waves from two sources, , 2l D, d, , 2 ´ 500 ´ 10-9 ´ 1, = 0.25 mm, 4 ´ 10-3, , Example 5 Fringe width in a particular YDSE is, , measured to be b. What will be the fringe width, if, wavelength of the light is doubled, separation, between the slits is halved and separation between, the screen and slits is tripled?, , Sol. We know that, b =, , 37°, r, , Sol. Given, I1 = I and I2 = 4I, , (iii) At f = p,, Intensity, IR = I + 4I + 2 I × 4I cos( p ) = I = Imin, , r, , i, , superimpose with a phase difference of (i) zero, (ii), p/ 2, (iii) p and (iv) ratio of maximum and minimum, intensities., , lD, d, , Now, the changed values are l ¢= 2 l, d ¢=, , d, and D¢= 3D., 2, , The new fringe width will be, l¢ D¢, b ¢=, d¢, ( 2l )( 3D), lD, Þ, b ¢=, = 12, = 12b, æ dö, d, ç ÷, è 2ø, The fringe width increases to 12 times of the original., , Example 6 In Young’s double slit experiment,, separation between slits is 1mm, distance of screen, from slits is 2m. If wavelength of incident light is, 500 nm. determine, (i) fringe width,, (ii) angular fringe width,

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50, , CBSE Term II Physics XII, , (iii) distance between 4th bright fringe and 3rd dark, fringe and, (iv) if whole arrangement is immersed in water, (m w = 4/ 3), new angular fringe width., , Sol. Given, d = 1 mm, D = 2m, l = 500 nm = 500 ´ 10-9 m, , (i) Fringe width,, Dl 2 ´ 500 ´ 10-9, b=, =, = 10-3 m = 1 mm, d, 1 ´ 10-3, (ii) Angular fringe width,, l 500 ´ 10-9, q0 = =, = 5 ´ 10-4 rad, d, 1 ´ 10-3, (iii) Distance of 4th bright fringe from centre,, nDl 4Dl, y4 =, =, d, d, Distance of 3rd dark fringe from centre,, ( 2n + 1 ) Dl, y3¢ =, d, ( 2 ´ 2 + 1 ) Dl, =, 2d, 5 Dl, =, 2d, Distance between 4th bright fringe and 3rd dark fringe,, 4Dl 5 Dl 3Dl, y4 - y3 ¢=, =, d, 2d, 2d, 3, 3, 3, = b = ´ 1 = mm, 2, 2, 2, l, l, 3, (iv) Wavelength in water, l w =, =, = l, m w 4/ 3 4, q0 5 ´ 10-4, =, mw, 4/ 3, 15, =, ´ 10-4 rad, 4, , ( q0 ) w =, , Example 7 A slit of width d is illuminated by red light, of wavelength 6500Å. For what value of d, will the, first maximum fall at angle of diffraction of 30°?, Sol. For first secondary maximum of the diffraction pattern,, 3l, d sin q =, 2, , \, , d=, , 3l, 3 ´ 6500 ´ 10- 10 m, =, 2sin q, 2 ´ sin 30°, , = 1.95 ´ 10-6 m, , Example 8 A beam of light of wavelength 600 nm from, a distant source falls on a single slit 1.0 mm wide, and the resulting diffraction pattern is observed on, a screen 2m away. What is the distance between, the second bright fringe on either side of the, central bright fringe?, Sol. For the diffraction at a single slit, the position of maxima is, given by, l, asin q = ( 2n + 1 ), 2, y, For small value of q, sin q » q =, D, For second bright fringe, n = 2., y 5l, 5l, or y =, \, a´ =, D, D, 2, 2a, Substituting the values, we have, 5 ´ 2 ´ 6 ´ 10-7, y=, 2 ´ 1 ´ 10-3, = 3.0 ´ 10-3 m = 3.0 mm, Distance between second maxima on either side of central, maxima = 2y = 6.0 mm., , Example 9 Estimate how large can be the aperture, opening to work with laws of ray optics using a, monochromatic light of wavelength 450 nm to a, distance of around 20 m., Sol. Now, here we are given Fresnel distance equal to 20 m and, l = 450nm, we have to estimate value of a., a2, Putting z F = , a = z F l, l, a = 20 ´ 450 ´ 10- 9, = 9000 ´ 10- 9, = 9 ´ 10- 6, = 3 ´ 10-3 m or 3 mm

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51, , CBSE Term II Physics XII, , Chapter, Practice, PART 1, Objective Questions, l, , Multiple Choice Questions, 1. Rays diverging from a point source produce a, wavefront of which shape?, (a) Cylindrical, (c) Plane, , (b) Spherical, (d) Cubical, , 2. According to Huygens’ principle, each point of the, wavefront is the source of, (a) secondary disturbance, (c) third disturbance, , (b) primary disturbance, (d) fourth disturbance, , 3. For light diverging from a point source, the, wavefront is spherical and the intensity ……… in, proportion to the distance squared., (a) increases, (c) remains the same, , When i > i C (critical angle of incidence), then, wavefront EC is, (a) formed further deep in medium 2, (b) formed closer to surface line AC, (c) formed perpendicular to AC, (d) formed in medium 1 (on same side of AB), , 7. Which of the following statement is incorrect, conditions for producing sustained interference?, (a) Phase difference between interferring waves remains, constant with time., (b) Interferring waves have nearly same amplitude levels., (c) Interferring waves are of same frequency., (d) Interferring waves are moving only in opposite, directions., , 8. In Young’s double slit experiment, if source S is, shifted by an angle f as shown in figure. Then,, central bright fringe will be shifted by angle f, towards, [Delhi 2020], A, , (b) decreases, (d) None of these, , S¢, , 4. The direction of wavefront of a wave with wave, , S, , S2, , (b) perpendicular, (d) at an angle of q, , 5. Light waves travel in vacuum along the Y-axis., Which of the following may represent the, wavefront?, (a) y = constant, (c) z = constant, , (b) x = constant, (d) x + y + z = constant, , 6. In given figure, light passes from denser medium 1, , O, , Q, , motion is, (a) parallel, (c) opposite, , S1, , f, , B, , (a) end A of screen, (b) end B of screen, (c) does not shift at all, (d) Either ends A and B depending on extra phase difference, caused by shifting of source, , 9. The correct curve between fringe width b and, distance between the slits d in figure is, , to rarer medium 2., Incident wavefront, , (a), B, , Medium 1, v1, , i A, , Medium 2, , (b), , v1t, , d, i, r, , r, , d, , C, , v2t, , v2, , (c), v2 > v1, E, , (d), , Refracted wavefront, , d, , d

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52, , CBSE Term II Physics XII, , 10. A thin film of oil is spread on the surface of water., , 17. A parallel beam of fast moving electrons is incident, , The beautiful colours exhibited in the light of sun is, due to, , normally on a narrow slit. A screen is placed at a, large distance from the slit. If the speed of the, electrons is increased, which of the following, statement is correct?, , (a) dispersion of light, (b) polarisation of light, (c) interference of light, (d) diffraction of light, , (a) Diffraction pattern is not observed on the screen in the, case of electrons., (b) The angular width of the central maximum of the, diffraction pattern will increase., (c) The angular width of the central maximum will decrease., (d) The angular width of the central maximum will remains, the same., , 11. The phase difference between the two light waves, reaching at a point P is 100p. Their path difference, is equal to, (a) 10 l, (c) 50l, , (b) 25 l, (d) 100l, , 18. In diffraction from a single slit the angular width of, the central maxima does not depends on, , 12. In the phenomenon of interference, energy is, , (a) l of light used, (b) width of slit, (c) distance of slits from the screen D, (d) ratio of l and slit width, , (a) destroyed at destructive interference, (b) created at constructive interference, (c) conserved but it is redistributed, (d) same at all points, , 13. Two sources S1 and S2 of intensity I1 and I 2 are, , 19. What should be the slit width to obtain 10 maxima, , placed infront of a screen Fig (a). The pattern of, intensity distribution seen in the central portion is, given by Fig (b)., [NCERT Examplar], , of the double slit pattern within the central maxima, of the single slit pattern of slit width 0.4 mm?, (a) 0.4 mm, , (c) 0.6 mm, , (d) 0.8 mm, , 20. In a single diffraction pattern observed on a screen, , S1, , placed at D m distance from the slit of width d m,, the ratio of the width of the central maxima to the, width of other secondary maxima is, , x, S1, (a), , x, (b), , (a) 2 : 1, , In this case, which of the following is/are incorrect, for S1 and S2 ?, (a) Same intensities, (b) Constant phase difference, (c) Same phase, (d) Same wavelength, , 14. Two identical and independent sodium lamps act as, (a) coherent sources, (b) incoherent sources, (c) Either (a) and (b), (d) None of the above, , 15. In Young’s double slit experiment, distance, between slits is kept 1 mm and a screen is kept 1 m, apart from slits. If wavelength of light used is, 500 nm, then fringe spacing is, (a) 0.5 mm, (c) 0.25 mm, , (b) 0.2 mm, , (b) 0.5 cm, (d) 0.25 cm, , 16. In a Young’s double slit experiment, the source is, white light. One of the holes is covered by a red, filter and another by a blue filter. In this case, there, shall be, [NCERT Exemplar], (a) alternate interference patterns of red and blue, (b) an interference pattern for red distinct from that for blue, (c) no interference fringes, (d) an interference pattern for red mixing with one for blue, , l, , (b) 1 : 2, , (c) 1 : 1, , (d) 3 : 1, , Assertion-Reasoning MCQs, Direction (Q.Nos. 21-29) Each of these questions, contains two statements Assertion (A) and Reason (R)., Each of these questions also has four alternative choices,, any one of which is the correct answer. You have to select, one of the codes (a), (b), (c) and (d) given below., (a) Both A and R are true and R is the correct, explanation of A., (b) Both A and R are true, but R is not the correct, explanation of A., (c) A is true, but R is false., (d) A is false and R is also false., , 21. Assertion Speed of light is independent of its, colour only in vacuum., Reason Red colour light travels slower than violet, colour light in glass., , 22. Assertion Increase in the wavelength of light due, to Doppler’s effect is red shift., Reason When the wavelength increases, then, wavelength in the middle of the visible region of, the spectrum will move towards the red end of the, spectrum.

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53, , CBSE Term II Physics XII, , 23. Assertion If we have a point source emitting waves, uniformly in all directions, the locus of point which, have the same amplitude and vibrate in the same, phase are spheres., Reason Each point of the wavefront is the source of, a secondary disturbance and the wavelets, emanating from these points spread out in all, directions with the speed of the wave., , 24. Assertion Two coherent source transmit waves of, equal intensity I 0 . Resultant intensity at a point,, l, where path difference is is also I 0 ., 3, Reason In interference, resultant intensity at any, point is the average intensity of two individual, intensities., , 25. Assertion In Young’s double slit experiment, ratio, of, , I max, is infinite., I min, , Reason If width of any one of the slits is slightly, increased, then this ratio will decrease., , 26. Assertion No interference pattern is detected,, when two coherent sources are infinitely close to, each other., Reason The fringe width is inversely proportional, to the distance between the two slits., , 27. Assertion In Young’s double slit experiment, the, fringe width for dark fringes is same as that for, white fringes., Reason In Young’s double slit experiment, when, the fringes are formed with a source of white light,, then only dark and bright fringes are observed., , 28. Assertion Diffraction does not determines the, , wavefront. Light travels in the form of waves., A wavefront is the locus of points (wavelets) having, the same phase (a surface of constant phase) of, oscillations. A wavelet is the point of disturbance, due to propagation of light. Wavefront may also be, defined as the hypothetical surface on which the, light waves are in the same phase., (i) Huygens’ original theory of light assumed that,, light propagates in the form of, (a) minute elastic particles, (b) transverse electromagnetic wave, (c) transverse mechanical wave, (d) longitudinal mechanical wave, , (ii) A wave normal, (a) is parallel to a surface at the point of incidence of a, wavefront, (b) is the line joining the source of light and an observer, (c) gives the direction of propagation of a wave at a given, point, (d) is the envelope that is tangential to the secondary, wavelets, , (iii) Rays diverging from a linear source form a, wavefront that is, (a) cylindrical, (b) spherical, (c) plane, (d) cubical, , (iv) According to Huygens’ principle, a wavefront, propagates through a medium by, (a) pushing medium particles, (b) propagating through medium with speed of light, (c) carrying particles of same phase along with it, (d) creating secondary wavelets which forms a new, wavefront, , (v) In case of reflection of a wavefront from a reflecting, surface,, , limitations of the concepts of light rays., Reason A beam of width a starts to spread out due to, diffraction after it has travelled a distance ( 2a 2 / l )., , l, , Incident, wavefront, E, , 29. Assertion In interference, different maximas have, , i, , same intensities., Reason In diffraction phenomenon, different, maximas have different intensities., , M, , Case Based MCQs, Direction Read the following passage and answer the, questions that follows, , 30. Wavefront, In 1678, a Dutch scientist, Christian Huygens’, propounded the wave theory of light. According to, him, wave theory introduced the concepts of, , A, , Reflected, wavefront, , B, i, , r, , C, , N, , I. points A and E are in same phase., II. points A and C are in same phase., III. points A and B are in same phase., IV. points C and E are in same phase., Which of the following is/are correct?, (a), (b), (c), (d), , Both I and II, Both II and III, Both III and IV, Both I and IV

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54, , CBSE Term II Physics XII, , PART 2, Subjective Questions, l, , Short Answer (SA) Type Questions, 1. What is the shape of the wavefront on earth for, sunlight?, , [NCERT Exemplar], , 2. Discuss Doppler’s effect in the electromagnetic, waves., 3. Answer the following questions., (i) When a low flying aircraft passes overhead, we, sometimes notice a slight shaking of the picture, on our TV screen. Suggest a possible, explanation., (ii) As, you have learnt in the text, the principle of, linear superposition of wave displacement is, basic to understanding intensity distributions in, diffraction and interference patterns., What is the justification of this principle? [NCERT], , 4. The 6563 Å H a -line emitted by hydrogen in a star, is found to be red shifted by 15 Å. Estimate the, speed with which the star is receding from the, earth., [NCERT], , 5. In Young’s double slit experiment, the two slits are, illuminated by two different lamps having same, wavelength of light. Explain with reason, whether, interference pattern will be observed on the screen, or not., [All India 2017], , 6. For a single slit of width a, the first minimum of the, interference pattern of a monochromatic light of, l, wavelength l occurs at an angle of . At the same, a, l, angle of , we get a maximum for two narrow slits, a, separated by a distance a . Explain., [Delhi 2014], , 7. Draw the intensity pattern for single slit diffraction, and double slit interference. Hence, state two, differences between interference and diffraction, patterns., [All India 2017], , 8. State briefly two features which can distinguish the, characteristic features of an interference pattern, from those observed in the diffraction pattern due, to a single slit., , 9. Why is the diffraction of sound waves more evident, in daily experience than that of light wave?, [NCERT Exemplar], , 10. In deriving the single slit diffraction, pattern, it was stated that the intensity is zero at, , nl, . Justify this by suitably dividing the slit, d, to bring out the cancellation., [NCERT], angles, , 11. Two wavelengths of sodium light of 590 nm and, 596 nm are used in turn to study the diffraction, taking place at a single slit of aperture 2 ´ 10 -6 m. If, the distance between the slit and the screen is 1.5, m, calculate the separation between the positions of, the second maxima of diffraction pattern obtained, in the two cases., [CBSE 2019], , 12. In the diffraction due to a single slit experiment,, the aperture of the slit is 3 mm. If monochromatic, light of wavelength 620 nm is incident normally on, the slit, calculate the separation between the first, order minima and the third order maxima on one, side of the screen. The distance between the slit and, the screen is 1.5 m., [CBSE 2019], , 13. (i) In a single slit diffraction experiment, the width, of the slit is made double the original width., How does this affect the size and intensity of the, central diffraction band? Explain., (ii) When a tiny circular obstacle is placed in the, path of light from a distant source, a bright spot, is seen at the centre of the obstacle. Explain,, why., [CBSE 2011], , 14. For the same objective, find the ratio of the least, separation between two points to be distinguished, by a microscope for light of 5000 Å and electrons, accelerated through 100V used as the illuminating, substance., [NCERT Exemplar], , 15. (i) Why are coherent sources necessary to produce, a sustained interference pattern?, (ii) In Young’s double slit experiment using, monochromatic light of wavelength l , the, intensity of light at a point on the screen, where, path difference is l , is K unit. Find out the, intensity of light at a point, where path, difference is l / 3., [Delhi 2012, NCERT], , 16. Two coherent light waves of intensity 5 ´ 10 -2 Wm -2, each superimpose and produce the interference, pattern on a screen. At a point where the path, l, difference between the waves is , l being, 6, wavelength of the wave, find the, (i) phase difference between the waves,, (ii) resultant intensity at the point and, (iii) resultant intensity in terms of the intensity at the, maximum., [All India 2020]

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55, , CBSE Term II Physics XII, , 17. Consider a two slit interference arrangements, (figure), such that the distance of the screen from, the slits is half the distance between the slits., Obtain the value of D in terms of l, such that the, first minima on the screen falls at a distance D from, the centre O., [NCERT Exemplar], T1, P, O, , S1, Source, C, , S, S2, , T2, Screen, , 18. Answer the following questions:, (i) In a double slit experiment using light of, wavelength 600 nm, the angular width of the, fringe formed on a distant screen is 0. 1°. Find the, spacing between the two slits., (ii) Light of wavelength 500 Å propagating in air, gets partly reflected from the surface of water., How will the wavelengths and frequencies of the, reflected and refracted light be affected?, [Delhi 2020, 15], , 19. A beam of light consisting of two wavelengths, 650 nm and 520 nm is used to obtain interference, fringes in a Young’s double slit experiment., (i) Find the distance of the third bright fringe on, the screen from the central maximum for, wavelength 650 nm., (ii) What is the least distance from the central, maximum, where the bright fringes due to both, the wavelengths coincide?, [NCERT], , 20. A small transparent slab containing material of, m = 15, . is placed along AS2 (figure). What will be, the distance from O of the principal maxima and of, the first minima on either side of the principal, maxima obtained in the absence of the glass slab?, [NCERT Exemplar], S1, A, L = d/4, , P1, q, , C, S2, , O, , Screen, , (AC = CO = D, S1C = S2C = d << D), l, , Long Answer (LA) Type Questions, 21. (i) State Huygens’ principle. Using this principle,, draw a diagram to show how a plane wavefront, incident at the interference of the two media gets, refracted when it propagates from a rarer to a, denser medium., Hence, verify Snell’s law of refraction., , (ii) Is the frequency of reflected and refracted light, same as the frequency of incident light?, [Delhi 2020, 13], , 22. (i) Using Huygens’ construction of secondary, wavelets explain how a diffraction pattern is, obtained on a screen due to a narrow slit on, which a monochromatic beam of light is incident, normally., (ii) Show that the angular width of the first diffraction, fringe is half that of the central fringe., , 23. (i) What is the effect on the interference fringes to a, Young’s double slit experiment, when, (a) the width of the source slit is increased and, (b) the monochromatic source is replaced by a source, of white light? Justify your answer in each case., (ii) The intensity at the central maxima in Young’s, double slit experiment set-up is I 0 . Show that the, intensity at a point, where the path difference is, [Foreign 2012], l / 3 is I 0 / 4., , 24. Consider two coherent sources S1 and S2, producing monochromatic waves to produce, interference pattern., Let the displacement of the wave produced by S1, be given by y1 = a cos wt and the displacement by, S2 be y 2 = a cos(wt + f )., Find out the expression for the amplitude of the, resultant displacement at a point and show that the, intensity at that point will be, f, I = 4a 2 cos 2, 2, Hence, establish the conditions for constructive, and destructive interference., [Delhi 2015], , 25. (i) In Young’s double slit experiment, derive the, condition for, (a) constructive interference and, (b) destructive interference at a point on the screen., (ii) A beam of light consisting of two wavelengths,, 800 nm and 600 nm is used to obtain the, interference fringes on a screen placed 1.4 m, away in a Young’s double slit experiment. If the, two slits are separated by 0.28 mm, calculate the, least distance from the central bright maximum, where the bright fringes of the two wavelengths, coincide., [All India 2012], , 26. Four identical monochromatic sources A, B, C, D as, shown in the (figure) produce waves of the same, wavelength l and are coherent. Two receiver R1, and R 2 are at great but equal distances from B.

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56, , CBSE Term II Physics XII, , (i) Which of the two receivers picks up the larger, signal?, (ii) Which of the two receivers picks up the larger, signal, when B is turned off?, (iii) Which of the two receivers picks up the larger, signal, when D is turned off?, (iv) Which of the two receivers can distinguish, which of the sources B or D has been turned off?, (R1 B = d = R 2 B, [NCERT Exemplar], AB = BC = BD = l / 2), , l/2, B, , C, , l/2, , D, l, , Dark, , Monochromatic, light source, , l/2, A, , Bright, , Bright, , R2, , R1, , beams overlap. Alternate bright and dark equally, spaced vertical bands (interference fringes) can be, observed on a screen placed at same distance from, the slits. If either of S1 or S2 is covered, then the, fringes disappear., , Case Based Questions, Direction Read the following passage and answer the, questions that follows, , 27. Young’s Experiment, In 1801, Thomas Young was the first who, demonstrated the interference of light., In his experimental arrangement (as shown in, figure), monochromatic light (single wavelength), from a narrow vertical slit S falls on two other, narrow slits S1 and S2 which are very close together, and equidistant to S. S1 and S2 act as two coherent, sources (both being derived from S ). The emerging, beams spread into the region beyond the slits., Superposition occurs in the shaded area, where the, , S1, S0, S1, , Dark, , }, , Interference, effects in, Bright, region, where, Dark beams overlap, , Bright, Single, slit, , Double, slits, , Dark, Bright, Viewing, screen, , Young’s experimental arrangement to produce interference pattern, , (i) Suppose while performing double slit, experiment, the space between the slits and the, screen is filled with water. How does the, interference pattern change?, (ii) In Young’s double slit experiment, if the distance, between the slits is halved, what changes in the, fringe width will take place?, (iii) The ratio of the widths of two slits in Young’s, double slit experiment is 4:1., Evaluate the ratio of intensities at maxima and, minima in the interference pattern. [Delhi 2015]

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Chapter Test, Multiple Choice Questions, , 9. Whose fringe width will be larger, the one for red, , 1. A monochromatic light refracts by the medium of, refractive index 1.5 in vacuum. The wavelength of, refracted wave will be, , Long Answer Type Questions, , 2. Two distinct light bulbs as sources, (a) can produce an interference pattern, (b) cannot produce a sustained interference pattern, (c) can produce an interference pattern, if they produce light of, same frequency, (d) can produce an interference pattern only when the light, produced by them is monochromatic in nature, , 3. From a single slit, the first diffraction minima is obtained, at 30° for a light of 6500 Å wavelength. The width of the, slit is, (b) 1.3 m, , (c) 5. 4 ´ 10 - 4 km, , (d) 12, . ´ 10 - 2 cm, , 4. In Young’s double slit experiment, two disturbance, arriving at a point P have phase difference of p / 2. The, intensity of this point expressed as a fraction of, maximum intensity I0 is, 3, (a) I 0, 2, 4, (c) I 0, 3, , 10. What do you understand by optical path? Prove that, optical path length d o = nd m, where d m is the, distance travelled in medium having refractive, index n ., , (a) equal, (b) increase, (c) decrease, (d) depend upon the intensity of refracted light, , (a) 3250 Å, , light or the one for yellow light, all other things be, the same?, , 1, (b) I 0, 2, 3, (d) I 0, 4, , 5. If the width of slit is decreased in a single slit diffraction,, then the width of central maxima will, , 11. Explain the following giving reasons:, (i) When monochromatic light is incident on a, surface separating two media, then both reflected, and refracted light have the same frequency as the, incident frequency., (ii) When light travels from a rarer to a denser, medium, then speed decreases. Does this decrease, in speed imply a reduction in the energy carried, by the wave?, (iii) In the photon picture of light, intensity of light is, determined by the square of the amplitude of the, wave. What determines the intensity in the photon, picture of light?, , 12. A light of 2000 - 8000 Å range is allowed to fall on, two slits having separation of 2 mm between them. A, screen is placed at 2.5 m away from the slit., Find the wavelength in the visible region that will be, present on the screen at 10 -3 m from the central, maxima. Also, find the wavelength that will be, present at that point of screen in the infrared as well, as in the ultraviolet region., [Ans. 8000 Å (infrared), 4000 Å (visible),, 2666 Å (ultraviolet), 2000 Å (ultraviolet)], , (a) increase, (b) decrease, (c) remain unchanged, (d) not depend on the width of slit., , 13. (i) Explain two features to distinguish between the, , Short Answer Type Questions, , 6. Use Huygens’ geometrical construction to show the, behaviour of a plane wavefront,, (i) passing through a biconvex lens and, (ii) reflected by a concave mirror., , 7. The ratio of maximum and minimum intensities of two, sources is 4 : 1. Find the ratio of their amplitudes., (Ans. 3 : 1), , 8. In a single slit diffraction experiment, the width of the slit, is decreased. How will the (a) size (b) and intensity of the, central bright band be affected? Justify your answer., , interference pattern in Young's double slit, experiment with the diffraction pattern obtained, due to a single slit., (ii) A monochromatic light of wavelength 500 nm is, incident normally on a single slit of width 0.2 mm, to produce a diffraction pattern. Find the angular, width of the central maximum obtained on the, screen., Estimate the number of fringes obtained in, Young's double slit experiment with fringe width, 0.5 mm, which can be accommodated within the, region of total angular spread of the central, maximum due to single slit., [Ans. 5), , Answers, Multiple Choice Questions, 1. (c), , 2. (b), , 3. (b), , 4. (b), , 5. (a), , For Detailed Solutions, Scan the code

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58, , CBSE Term II Physics XII, , EXPLANATIONS, PART 1, 1. (b) Wavefronts emitting from a point source are spherical, wavefronts., , Rays are, perpendicular, to wavefronts, , Point, source, , Wavefronts, are concentric, spheres, Light rays, , 2. (a) According to Huygens’ principle, each point of the, wavefront is the source of a secondary disturbance and the, wavelength emanating from these points spread out in all, directions with the speed of the wave., 3. (b) Consider the diagram in which light diverges from a, point source O., , We define critical angle iC by the formula,, n, sin iC = 2, n1, , ...(ii), , Thus, if i = i C , then from Eqs. (i) and (ii), sin r =1 and r = 90°., Obviously for i > i C , there cannot be any refracted wave. So,, for all angles of incidence greater than the critical angle, we, will not have any wavefront in medium 2. Hence, it is formed, in medium 1 (on same side of AB)., 7. (d) The statement given in option (d) is incorrect and it can, be corrected as,, The motion of interferring waves can be in same or in, opposite directions., Rest statements are correct., 8. (b) Since, when the source S is on the perpendicular bisector,, then the central bright fringe occurs at O, which is also on, the perpendicular bisector., If S is shifted by an angle f to point S¢, then the central fringe, appears at a point O¢ at an angle -f (shown in figure below),, which means that it is shifted by the same angle on the other, side of the bisector, i.e. towards end B of screen., A, , S', S, , O, r, , S1, , f, Q, , S2, , f, , O, , Spherical, , Due to the point source, light propagates in all directions, symmetrically and hence, wavefront will be spherical as, shown in the diagram., If power of the source is P, then intensity of the source will be, P, I=, 4pr 2, where, r is radius of the wavefront at any time., Hence, intensity will decreases in proportion to the distance, squared distance., 4. (b) The speed with which the wavefront moves outwards, from the source is called the speed of the wave. The, direction of the wave travels in a direction perpendicular to, the wavefront (as shown below), , Light rays, , Plane wavefront, , 5. (a) As, velocity of light is perpendicular to the wavefront and, light wave is travelling in vacuum along the Y-axis, therefore, the wavefront is represented by y = constant., 6. (d) As we know, for given figure from Snell’s law, sin i n 2, ...(i), =, sin r n1, where, n1 and n 2 = refractive indices of medium 1 and, medium 2, respectively., , O', B, , 9. (b) As, b =, , lD, d, , 1, d, So, the correct graph is represented by option (b)., 10. (c) The light reflected from the oil film produced two, coherent waves and these waves are superposed, (interference) and produce beautiful colours., 11. (c) Given, Df = 100 p, We know that, change in phase difference,, 2p, i.e., Df =, ´ Dx, l, where, Dx = path difference., l, l, Þ, Dx = D f ´, = 100p ´, = 50l, 2p, 2p, 12. (c) In the phenomenon of interference, energy is conserved, but it is redistributed., 13. (c) Consider the pattern of the intensity shown in the figure, given in question., (i) As intensities of all successive minima is zero, hence, we can say that two sources S1 and S2 are having same, intensities., (ii) Regular pattern shows constant phase difference, between sources S1 and S2 but these sources are not in, same phase., \ bµ

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59, , CBSE Term II Physics XII, , (iii) We are using monochromatic light in YDSE to avoid, overlapping and to have very clear pattern on the, screen, so wavelength of S1 and S2 will be same., Thus, the statement given option (c) is incorrect, rest are, correct., 14. (b) Two identical and independent sodium lamps (i.e. two, independent sources of light) can never be coherent., So, sodium lamps will be incoherent., 15. (a) Fringe spacing,, b=, , Dl 1 ´ 5 ´ 10-7, =, m, d, 1 ´ 10-3, , (1 nm = 10-9 m), , = 5 ´ 10-4 m = 0.5 mm, 16. (c) In this case due to filteration, only red and blue lights are, present. In YDSE, monochromatic light is used for the, formation of fringes on the screen. Hence, in this case there, shall be no interference fringes., 17. (c) Angular fringe width is the ratio of fringe width to, distance D of screen from the source. i.e., b, q=, D, As, D is taken large, hence angular fringe width of the, central maximum will decrease., 18. (c) Angular width of central maxima,, 2q = 2l / e, Thus, q does not depend on D, i.e. distance between the slit, and the screen., 19. (b) As, the path difference aq is l, then, l, q=, a, 10l 2l, Þ, =, d, a, d 10, Þ, a= =, = 0.2 mm, 5 5, So, the width of each slit is 0.2 mm., 20. (a) Width of central maxima = 2lD/ e, Width of other secondary maxima = lD / e, \ Width of central maxima : Width of other secondary, maxima = 2:1., 21. (c) Speed of light is independent of its colour only in, vacuum., As wavelength of red colour ( l1 ) is higher than wavelength, of blue colour (l 2 )., v, l, For light travelling from medium 1 to medium 2, 1 = 1, v2 l 2, where, v1 and v2 are velocities., So, red colour light travels faster than violet colour light in, glass., Therefore, A is true but R is false., 22. (a) Increase in wavelength of light, when the source move away, from the observer due to Doppler’s effect is called red shift., The visible regions shifts towards red end of, electromagnetic spectrum and hence called red shift., Therefore, both A and R are true and R is the correct, explanation of A., , 23. (a) According to Huygens’ principle, each point of the, wavefront is the source of a secondary disturbance and the, wavelets emanating from these point spread out in all, directions with the speed of wave., These wavelets emanating from the wavefront are usually, referred to as secondary wavelets and if we draw a common, tangent to all these spheres, we obtain the new position of the, wavefront at a later time., So, for a point source emitting waves uniformly in all, directions, the locus of points which have the same amplitude, and vibrate in the same phase are spheres and we have a, spherical wave., Therefore, both A and R are true and R is the correct, explanation of A., 2p, 2p l 2p, 24. (c) Phase difference, f =, × Dx =, ´ =, l, l 3, 3, 1, 2 æ fö, and, I = 4I0 cos ç ÷ = 4I0 ´ = I0, 4, è 2ø, Therefore, A is true but R is false., 25. (b) Imax > 4I0 and Imin = 0, Imax, = infinite, Imin, , 26., , 27., , 28., , 29., , 30., , If width of one slit is slightly increased Imin > 0, then this, ratio will be less than infinite., Therefore, both A and R are true but R is not the correct, explanation of A., (a) When distance between slits (d) is negligibly small,, 1, æ lD ö, fringe width b ç =, ÷ which is proportional to may, d, è d ø, become too large. Even a single fringe may occupy the, whole screen, hence the pattern cannot be detected., Therefore, both A and R are true and R is the correct, explanation of A., (c) In Young’s double slit experiment, fringe width for dark, and white fringes are same, while in the same experiment,, when a white light (as source) is used, the central fringe is, white around which few coloured fringes are observed on, either side., Therefore, A is true but R is false., (d) Diffraction determines the limitations of the concept of, light rays., a2, A beam of width a travels a distance , called the Fresnel, l, distance, before it starts to spread out due to diffraction., Therefore, A is false and R is also false., (c) All bands of bright interference receive equal amount of, energy collectively. Thus, the different maximas have same, intensities., But in diffraction, only central maxima receives most of the, energy from the slit and hence the remaining maxima’s, intensity falls down significantly., Therefore, A is true but R is false., (i) (d) Huygens’ original wave theory of light assumes that,, light propagates in the form of longitudinal mechanical wave., (ii) (c) A wave normal is a line perpendicular to the, wavefront. It gives the direction of a moving wave.

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60, , CBSE Term II Physics XII, , (iii) (a) Wavefronts emanating from a linear source are, cylindrical wavefronts., (iv) (d) According to Huygens’ principle, each point of the, wavefront is the source of a secondary disturbance and, the wavelets emanating from these points spread out in, all directions with the speed of the wave., (v) (c) Figure shows AB as incident wavefront, so A and B are, in same phase., B, A, E, A, , B, C, , By the time B reaches C, secondary wavelet from A, reaches E., So, points C and E are same time intervals apart as they, are in same phase., Hence, option (c) is correct., , PART 2, 1. We know that, the sun is at very large distance from the, earth. Assuming sun as spherical, it can be considered as, point source situated at infinity., Due to the large distance, the radius of wavefront can be, considered as large (infinity) and hence wavefront is almost, plane., , Sun, , 4. Given, l = 6563 Å, Dl = + 15 Å and c = 3 ´ 108 ms-1, Since, the star is receding away, hence its velocity v is, negative., vl, \, Dl = c, c Dl, or, v=l, 3 ´ 108 ´ 15, =6563, = - 6.86 ´ 105 ms -1, Negative sign shows that star is receding away from the, earth., 5. No interference pattern will be observed on the screen. This, is because, the sources will serve as incoherent sources., Here, two different lamps will act as different sources of, light which do not maintain constant phase difference., Dx, 6. In diffraction, angular position, q =, a, For first minima, Dx = l, l, \, q=, a, In interference, d = a (given) and angular position,, Dx, q=, a, \ Angular position of first maxima ( Dx = l ), q = l / a, 7. Intensity pattern for single slit diffraction is shown below, Intensity, Imax, , Almost, plane, wavefront, , 2. According to the Doppler’s effect, whenever there is a, relative motion between a source of light and observer, the, apparent frequency of light received by observer is different, from the true frequency of light emitted from the source of, light., The fractional change in frequency is given, by, Dn, v, = - radial, n, c, where, vradial is the component of the source velocity along, the line joining the observer to the source relative to the, observer and vradial is considered positive, when the source, moves away from the observer., 3. (i) We notice a slight shaking of the picture on our TV, screen because a low flying aircraft reflects the TV signal, and there may be an interference between the, direct signal and the reflected signal., (ii) The superposition principle follows the linear character of, the differential equation governing wave motion. If y1 and, y2 be the solutions of wave equation, so there can be any, linear combination of y1 and y2 . When the amplitudes are, large and non-linear effects are important, then the, situation is more complicated., , q, Angular, position, , l, a, , –l, a, , O Earth, , Intensity pattern for double slit interference pattern is as, shown below, Intensity, 4I0, , –l, , –l 0, 2, , l, 2, , l, , Path, difference, , Difference between diffraction and interference pattern are, (i) In interference pattern, all maxima and all minima are of, same width but in diffraction pattern, width of central, maxima is maximum and for successive maxima, it goes, on decreasing., (ii) In interference pattern, each maxima have same, intensity while in diffraction pattern, intensity of central, maxima is largest and it decreases rapidly for successive, maxima., 8. In case of single slit, the diffraction pattern obtained on the, screen consists of a central bright band having alternate, dark and weak bright band of decreasing intensity on both, sides. The diffraction pattern can be graphically represented, as

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61, , CBSE Term II Physics XII, , The distane of first order minima from central maxima,, nlD lD, x1 =, =, ( n = 1), b, b, 620 ´ 10-9 ´ 1.5, =, 3 ´ 10-3, , Y, Intensity(I), , XN, , = 310 ´ 10-6 m, , X, –3l, , C, –2l, –l, l, 2l, Path difference ( d sin q), , 3l, , Points to compare the intensity distribution between, interference and diffraction, (i) In interference, all bright fringes have same intensity,, but in diffraction all the bright fringes are not of same, intensity., (ii) In interference, the widths of all the fringes are same but, in diffraction fringes are of different widths., The point C corresponds to the position of central maxima, and the position -3l, -2l, - l, l, 2l, 3l… are secondary, minima. The above conditions for diffraction maxima and, minima are exactly reverse of mathematical conditions for, interference maxima and minima., 9. As we know that, the frequencies of sound waves lie, between 20 Hz to 20 kHz, so their wavelength ranges, between 15 m to 15 mm. The diffraction occurs, if the, wavelength of waves is nearly equal to slit width., As, the wavelength of light waves is 7000 ´ 10-10 m to, 4000 ´ 10-10 m, the slit width is very near to the wavelength, of sound waves as compared to light waves. Thus, the, diffraction of sound waves is more evident in daily life than, that of light waves., 10. Let the slit be divided into n smaller slits each of width,, d, d¢ =, n, nl nl, l, Angle,, q=, =, =, d, d ¢n d ¢, Therefore, each of the smaller slit would send zero intensity, in the direction of q. Hence, for the entire single slit, intensity, nl, at angle, would be zero., d, 11. Given, l1 = 590 nm, l 2 = 596 nm,, d = 2 ´ 10-6 m , D = 1.5 m, Distance of secondary maxima from centre,, 3 Dl, x=, 2 d, Spacing between the first two maxima of sodium light, 3D, Þ, x 2 - x1 =, ( l 2 - l1 ), 2d, 3 ´ 1.5, =, ( 596 - 590) ´ 10-9, 2 ´ 2 ´ 10-6, = 6.75 ´ 10-3 m = 6.75 mm, 12. Given, l = 620 nm = 620 ´ 10-9 m, Aperture of slit, b = 3 mm = 3 ´ 10-3 m, Distance between source and screen,, D = 1.5 m, , The distance at third order maxima from central maxima,, ( 2n + 1 ) l D 7l D, x3 ¢ =, =, 2b, 2b, =, , 7 ´ 620 ´ 10-9 ´ 1.5, 2 ´ 3 ´ 10-3, , = 1085 ´ 10-6 m, Thus, distance between x1 and x 3 is, x = x 3 - x1, = ( 1085 - 310) ´ 10-6, = 775 ´ 10-6 m, 13. (i) We know that, width of central maximum is given as, 2Dl, 2y =, a, where, a = width of slit., When a¢ = 2a, then width of central maximum, 2Dl lD, =, =, 2a, a, Thus, the width of central maximum became half. But, in case of diffraction, intensity of central maxima does, not changes with slit width. Thus, the intensity remains, same in both cases., (ii) When a tiny circular obstacle is placed in the path of, light from a distant source, a bright spot is seen at the, centre of the obstacle because the waves diffracted, from the edge of circular obstacle interfere, constructively at the centre of the shadow resulting in, the formation of a bright spot., 14. We know that,, 1 2sin b, Resolving power = =, d 1 .22 l, 1.22 l, Þ, dmin =, 2 sin b, where, l is the wavelength of light and b is the angle, subtended by the objective at the object., For the light of wavelength 5500 Å,, dmin =, , 1.22 ´ 5500 ´ 10-10, 2 sin b, , For electrons accelerated through 100 V, the de-Broglie, wavelength,, 12. 27, l=, V, 12.27, =, = 0.12 ´ 10-9 m, 100, 1.22 ´ 012, . ´ 10-9, ¢ =, dmin, 2 sin b, , ... (i)

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62, , CBSE Term II Physics XII, , Ratio of the least separation,, ¢, dmin, 012, . ´ 10-9, =, = 0.2 ´ 10-3, dmin 5500 ´ 10-10, , 17. From the given figure of two slit interference arrangements,, we can write, , 15. (i) In order to obtain a well defined observable, interference pattern, the intensity at points of, constructive and destructive interference must be, maintained maximum and almost zero, respectively., The two sources producing interference must be, coherent in nature., (ii) Intensity of light at a point on the screen is given by, IR = I1 + I2 + 2 I1I2 cos f, For the path difference l, phase difference is 2p., As, sources are coherent and taken out of the same source, in Young’s double slit experiment,, I1 = I2 = I, Þ, IR = 2I + 2I cos 2p, Þ, IR = 4I, …(i), Þ, 4I = K unit, l, For the path difference,, corresponding to phase, 3, 2p, difference of, ., 3, 2p, …(ii), IR = 2I + 2I cos, = 2I - I = I, 3, From Eqs. (i) and (ii), we conclude, K, IR = unit, 4, 16. Given, intensity, I = 5 ´ 10-2 Wm -2, l, Path difference, Dx =, 6, (i) As, path difference between the interfering waves is, l, given as Dx = f, 2p, where, f = phase difference., l, l, p, Þ, f= Þ f=, 2p, 6, 3, (ii) Resultant intensity,, IR = I1 + I2 + 2 I1I2 cos f, Here, I1 = I2 = I = 5 ´ 10-2 Wm -2, p, Þ, IR = 2I + 2I cos, 3, = 3I = 3 ´ 5 ´ 10-2, , Source, , (iii) As, IR = 3I, …(i), Maximum intensity is obtained, when phase difference is, zero or even integral multiple of 2p., Þ Imax = I + I + 2 I ´ I cos 2np, (Q cos 2np = 1 ), …(ii), , x, C, , S, S2, , T1, PD, O, D, T2, , Screen, , T2 P = T2 O + OP = D + x, T1P = T1O - OP = D - x, , and, , 2, 2, 2, 2, SP, 1 = ( ST, 1 1 ) + ( PT1 ) = D + ( D - x ), , S2 P = ( S2T2 ) 2 + ( T2 P ) 2 = D2 + ( D + x ) 2, , and, , The minima will occur when S2 P - SP, 1 = ( 2n - 1 ), , l, 2, , l, 2, (for first minima, n = 1), , [ D2 + ( D + x ) 2 ]1/ 2 - [ D2 + ( D - x ) 2 ]1/ 2 =, , i.e., , If x = D, we can write, ( D2 + 4D2 )1/ 2 - ( D2 + 0)1/ 2 =, Þ, , l, 2, l, 5D - D =, 2, , Þ, Þ, , D ( 5 - 1) = l / 2, l, 2 ( 5 - 1), , or, , D=, , Putting, , 5 = 2 . 236, , Þ, \, , l, 2, , ( 5 D2 )1/ 2 - ( D2 )1/ 2 =, , 5 - 1 = 2 . 236 - 1 = 1 . 236, l, D=, = 0.404 l, 2( 1.236), , 18. (i) Angular width, q =, Given,, , l, l, or d =, d, q, , l = 600 nm = 6 ´ 10-7 m, 0.1 p, p, q=, rad =, rad, 180, 1800, , 6 ´10-7 ´1800, = 3. 44 ´10-4 m, p, (ii) Frequency of a light depends on its source only, so the, frequencies of reflected and refracted light will be same, as that of incident light., Reflected light is in the same medium (air), so its, wavelength remains same as 500 Å., Wavelength of refracted light, l r = l / m w, where, m w = refractive index of water., So, wavelength of refracted wave will be decreased., 19. Given, wavelength, l1 = 650 nm = 650 ´ 10- 9 m, \, , = 15 ´ 10-2 Wm -2, , = 2I + 2I, = 4I, Dividing Eq. (i) by Eq. (ii), we get, IR, 3I, =, Imax 4I, 3, or, IR = Imax, 4, , D, , S1, , and, , d=, , l 2 = 520 nm = 520 ´ 10- 9 m, , (i) For third bright fringe, n = 3, The distance of third bright fringe from central, maximum.

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63, , CBSE Term II Physics XII, nlD, D, = 3 ´ 650 ´ 10- 9 ´ m, d, d, 3 ´ 650 ´ 10- 9 ´ 1.2, = 1.17 ´ 10- 3 m, =, 2 ´ 10-3, (ii) Let nth bright fringe due to wavelength l 2 = 520 nm,, coincide with ( n + 1)th bright fringe due to wavelength, l1 = 650 nm., D, D, i.e., nl 2 = ( n - 1 ) l1, d, d, n ´ 520 ´ 10- 9 = ( n - 1 ) 650 ´ 10- 9, or, 4n = 5 n - 5, or, n =5, Thus, the least distance,, D, D, x = nl 2 = 5 ´ 520 ´ 10- 9, d, d, D, -9, x = 2600 ´ 10 m, d, x=, , 1.2 ´ 10-9, m, 2 ´ 10- 3, = 1.56 ´ 10- 3 m = 1.56 mm, = 2600 ´, , 20. In case of transparent glass slab of refractive index m, the, path difference will be calculated as, Dx = 2d sin q + ( m - 1 ) L, For the principal maxima (path difference is zero),, i.e., 2d sin q0 + ( m - 1 ) L = 0, L ( m - 1 ) - L ( 05, . ), or, (Q L = d / 4 ), sin q0 = =, 2d, 2d, -1, or, sin q0 =, 16, -D, \, OP = D tan q0 » D sin q0 =, 16, l, For the first minima, the path difference is ± ., 2, l, \, 2d sin q1 + 05, . L=±, 2, or, , sin q1 =, , ± l / 2 - 05, . L ± l/ 2 - d / 8, =, 2d, 2d, , ± l/ 2 - l/ 8, 1 1, =± 2l, 4 16, [Q The diffraction occurs, if the wavelength of waves is nearly, equal to the side width (d)], 1 1, 3, +, On the positive side sin q¢1 = + =, 4 16 16, 1 1, 5, On the negative side sin q¢¢1- = - =4 16, 16, The first principal maxima on the positive side is at distance, sin q¢1+, 3, 3D, above, =D, D tan q¢1+ = D, =, 2, 2, 2, 247, 1 - sin q¢1, 16 - 3, =, , point O., The first principal minima on the negative side is at distance, 5D, 5D, below point O., D tan q¢¢1 =, =, 2, 2, 231, 16 - 5, , 21. (i) Huygens’ Principle, Each point on the given wavefront (called primary, wavefront) is the source of a secondary disturbance, (called secondary wavelets) and the wavelets, emanating from these points spread out in all, directions with the speed of the wave. Consider a, plane wavefront AB incident on a plane surface XY,, separating two media 1 and 2, as shown in figure. Let v1, and v2 be the velocities of light in two media, with, v2 < v1., N, , S, X, , Incident wavefront, B, v1t, , i, i, , r, , A, r v2t, , C, , Rarer-v1, Y, Denser-v2, , Refracted wavefront, D, , The wavefront first strikes at point A and then at the, successive points towards C. According to Huygens’, principle from each point on AC, the secondary, wavelets starts growing in the second medium with, speed v2 . Let the distrubance takes time t to travel, from B to C, then BC = v1t., During the time the disturbance from B reaches the, point C, the secondary wavelets from point A must, have spread over a hemisphere of radius AD = v2 t in, the second medium. The tangent plane CD drawn from, point C over this hemisphere of radius v2 t will be the, new refracted wavefront. Let the angles of incidence, and refraction be i and r, respectively., BC, From right DABC, we have, sin ÐBAC = sin i =, AC, AD, From right DADC , we have sin ÐDCA = sin r =, AC, sin i BC v1t, sin i v1, or, \, =, =, =, = m 21, sin r AD v2 t, sin r v2, This proves Snell’s law of refraction. The constant m 21, is called the refractive index of the second medium, with respect to first medium., Further, since the incident ray SA, the normal AN and, the refracted ray AD are respectively perpendicular to, the incident wavefront AB, the dividing surfae XY and, the refracted wavefron CD (all perpendicular to the, plane of the paper), therefore they all lie in the plane, of the paper, i.e. in the same plane. This proves, another law of refraction., (ii) The reflection and refraction phenomenon occur due to, interaction of corpuscles of incident light and the, atoms of matter on receiving light energy, the atoms, are forced to oscillate about their mean positions with, the same frequency as incident light. According to, Maxwell’s classical theory, the frequency of light, emitted by a charged oscillator is same as its frequency, of oscillation. Thus, the frequency of reflected and, refracted light is same as the incident frequency., 22. (i) Consider a parallel beam of light from a lens falling on a, slit AB. As diffraction occurs, the pattern is focused on, the screen with the help of lens L 2 . We will obtain a, diffraction pattern that is central maximum at the centre

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64, , CBSE Term II Physics XII, , O, flanked by a number of dark and bright fringes called, secondary maxima and minima., a, , n, Perpe, q, A, q, C, B, , S, , P, , L2, , Slit, L1, , (ii) Given, OP = yn, , r, , dicula, , y, , q, , yn, , S1, , P, , O, , d, S2, , O, , q, , Plane, wavefront, , D, , The distance OP equals one-third of fringe width of the, pattern., b, 1 æ Dl ö Dl, i.e., yn =, = ç, ÷=, 3 3 è d ø 3d, dyn l, Þ, =, D, 3, dyn l, Path difference = S2 P - SP, =, 1 =, D, 3, 2p, \ Phase difference, f =, ´ path difference, l, 2p l 2p, =, ´ =, l 3, 3, If intensity at central fringe is I0 , then intensity at a point, P, where phase difference f, is given by, I = I0 cos 2 f, , D, , Each point on the plane wavefront AB sends out the, secondary wavelets in all directions. The waves from, points equidistant from the centre C , lying on the, upper and lower half, reach point O with zero path, difference and hence reinforce each other producing, maximum intensity at O., (ii) Let l and a be the wavelength and slit width of, diffracting system, respectively. Let O be the position, of central maximum., Condition for the first minimum is given by, ... (i), asin q = ml, Let q be the angle of diffraction., As, diffraction angle is small, \, sin q » q, For first diffraction minimum, q = q1 (let), , 2p ö, æ, I = I0 ç cos, ÷, 3ø, è, , Þ, , pö, æ, = I0 ç - cos ÷, 3ø, è, , A, q1, q, , B, , 2, , O, , C, , 2, , 2, , I, æ 1ö, = I0 ç - ÷ = 0, 4, è 2ø, I0, ., 4, 24. Given, the displacements of two coherent sources y1 = a cos, w t and y2 = a cos( wt + f)., Hence, the intensity at point P would be, , For the first minimum, take m = 1, l, aq1 = l Þ q1 =, a, Now, angular width, AB = q1, Angular width, BC = q1, Angular width, AC = 2 q1, Hence, the angular width of the first diffraction fringe, is half that of the central fringe., s l, 23. (i) (a) For interference fringes to be seen, £ condition, S d, should be satisfied, where s = size of the source and, d = distance of the source from the plane of two slits., As, the source slit width increases, fringe pattern gets, less and less sharp. When the source slit is too wide,, the above condition does not get satisfied and the, interference pattern disappears., (b) The interference pattern due to the different colour, components of white light overlap. The central bright, fringes for different colours are at the same position., Therefore, central fringe is white. And on the either, side of the central fringe (i.e. central maxima),, coloured bands will appear. The fringe close to either, side of central white fringe is red and the farthest will, be blue., , By principle of superposition,, y = y1 + y2 = a cos wt + a cos( wt + f), y = a cos wt + a cos wt cos f - asin wt sin f, y = a( 1 + cos f) cos wt + ( - asin f)sin wt, Let a( 1 + cos f) = A cos q, …(i), and, …(ii), asin f = A sin q, \, y = A cos q cos wt - A sin qsin wt, …(iii), Þ, y = A cos( wt + q), Squaring and adding Eqs. (i) and (ii), we get, ( A cos q) 2 + ( A sin q) 2 = a2 ( 1 + cos f) 2 + ( asin f)2, A 2 (cos 2 q + sin 2 q), 2, , = a ( 1 + cos f + 2 cos f) + a2 sin 2 f, Þ, Þ, , 2, , A 2 ´ 1 = a2 + a2 + 2a2 cos f = 2a2 ( 1 + cos f), fö, æ, æ fö, A 2 = 2a2 ç 2 cos 2 ÷ = 4a2 cos 2 ç ÷, 2ø, è, è 2ø, , f, If I is the resultant intensity, then I = 4a2 cos 2 ., 2

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65, , CBSE Term II Physics XII, f, From constructive interference, cos = ± 1, 2, f, Þ, = np Þ f = 2np, 2, For destructive interference,, f, cos = 0, 2, f, p, Þ, = ( 2n + 1 ), 2, 2, f = ( 2n + 1 ) p, 25. (i) Let two coherent sources of light, S1 and S2 (narrow, slits) are derived from a source S. The two slits S1 and S2, are equidistant from source, S., Now, suppose S1 and S2 are separated by distance d. The, slits and screen are distance D apart., P, , S1, , A, q, , S, , C, , yn, , d/2, , q, , O d, , M, , d/2, , S2, D, , B, Screen, , Considering any arbitrary point P on the screen at a, distance yn from the centre O. The path difference, between interfering waves is given by S2 P - S1 P., i.e. Path difference = S2 P - S1 P = S2 M, S2 P - S1 P = d sin q, where,, S1 M ^ S2 P, [Q ÐS2 S1 M = ÐOCP (by geometry), Þ S1 P = PM Þ S2 P = S2 M ], If q is small, then sin q » q » tan q, \ Path difference,, S2 P - S1 P = S2 M = dsin q » d tan q, æy ö, Path difference = d ç n ÷, …(i), è Dø, OP yn, (Q in DPCO, tan q =, = ), CO D, For constructive interference, Path difference = nl, where n = 0, 1, 2, ... [from Eq. (i)], Dnl, Þ, yn =, d, D( n + 1 ) l, Þ, yn+ 1 =, d, Q Fringe width of dark fringe = yn + 1 - yn, (Q dark fringe exist between two bright fringes), Dl, Dnl, b=, (n + 1) d, d, Dl, Dl, =, ( n + 1 - n) =, d, d, Dl, Fringe width of dark fringe, b =, …(ii), d, , For destructive interference, l, Path difference = ( 2n - 1 ) , where n =1, 2, 3,..., 2, yn¢ d, l, [from Eq. (i)], Þ, = ( 2n - 1 ), D, 2, ( 2n - 1 ) D l, Þ, yn¢ =, 2d, where, yn¢ is the separation of nth order dark fringe from, central fringe., Dl, yn¢ + 1 = ( 2n + 1 ), \, 2d, \ Fringe width of bright fringe = Separation between, ( n + 1 )th and nth order dark fringe from centred fringe,, Þ, b = yn¢ + 1 - yn¢, ( 2n + 1 ) Dl ( 2n - 1 ) Dl, or, b=, 2d, 2d, Dl, =, ( 2n + 1 - 2n + 1 ), 2d, Dl, =, 2d, Fringe width of bright fringe,, Dl, …(iii), b=, d, From Eqs. (ii) and (iii), we can see that,, Fringe width of dark fringe = Fringe width of bright, fringe, Dl, b=, d, (ii) Given, l1 = 800 nm, l 2 = 600 nm, D = 1.4 m and d = 0.28 mm = 2.8 ´ 10-4 m, Let nth order bright fringe of l = 800 nm coincide with, ( n + 1 )th order 600 nm wavelength., Dn l1 D( n + 1 ) l 2, \, =, d, d, Þ, nl1 = ( n + 1 ) l 2, n ´ 800 ´ 10- 9 = ( n + 1 ) ´ 600 ´ 10- 9, n+1 4, =, n, 3, 1 4, 1, Þ, = -1 =, n 3, 3, n=3, Dn l 1, \ Least distance from central fringe, yn =, d, 1.4 ´ 3 ´ 800 ´ 10- 9, -3, = 12 ´ 10 m, yn =, 2.8 ´ 10- 4, yn = 12 mm, 26. Consider the disturbances at the receiver R1, which is at a, distance d from B., Let the wave at R1 because of A be yA = a cos wt. The path, difference of the signal from A with that from B is l / 2 and, hence the phase difference is p., Thus, the wave at R1 because of B is, yB = a cos ( wt - p) = - a cos wt, The path difference of the signal from C with that from A is l, and hence the phase difference is 2p.

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67, , CBSE Term II Physics XII, , CHAPTER 04, , Dual Nature of, Radiation and, Matter, In this Chapter..., l, , Electron Emission, , l, , Photoelectric Effect, , l, , Einstein's Photoelectric Equation, , Electron Emission, The electrons in the outer shell (valence electrons), which are, free to move easily within the metal surface but cannot leave, it, are called free electrons., These electrons can be emitted from the metals, if they have, sufficient energy to overcome the attractive pull of metal, surface. So, the phenomenon of emission of electrons from, the surface of a metal is called electron emission., The minimum energy required for the electron emission from, the metal surface can be supplied to the free electrons by any, of the following physical processes, (i) Thermionic Emission Sufficient thermal energy can, be imparted to the free electrons by suitable heating, so, that they can come out of the metal., (ii) Field Emission or Cold Cathode Emission It is the, phenomenon of emission of electrons from the surface of, a metal by applying a very strong electric field, ( ~ 10 8 Vm -1 ) to a metal. One of the examples of cold, emission is spark plug., (iii) Photoelectric Emission It is the phenomenon of, emission of electrons from the surface of metal, when, light radiations of suitable frequency fall on it. Here,, the energy to the free electrons for their emission is, being supplied by light photons., , l, , Particle Nature of Light : The Photon, , l, , Photocell, , l, , Wave Nature of Particles :, de-Broglie Hypothesis, , Work Function, For electron emission to take place, a certain minimum, amount of energy is required for an electron to pull it out, from the surface of the metal. This minimum energy required, by an electron to escape from the metal surface is called the, work function of the metal. It is generally denoted by f0 or, W0 and measured in eV (electron volt)., One electron volt (1 eV) is the energy gained by an electron,, when it has been accelerated by a potential difference of one, volt (1 V), so that 1 eV = 1.602´ 10 -19 J. It depends on the, properties of the metal and the nature of its surface. It, decreases with the increase in temperature., , Photoelectric Effect, It is the phenomenon of emission of electrons from the, surface of metal, when radiations of suitable frequency fall on, it. The emitted electrons are called photoelectrons and the, current, so produced is called photoelectric current., Note Alkali metals like lithium, sodium, etc., show photoelectric effect, with visible light, whereas the metals like zinc, cadmium, etc.,, are sensitive only to ultraviolet light., , Hertz’s Observations, He observed that high voltage sparks across the detector loop, were enhanced, when the emitter plate was illuminated by, ultraviolet light from an arc lamp.

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68, , CBSE Term II Physics XII, , Lenard observed that, if a potential difference is applied, across the two metal plates enclosed in an evacuated tube,, then there is no flow of current in the circuit., However, when one plate (called emitter plate) enclosed in, the evacuated tube, kept at negative potential is exposed with, ultraviolet radiations, current begins to flow in the circuit., As soon as ultraviolet radiations falling on the emitter plate, are stopped, the current flowing is also stopped., Hallwachs and Lenard also observed that when ultraviolet, light fall on the emitter plate, no electrons were emitted,, until the frequency of the incident light was smaller than a, certain minimum value, called the threshold frequency., This minimum frequency depends on the nature of the, material of the emitter plate., , On the basis of the experimental arrangement used for, studying photoelectric effect, the variations of photocurrent, with intensity of radiation, frequency of radiation and the, potential difference between the plates A and C are as follows, Source, Evacuated, glass tube, Electrons, , C, , A, Reversing key, , mA, , V, , K, , B, , Experimental arrangement for the study of, photoelectric effect, , Photoelectric, current, , Effect of Intensity of Light on, Photoelectric Current, For a fixed frequency of incident radiation and accelerating, potential, the photoelectric current increases linearly with, increase in intensity of incident light which is as shown, below, , O Intensity of light, , Effect of Potential on Photoelectric Current, Variation of photoelectric current with potential for different, intensities but constant frequency is as shown in the graph, below, I 3 > I2 > I1, , I3, I2, I1, , Stopping, potential, O, –V0, Retarding potential, , Experimental Study of, Photoelectric Effect, , Quartz, window, Photosensitive, plate, , As, the photoelectric current is directly proportional to the, number of photoelectrons emitted per second. So, the, number of photoelectrons emitted per second is directly, proportional to the intensity of the incident radiation., , Photoelectric, current, , Hallwachs’ and Lenard’s Observations, , Collector plate, potential, , From the above graph, we can observe that, (i) After a certain value of accelerating potential, when all, photoelectrons reach the plate A and the photocurrent, ceases. On increasing the value of accelerating, potential, this maximum value of photoelectric current, is called saturation current., (ii) When the potential is decreased, the current decreases, but does not become zero at zero potential. This shows, that even in the absence of accelerating potential, few, photoelectrons manages to reach the collector plate on, their own due to their kinetic energy., (iii) For a particular frequency of incident radiation, when, minimum negative potential V0 is applied to the, collector plate w.r.t. emitter plate, photoelectric current, becomes zero at a particular value of negative potential, V0 called stopping potential or cut-off potential., In this condition, the stopping potential is sufficient to repel, even the most energetic photoelectron with maximum kinetic, energy K max , which is given as, 1, 2, K max = eV0 = mv max, 2, where, m is the mass of photoelectron and v max is the, maximum velocity of emitted photoelectron., For the radiation of a given frequency and material of emitter, plate, the value of stopping potential V0 is independent of, the intensity of the incident radiation. It means that, the, maximum kinetic energy of emitted photoelectron depends, on the light source and the emitter plate material but is, independent of intensity of incident radiation., , Effect of Frequency of Incident Radiation on, Stopping Potential, Variation of photoelectric current with potential for different, frequencies but constant intensity of incident radiation is as, shown in the graph below

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69, , CBSE Term II Physics XII, , Photoelectric current, n3 > n2 > n1, n3, , –V03, , n2, , Saturation, current, , n1, , –V02 –V01, , 0, , Collector plate, potential, , Retarding potential, , From the above graph, we observe that, (i) The value of stopping potential is different for radiations, of different frequencies but the value of saturation, current (for a given intensity) remains constant., (ii) The value of stopping potential is more negative for, incident radiation of higher frequency. This means, that, the energy of the emitted electrons depends on, the frequency of incident radiations., So, greater the frequency of incident radiation, greater, is the maximum kinetic energy of photoelectrons., Consequently, greater retarding potential or stopping, potential is required to stop them completely., (iii) The value of saturation current depends upon the, intensity of incident radiation but is independent of the, frequency of the incident radiation., The graph between stopping potential and the frequency of, the incident radiation for two different metals A and B, is as, shown below, Metal A, Stopping, potential, (V0), 0, , Metal B, n > n0, n0, , n0¢, , n > n0N, , Frequency of incident, radiation (n), , From the graph, we observe that, (i) The stopping potential V0 varies linearly with the, frequency of incident radiation for a given, photosensitive material., (ii) There exists a certain minimum cut-off frequency n 0 for, which the stopping potential is zero. This frequency is, called threshold or cut-off frequency. For a frequency, lower than cut-off frequency, no photoelectric emission, is possible even, if the intensity is large., If frequency of incident radiation is more than, the threshold frequency, the photoelectric emission, starts instantaneously without any apparent time lag, ( -10 -9 s or less) even, when the incident radiation is, very dim., , Laws of Photoelectric Emission, The laws of photoelectric emission are as follows, (i) For a given material and a given frequency of incident, radiation, the photoelectric current or number of, , photoelectrons ejected per second is directly, proportional to the intensity of the incident light., (ii) For a given material and frequency of incident, radiation, saturation current is found to be proportional, to the intensity of incident radiation, whereas the, stopping potential is independent of its intensity., (iii) For a given material, there exists a certain minimum, frequency of the incident radiation below which no, emission of photoelectrons takes place. This frequency, is called threshold frequency., Above the threshold frequency, the maximum kinetic, energy of the emitted photoelectrons or equivalent, stopping potential is independent of the intensity of the, incident light but depends upon only the frequency (or, wavelength) of the incident light., (iv) The photoelectric emission is an instantaneous process., The time lag between the incidence of radiations and, emission of photoelectrons is very small, less than even, 10 -9 s., , Photoelectric Effect and Wave, Theory of Light, Huygens’ wave theory of light could not explain the, photoelectric emission due to the following main reasons, (i) According to the wave nature of light, free electrons at, the surface of the metal absorb the radiant energy, continuously., The greater the intensity of radiation, the greater, should be the energy absorbed by each electron. The, maximum kinetic energy of the photoelectrons on the, surface is then expected to increase with increase in, intensity., But according to experimental facts, the maximum, kinetic energy of ejected photoelectrons is, independent of intensity of incident radiation., (ii) According to wave theory of light, no matter what the, frequency of radiation is, a sufficiently intense beam of, radiation should be able to impart enough energy to, the electrons, so that they exceed the minimum energy, needed to escape from metal surface., A threshold frequency, therefore should not exist, which contradicts the experimental fact that, no, photoelectric emission takes place below that, threshold frequency, no matter whatsoever may be its, intensity., (iii) According to the wave theory of light, the absorption of, energy by electron takes place continuously over the, entire wavefront of the radiation. Since, a large number, of electrons absorb energy, the energy absorbed per, electron per unit time turns out to be small., Hence, it will take hours or more for a single electron, to come out of the metal which contradicts the, experimental fact that photoelectron emission is, instantaneous.

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70, , CBSE Term II Physics XII, , Þ, , eV0 = h (n - n 0 ), , In terms of threshold wavelength, photoelectric equation is, given as, æ1 1 ö, ÷÷, eV0 = hc çç è l l0 ø, where,, and, , l = wavelength of the incident radiation,, l 0 = threshold wavelength of the metal surface, c = velocity of light., , Graphs Related to Photoelectric Effect, From Einstein Photoelectric Equation, The important graphs related to photoelectric effect are as, follows, (i) Frequency n and stopping potential V0 graph, We know that, eV0 = hn - f0, hn f0, Þ, V0 =, e, e, So,, V0 µ n, Stopping potential, , 0, , n0, , tan q =, , Maximum KE, , Kmax, , q, n0, , tan q = h, Frequency, , n, , – f0, , (iii) Frequency n and photoelectric current I graph, The graph given below shows that, the photoelectric, current I is independent of frequency of the incident, light, till intensity remains constant., I, , n0, , Frequency n, , (iv) Intensity and stopping potential V0 graph, V0, , Intensity, , (v) Photoelectric current I and time lag t graph, I, , 0 10 –9 s, , t, Time, , Particle Nature of Light : The Photon, Photoelectric effect gave evidence that light consists of, packets of energy. These packets of energy were called light, quantum that are associated with particles named as, photons. So, photons confirm the particle nature of light., , V0, , q, , K max µ n, , Photoelectric, current, , On the basis of this, Albert Einstein explained photoelectric, effect and the following photoelectric equation, 1, 2, K max = mv max, = hn - f0, 2, where, f0 is work function and K max is the maximum kinetic, energy of emitted electrons., Also,, K max = eV0, , Þ, , Stopping, potential, , According to Planck’s quantum theory, the energy of an, electromagnetic wave is not continuously distributed over, the entire wavefront of waves., Instead of this, these waves travel in the form of discrete, packets or bundels of energy called quanta of energy of, radiation. Each quantum of energy radiate an energy, which, is given by, E = hn, where, h is Planck’s constant and n is the frequency of light, radiation., , (ii) Frequency n and maximum kinetic energy K max, graph, As,, K max = hn - f0, , Photoelectric, current, , Einstein’s Photoelectric Equation, , h, = Slope, e, n, Frequency, , – f0, e, , It could be seen that, V0 versus n curve is a straight line, with slope = h/e and is independent of the nature of, material., , Characteristic Properties of Photons, Different characteristic properties of photons which are as, given below, (i) In interaction of radiation with matter, radiation, behaves as if it is made up of particles called photons., (ii) A photon travels at a speed of light c in vacuum,, i. e. 3 ´ 10 8 m / s., (iii) It has zero rest mass, i.e. the photon cannot exist at rest.

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71, , CBSE Term II Physics XII, , (iv) The inertial mass of a photon is given by, E, h hv, m= 2 =, =, cl c 2, c, (v) Photons travel in a straight line., (vi) Irrespective of the intensity of radiation, all the photons, of a particular frequency n or wavelength l have the same, hc ö, æ, æ hn h ö, energy E ç = hn = ÷ and momentum, p ç =, = ÷., lø, è, è c lø, (vii) Photons are not deflected by electric and magnetic, fields. This shows that, photons are electrically, neutral., (viii) In a photon-particle collision (such as photoelectron, collision), the energy and momentum are conserved., However, the number of photons may not be, conserved in a collision., (ix) Photons may show diffraction under given conditions., , Photocell, It is a device which converts light energy into electrical, energy. It is also called an electric eye. As, the photoelectric, current sets up in the photoelectric cell corresponding to, incident light, it provides the information about the objects, as seen by our eye in the presence of light., Incident, light, C, , Evacuated, glass bulb, , Emitter, (Cathode), , B, , (i) Used in television camera for telecasting scenes and in, photo telegraphy., (ii) Reproduction of sound in cinema film., (iii) Used in counting devices., (iv) Used in burglar alarm and fire alarm., (v) To measure the temperature of stars., (vi) Used for the determination of Planck’s constant., , Wave Nature of Particles :, de-Broglie Hypothesis, Louis victor de-Broglie, put forward a hypothesis that, moving, material particles of matter should display wave like, properties under suitable condition. Thus, he proposed that, the wavelength l associated with moving particle is given as, h h, l=, =, mv p, where, m, v and p are the mass, velocity and momentum of the, particle respectively and h is Planck’s constant., , Relation between de-Broglie Wavelength (l), and Temperature (T ), For a thermal neutron at temperature T, de-Broglie, wavelength is given as, h, h, l= =, p, 3 mKT, , Collector (Anode), , A, , –, , Applications of Photocell, , +, , Photocell, , mA, , de-Broglie Wavelength of Charged Particles, The wavelength associated with charged particles accelerated, through a potential is given as, h, h, l= =, p, 2mqV, where, q is the charge on the particle., 1 . 227, Specifically, for electrons, l =, nm., V

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72, , CBSE Term II Physics XII, , Solved Examples, Example 1. The photoelectric cut-off voltage in a, certain experiment is 1.5 V. What is the maximum, kinetic energy of photoelectrons emitted?, Sol. Maximum kinetic energy of photoelectrons, K max = eV0, where, e = 1.6 ´10-19 C, V0 = 1.5 V, \ K max = 1 .6 ´ 10-19 ´ 1 .5 = 2.4 ´ 10-19 J, , \ Threshold frequency, n 0, 91, . ´ 10-31 ´ ( 6.0 ´ 105 ) 2, = 7.21 ´ 1014 2 ´ 6.62 ´ 10-34, = 7.21 ´ 1014 - 2.47 ´ 1014, = 4.74 ´ 1014 Hz, , Example 5. A student performs an experiment on, , Example 2. In an experimental study of photoelectric, effect, the stopping potential for a metallic surface, is 5.68 V. What will be the maximum velocity of the, electrons?, , photoelectric effect, using two materials A and B. A, plot of Vstop versus n is given in the figure., Vstop, 3, 2.5, 2, 1.5, 1, , Sol. Stopping potential, V0 = 5.68 V, m e = 91, . ´ 10-31 kg, 1, 2, Maximum KE, K max Þ mvmax, = eV0, 2, Þ, , 2 ´ 1 .6 ´ 10-19 ´ 5 .68, 91, . ´ 10-31, 6, -1, = 1 .414 ´ 10 ms, , vmax =, , the slope of the cut-off voltage versus frequency of, incident light is found to be 4 .12 ´ 10 -15 V-s., Calculate the value of Planck’s constant., Sol. We know that, eV0 = hn - W, \, \, \, , W, æ hö, V0 = ç ÷ n e, è eø, h, h, Slope = or 4.12 ´ 10-15 =, e, e, h = 412, . ´ 10-15 ´ e, , B, , 5, , 10, 15, (× 1014 Hz), Frequency (Hz), , 2eV0, =, m, , Example 3. In an experiment on photoelectric effect,, , A, , (i) Which material A or B has a higher work, function?, (ii) Given the electric charge of an electron, = 1.6 ´ 10 -19 C, find the value of h obtained from, the experiment for both A and B., Comment on whether it is consistent with the, Einstein’s theory., Sol. (i) Given, threshold frequency of A is given by, n OA = 5 ´ 104 Hz, and for B, n OB = 10 ´ 1014 Hz, , = 412, . ´ 10-15 ´ 1 .6 ´ 10-19 Js-1, = 6.592 ´ 10-34 Js -1, 14, , Example 4. Light of frequency 7.21 ´ 10 Hz is, incident on a metal surface. Electrons with a, maximum speed of 6.0 ´ 105 m/ s are ejected from, the surface. What is the threshold frequency for, photoemission of electrons?, Sol. Using Einstein’s photoelectric equation,, 1, 2, hn - hn 0 = mvmax, 2, 1, 2, We get hn 0 = hn - mvmax, 2, 2, mvmax, \, n0 = n 2h, Given, n = 7.21 ´ 1014 Hz, vmax = 6.0 ´ 105 m/s, m = mass of electron = 91, . ´ 10- 31 kg, , We know that, work function,, f = hn 0 or f0 µ n 0, fOA, 5 ´ 1014, So,, =, <1 Þ, fOB 10 ´ 1014, , fOA < fOB, , Thus, work function B is higher than A., h, 2, (ii) For metal A, slope = =, e ( 10 - 5 ) 1014, 2 ´e, 2 ´ 1 .6 ´ 10-19, =, 14, 5 ´ 10, 5 ´ 1014, = 6.4 ´ 10-34 J-s, h, 2 .5, For metal B, slope = =, e ( 15 - 10) 1014, 25, . ´e, or, h=, 5 ´ 1014, 2 .5 ´ 1 .6 ´ 10-19, =, = 8 ´ 10-34 J-s, 5 ´ 1014, Since, the value of h from experiment for metals A and B is, different. Hence, experiment is not consistent with theory., or, , h=

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73, , CBSE Term II Physics XII, , Example 6. The momentum of photon of, , electromagnetic radiation is 3.3 ´ 10 -29 kg -ms -1 ., Find out the frequency and wavelength of the wave, associated with it., , Sol. (i) Given, h = 6.63 ´ 10-34 J/s, c = 3 ´ 108 ms -1, and p = 3.3 ´ 10-29 kg- ms -1, hn, Momentum, p =, c, pc, 3.3 ´ 10-29 ´ 3 ´ 108, Þ, n=, =, = 1.5 ´ 1013 Hz, h, 6.63 ´ 10-34, c, 3 ´ 108, (ii), l= =, = 2 ´ 10-5 m, n 1 .5 ´ 1013, , Example 7. The wavelength of a photon is 1.4 Å. It, , 6.63 ´ 10- 34 ´ 3 ´ 108 6.63 ´ 10- 34 ´ 3 ´ 108, =, - 4.26 ´ 10-16, l2, 1.4 ´ 10- 9, 6.63 ´ 10- 34 ´ 3 ´ 108, = 14.21 ´ 10-16 - 4.26 ´ 10-16, l2, l 2 = 2 ´ 10-10 m, = 2.0 Å, , Example 8. What is the, (a) momentum,, (b) speed and, (c) de-Broglie wavelength of an electron with, kinetic energy of 120 eV?, Sol. (a) Using the relation, E =, , collides with an electron. The energy of the, scattered electron is 4.26 ´ 10 - 16 J. Find the, wavelength of photon after collision., (Take, h = 6.63 ´ 10 - 34 J-s), Sol. Initial wavelength of photon is l1 and final wavelength of, photon is l 2 ., hc hc, 4.26 ´10-16 =, \, l1 l 2, hc hc, =, - 4.26 ´ 10-16, l 2 l1, , p = 2mE, =, , p2, , we get, 2m, , 2 ´ 91, . ´ 10-31 ´ 120 ´ 1 .6 ´ 10-19, , = 5.91 ´ 10-24 kg - ms -1, (b) Speed, v =, , p 5.91 ´ 10-24, =, = 6.5 ´ 106 ms -1, m 9.1 ´ 10-31, , (c) de-Broglie wavelength is given by, l=, , h, 6.62 ´ 10-34, =, = 1.12 ´ 10-10 m, p, 5 .91 ´ 10-24

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74, , CBSE Term II Physics XII, , Chapter, Practice, PART 1, Objective Questions, l, , Multiple Choice Questions, 1. Work-function f0 for a metal will change, if it is, (a), (b), (c), (d), , heated, cooled, coated with some other metal, All of the above, , 2. Hallwach connected an uncharged zinc plate to an, electroscope as shown below, Uncharged, zinc plate, , 4. In an experiment on photoelectric effect with the, increase in potential difference of emitter and, collector plate, keeping the frequency and intensity, fixed of the incident light, the photoelectric current, (a) increases, (b) decreases, (c) remains constant, (d) increases initially and then becomes constant, , 5. The cathode of a photoelectric cell is changed, such, that the work function changes from W1 to W2, ( W2 > W1 ). If the current before and after are i1 and, i 2 , respectively and all other conditions remaining, unchanged, then, [Delhi 2020], (a) i1 = i2, , (b) i1 < i2, , (c) i1 > i2, , (d), , W1, i, = 1, W2, i2, , 6. In photoelectric effect experiment, collector plate, , This plate is then irradiated by UV-light. The result, of this is correctly shown in, , is made negative with respect to emitter plate as, shown in figure below till it reach a certain, potential V 0 , when photocurrent is zero., Light, , +, , (a), , P, , Collector, plate, -, , (b), Emitter plate, V0, , (c), , (d), , If K indicates kinetic energy of an emitted, photoelectron, then at point P, (a) K > eV0, (c) K = eV0, , (b) K < eV0, (d) 0 £ K £ eV0, , 7. Consider a beam of electrons (each electron with, , 3. In the phenomenon of electric discharge through, gases at low pressure, the coloured glow in the tube, appears as a result of, (a) excitation of electrons in the atoms, (b) collision between the atoms of the gas, (c) collision between the charged particles emitted from the, cathode and the atoms of the gas, (d) collision between different electrons of the atoms of the, gas, , energy E 0 ) incident on a metal surface kept in an, evacuated chamber, then, [NCERT Exemplar], (a) no electrons will be emitted as only photons can emit, electrons, (b) electrons can be emitted but all with an energy, E0, (c) electrons can be emitted with any energy, with a, maximum of E0 - f, (f is the work function), (d) electrons can be emitted with any energy, with a, maximum of E0

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75, , CBSE Term II Physics XII, , 8. The formula for kinetic mass of a moving photon is, (where, h is Planck constant and n, l, c are, frequency, wavelength and speed of photon,, respectively), (b) hl / e, , (c) hn/ e, , (d) h / cl, , 9. The wavelength of a photon needed to remove a, proton from a nucleus which is bound to the nucleus, with 1 MeV energy is nearly, [NCERT Exemplar], (b) 1 . 2 ´ 10-3 nm, (d) 1 . 2 ´ 10 nm, , (a) 1 . 2 nm, (c) 1 . 2 ´ 10-6 nm, , 13 The value of stopping potential V 0 from the given, graph is, , [All India 2020], Photoelectric, current, , (a) hn / l, , (a) Curves a and b represent incident radiations of different, frequencies and different intensities., (b) Curves a and b represent incident radiations of same, frequency but of different intensities., (c) Curves b and c represent incident radiations of different, frequencies and different intensities., (d) Curves b and c represent incident radiations of same, frequency having same intensity., , 10. A point source of light is used in an experiment on, photoelectric effect. Which of the following curves, best represents the variation of photoelectric, current i with distance s of the source from the, emitter?, i, Photoelectric, current, , a, , 0.2 V Collector plate, potential, , O, , – 0.54 V, , b, , (a) - 0.54 V, (c) 0.2 V, , c, d, , (a) a, (c) c, , 14. A student plot a graph while performing an, , s, , Distance, , (b) 0.54 V, (d) - 0.2 V, , (b) b, (d) d, , 11. Variation of photoelectric current with collector, , experiment on photoelectric effect using an, evacuated glass tube, for light radiation of same, intensity at various frequencies as shown below, , I1, I2, , (b), , I1, , (d), , -V0, , O Collector plate, potential, , I1, I2, , O Collector plate, potential, , 12. The figure shows a plot of photocurrent versus, anode potential for a photosensitive surface for, three different radiations. Which one of the, following is a correct statement?, Photocurrent, b, c, , Retarding potential, , a, Anode potential, , n1, X, , O, , V0 Collector plate, potential, , Photoelectric, current, , Photoelectric, current, -V0, , I2, I1, , Z, n2, I2, , -V01 -V02, , V0 Collector plate, potential, , (c), , Y, , Photoelectric, current, , (a), , Photoelectric, current, , plate potential for different intensities I1 and I 2, (such that I1 > I 2 ) at a fixed frequency is, , Here,, (a) X ® Photoelectric current; Y ® Retarding potential; Z ®, Stopping potential, (b) X ® Retarding plate potential; Y ® Photocurrent; Z ®, Stopping potential, (c) X ® Collector plate potential; Y ® Photocurrent; Z ®, Saturation current, (d) X ® Retarding plate potential; Y ® Photocurrent; Z ®, Saturation current, , 15. The photoelectric threshold wavelength for silver is, l 0 . The energy of the electron ejected from the, surface of silver by an incident wavelength, [All India 2020], l ( l < l 0 ) will be, æ l - lö, (a) hc ç 0, ÷, è ll 0 ø, (c), , hc, l0 - l, , (b), , h æ l0 - l ö, ç, ÷, c è ll 0 ø, , (d) hc ( l 0 - l )

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76, , CBSE Term II Physics XII, , K max . When the ultraviolet light is replaced by, X-rays, both V 0 and K max increase., Reason Photoelectrons are emitted with speeds, ranging from zero to a maximum value because of the, range of frequencies present in the incident light., , 16. If all the following particles are moving with the, same velocity, then the particle of maximum, momentum will be, (a) b-particle, (c) a-particle, , (b) proton, (d) neutron, , 17. An electron is moving with an initial velocity, , 23. Assertion The photocells inserted in the door, , v = v 0 $i and is in a magnetic field B = B 0 $j. Then, its, de-Broglie wavelength, [NCERT Exemplar], , light electric circuit are used as automatic door, opener., Reason Abrupt change in photocurrent, helps to, open the door., , (a) remains constant, (b) increases with time, (c) decreases with time, (d) increases and decreases periodically, , 24. Assertion A particle at rest breaks into two, particles of different masses. They fly off in, different directions and their de-Broglie, wavelengths will be same., Reason Their speed will be same., , 18. The de-Broglie wavelength of a particle of kinetic, energy K is l. What will be the wavelength of the, K, particle, if its kinetic energy is ?, 9, (a) l, (c) 3l, , (b) 2l, (d) 4l, , 19. A proton, a neutron, an electron and an a-particle, have same energy. Then, their de-Broglie, wavelengths compare as, [NCERT Exemplar], (a) l p = l n > l e > l a, (c) l e < l p = l n > l a, l, , (b) l a < l p = l n < l e, (d) l e = l p = l n = l a, , Assertion-Reasoning MCQs, Direction (Q. Nos. 20-24) Each of these questions, contains two statements Assertion (A) and Reason (R)., Each of these questions also has four alternative, choices, any one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given, below., (a) Both A and R are true and R is the correct, explanation of A., (b) Both A and R are true, but R is not the correct, explanation of A., (c) A is true, but R is false., (d) A is false and R is also false., , 20. Assertion Heinrich Hertz observed that, high, voltage spark across detector loop were enhanced, when the emitter plate was illuminated by, UV-light., , Reason Light shining on the metal surface, facilitate the escape of free electrons., , 21. Assertion If distance of the point source is, increased from the photoelectric plate, then, stopping potential will remain unchanged., Reason Saturation current will decrease., , 22. Assertion When ultraviolet light is incident on a, photocell, its stopping potential is V 0 and the, maximum kinetic energy of the photoelectrons is, , l, , Case Based MCQs, Direction Read the following passage and answer the, questions that follows, , 25. Photocell, Photocell is a device which converts light energy, into electrical energy. It is also called an electric, eye. As, the photoelectric current sets up in the, photoelectric cell corresponding to incident light, it, provides the information about the objects as has, been seen by our eye in the presence of light., Incident, light, , C, , Collector (Anode), , A, , Evacuated, glass bulb, , Emitter, (Cathode), –, , B, , +, , mA, , A photocell consists of a semi-cylindrical, photosensitive metal plate C (emitter) and a wire, loop A (collector) supported in an evacuated glass, or quartz bulb. When light of suitable wavelength, falls on the emitter C, photoelectrons are emitted., (i) It is observed that no electrons are emitted, when, frequency of light is less than a certain minimum, frequency., This minimum frequency depends on, (a) potential difference of emitter and collector plates, (b) distance between collector and the emitter plate, (c) size (area) of the emitter plate, (d) material of the emitter plate

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77, , CBSE Term II Physics XII, , (ii) The work function of a metal used in photocell is, hc/ l 0 . If light of wavelength l is incident on its, surface, then the essential condition for the electron, to come out from the metal surface is, (a) l ³ l 0, , (b) l ³ 2l 0, , (c) l £ l 0, , (d) l £ l 0 / 2, , (iii) Variation of photoelectric current with intensity of, light for a photocell is, , 3. Two monochromatic radiations, blue and violet,, of the same intensity are incident on a, photosensitive surface and cause photoelectric, emission. Would, (i) the number of electrons emitted per second and, (ii) the maximum kinetic energy of the electrons be, equal in the two cases? Justify your answer., , (b), , Photoelectric, current, , (a), , Photoelectric, current, , [Delhi 2010], , Photoelectric, current, , Photoelectric, current, , (d), , Intensity of light, , potential with the frequency of radiation incident on, a metal plate. How can the value of Planck’s, constant be determined from this graph?, Intensity of light, , Intensity of light, , (c), , 4. Draw a graph to show the variation of stopping, , of the metals given below, which ones will show, photoelectric emission and why?, [CBSE 2018], , Intensity of light, , (iv) A photon of energy 3.4 eV is incident on a metal, surface of a photocell whose work function is 2 eV., Maximum kinetic energy of the photoelectron, emitted by the metal surface will be, (a) 1.4 eV, (b) 1.7 eV, (c) 5.4 eV, (d) 6.8 eV, , (v) Photocell is based on photoelectric effect that gave, evidence that light in interaction with matter,, (a), (b), (c), (d), , is converted into particles of same size, is converted into particles of same energy, is converted into mass following E = mc 2, behaves as if it was made of packets of energy, each of, energy hn, , PART 2, Subjective Questions, l, , 5. If light of wavelength 412.5 nm is incident on each, , Short Answer (SA) Type Questions, 1. There are materials which absorb photons of, shorter wavelength and emit photons of longer, wavelength. Can there be stable substances which, absorb photons of larger wavelength and emit light, of shorter wavelength?, [NCERT Exemplar], , 2. Two monochromatic beams A and B of equal, intensity I, hit a screen. The number of photons, hitting the screen by beam A is twice that by beam, B, then what inference can you make about their, frequencies?, [NCERT Exemplar], , Metal, , Work Function (eV), , Na, , 1.92, , K, , 2.15, , Ca, , 3.20, , Mo, , 4.17, , 6. The work function of Cs is 2.14 eV. Find, (i) threshold frequency for Cs and, (ii) wavelength of incident light, if the photocurrent, is brought to zero by stopping potential of 0.6 V., , 7. The threshold frequency for a certain metal is, , 3.3 ´ 1014 Hz. If light of frequency 8.2 ´ 1014 Hz is, incident on the metal, predict the cut-off voltage, for the photoelectric emission., [NCERT], , 8. Find the frequency of light which ejects electrons, from a metal surface, fully stopped by a retarding, potential of 3.3 V. If photoelectric emission begins, in this metal at a frequency of 8 ´ 1014 Hz, calculate, the work function (in eV) for this metal., [All India 2018C], , 9. Consider a metal exposed to light of wavelength, 600 nm. The maximum energy of the electron, doubles, when light of wavelength 400 nm is used., Find the work function in eV., [NCERT Exemplar], , 10. Light of same wavelength is incident on three, photosensitive surfaces A, B and C. The following, observations are recorded., (i) From surface A, photoelectrons are not emitted., (ii) From surface B, photoelectrons are just emitted., (iii) From surface C, photoelectrons with some, kinetic energy are emitted., Compare the threshold frequencies of the three, surfaces and justify your answer., [Delhi 2020]

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78, , CBSE Term II Physics XII, , 11. Figure shows the stopping potential V 0 for the, 1, graph, for two metals A and, l, B, l being the wavelength of incident light., , photoelectron versus, , O, , radiation is equal to the de-Broglie wavelength of, its quantum (photon)., [NCERT], , 19. A proton and an a-particle are accelerated through, the same potential. Which one of the two has, (i) greater value of de-Broglie wavelength, associated with it and, (ii) less kinetic energy?, Give reasons to justify your answer., [Delhi 2014], , V0, A, , 18. Show that the wavelength of electromagnetic, , B, 1/l, , (i) How is the value of Planck’s constant, determined from the graph?, (ii) If the distance between the light source and the, surface of metal A is increased, how will the, stopping potential for the electrons emitted from, it be effected? Justify your answer., [Delhi 2020], , 12. Explain with the help of Einstein’s photoelectric, equation any two observed features in, photoelectric effect with cannot be explained by, wave theory., [Delhi 2019], , 13. Why is wave theory of electromagnetic radiation, not able to explain photoelectric effect ? How does, photon picture resolve this problem?, [Delhi 2019], , 14. In the wave picture of light, intensity of light is, determined by the square of the amplitude of the, wave. What determines the intensity in the photon, picture of light?, [All India 2016], , 15. If the frequency of light incident on the cathode of, a photocell is increased, how will the following be, affected? Justify your answer., (i) Energy of the photoelectrons, (ii) Photocurrent, [Delhi 2020], , 16. A 100 W sodium lamp radiates energy uniformly in, all directions. The lamp is located at the centre of a, large sphere that absorbs all the sodium light which, is incident on it., The wavelength of the sodium light is 589 nm., (i) What is the energy per photon associated with, the sodium light?, (ii) At what rate are the photons delivered to the, sphere?, [NCERT], , 17. The work functions for the following metals are, given, Na = 2.75 eV, K = 2.30 eV, Mo = 4.17 eV,, Ni = 5.15 eV. Which of these metals will not give, photoelectric emission for a radiation of, wavelength 3300 Å from a He-Cd laser placed 1 m, away from the photocell? What happens, if the, laser is brought nearer and placed 50 cm away?, [NCERT], , 20. The two lines marked A and B in the given figure,, show a plot of de-Broglie wavelength l versus, , 1, , ,, V, where V is the accelerating potential for two nuclei, 2, 3, 1 H and 1 H., A, l, , B, , 1, ÖV, , (i) What does the slope of the lines represent?, (ii) Identify, which of the lines corresponded to, these nuclei., [All India 2010], , 21. Assuming an electron is confined to a 1 nm wide, region, find the uncertainty in momentum using, Heisenberg uncertainty principle (Dx ´ Dp » h). You, can assume, the uncertainty in position Dx as 1 nm., Assuming p » Dp, find the energy of the electron, in eV., [NCERT Exemplar], , 22. A particle is moving three times as fast as an, electron. The ratio of the de-Broglie wavelength of, the particle to that of the electron is 1.813 ´ 10 -4 ., Calculate the particle’s mass and identify the, particle., [All India 2011], , 23. (i) For what kinetic energy of a neutron will associated, de-Broglie wavelength be 1.40 ´10 -10 m?, , (ii) Also, find the de-Broglie wavelength of a neutron,, in thermal equilibrium with matter, having an, average kinetic energy of (3/2) kT and, temperature is 300 K., [NCERT], , 24. Electrons are emitted from the cathode of a, photocell of negligible work function, when, photons of wavelength l are incident on it. Derive, an expression for the de-Broglie wavelength of the, electrons emitted in terms of the wavelength of the, incident light., [All India 2017 C]

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79, , CBSE Term II Physics XII, , l, , Long Answer (LA) Type Questions, 25. Sketch the graphs showing variation of stopping, potential with frequencies of incident radiations for, two photosensitive materials A and B having, threshold frequencies n A > n B ., (i) In which case is the stopping potential more and, why?, (ii) Does the slope of the graph depend on the nature, of the material used? Explain., [All India 2016], , 26. Define the term cut-off frequency in photoelectric, emission. The threshold frequency of a metal is n., When the light of frequency 2n is incident on the, metal plate, the maximum velocity of photoelectron, is v1 . When the frequency of the incident radiation, is increased to 5n, the maximum velocity of, photoelectrons is v 2 . Find the ratio v1 : v 2 ., , [Foreign 2016], , 27. (i) State two important features of Einstein's, photoelectric equation., (ii) Radiation of frequency 1015 Hz is incident on, two photosensitive surfaces P and Q. There is no, photoemission from surface P. Photoemission, occurs from surface Q but photoelectrons have, zero kinetic energy. Explain these observations, and find the value of work function for surface Q., [Delhi 2017], , 28. (i) Write the important properties of photons which, are used to establish Einstein’s photoelectric, equation., (ii) Use this equation to explain the concept of, (a) threshold frequency and, (b) stopping potential., [Delhi 2015], , 29. Write Einstein’s photoelectric equation. The, maximum kinetic energy of the photoelectrons gets, doubled when the wavelength of light incident on, the surface changes from l1 to l 2 . Derive an, expressions for the threshold wavelength l 0 and, work function for the metal surface. [All India 2015], , 30. In the study of a photoelectric effect, the graph, between the stopping potential V and frequency n, of the incident radiation on two different metals P, and Q is shown below, P, Q, , 4, 2, , V, (volt) 0, –2, , 4, 6, n (× 1014) Hz, , (i) Which one of the two metals has higher, threshold frequency?, (ii) Determine the work function of the metal which, has greater value., (iii) Find the maximum kinetic energy of electron, emitted by light of frequency 8 ´ 1014 Hz for this, metal., [Delhi 2017], , 31. (i) Describe briefly three experimentally observed, features in the phenomenon of photoelectric, effect., (ii) Discuss briefly how wave theory of light cannot, explain these features., [Delhi 2015, 16], , 32. What is the de-Broglie wavelength of, (i) a bullet of mass 0.040 kg travelling at the speed, of 1.0 km/s,, (ii) a ball of mass 0.060 kg moving at a speed of, 1.0 m/s and, (iii) a dust particle of mass 1.0 ´ 10 –9 kg drifting with, a speed of 2.2 m/s?, [NCERT], , 33. A mercury lamp is a convenient source for studying, frequency dependence of photoelectric emission,, since it gives a number of spectral lines ranging, from the UV to the red end of the visible spectrum., In our experiment with rubidium photocell, the, following lines from a mercury source were used, l1 = 3650 Å, l 2 = 4047 Å, l 3 = 4358 Å,, l 4 = 5461 Å , l5 = 6907 Å, The stopping voltages respectively were measured, to be, V 01 = 1.28 V, V 02 = 0.95 V,, V 03 = 0.74 V, V 04 = 0.16 V, V 05 = 0, Determine the value of Planck’s constant h, the, threshold frequency and work function for the, material., [NCERT], , 34. Light of intensity 10 -5 Wm -2 falls on a sodium, photocell of surface area 2 cm 2 . Assuming that, the, top 5 layers of sodium absorb the incident energy,, estimate the time required for photoelectric, emission in the wave picture of radiation. The work, function of the metal is given to be about 2 eV., What is the implication of your answer?, Effective atomic area = 10 -20 m 2 ., [NCERT], , 35. An electron, a-particle and a proton have the same, de-Broglie wavelengths. Which of these particle, has, (i) minimum kinetic energy?, (ii) maximum kinetic energy and why?

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80, , CBSE Term II Physics XII, , In what way has the wave nature of electron beam, exploited in electron microscope?, l, , Case Based Questions, Direction Read the following passage and answer the, questions that follows, , 36. Observation of Photoelectric Effect, The phenomenon of photoelectric emission was, discovered in 1887 by Heinrich Hertz during his, electromagnetic waves by means of spark across the, detector loop were enhanced., Light, , Metal, surface, , Detector, Electrons, Ammeter, , +, , –, Vacuum chamber, , –, +, Battery, , When the emitter plate was illuminated by, ultraviolet light from an arc lamp. According to this, effect, there is emission of electrons from the, surface of metal when a light beam of suitable, frequency is incident on it., (i) Light of wavelength 2500 Å falls on a metal, surface of work function 3.5 eV. What is the, kinetic energy (in eV) of, (a) the fastest and, (b) the slowest electron emitted from the surface?, If the same light falls on another surface of work, function 5.5 eV, what will be the energy of, emitted electrons?, (ii) An electron is accelerated through a potential, difference of 64 V. What is the de-Broglie, wavelength associated with it? To which part of, the electromagnetic spectrum this wavelength, correspond?, [Delhi 2010]

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Chapter Test, 8. Ultraviolet light of wavelength 200 nm is incident on, , Multiple Choice Questions, , 1. If photons of frequency n are incident on the surfaces, n, n, of metals A and B of threshold frequencies and ,, 2, 3, respectively, the ratio of the maximum kinetic energy, of electrons emitted from A to that from B is, (a) 2 : 3, , (b) 3 : 4, , (c) 1 : 3, , (d) 3 : 2, , 2. The kinetic energy of a proton and that of an, a-particle are 4 eV and 1 eV, respectively. The ratio of, the de-Broglie wavelengths associated with them, will, be, (a) 2 : 1, , (b) 1 : 1, , (c) 1 : 2, , (d) 4 : 1, , 3. A photocell connected in an electrical circuit is placed, at a distance d from a source of light. As a result,, current I flows in the circuit. What will be the current, in the circuit when the distance is reduced to d/2?, (a) I, (c) 4I, , 4., , (b) 2I, (d) I / 2, , The de-Broglie wavelength of a neutron at 27°C is l., What will be its wavelength at 927°C?, l, (a), 2, l, (c), 4, , l, (b), 3, l, (d), 9, , 5. Which one of the following statements regarding, photoemission of electrons is correct?, (a) Kinetic energy of electrons increases with the intensity of, incident light., (b) Electrons are emitted when the wavelength of the, incident light is above a certain threshold wavelength., (c) Photoelectric emission is instantaneous with the incidence, of light., (d) Photoelectrons are emitted whenever a gas is irradiated, with ultraviolet light., , Short Answer Type Questions, , 6. What are the energies of photons at the (i) violet and, (ii) red ends of the visible spectrum? The wavelength, of light is about 390 nm for violet and about 760 nm, for red., [Ans. (i) 3.17 eV and (ii) 1.63 eV], , 7. The de-Broglie wavelength of a particle of kinetic, energy K is l. What would be the wavelength of the, K, particle, if its kinetic energy were ?, 4, (Ans. 1 : 2), , polished surface of iron. Work function of the surface is, 4.71 eV. Calculate its stopping potential., (Ans. 1.50 V), , 9. Light of wavelength 488 nm is produced by an argon, laser, which is used in the photoelectric effect. When, light from this spectral line is incident on the emitter,, the stopping (cut-off) potential of photoelectrons is, 0.38 V. Find the work function of the material from, which the emitter is made., (Ans. 2.17 eV), , 10. Find the, (i) maximum frequency and, (ii) minimum wavelength of X-rays produced by 30 kV, electrons., [Ans. (i) 7.24 ´ 10 18 Hz and (ii) 0.0414 nm], , 11. The energy flux of sunlight reaching the surface of the, earth is 1.388 × 10 3 W/m 2 . How many photons (nearly), per square metre are incident on the earth per second?, Assume that the photons in the sunlight have an, average wavelength of 550 nm., (Ans. 3.838 ´ 10 2 ), , Long Answer Type Questions, , 12. What is the de-Broglie wavelength of a nitrogen, molecule in air at 300 K? Assume that, the molecule is, moving with the root-mean-square speed of molecules, at this temperature., (Take, atomic mass of nitrogen = 14.0076 u), (Ans. 0.028 nm), , 13. The work function of caesium metal is 2.14 eV. When, , light of frequency 6 ×10 14 Hz is incident on the metal, surface, photoemission of electrons occurs. What is the, , (i) maximum kinetic energy of the emitted electrons,, (ii) stopping potential and, (iii) maximum speed of the emitted photoelectrons?, , [Ans. (i) 0.35 eV, (ii) 0.35 V and (iii) 350.7 km/s], , 14. Monochromatic light of wavelength 632.8 nm is, produced by a helium-neon laser. The power emitted is, 9.42 mW., (i) Find the energy and momentum of each photon in the light, beam., (ii) How many photons per second, on the average, arrive at a, target irradiated by this beam? (Assume the beam to have, uniform cross-section, which is less than the target area.), (iii) How fast does a hydrogen atom have to travel in order to, have the same momentum as that of the photon?, , [Ans. (i) 3.14 ´ 10 - 19 J, 1. 05 ´ 10 -27 kg-m/s, (ii) 3 ´ 10 16 photons/s, (iii) 0.63 m/s], , Answers, Multiple Choice Questions, 1. (b), , 2. (b), , 3. (c), , 4. (a), , 5. (c), , For Detailed Solutions, Scan the code

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82, , CBSE Term II Physics XII, , EXPLANATIONS, , Thus, at point P (as shown in the given figure), it can be seen, clearly that the electron must have been repelled, so, maximum kinetic energy, K = eV0 ., 7. (d) When a beam of electrons of energy E0 is incident on a, metal surface kept in an evacuated chamber electrons can, be emitted with maximum energy E0 (due to elastic, collision) and with any energy less than E0 , when part of, incident energy of electron is used in liberating the, electrons from the surface of metal., 8. (d) We know that, E = hn and, E = mc 2 (Einstein mass-energy equation), , m = hn/ c 2, ( hc / l ) h, Kinetic mass, m =, =, cl, c2, 9. (b) Given,, Energy of a photon, E = 1 MeV = 106 eV, Þ, , hc = 1240 eV-nm, hc, E=, l, hc 1240 eV- nm, l=, =, E, 106 eV, = 1 . 2 ´ 10-3 nm, , Now,, Now,, Þ, , 10. (d) As the distance of source from the surface increases,, intensity of radiation decreases., 1, Intensity I µ, Q, (distance d) 2, Since, photoelectric current (i) µ intensity (I), Thus, the variation of i versus s is correctly depicted by the, curve d in the given figure., 11. (d) Here for greater intensity I1, more photoelectrons are, emitted and hence saturation current is more. Thus, graph, corresponding to I1 will be above than that of I2 . Since, the, stopping potential is independent of intensity, hence the, graphs converge at same value of stopping potential V0 for, both the intensities I1 and I2 ., Therefore, for fixed frequency and intensity of incident, light, photoelectric current increases with increase in, potential applied to the collector as shown in the graph, Photoelectric, current, , 1. (d) The work function f 0 depends on the properties of the, metal and the nature of its surface., So, its value will vary with the presence of surface, impurities and with increase/decrease in temperature of the, surface., Thus, the value of f 0 for a metal will change, if it is cooled, or heated or coated with some other metal., 2. (a) Hallwach observed that when the uncharged zinc plate is, irradiated by UV-rays, then it becomes, positively charged. Positive charge on this plate is further, enhanced when it was continuously illuminated by, UV-light.This will make the leaves to move apart from each, other, as same type of charge flows through the leaves of an, electroscope, making it repel each other., Thus, it can be concluded that, negatively charged particles, are emitted from zinc plate under the action of UV-light,, making the leaves to move apart from each other., Hence, option (a) is correct., 3. (c) In the phenomenon of electric discharge through gases at, low pressure, as the charged particles emitted from the, cathode collides with the atoms of the gas, coloured glow, appears in the tube., 4. (d) Keeping the frequency n of the light and its intensity I, fixed, firstly with the increase in the positive potential of, collector plate, the photoelectric current increases. At some, stage, for a certain positive potential of collector plate, the, photoelectric current becomes maximum or saturates. If we, further increase the potential of collector plate, the, photocurrent does not increase., 5. (a) We know that, for a given photosensitive material and, frequency of incident radiation (above the threshold, frequency), the photoelectric current is directly, proportional to the intensity of incident light only. Since, in, the given case, only the work function has been changed, from W1 to W2 keeping the intensity of the incident light, same. This means, the value of photocurrent in both the, cases will also remain same. i.e. i1 = i2 ., 6. (c) In the given condition, photocurrent is zero when the, stopping potential is sufficient to repel even the most, energetic photoelectrons, with the maximum kinetic energy, K, so that K = eV0 ., , and, , mc 2 = hn, , \, , PART 1, , -V0, , I1, I2, , O Collector plate, potential, , 12. (b) Since in the graph retarding potential is same in graphs a, and b and photocurrent is different, so for curves they have, same frequency but different intensity of light., Photocurrent, , b, , a, , c, Retarding potential Anode potential, , 13. (a) For a particular frequency of incident radiation, the, minimum negative (retarding) potential V0 given to the

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83, , CBSE Term II Physics XII, , collector plate for which the photoelectric current stops, or becomes zero is called the cut-off or stopping potential., From given the graph, the value of stopping potential,, V0 = - 054, . V., 14. (c) For incident radiation at same intensity and at varying, frequencies, the given graph represents the variation of, photocurrent with collector plate potential. This can be as, shown in graph below, Photocurrent Y, , Saturation current Z, n2, , n1, , –V02 –V01, O, Retarding, potential, , Collector plate, potential X, , Thus, X ® collector plate potential,, Y ® photocurrent and Z ® saturation current., 15. (a) From the photoelectric equation,, E = W + KE, hc hc, Þ, KE = E - W =, l l0, æ1, æ l - lö, 1 ö, = hc ç ÷ = hc ç 0, ÷, è l l0 ø, è ll 0 ø, 16. (c) Momentum, p = m ´ c Þ p µ m, Because mass of a-particle is more in comparison to other, particles, so the momentum of a-particle will be highest., 17. (a) Given, v = v0 i$ Þ B = B0 $j, Force on moving electron due to magnetic field,, F = - e( v ´ B ), = - e[ v0 $i ´ B0 $j], Þ, = - ev B k$, 0, , 0, , As this force is perpendicular to v and B, so the magnitude of, v will not change, i.e. momentum ( = mv) will remain, constant in magnitude. Hence, de-Broglie wavelength, h, remains constant., l=, mv, h, ...(i), 18. (c) de-Broglie wavelength, l =, 2mK, K, When the kinetic energy is , then, 9, h, 3h, [using Eq. (i)], l¢ =, =, = 3l, 2, mK, æKö, 2m ç ÷, è 9ø, 19. (b) We know that, the relation between l and K is given by, h, l=, 2mK, 1, or, lµ, m, Since, m p = m n, hence l p = l n, As,, m a > m p , therefore l a < l p, , As,, m e < m n, therefore l e > l n, Hence, l a < l p = l n < l e, 20. (a) In a detector loop, when the light falls on the metal, surface, i.e. the emitter plate, some electrons near the, surface absorb enough energy from incident radiations, i.e., from UV-light, to overcome the attraction of the positive, ions in the material of the surface., After gaining sufficient energy from UV-light, the electrons, escape from the emitter plate into the surrounding space,, thus enhancing the high voltage spark across detector loop., Therefore, both A and R are true and R is the correct, explanation of A., 21. (b) Stopping potential is dependent on frequency, but, independent of the intensity. So, increasing the distance,, affects the intensity as., 1, Intensity µ, (Distance) 2, So, the stopping potential will not change. But value of, saturation current depends on the intensity of incident, radiation. So, more the distance, the intensity will decrease., Hence, the saturation current will also decrease., Therefore, both A and R are true but R is not the correct, explanation of A., 22. (c) Since the frequency of ultraviolet light is less than the, frequency of X-rays, the energy of each incident photon will, be more for X-rays, KE photoelectron = hn - f, Stopping potential is used to stop the fastest photoelectron,, hn f, V0 =, e, e, So, KE max and V0 both increase., But KE ranges from zero to KE max because of loss of energy, due to subsequent collisions before getting ejected and not, due to range of frequencies in the incident light., Therefore, A is true but R is false., 23. A person approaching a doorway may intrupt a light beam, which is incident on photocell., This interuption will leads to abrupt change in the amount, of photocurrent. Thus, this change in photocurrent helps to, start a motor which is fitted in the doorway which opens the, door., Therefore, both A and R are true and R is the correct, explanation of A., h, 24. (c) de-Broglie wavelength, l =, p, Since, their momenta are same due to conservation of linear, momentum., Hence, their wavelengths are same but their speed will be, different., Therefore, A is true but R is false., 25. (i) (d) According to photoelectric effect, no electrons were, emitted at all when the frequency of the incident light, was smaller than a certain minimum value, called the, threshold frequency. This minimum frequency depends, on the nature of the material of the emitter plate., (ii) (c) When the wavelength of incident light is l £ l 0 , then, the electrons will come out of the metal surface.

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84, , CBSE Term II Physics XII, , (iii) (d) Photocurrent varies linearly with intensity. The, photocurrent is directly proportional to the number of, photoelectrons emitted per second. This implies that, it, is a straight line passing through origin., Hence, option (d) is correct., (iv) (a) Given, work function = 2eV, Energy of incident photon = 3.4 eV, From Einstein’s equation of photoelectric effect,, hn = hn 0 + K, 3.4 eV = 2 eV + K, K = 3 .4 eV -2.0 eV = 1.4 eV, (v) (d) According to photoelectric effect, light consists of, packets of energy., In interaction of radiation with matter, radiation behaves, as, if it is made up of particles called photons. Thus,, photons confirms the particle nature of light., Each photon have energy, E = hn., , PART 2, 1. According to first statement, when the materials which, absorb photons of shorter wavelength has high energy of the, incident photon on the material and low energy of emitted, photon of longer wavelength., But in second statement, the energy of the incident photon, is low for the substances which has to absorb photons of, larger wavelength and energy of emitted photon is high to, emit light of shorter wavelength. This means in this, statement material has to supply the energy for the emission, of photons. But this is not possible for a stable substances., 2. The number of photons of beam A = n A, The number of photons of beam B = n B, According to the question, n A = 2n B, Let n A be the frequency of beam A and n B be the frequency, of beam B., Intensity µ Energy of photons, \, Þ, I µ hn ´ Number of photons, IA n A n A, \, =, IB n Bn B, According to the question, IA = IB, \, n A n A = n Bn B, n A nB 1, or, =, =, n B nA 2, So,, nB = 2 nA, 3. The intensities for both the monochromatic radiations are, same but their frequencies are different. It represents, (i) the number of electrons ejected in two cases are same, because it depends on the number of incident photons., (ii) As, KE max = hn - f 0 = hc / l - f 0, (Einstein’s photoelectric equation), Since, wavelength l of violet colour is minimum. So,, the KE max of violet radiation will be more., \ The KE max of violet radiation will be more., 4. The variation of stopping potential with the frequency of, radiation, incident on a metal plate is a straight line AB as, shown in the figure., , Take, two points C and D on the graph., Stopping V2, potential (V), V1, , D, , B, , C, , A, n0, n1 n2, Frequency of radiation (n), , The corresponding frequency of radiation is n1, n 2 and, stopping potential is V1, V2 ., Then,, eV1 = hn1 - f 0 and eV2 = hn 2 - f 0, \, e ( V2 - V1 ) = h ( n 2 - n1 ), e ( V2 - V1 ), or, h=, = e ´ slope of ( V - n ) graph, n 2 - n1, Thus, Planck’s constant can be determined using graph., 5. Given, l = 4125, . nm = 412.5 ´ 10-9 m, \, , E=, , hc, 6.63 ´ 10-34 ´ 3 ´ 108, eV = 3.01 eV, =, l 412.5 ´ 10-9 ´ 1 . 6 ´ 10-19, , From the given question, work function f of the following, metals are given as, Na ® 1.92 eV, K ® 2.15 eV, Ca ® 3.20 eV, Mo ® 4.17 eV, As the given energy is greater than the work function of Na, and K only, hence these metals shows photoelectric, emission., 6. Given, f = 214, . eV = 214, . ´ 1.6 ´ 10-19 J, and, Vstopping = 0.6 V, (i) Work function, f = hn 0, So, threshold frequency, n 0 =, Þ, , f 214, . ´ 1.6 ´ 10-19, =, h, 6.62 ´ 10-34, , n 0 = 5.17 ´ 1014 Hz, , (ii) As, K max = eV0 = 0.6 eV, Energy of photon, E = K max + f, = 0.6 eV + 2.14 eV, = 2.74 eV, hc, Hence, wavelength of photon, l =, E, 8, -34, 6.62 ´ 10 ´ 3 ´ 10, (Qc = 3 ´ 108 m/s), =, 2.74 ´ 1.6 ´ 10-19, = 4530 Å, , h( n - n0 ), e, 6.63 ´ 10-34 ( 8.2 ´ 1014 - 3.3 ´ 1014 ), =, 1.6 ´ 10–19, = 2.03 V, , 7. Cut-off voltage , V0 =, , 8. Given, V = 3.3 V, and frequency of photons, n = 8 ´ 1014 Hz, As we know,, eV0 = hn - f, So,, f = hn - eV0

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85, , CBSE Term II Physics XII, 6.63 ´ 10-34 ´ 8 ´ 1014, eV - 3.3 eV, 1.6 ´ 10-19, = ( 3.31 - 3.3) eV, f = 0.01 eV, , Þ, , Þ f ( eV ) =, Þ, , or, , 9. Given, for the first condition, l = 600nm, For the second condition, l ¢ = 400nm, K max, ¢ = 2K max, hc, Here,, K max, -f, ¢ =, l, hc, Þ, 2K max =, - f0, l¢, æ 1240, ö æ 1240, ö, Þ, 2ç, - f÷ » ç, - f÷, (Qhc » 1240 eV -nm ), è 600, ø è 400, ø, 1240, = 1 .03 eV, 1200, 10. From Einstein’s photoelectric equation,, 1, 2, K max = mvmax, = h( n - n 0 ), 2, where, h = Planck’s constant,, n = frequency of incident light, and n 0 = threshold frequency of the photosensitive surface., So, for photoemission to takes place, n > n 0 ., As the wavelength of light incident is same for all the three, surfaces, so, (i) threshold frequency of surface A is higher than the, frequency of incident light, as no emission takes, place., (ii) threshold frequency of surface B is equal to the, frequency of incident light, as photoelectrons are just, emitted., (iii) threshold frequency of surface C is lower than the, frequency of incident light, as the emitted, photoelectrons have some kinetic energy., \, ( n0 )A > ( n0 )B > ( n0 )C, 11. (i) The variation of stopping potential V0 for the, 1, photoelectron versus graph is as shown below, l, Þ, , f=, , V0, V2, V1, O, , B, D, C, , 1/l1 1/l2, , 1/l, , Take, any two points C and D on the graph as shown, above., According to Einstein’s photoelectric equation, we can, hc, write, eV1 =, …(i), - f0, l1, where, f 0 is the work function of metal A., hc, and eV2 =, - f0, l2, Subtracting Eq. (i) from Eq. (ii), we get, , …(ii), , æ 1, 1ö, e( V2 - V1 ) = hc ç, - ÷, è l 2 l1 ø, e( V2 - V1 ), e( V2 - V1 ) l1l 2, h=, =, æ 1, c( l1 - l 2 ), 1ö, - ÷, cç, è l 2 l1 ø, , Thus, Planck’s constant can be determined from graph., Note Since, h is a constant, so it will be same for both, metals A and B., (ii) Stopping potential V0 for the electrons emitted will not, be affected by the increase in distance between light, source and the metal surface A., This is because, V0 is independent of the intensity of the, incident light but depends only upon the frequency (or, wavelength) of incident light., So, increase in the given distance affects only the, intensity of the light but not the frequency. Thus, V0, remains same., 12. Two salient features observed in photoelectric effect and, their explanation on the basis of Einstein’s photoelectric, equation is given as below, (i) Threshold Frequency For KE max ³ 0., Þ, n ³ n0, i.e. The phenomenon of photoelectric effect takes place,, when incident frequency is greater or equal to a, minimum frequency (threshold frequency) n 0 fixed for, given metal., (ii) Effect of Intensity of Incident Light The number of, photons incident per unit time per unit area increases, with the increase of intensity of incident light. More, number of photons facilitates ejection of more number, of photoelectrons from metal surface leads to further, increase of photocurrent till its saturation value is, reached., 13. The wave theory of light is not able to explain the observed, features of photoelectric current because of following, reasons, (i) According to wave nature of light, the free electrons at, the surface of the metal absorb the radiant energy, continuously., The greater the intensity of radiation, the greater should, be the energy obtained by each electrons. The maximum, kinetic energy of the photoelectrons on the surface is, then expected to increase with increase in intensity. But, according to experimental facts, the maximum kinetic, energy of ejected photoelectrons is independent of, intensity of incident radiation., (ii) Wave theory states that, energy carried by wave is, independent of frequency of light wave and hence wave of, high intensity and low frequency (less than threshold, frequency) should stimulate photoelectric emission but, practically, it does not happen., Considering the following few properties of photon, the, above problem was resolved, (i) In interaction of radiation with radiation behaves as, if it, is made up of particle called photon., (ii) Energy of a photon is directly proportional to the, frequency of the incident light., 14. For a given frequency, intensity of light in the photon, picture is determined by

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86, , CBSE Term II Physics XII, , I=, , energy of photons n ´ hn, =, area ´ time, A´t, , where, n is the number of photons incident normally on, cross-sectional area A in time t., 15. (i) The energy of photoelectrons in a photocell is given by, hc, E=, = hn Þ E µ n, l, So, if the frequency of light incident on the cathode is, increased, the energy of photoelectrons increases, linearly., (ii) As, photoelectric current/photocurrent of the photocell is, independent of frequency of the incident light, till, intensity remains constant. So, when the frequency of, light incident on the cathode of photocell is increased, keeping other factors same, the photoelectric current, remains the same., 16. Given, power of lamp, P = 100 W, , Thus, wavelength of electromagnetic radiation is equal to the, de-Broglie wavelength., 19. (i) The de-Broglie wavelength of a particle is given by, h, l=, 2mV0 q, Since, a-particle and proton both are accelerated, through the same potential V0 ., 1, \, lµ, mq, , (i) Energy of each photon,, hc 6.63 ´ 10- 34 ´ 3 ´ 108, E=, =, l, 589 ´ 10- 9, , =, , 3.38 ´ 10-19, eV = 211, . eV, 1.6 ´ 10-19, , (ii) Let n photons are delivered per second., Power, (from P = En ), \, n=, Energy of each photon, 100, =, = 3 ´ 1020 photons/s, 3.38 ´ 10-19, = 3 ´ 1020 photons/s, 17. Energy of the incident radiation of wavelength l,, hc ( 6.63 ´ 10-34 ) ´ ( 3 ´ 108 ), E=, =, l, 3300 ´ 10-10 ´ 1 .6 ´ 10-19, = 3.76 eV, This energy of the incident radiation is greater than the, work function of Na and K but less than those of Mo and Ni., So, photoelectric emission will occur only in Na and K, metals and not in Mo and Ni., If the laser is brought closer, the intensity of incident, radiation increases. This does not affect the result regarding, Mo and Ni metals, while photoelectric current from Na and K, will increase in proportion to intensity., 18. The momentum of an electromagnetic wave of frequency n,, wavelength l is given by, hn h, p=, =, c, l, h, or, l=, p, de-Broglie wavelength of photon,, h, l=, p, , =, , mp qp, ma qa, , Mass of a-particle = 4 ´ mass of proton, mp 1, ma = 4 ´ mp Þ, =, ma 4, , Planck’s constant, h = 6.63 ´ 10- 34 J-s, , = 3.38 ´ 10-19 J, , lp, , As, charge on a-particle = 2 ´ charge on proton, qp 1, =, q a = 2q p Þ, qa 2, , Wavelength of the sodium light, l = 589 nm = 589 ´ 10- 9 m, , (Q c = 3 ´ 108 m / s ), , la, , or, , la, 1 1, 1, =, × =, lp, 4 2 2 2, , \, Þ, , lp = 2 2la, , i. e. Proton has greater de-Broglie wavelength than that of, a-particle., (ii) KE µ q (for same accelerating potential), The charge of an a-particle is more as compared to a, proton, so it will have a greater value of KE . Hence,, proton will have lesser KE., 20. de-Broglie wavelength of accelerating charged particle is, given by, h, l=, 2mqV, h, = constant, 2mq, h, (i) The slope of the lines represents, 2mq, , Þ, , l V =, , where, h = Planck’s constant, q = charge and, m = mass of charged particle., (ii) 1H2 and 1H3 carry same charge (as they have same atomic, number)., 1, \, l V µ, m, The lighter mass, i.e. 1H2 is represented by line of greater, slope, i.e. A and similarly, 1H3 by line B., 21. Given, Dx = 1 nm = 10-9 m, As, DxDp » h, h, h, \, Dp =, =, Dx 2pDx, 6.6 ´ 10-34 J-s, =, 2 ´ ( 22 / 7) 10-9 m, = 1 .05 ´ 10- 25 kg -m / s, \, , Energy, E =, , p2 ( Dp) 2, =, 2m, 2m, , (Q p » Dp)

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87, , CBSE Term II Physics XII, , =, , ( 1 .05 ´ 10-25 ) 2, J, 2 ´ 9.1 ´ 10-31, , ( 1.05 ´ 10-25 ) 2, eV, 2 ´ 9.1 ´ 10-31 ´ 1.6 ´ 10-19, = 3.8 ´ 10-2 eV, = 3 velectron, =, , 22. Given, vparticle, , l particle = 1.813 ´ 10, , and, , m particle, , Þ, , m electron, =, , \, , =, , ...(i), , l electron, , h, mv, , l=, , As,, , -4, , V, , (de-Broglie equation), , l electron ´ velectron, l particle ´ vparticle, , Stopping, potential (V), , l electron ´ velectron, 1.813 ´ 10- 4 ´ l electron ´ 3velectron, = 1839 ´ 9.1 ´ 10-31, = 1.673 ´ 10-27 kg, , \ Particle is either a proton or a neutron., 23. (i) de-Broglie wavelength, l = 1.40 ´ 10-10 m, Mass of neutron, m n = 1.675 ´ 10- 27 kg, Using the formula, wavelength associated with kinetic, energy,, h, l=, 2m KE, KE =, =, , h2, 2l2 m n, ( 6.63 ´ 10- 34 ) 2, 2 ´ ( 1.40 ´ 10-10 ) 2 ´ 1.675 ´ 10- 27, , = 6.686 ´ 10- 21 J, (ii) Kinetic energy associated with temperature,, 3, 3, KE = kT = (1.38 ´ 10-23 ) ´ 300, 2, 2, = 6.21 ´ 10-21 J, (Q absolute temperature, T = 300 K and, Boltzmann’s constant, k = 1.38 ´ 10-23 J / K), de-Broglie wavelength associated with kinetic energy,, h, l=, 2m n KE, =, , 6.63 ´ 10-34, 2 ´ 1.675 ´ 10-27 ´ 6.21 ´ 10-21, , = 1.45 ´ 10-10 m, = 1.45 Å, 24. We know that,, hc hc 1, =, + mv2, l l0 2, Neglecting the wave equation, we get, hc 1, = mv2, l 2, de-Broglie wavelength is given by, , B, A, , A, nB, nA n, Frequency of radiation (n), , [from (Eq. (i)], , m particle = 1839 m electron, , or, , h, mv, h l, hl, le =, =, 2me, 2mhe, 25. We know that, K max = eV = h( n - n 0 ), h, h, or, V = n - n0, e, e, le =, , n, , (i) From the graph for the same value of n , stopping, potential is more for material B., h, As, V = ( n - n 0 ), e, \ V is higher for lower value of n 0 . Here n B < n A , so, VB > VA ., h, (ii) Slope of the graph is given by which is constant for all, e, the materials. Hence, slope of the graph does not depend, on the nature of the material used., 26. For a given material, there exists a certain minimum, frequency of the incident radiation below which no, emissions of photoelectrons takes place. This frequency is, called threshold frequency or cut-off frequency of that, material. Above the threshold frequency, the maximum, kinetic energy of the emitted photoelectrons or equivalent, stopping potential is independent of intensity of incident, light but depends only upon the frequency (or wavelength), of the incident light., Given that, threshold frequency of metal is n and frequency, of light is 2n. Using Einstein’s equation for photoelectric, effect, we can write, 1, … (i), h ( 2n - n ) = mv12, 2, Similarly, for light having frequency 5 f , we have, 1, … (ii), h ( 5 n - n ) = mv22, 2, Using Eqs. (i) and (ii), we get, n, v2, = 12, 4n v2, Þ, , v1, 1, v, 1, =, Þ 1 =, v2, 4, v2 2, , Hence, the ratio is 1 : 2., 27. (i) Refer to Q-12 (SA) part-2., (ii) Energy of incident photon is less than work function of P, but just equal to that of Q., For surface Q,, hn, Work function, f 0 =, (eV), e

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88, , CBSE Term II Physics XII, , 6.6 ´ 10-34 ´ 1015, = 41, . eV, 1.6 ´ 10-19, 28. (i) Important properties of photons which are used to, establish Einstein’s photoelectric equations as given, below, (a) In interaction of radiation with matter, radiation, behaves as, if it is made up of particles called photons., (b) All photons of light of a particular frequency n or, wavelength l have the same energy E ( = hn = hc / l ), and momentum p ( = hn / c = h / l ), whatever the, intensity of radiation may be., (c) Photons are not deflected by electric and magnetic, fields. This shows that, photons are electrically, neutral., (ii) Since, Einstein’s photoelectric equation is given by, 1, 2, KE max = mvmax, = hn - hn 0 = eV0, 2, (a) For a given material, there exist a certain minimum, frequency of the incident radiation, below which no, emission of photoelectron takes place. This frequency, is called threshold frequency ( n 0 ). Above threshold, frequency, the maximum kinetic energy of the, emitted photoelectron or equivalent stopping, potential is independent of the intensity of the, incident light but depends only upon the frequency of, the incident light., (b) If the collecting plate in the photoelectric apparatus, is made at high negative potential, then most of the, high energetic electrons get repelled back along the, same path and the photoelectric current in the circuit, becomes zero. So, for a particular frequency of, incident radiation, the minimum negative potential, for which the electric current becomes zero is called, cut-off or stopping potential ( V0 )., =, , 29. Einstein’s photoelectric equation, 1, 2, K max = mvmax, = hn - f 0 = hn - hn 0, 2, According to the question,, hc, K max =, - f0, l1, , ...(i), , Let the maximum kinetic energy for the incident radiation, (of wavelength l 2 ) be K¢max ., hc, ...(ii), Þ, K ¢max =, - f0, l2, From Eqs. (i) and (ii), we get, æ hc, ö, hc, - f0 = 2 ç, - f0 ÷, è l1, ø, l2, Þ, , æ2, 1 ö, f 0 = hc ç ÷, è l1 l 2 ø, , Þ, , æ2, 1 ö, hn 0 = hc ç ÷, è l1 l 2 ø, æ2, 1 ö, c, =cç ÷, è l1 l 2 ø, l0, , ¢ = 2 K max ), (QK max, , Þ, , æ2, 1, 1 ö, =ç ÷, l 0 è l1 l 2 ø, , Þ, , æ l1l 2 ö, l0 = ç, ÷, è 2l 2 - l1 ø, , 30. (i) Since, Q has greater negative intercept, it will have, greater f (work function) and hence higher threshold, frequency., (ii) To know work function of Q, we put, V = 0 in the following equation, hv f, V =, e, e, hn f, Þ, 0=, - Þ f = hn, e, e, \, f = 6.6 ´ 10-34 ´ 6 ´ 1014 J, 6.6 ´ 6 ´ 10-20, eV = 2.5 eV, 1.6 ´ 10-19, (iii) From the equation, nl = c, c 3 ´ 108 30, Þ, l= =, =, ´ 10-7 m, n 8 ´ 1014 8, 30, =, ´ 103 ´ 10-10 m, 8, 30, =, ´ 103 Å = 3750 Å, 8, 12375 12375, Energy =, eV = 3.3 eV, =, l(Å), 3750, =, , \ Maximum KE of emitted electron = 3.3 - 2.5 eV, = 0.8 eV, 31. (i) Three experimentally observed features in the, phenomenon of photoelectric effect is, (a) Intensity When intensity of incident light increases, as one photon ejects one electron, the increase in, intensity will increase the number of ejected, electrons. Frequency has no effect on photoelectron., (b) Frequency When the frequency of incident photon, increases, the kinetic energy of the emitted, electrons increases. Intensity has no effect on, kinetic energy of photoelectron., (c) No Time Lag When energy of incident photon is, greater than the work function, the photoelectron is, immediately ejected. Thus, there is no time lag, between the incidence of light and emission of, photoelectron., (ii) Also, according to the wave theory, the absorption of, energy by electron takes place continuously over the, entire wavefront of the radiation., Hence, it will take hours or more for a single electron to, come out of the metal which contradicts the, experimental fact that photoelectron emission is, instantaneous., 32. (i) Given, mass of bullet, m = 0.040 kg, Speed of bullet, v = 1000 m/s, h, 6.63 ´ 10-34, de-Broglie wavelength, l =, =, mv 0.040 ´ 1 ´ 103, = 1 .66 ´ 10-35 m, (ii) Mass of the ball, m = 0.060 kg, and speed of the ball, v = 1 m/s

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89, , CBSE Term II Physics XII, h 6.63 ´ 10-34, =, mv, 0.060 ´ 1, = 1.1 ´ 10-32 m, (iii) Mass of a dust particle, m = 1 ´ 10-9 kg, , de-Broglie wavelength, l =, , and speed of the dust particle, v = 2.2 m/s, h, 6.63 ´ 10-34, de-Broglie wavelength, l =, =, mv 1 ´ 10-9 ´ 2.2, = 3.0 ´ 10-25 m, 33. Given, the following wavelengths from a mercury source, were used, l1 = 3650 Å = 3650 ´ 10-10 m, l 2 = 4047 Å = 4047 ´ 10-10 m, l 3 = 4358 Å = 4358 ´ 10-10 m, l 4 = 5461 Å = 5461 ´ 10-10 m, , 2 ´ 1.6 ´ 10-19, = 1.6 ´ 107 s, 2 ´ 10-26, hc hc 1 2, 35. We know that,, =, + mv, l l0 2, , The stopping voltages are as follows, V01 = 1.28 V, V02 = 0.95 V, V03 = 0.74 V, V04 = 0.16 V and V05 = 0, Frequencies corresponding to wavelengths,, c, 3 ´ 108, = 8.219 ´ 1014 Hz, =, l1 3650 ´ 10-10, , Similarly,, n 2 = 7.412 ´ 1014 Hz,, n 4 = 5.493 ´ 1014 Hz ,, , n 3 = 6.884 ´ 1014 Hz, n5 = 4.343 ´ 1014 Hz, , As we know that, eV0 = hn - f 0, hn f 0, V0 =, e, e, As the graph between V0 and frequency n is a straight line., h, The slope of this graph gives the values of ., e, h V01 - V04, 1 .28 - 016, ., \, =, =, e, n1 - n 4, ( 8.219 - 5 .493) ´ 1014, 1.12, =, 2.726 ´ 1014, 1 .12 ´ 1 .6 ´ 10-19, = 6.573 ´ 10-34 J-s, h=, 2 .726 ´ 1014, As, n average = 5 ´ 1014 Hz, , 2 ´ 10-9, = 2 ´ 10-26 W, 1017, \ Time required for photoelectric emission will be,, Energy required per electron for ejection, t=, Energy absorbed per second per atom, =, , =, , l5 = 6907 Å = 6907 ´ 10-10 m, , n1 =, , Incident power, P = Intensity ´ Area, = 10-5 ´ 2 ´ 10- 4, = 2 ´ 10-9 W, According to wave picture, the incident power is uniformly, absorbed by all the electrons continuously., Hence, energy absorbed per second per electron, incident power, =, number of electrons of five layers, , (given), , \Work function, f 0 = hn 0 = 6.573 ´ 10-34 ´ 5 ´ 1014, = 32.865 ´ 10- 20 J, = 2.05 eV, 34. Here, I = 10-5 Wm -2 , A = 2 ´ 10- 4 m 2, n = 5, t = ?,, f 0 = 2 eV = 2 ´ 1.6 ´ 10- 19 J, Sodium has one conduction electron per atom and effective, atomic area = 10-20 m 2, Number of conduction electrons in five layers, 5 ´ Area of one layer, =, Effective atomic area, 5 ´ 2 ´ 10- 4, =, = 1017, 10-20, , Neglecting the work function, we get, hc 1, = mv 2, l 2, 2hc, Þ, v=, ml, de-Broglie wavelength is given by, h, le =, mv, h l, hl, \, le =, =, 2, mc, 2mhc, 36. (i) Wavelength of incident radiation, l = 2500 Å, Work function, f 0 = 3.5 eV, According to Einstein’s photoelectric equation,, hc, = f 0 + KE max, l, hc, Þ, KE max =, - f0, l, é ( 6.63 ´ 10- 34 ) ( 3 ´ 108 ), ù, 1, =ê, ´, - 3.5 ú eV, -10, -19, 2500, ´, 10, 1, ., 6, ´, 10, ë, û, = ( 4.97 - 35, . ) eV = 1.47 eV, (a) KE of fastest electron = 1.47 eV, (b) KE of slowest electron = 0 eV, If the same light (having energy 4.97 eV) falls on the, surface (of work function 5.5 eV), then no photoelectron, will emit., (ii) Given, V = 64 V, Now, from de-Broglie equation,, 12.27, l=, Å, V, 12.27, =, Å, 64, 12.27, =, Å, 8, nm, = 0153, ., This wavelength belongs to the X-ray part of the, electromagnetic spectrum.

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90, , CBSE Term II Physics XII, , CHAPTER 05, , Atoms, In this Chapter..., l, , a-particles Scattering, Experiment by Rutherford, , l Electron Orbits, l Bohr's Model of Hydrogen Atom, , l Rutherford's Model of Atom, , Atoms are the basic units of matter and defining structure of, elements. In 1898, JJ Thomson for the first time proposed, that the physical structure of atom and named it as, Plum-Pudding model., Then, Rutherford performed an experiment on a-particle, scattering in early 90’s to investigate the atomic structure., But his work was rejected on the basis of classical theory. His, shortcomings was rectified by Neils Bohr through his atomic, model., , a-Particles Scattering, Experiment by Rutherford, This experiment was suggested by Rutherford in 1911 as, given in the figure below, Radioactive, source, , Most, a-particles, pass in, straight line, , Gold foil, (10, a, , S, Lead cavity, , a, , –8, , m thick), , q, ZnS, screen, , Collimator, , About one a-particle in 8000, a-particles is reflected back, , Microscope, detector, , Experimental arrangement for Rutherford’s, a-particle scattering experiment, , H Geiger and E Marsden took a collimated beam of, a-particles of energy 5.5 MeV and made it fall on, 2.1 ´ 10 -7 m thick gold foil. The a-particles were observed, , l Hydrogen Spectrum, , through a rotatable detector consisting of a zinc sulphide, screen and microscope and it was found that a-particles got, scattered., Observations, Rutherford made the following observations from his, experiment that are given below, (i) Most of the a-particles passed through the gold foil, without any appreciable deflection., (ii) Only about 0.14% of the incident a-particles scattered, by more than 1°., (iii) About one a-particle in every 8000 a-particles, deflected by more than 90°., (iv) The number of a-particles scattered per unit area, N( q) at scattering angle q varies inversely as sin 4 q/ 2 ., 1, N( q) µ 4, sin q/ 2, (v) The force between a-particles and nucleus is given by, 1 ( 2e) ( Ze), F=, ×, 4 pe 0, r2, where, r is the distance between the a-particles and, the nucleus., Conclusions, On the basis of his experiment, Rutherford concluded that, (i) Atom has a lot of empty space and practically the, entire mass of the atom is confined to an extremely, small central core called nucleus, whose size is of the, order from 10 - 15 m to 10 - 14 m., (ii) Scattering of a-particles (positively charged) is due to, the Coulomb’s law for electrostatic force of repulsion, between the positive charge of nucleus and a-particles.

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91, , CBSE Term II Physics XII, , (iii) Distance between electron and nucleus is from 104 to, 105 times the size of the nucleus itself., (iv) More is the distance of the velocity vector of an, a-particle from the central line of the nucleus, lesser, is the angle of scattering., , Rutherford’s Model of Atom, The essential features of Rutherford’s nuclear model of the, atom or planetary model of the atom are given below, (i) Every atom consists of a central core, called the, atomic nucleus, in which the entire positive charge, and almost entire mass of the atom is concentrated., (ii) The size of nucleus is of the order of 10 -15 m, which is, very small as compared to the size of the atom which, is of the order of 10 -10 m., (iii) The atomic nucleus is surrounded by certain number, of electrons. As atom on the whole is electrically, neutral, the total negative charge of electrons, surrounding the nucleus is equal to total positive, charge on the nucleus., (iv) These electrons revolve around the nucleus in various, circular orbits as the planets do around the sun., The centripetal force required by electrons for, revolution is provided by the electrostatic force of, attraction between the electrons and nucleus., Distance of Closest Approach, At a certain distance r0 from the nucleus, whole of the KE of, a-particle converts into electrostatic potential energy and, a-particles cannot go further close to nucleus, this distance, ( r0 ) is called distance of closest approach., It is given as, \, , 1 2Ze 2, r0 =, ×, 4 pe0, K, , or, , r0 =, , 1, 2Ze 2, ×, 4 pe0 æ 1, 2ö, ç mv ÷, 2, è, ø, , where, m = mass of a-particle and v = initial velocity of, a-particle., Angle of Scattering (q), Angle by which a-particle gets deviated from its original path, around the nucleus is called angle of scattering., Impact Parameter (b), Perpendicular distance of the velocity vector of a-particle, from the central line of the nucleus of the atom is called, impact parameter., q, Ze 2 cot, 1, 2, b=, ×, 4 pe 0, KE, where, b = impact parameter,, q = angle of scattering, , 1, mv 2 ., 2, In case of head-on-collision, the impact parameter is, minimum and the a-particle rebounds back (q = p)., For a large impact parameter, the a-particle goes nearly, undeviated and has a small deflection ( q = 0 ° )., , and KE = kinetic energy of a-particle =, , q, , —, b, —, , Target nucleus, , Trajectory of a-particles in the Coulombic field of, a target nucleus, , Electron Orbits, The Rutherford nuclear model of the atom pictures the atom, as an electrically neutral sphere consisting of a very small,, massive and positively charged nucleus at the centre, surrounded by the revolving electrons in their respective, dynamically stable orbits., The electrostatic force of attraction Fe between the, revolving electrons and centripetal force Fc keep them in, their orbits., \, Fc = Fe, mv 2, 1 Ze 2, Þ, =, ×, r, 4 pe0 r 2, Thus, the relation between the orbit radius and the electron, velocity is, Ze 2, r=, 4pe0 mv 2, For hydrogen, Z = 1, e2, Therefore, r =, 4pe 0 mv 2, The kinetic energy K and electrostatic potential energy U of, the electron in H-atom are, 1, e2 æ, e2 ö, çQ mv 2 =, ÷, K = mv 2 =, 2, 8pe0 r çè, 4 pe 0 r ÷ø, and, , U =-, , e2, 4pe0 r, , Thus, the total mechanical energy E of the electron in a, H-atom is, e2, E=8pe 0 r, The total energy of the electron is negative. This implies the, fact that the electron is bound to the nucleus.

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92, , CBSE Term II Physics XII, , Drawbacks of Rutherford’s Model, , Bohr’s Theory, , Rutherford’s model suffers two major drawbacks, (i) According to classical electromagnetic theory, the, electrons must radiate energy in the form of, electromagnetic wave., Due to this continuous loss of energy, radii of their, orbits should be continuously decreasing and, ultimately the electrons should fall in the nucleus., Thus, atom cannot remain stable., (ii) Due to continuous decrease in radii of electron’s, orbit, the frequency of revolution of electron will also, change. According to classical theory of, electromagnetism, frequency of EM wave emitted by, electron is equal to frequency of revolution of, electron., So, due to continuous change in frequency of, revolution of electron, it will radiate EM waves of all, frequencies, i.e. the spectrum of these waves will be, continuous in nature. But, this is not the case,, experimentally we get line spectrum. Rutherford, model was unable to explain line spectrum., , Bohr’s model is valid for all one electron atoms or ions which, consists of a tiny positively charged nucleus and an electron, revolving in a stable circular orbit around the nucleus. These, one electron atoms or ions can be called hydrogen like, atoms. e.g. Singly ionised helium (He + ) and doubly ionised, lithium (Li 2+ ), v, , Bohr’s Model of Hydrogen Atom, , Þ, , Bohr combined classical and early quantum concepts and, gave his theory in the form of three postulates, (i) Bohr’s first postulate states that, an electron in an, atom could revolve in certain stable orbits without the, emission of radiant energy, contrary to the predictions, of electromagnetic theory. According to this postulate,, each atom has certain definite stable states in which it, can exist and each possible state has definite total, energy. These are called the stationary states of the, atom., (ii) Bohr’s second postulate states that, the electron, revolves around the nucleus only in those orbits for, which the angular momentum is some integral, multiple of h / 2p, where h is the Planck’s constant, ( = 6.63 ´ 10 -34 J-s) ., nh, , where n = 1, 2, 3, ...., mvr =, 2p, It is also called principal quantum number., (iii) Bohr’s third postulate states that, an electron might, make a transition from one of its specified, non-radiating orbits to another of lower energy., When it does so, a photon is emitted having energy, equal to the energy difference between the initial and, final states. The frequency of the emitted photon is, given by, hn = E i - E f, , From the second postulate, the angular momentum of the, electron is, h, ...(ii), mvr = n, 2p, where, n (= 1, 2, 3, ...) is principal quantum number., From Eqs. (i) and (ii), we get, h 2 e0, ...(iii), r = n2, pmZe 2, , where, E i and E f are the energies of the initial and, final states and E i > E f ., , Electron (– e), Nucleus, +Ze, , r, , Let e, m and v be respectively the charge, mass and velocity, of the electron and r be the radius of the orbit. The positive, charge on the nucleus is Ze, where Z is the atomic number, (in case of H-atom, Z = 1). As, the centripetal force is, provided by the electrostatic force of attraction, we have, mv 2, 1 ( Ze) ´ e, =, ×, r, 4 pe 0, r2, mv 2 =, , Ze 2, 4 pe 0 r, , ...(i), , This is the equation for the radii of the permitted orbits., According to this equation,, rn µ n 2, Since, n = 1, 2, 3, ... it follows that the radii of the, permitted orbits increase in the ratio 1 : 4 : 9 : 16 : ..., from, the first orbit. Clearly, the stationary orbits are not equally, spaced., , Bohr Radius, The radius of the first orbit (n = 1) of H-atom (Z = 1) will be, h 2 e0, r1 =, pme 2, This is called Bohr radius and its value is 0.53 Å. Since,, r µ n 2 , the radius of the second orbit of H-atom will be, (4 ´ 0.53) Å and that of the third orbit (9 ´ 0.53) Å., Velocity of Electron in Stationary Orbits, Velocity of electrons in permitted orbits,, Ze 2 1, v=, ×, 2h e 0 n

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93, , CBSE Term II Physics XII, , where, principal quantum number, n = 1, 2, 3, … ., 1, Thus,, vµ, n, The velocity of electron in the first orbit (n = 1) of H-atom, ( Z = 1) is, e2, c, v1 =, =, (Q c = 3 ´ 10 8 m /s), 2h e 0 137, , Frequency of Electron in a Stationary Orbit, It is the number of revolutions completed per second by the, electron in a stationary orbit around the nucleus., æ, kZe 2, 1 ö, n=, ççQ k =, ÷, nhr, 4 pe 0 ÷ø, è, , Energy of Electron in Stationary Orbits, The energy E of an electron in an orbit is the sum of kinetic, and potential energies., The kinetic energy of the electron,, 1, Ze 2, KE = mv 2 =, 2, 8pe 0 r, The potential energy of the electron in an orbit of radius r, due to the electrostatic attraction by the nucleus is given by, 1 ( Ze)( -e), PE =, ×, 4p e 0, r, =-, , 1 Ze 2, ×, 4 pe0, r, , In terms of Rydberg constant R, its simplified form is, 2Rhc, PE = - 2, n, The total energy of the electron,, Ze 2, Ze 2, E = KE + PE =, 8pe 0 r 4 pe 0 r, =-, , Ze 2, Rhc, =- 2, 8pe 0 r, n, , Substituting for r from Eq. (iii), we get, E=-, , mZ 2 e4 æ 1 ö, ç 2÷, 2, 8e 0 h 2 è n ø, , where, n = 1, 2, 3, ... ., For hydrogen atom, Z = 1, -13 . 6, En =, eV, n2, , Energy Levels, The lowest state of the atom is called the ground state. The, energy of this state is -13.6eV. Therefore, the minimum, energy required to free the electron from the ground state of, the H-atom is -13.6 eV. It is called ionisation energy of the, , H-atom. The atom may acquire sufficient energy to raise, electron to higher energy state. In this condition, the atom is, said to be in excited state., According to Bohr’s third postulate, the frequency n of the, emitted electromagnetic wave (photon),, E - E1 mZ 2 e4 æ 1, 1 ö, n= 2, = 2 3ç 2 - 2÷, h, 8 e 0 h çè n1, n 2 ÷ø, The corresponding wavelength l of the emitted radiation is, given by, 1 n mZ 2 e4 æç 1, 1 ö, = =, - 2 ÷ (Qnl = c), 2, 2, 3, ç, l c 8 e 0 ch è n1, n 2 ÷ø, Þ, where,, and, , æ 1, 1, 1, = Z 2 Rç 2 - 2, çn, l, n2, è 1, , ö, ÷, ÷, ø, , 1, = wave number (number of waves per unit length), l, me4, R=, 2, 8e 0 ch 3, , R is called Rydberg constant and its value is 1.097 ´ 10 7 m -1 ., , Hydrogen Spectrum or Line, Spectra of Hydrogen Atom, Hydrogen spectrum consists of discrete bright lines in a dark, background and it is specifically known as hydrogen, emission spectrum., There is one more type of hydrogen spectrum where we get, dark lines on the bright background, it is known as, absorption spectrum., Balmer found an empirical formula by the observation of a, small part of this spectrum and it is represented by, 1, 1, 1, = R æç 2 - 2 ö÷ , where n = 3, 4, 5 , K, l, n ø, è2, where, R is a constant called Rydberg constant and its value, is 1.097 ´ 10 7 m -1 ., 1, So,, = 1.522 ´ 10 6 m -1 = 656.3 nm for n = 3, l, Other series of spectra for hydrogen were subsequently, discovered and known by the name of their discoverers. The, lines of Balmer series are found in the visible part of the, spectrum. Other series were found in the invisible parts of, the spectrum., e.g. Lyman series in the ultraviolet region and Paschen,, Brackett and Pfund in the infrared region., The wavelengths of line in these series can be expressed by, the given formulae, (i) For Lyman series (in ultraviolet region), 1, 1, 1, = R æç 2 - 2 ö÷ , where n = 2, 3, 4, ..., l, n ø, è1

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94, , CBSE Term II Physics XII, , (ii) For Balmer series (in visible region), 1, 1, 1, = R æç 2 - 2 ö÷ , where n = 3, 4, 5 , ..., l, n ø, è2, (iii) For Paschen series (in infrared region), 1, 1, 1, = R æç 2 - 2 ö÷ , where n = 4, 5 , 6, ..., l, n ø, è3, (iv) For Brackett series (in infrared region), 1, 1, 1, = R æç 2 - 2 ö÷ , where n = 5 , 6, 7, ..., l, n ø, è4, , de-Broglie’s Comment on, Bohr’s Second Postulate, According to de-Broglie, a stationary orbit is that which contains, an integral number of de-Broglie standing waves associated with, the revolving electron., h, l=, mv n, where, v n is speed of electron revolving in nth orbit., l, , (v) For Pfund series (in infrared region), 1, 1, 1, = R æç 2 - 2 ö÷ , where n = 6, 7, 8, ..., l, n ø, è5, Series limit, , Energy level (eV), , Ionised atom, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, , Paschen, series, , r, +, Nucleus, , Brackett, series, , Balmer, series, , A standing wave is shown on a circular orbit, n=7, n=6, n=5, n=4, n=3, n=2, , Pfund, series, , n=1, , Lyman, series, , Line spectra of the H-atom, , \, , 2prn =, , nh, nh, or mv n rn =, = n ( h / 2p), mv n, 2p, , i.e. Angular momentum of electron revolving in nth orbit must, be an integral multiple of h / 2p, which is the quantum condition, proposed by Bohr in his second postulate., Limitations of Bohr’s Model, There are following limitations of Bohr’s model as given below, (i) This model is applicable only to a simple atom like, hydrogen having Z = 1 . This theory fails, if Z > 1., (ii) It does not explain the fine structure of spectral lines in, H-atom., (iii) This model does not explain why orbits of electrons are, taken as circular whereas elliptical orbits are also possible.

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95, , CBSE Term II Physics XII, , Solved Examples, Example 1. In Rutherford’s a-particle experiment, , with thin gold foil, 8100 scattered a-particles per, unit area per minute were observed at an angle of, 60°. Find the number of scattered a-particles per, unit area per minute at an angle of 120°., , Sol. It is given that, q1 = 60°, q2 = 120°, N1 = 8100, Number of a-particles scattered at an angle q,, 1, Nµ, 4 æ qö, sin ç ÷, è 2ø, æ 4 q2 ö, çsin, ÷, N1 è, 2ø, Þ, =, N2, æ 4 q1 ö, çsin, ÷, 2ø, è, æ 4 120° ö, çsin, ÷, 8100 è, 2 ø, =, Þ, N2, æ 4 60° ö, çsin, ÷, 2 ø, è, 4, , Þ, , Þ, , 8100 sin 60°, =, N2, sin 4 30°, 1, 8100 ´, 16 = 900, N2 =, 9, 16, , Example 2. In Rutherford scattering experiment, what, will be the correct angle for a scattering for an, impact parameter, b = 0?, , q, Sol. We know that, impact parameter, b µ cot, 2, q, Þ, b = K cot, 2, where, K is a constant., Here, if, b = 0., K cot q/ 2 = 0, Þ, cot q/ 2 = 0, q p, Þ, =, 2 2, Hence,, q = p = 180°, , Example 3. It is found experimentally that, 13.6 eV, energy is required to separate a hydrogen atom into, a proton and an electron. Calculate the velocity of, the electron in hydrogen atom., Sol. Total energy of the electron in hydrogen atom is, - 13.6 eV = - 13.6 ´ 1.6 ´ 10-19 J = - 2.2 ´ 10-18 J, Total energy of the electron in hydrogen atom,, e2, E=8pe0 r, , \, , -, , e2, = - 2.2 ´ 10-18, 8pe0 r, , Orbital radius, r = -, , e2, 8pe0 E, , (9 ´ 109 Nm 2C - 2 ) (1.6 ´ 10-19 C) 2, (2) ( - 2.2 ´ 10-18 J), , Þ, , r=, , Þ, , r = 5.3 ´ 10-11 m, , The velocity of the revolving electron is given by, e2, r=, 4pe0 mv2, e, Þ, v=, 4pe0 mr, Putting, m = 9.1 ´ 10-31 kg and r = 5.3 ´ 10-11 m, (1.6 ´ 10-19 ), We have v =, 4p e0 ´ 9.1 ´ 10-31 ´ 5.3 ´ 10-11, v = 2.2 ´ 106 m / s, , Example 4. Using known values for hydrogen atom,, calculate radius of third orbit for Li 2 + ., , Sol. As, Z = 3 for Li +2, n2, r1, Z, Substituting, n = 3, Z = 3 and r1 = 0.529Å, (3) 2, We have, r3 =, (0.529) Å = 1.587Å, (3), , Further we know that, rn =, , Example 5. Using known values for hydrogen atom,, , calculate speed of electron in fourth orbit of He + ., , Sol. As, Z = 2 for He+ ., Z, v1, n, Substituting n = 4, Z = 2 and v1 = 2.19 ´ 106 ms -1, we get, , Also we know that, vn =, , æ 2ö, v4 = ç ÷ (2.19 ´ 106 ) ms -1 = 1.095 ´ 106 ms -1, è 4ø, , Example 6. Find the ratio of product of velocity and, time period of electron orbiting in 2nd and 3rd, stable orbits., Sol. We know that, time period, Tn µ n 3 and velocity, vn µ 1 / n., Þ Time period, Tn ´ velocity, vn µ n 2, \, , T2 v2 22 4, =, =, T3v3 32 9, , Example 7. The ground state energy of hydrogen atom, is - 13.6 eV. What is the kinetic and potential, energies of the electron in this state?, , Sol. As we know, kinetic energy = - total energy, = - ( -13.6) eV = 13.6 eV

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96, , CBSE Term II Physics XII, , and potential energy = + 2 (total energy), = 2 ( -13.6) eV = - 27.2 eV, , \, , l max, , 1 ö, æ1, = 1.097 ´ 107 ç - ÷, è 16 25 ø, , l max, , = 0.0246 ´ 107 m, = 40.650 ´ 10-7 m = 40650 Å, , 1, , Example 8. Determine the wavelength of the radiation, required to excite the electron in Li ++ from the, first to the third Bohr orbit., , Sol. Given, for Li + + , Z = 3 and as the excitation is from first to, third Bohr orbit, so n1 = 1, n 2 = 3., Using the relation,, æ 1, 1, 1 ö, 1ö, æ1, = Z 2 R çç 2 - 2 ÷÷ = ( 3) 2 R ç 2 - 3 ÷ = 8R, l, n, n, 1, 3, è, ø, 1, 2, è, ø, 1, 1, =, 8R 8 ´ 1.097 ´ 107, = 0.114 ´ 10-7, = 11.4 nm, , Þ Wavelength, l =, , Example 9. Find the largest and shortest wavelengths, in the Brackett series for hydrogen. In what region, of the electromagnetic spectrum does each series, lie?, Sol. The transition equation for Brackett series is given by, 1, 1 ö, æ1, = R ç 2 - 2 ÷ , where n = 2, 3, ..., l, 4, n, ø, è, The largest wavelength is corresponding to n = 5., , \, , The shortest wavelength corresponds to n = ¥., 1, 1ö, æ1, = 1.097 ´ 107 ç 2 - ÷, \, l min, ¥ø, è4, or, , l min = 14.58 ´ 10-7 m = 14585 Å, , Both of these wavelengths lie in infrared region of, electromagnetic spectrum., , Example 10. An electron of a hydrogen like atom is in, excited state. If total energy of the electron is, -4.6 eV, then evaluate, (i) the kinetic energy and, (ii) the de-Broglie wavelength of the electron., Sol. Given, total energy of electron, E = - 4.6 eV, (i) Kinetic energy of electron, K = - (Total energy E), Þ, = - E = - ( -4.6) = 4.6 eV, (ii) de-Broglie wavelength,, h, 6.6 ´ 10-34, ld =, =, 2mK, 2 ´ 9.1 ´ 10-31 ´ 4.6 ´ 1.6 ´ 10-19, = 0.57 ´ 10-9 m = 0.57 nm

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97, , CBSE Term II Physics XII, , Chapter, Practice, PART 1, Objective Questions, l, , 6. A given beam of a-particles has a distribution of, impact parameter b, so that the beam is scattered in, various direction as shown below, , Multiple Choice Questions, 1. The existence of a positively charged nucleus in an, atom was first suggested by the experiment of, (a) J J Thomson, (c) Chadwick, , (b) E Rutherford, (d) Hahn and Strassman, , q, Target nucleus, , b, , 2. In the a-particle scattering experiment, the shape of, the trajectory of the scattered a-particles depend, upon, [CBSE 2020], (a) only on impact parameter, (b) only on the source of a-particles, (c) Both impact parameter and source of a-particles, (d) impact parameter and the screen material of the detector, , 3. For scattering of a-particles, Rutherford’s, suggested that, (a) mass of atom and its positive charge were concentrated, at centre of atom, (b) only mass of atom is concentrated at centre of atom, (c) only positive charge of atom is concentrated at centre of, atom, (d) mass of atom is uniformly distributed throughout its, volume, , Then,, (a) b = minimum at q @ 0, (b) b = maximum at q @ p, (c) b = minimum at q @ p, (d) b = maximum at q at all values of q, , 7. The graph which depicts the results of Rutherford, gold foil experiment with a-particles is, q = scattering angle and Y = number of scattered, a-particles detected, (plots are schematic and not to scale)., , (a), , Y, , 4. In Rutherford’s nuclear model of the atom, if Fe, indicates electrostatic force between electron and, nucleus and Fc indicates the centripetal force on, revolving electron, then, (a), (b), (c), (d), , Fe = Fc, Fe > Fc, Fe < Fc, Fe = ¥ and Fc = 0, , (b), 0, , q, , Y, , p, , 0, , q, , p, , 0, , q, , p, , (d) Y, , (c) Y, 0, , q, , p, , 8. What will be the angular momentum in 4th orbit, if, , 1, 5. An alpha nucleus of energy mv 2 bombards a, 2, heavy nuclear target of charge Ze, then the distance, of closest approach for the alpha nucleus will be, proportional to, (a) v2, , (b) 1 / m, , (c) 1 / v4, , (d) 1 / Ze, , L is the angular momentum of the electron in the, 2nd orbit of hydrogen atom?, (a) 2L, , 3, (b) L, 2, , 2, (c) L, 3, , (d), , L, 2, , 9. In which of the following systems, will the radius of, the first orbit (n = 1) be minimum?, (a) Hydrogen atom, (c) Singly ionised helium, , (b) Deuterium atom, (d) Doubly ionised lithium

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98, , CBSE Term II Physics XII, , 10. The angular momentum of an electron in hydrogen, , 18. The first member of the Balmer series of hydrogen, , atom in ground state is, h, (a), p, 2p, (c), h, , atom has a wavelength of 6561Å. The wavelength of, the second member of the Balmer series (in nm) is, , h, (b), 2p, p, (d), h, , 11. The ratio of speed of an electron in ground state in, Bohr’s first orbit of hydrogen atom to the velocity, of light in air is, (a), , e2, 2phc, , (b), , 2ep, hc, , (c), , e3, 2phck, , (d), , 2 pe 2 k, hc, , 12. In accordance with the Bohr’s model, find the, quantum number that characterises the earth’s, revolution around the sun in an orbit of radius, 1 .5 ´ 1011 m with orbital speed 3 ´ 10 4 ms -1 ., (Take, mass of earth = 6 . 0 ´ 10 24 kg), (a) 2 . 6 ´ 1074, , (b) 8. 5 ´ 1080, , (c) 4.34 ´ 10100, , (d) 3.2 ´ 108, , Li ++ ion in its ground state, on the basic of Bohr’s, model, will be about, [NCERT Exemplar], (b) 27 pm, , (c) 18 pm, , (d) 13 pm, , 14. If the orbital radius of the electron in a hydrogen, , atom is 4.7 ´ 10 -11 m, compute the kinetic energy of, the electron in hydrogen atom., , (a) 15.3 eV, (c) 1.3.6 eV, , (b) - 1.53 eV, (d) - 13.6 eV, , 15. A set of atoms in an excited state decays, [NCERT Exemplar], (a) in general to any of the states with lower energy, (b) into a lower state only when excited by an external, electric field, (c) all together simultaneously into a lower state, (d) to emit photons only when they collide, , 16. The transition from the state n = 3 to n = 1 in a, hydrogen like atom results in ultraviolet radiation., Infrared radiation will be obtained in the transition, from, (a) 2 ® 1, (c) 4 ® 2, , (b) 3 ® 2, (d) 4 ® 3, , 17. In Pfund series, ratio of maximum to minimum, wavelength of emitted spectral lines is, (a), (c), , l max, l min, , 4, =, 3, , l max 16, =, l min, 7, , l, 9, (b) max =, l min 5, (d), , (b) 684, , (c) 486, , (d) 492, , Assertion-Reasoning MCQs, Direction (Q. Nos. 19-26) Each of these questions, contains two statements Assertion (A) and Reason (R)., Each of these questions also has four alternative, choices, any one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given, below., (a) Both A and R are true and R is the correct, explanation of A, (b) Both A and R are true, but R is not the correct, explanation of A, (c) A is true, but R is false, (d) A is false and R is also false, , 19. Assertion An atom consists of electron, proton and, , 13. Taking the Bohr radius as a 0 = 53 pm, the radius of, (a) 53 pm, , l, , (a) 268, , l max 36, =, l min 11, , neutron. Electron revolves around nucleus., However, most of space in an atom is empty., Reason From Rutherford’s experiment, size of the, nucleus is 10 -10 m and from kinetic theory, size of, the atom is 10 -15 m., , 20. Assertion If the electrons in an atom were, stationary, then they would fall into the nucleus., Reason Electrostatic force of attraction acts, between negatively charged electrons and positive, nucleus., , 21. Assertion If total energy of an electron in a H-atom, is positive, then the electron will be bound to the, nucleus., Reason Bohr model is valid for both one electron, atom and multi-electron atom., , 22. Assertion Atom as a whole is electrically neutral., Reason Atom contains equal amount of positive, and negative charges., , 23. Assertion Atoms of each element are stable and, emit characteristic spectrum., Reason The spectrum provides useful information, about the atomic structure., , 24. Assertion According to electromagnetic theory, an, accelerated particle continuously emits radiation., Reason According to classical theory, the proposed, path of an electron in Rutherford atom model will, be parabolic.

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99, , CBSE Term II Physics XII, , 25. Assertion Highest energy state corresponds with, the electron revolving in an orbit closest to the, nucleus., Reason Energies of the excited states come closer, and closer together as n decreases., , 26. Assertion Second orbit circumference of hydrogen, atom is two times the de-Broglie wavelength of, electrons in that orbit., Reason de-Broglie wavelength of electron in, ground state is minimum., l, , Case Based MCQs, Direction Read the following passage and answer the, questions that follows, , 27. a-Particle Scattering Experiment, In this experiment, H Geiger and E Marsden took, radioactive source ( 214, 83 Bi) for a-particles., A collimated beam of a-particles of energy 5.5 MeV, was allowed to fall on 2.1 ´ 10 -7 m thick gold foil., The a-particles were observed through a rotatable, detector consisting of a zinc sulphide screen &, microscope and it was found that a-particles got, scattered., These scattered a-particles produced scintillations, on the zinc sulphide screen., Observations of this experiment are as follows, I. Many of the a-particles pass through the foil, without deflection., II. Only about 0.14% of the incident a-particles, scattered by more than 1°., III. Only about one a-particle in every 8000, a-particles deflected by more than 90°., Based on these observation, they were able to, proposed a nuclear model of atom, are called, planetary model, in which entire positive charge, and most of the mass of atom is concentrated in a, small volume called the nucleus with electron, revolving around the nucleus as planets revolve, around the sun., (i) Rutherford’s atomic model can be visualised as, (a), , (b), , (c), , (d), , (ii) Gold foil used in Geiger-Marsden experiment is, about 10 -8 m thick. This ensures, (a) gold foil’s gravitational pull is small or possible, (b) gold foil is deflected when a-particle stream is not, incident centrally over it, (c) gold foil provides no resistance to passage of a-particles, (d) most a-particle will not suffer more than 1° scattering, during passage through gold foil, , (iii) In scattering, the impact parameter b is defined as, the, (a) maximum kinetic energy of projectile scattered, (b) distance of closest approach between projectile and, target, (c) closest distance between projectile and target if there, were no deflection, (d) None of the above, , (iv) The distance of closest approach, when a 15.0 MeV, proton approaches gold nucleus ( Z = 79) is, (a) 758 fm, (b) 7.58 fm, (c) 75.8 fm, (d) 0.758 fm, , (v) The fact that only a small fraction of the number of, incident particles rebound back in Rutherford, scattering indicates that, (a) number of a-particles undergoing head-on-collision is, small, (b) mass of the atom is concentrated in a small volume, (c) mass of the atom is concentrated in a large volume, (d) Both (a) and (b), , 28. Hydrogen Spectrum, Hydrogen spectrum consists of discrete bright lines, in a dark background and it is specifically known as, hydrogen emission spectrum.There is one more, type of hydrogen spectrum that exists where we get, dark lines on the bright background, it is known as, absorption spectrum., Balmer found an empirical formula by the, observation of a small part of this spectrum and it is, represented by, 1, 1 ö, æ 1, =Rç, ÷ , where n = 3, 4, 5, K ., 2, l, n2 ø, è2, For Lyman series, the emission is from first state to, nth state, for Paschen series, it is from third state to, nth state, for Brackett series, it is from fourth state, to nth state and for Pfund series, it is from fifth, state to nth state., (i) Number of spectral lines in hydrogen atom is, (a) 8, (c) 15, , `, , (b) 6, (d) ¥

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100, , CBSE Term II Physics XII, , (ii) Which series of hydrogen spectrum corresponds to, ultraviolet region?, [CBSE 2020], (a), (b), (c), (d), , Balmer series, Brackett series, Paschen series, Lyman series, , C, B, A, l1 l2 l3 l4 l5 l6, , G, , (a) l1, l 2 , l 3, (b) l 4 , l5 , l 6, (c) l1, l 4 , l 6, (d) l1, l 2 , l 3, l 4 , l5 , l 6, , (b) 1 : 4 : 9, (d) 3 : 2 : 1, , (v) When an electron jumps from the orbit n = 2 to, n = 4, then wavelength of the radiations absorbed, will be (R is Rydberg’s constant), , replaced by the positively charged anti-particle of, the electron (called the positron which is as, massive as the electron). What would be the, ground state energy of positronium?, [NCERT Exemplar], , 6. The total energy of an electron in the first excited, state of hydrogen atom is about -3.36 eV., (i) What is the kinetic energy of the electron in first, excited state?, (ii) What is the potential energy of the electron in, the first excited state?, (iii) If zero potential energy reference is changed,, which of the above would change?, energy change when electron in hydrogen atom is, replaced by a particle of mass 200 times that of the, electron but having the same charge? [All India 2013], , 8. How many different wavelengths may be observed, in the spectrum from a hydrogen sample, if the, atoms are excited to states with principal quantum, number n?, , 9. The figure shows energy level diagram of hydrogen, atom, A, , 16, (b), 5R, 3R, (d), 16, , PART 2, Subjective Questions, l, , [Delhi 2019], , 7. Define ionisation energy. How would the ionisation, , (iv) The ratio of the wavelength for 2 ® 1 transition in, Li ++ , He + and H is, , 16, (a), 3R, 5R, (c), 16, , first excited state of hydrogen atom., , 5. Positronium is just like a H-atom with the proton, , (iii) The figure shows energy levels of an atom with six, transitions of wavelengths l 1 , l 2 , l 3 , l 4 , l 5 and, l 6 . The following wavelengths also occur in, absorption spectrum, [CBSE 2020], , (a) 1 : 2 : 3, (c) 4 : 9 : 36, , 4. Calculate the orbital period of the electron in the, , Short Answer (SA) Type Questions, 1. Explain briefly how Rutherford scattering of, a-particle by a target nucleus can provide, information on the size of the nucleus. [Delhi 2019], , 2. An a-particle moving with initial kinetic energy K, towards a nucleus of atomic number Z approaches, a distance d at which it reverses its direction., Obtain the expression for the distance of closest, approach d in terms of the kinetic energy K of, [All India 2016 C], a-particle., , 3. State Bohr postulate of hydrogen atom that gives, the relationship for the frequency of emitted, photon in a transition., [Foreign 2016], , B, , C, , n=4, D, , E, , n=3, F, , n=2, , n=1, , (i) Find out the transition which results in the, emission of a photon of wavelength 496 nm., (ii) Which transition corresponds to the emission of, radiation of maximum wavelength? Justify your, answer., [All India 2015C], 10. Calculate the shortest wavelength present in, Paschen series., , 11. The short wavelength limit for the Lyman series of, , the hydrogen spectrum is 913.4 Å. Calculate the, short wavelength limit for Balmer series of the, hydrogen spectrum., [Delhi 2016], , 12. Different elements have different atomic spectra., Comment., , 13. Find the ratio between the wavelengths of the 'most, energetic' spectral lines in the Balmer and Paschen, series of the hydrogen spectrum.

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101, , CBSE Term II Physics XII, , 14. Obtain the expression for the ratio of the, , 23. The ground state energy of hydrogen atom is, , de-Broglie wavelengths associated with the, electron orbiting in the second and third excited, states of hydrogen atom., [Delhi 2019], , – 13.6 eV. If an electron makes a transition from an, energy level – 0.85 eV to – 1.51 eV, calculate the, wavelength of the spectral line emitted. To which, series of hydrogen spectrum, does this wavelength, belong?, [All India 2012], , 15. State Bohr’s quantisation condition of angular, momentum. Calculate the shortest wavelength of, the Brackett series and state to which part of the, electromagnetic spectrum does it belong.[Delhi 2019], , 16. Using Bohr’s second postulate of quantisation of, orbital angular momentum, show that the, circumference of the electron in the nth orbital, state in hydrogen atom is n-times the de-Broglie, wavelength associated with it., [Delhi 2020], , 17. Why motion of planets around the sun cannot be, governed by Bohr’s postulate of quantisation of, angular momentum?, , 18. In a Geiger-Marsden experiment, calculate the, distance of closest approach to the nucleus of, Z = 80, when an a-particle of 8 MeV energy, impinges on it before it comes to momentarily rest, and reverses its direction., How will the distance of closest approach be, affected when the kinetic energy of the a-particle, is doubled?, [All India 2012], , 19. Using Rutherford model of the atom, derive the, expression for the total energy of the electron in, hydrogen atom. What is the significance of total, negative energy possessed by the electron?, [All India 2014], , 20. Define the distance of closest approach. An, a-particle of kinetic energy K is bombarded on a, thin gold foil. The distance of the closest approach is r., What will be the distance of closest approach for an, a-particle of double the kinetic energy?, Write two important limitations of Rutherford, nuclear model of the atom., [All India 2016], , 21. A photon emitted during the de-excitation of, electron from a state n to the first excited state in a, hydrogen atom, irradiates a metallic cathode of, work function 2eV, in a photocell, with a stopping, potential of 0.55 V. Obtain the value of the quantum, number of the state n., [All India 2019], , 22. A 12.5 eV electron beam is used to bombard, gaseous hydrogen at room temperature. Upto, which energy level, the hydrogen atoms would be, excited?, Calculate the wavelengths of the first member of, Lyman and first member of Balmer series., [Delhi 2014], , l, , Long Answer (LA) Type Questions, 24. (i) In H-atom, an electron undergoes transition, from second excited state to the first excited state, and then to the ground state. Identify the, spectral series to which these transitions belong., (ii) Find out the ratio of the wavelengths of the, emitted radiations in the two cases., , 25. Show that the first few frequencies of light that is, emitted when electrons fall to nth level from levels, higher than n, are approximate harmonics (i.e., in, the ratio 1 : 2 : 3 K) when n >> 1. [NCERT Exemplar], , 26. State any two postulates of Bohr’s theory of, H-atom. What is the maximum possible number of, spectral lines, when the H-atom is in its second, excited state? Justify your answer. Calculate the, ratio of the maximum and minimum wavelengths, of the radiations emitted in this process., [All India 2010], , 27. Obtain an expression for the frequency of radiation, emitted, when a hydrogen atom de-excites from, level n to level ( n -1 ) . For large n, show that this, frequency equals to the classical frequency of, revolution of the electron in the orbit., [NCERT], , 28. (i) The kinetic energy of the electron orbiting in the, first excited state of hydrogen atom is 3.4 eV., Determine the de-Broglie wavelength associated, with it., (ii) Calculate the de-Broglie wavelength associated, with the electron in the second excited state of, hydrogen atom. The ground state energy of the, hydrogen atom is 13.6 eV., [Delhi 2020], , 29. The stopping potential for the photoelectrons, emitted from a metal surface of work function, 1.7 eV is 10.4 V. Find the wavelength of the, radiation used. Also, identify the energy levels in, hydrogen atom which will emit this wavelength., (Take, h = 6.63 ´ 10 - 34 J-s, c = 3 ´ 10 8 ms -1 and, e = 1.6 ´ 10 - 19 ), , 30. (i) Using the Bohr’s model, calculate the speed of the, electron in a H-atom in the n = 1, 2 and 3 levels., (ii) Calculate the orbital period in each of these, levels., [NCERT]

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102, , CBSE Term II Physics XII, , 31. If a proton had a radius R and the charge was, , Total energy, E(eV), Unbound (ionised), atom, , uniformly distributed, calculate using Bohr, theory, the ground state energy of a H-atom when, (i) R = 0.1Å and (ii) R = 10Å., [NCERT Exemplar], , 0, , 32. The first four spectral in the Lyman series of a, H-atom are l = 1218Å, 1028Å, 974.3Å and 951.4Å., If instead of hydrogen, we consider deuterium,, calculate the shift in the wavelength of these, lines., [NCERT Exemplar], , –0.85, , n=5, n=4, , –0.51, , n=3 Excited, states, , –3.40, , n=2, , 33. Using Bohr’s postulates, derive the expression for, the frequency of radiation emitted when electron, in hydrogen atom undergoes transition from, higher energy state (quantum number n i ) to the, lower state, ( n f ). When electron in hydrogen, atom jumps from energy state n i = 4 to, n f = 3, 2, 1. Identify the spectral series to which, the emission lines belong., [All India 2018, 13], , 34. A hydrogen like atom (atomic number Z) is in a, higher excited state of quantum number n. It can, emit two photons of energies 10.20 eV and, 17.00 eV. When make transition to the first excited, state. But emits photons of energies 4.25 eV and, 5.95 eV, if make transition to second excited state., Determine the values of Z and n. [All India 2014 C], l, , Case Based Questions, Direction Read the following passage and answer, the questions that follows, , 35. Excited State of Atom, At room temperature, most of the H-atoms are in, ground state. When an atom receives some, energy (i.e. by electron collisions), the atom may, acquire sufficient energy to raise electron to, higher energy state. In this condition, the atom is, said to be in excited state. From the excited state,, the electron can fall back to a state of lower, energy, emitting a photon equal to the energy, difference of the orbit., , Ground state, n=1, , –13.6, ., , In a mixture of H—He + gas (He + is single ionised, He atom), H-atoms and He + ions are excited to their, respective first excited states. Subsequently, H-atoms, transfer their total excitation energy to He + ions (by, collisions)., (i) Calculate the wavelength of radiation emitted,, when an electron having total energy -15, . eV, makes a transition to the ground state., [Given, energy in the ground state, = - 13.6 eV and Rydberg’s constant, = 1.09 ´ 10 7 m -1 ], (ii) Hydrogen atom has only one electron but its, emission spectrum has many lines. Explain with, reason., (iii) Find the relation between the three wavelengths, l 1 , l 2 and l 3 from the energy level diagram, shown below., C, l3, , l1, l2, , B, A

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Chapter Test, Multiple Choice Questions, , 1., , 8. Calculate the wavelength of Ha -line in Balmer series of, , Atoms consist of a positively charged nucleus is, obvious from the following observation of, Geiger-Marsden experiment, (a) most of a-particles do not pass straight through the, gold foil, (b) many of a-particles are scattered through the acute, angles, (c) very large number of a-particles are deflected by large, angles, (d) None of the above, , 2. The simple Bohr model cannot be directly applied to, calculate the energy levels of an atom with many, electrons. This is because, (a), (b), (c), (d), , of the electrons not being subject to a central force, of the electrons colliding with each other, of screening effects, the force between the nucleus and an electron will no, longer be given by Coulomb’s law, , 3. The number of spectral lines produced due to, transition among three energy levels will be, (a) 10, (b) 8, (c) 6, (d) 3, , hydrogen atom., (Take, Rydberg constant, R = 1.097 ´ 10 7 m -1 ), , (Ans. 6.56 ´ 10 - 7 m), , 9. Explain, in brief, why Rutherford’s model cannot account, for the stability of an atom., , 10. The value of ground state energy of hydrogen atom is, -13.6 eV., (i) Find the energy required to move an electron from the, ground state to the first excited state of the atom., (ii) Determine (a) the kinetic energy and (b) orbital radius, in the first excited state of the atom., (Given, the value of Bohr's radius= 0.53 Å), [Ans. (i) 10.2eV, (ii) (a) 3.4 eV (b) 2.12 Å], , 11. Assume that, there is no repulsive force between the, electrons in an atom but the force between positive and, negative charges is given by Coulomb’s law as usual., Under such circumstances, calculate the ground state, energy of a He-atom., (Ans. - 54. 4 eV), , Long Answers Type Questions, , 12. (i) State Bohr’s postulate to define stable orbits in, , 4. The de-Broglie wavelength of an electron in first, Bohr’s orbit is equal to, , hydrogen atom. How does de-Broglie’s hypothesis, explain the stability of these orbits?, (ii) A hydrogen atom initially in the ground state absorbs, a photon which excites it to the n = 4 level. Estimate, the frequency of the photon., [Ans. (ii) 3. 1 ´ 10 15 Hz], , 13. The wavelength of light from the spectral emission line of, , 1, of circumference of orbit, 4, 1, (b) of circumference of orbit, 2, (c) twice of circumference of orbit, (d) the circumference of orbit, , (a), , sodium is 589 nm. Find the kinetic energy at which, (i) an electron and, (ii) a neutron, would have the same de-Broglie, wavelength. [Ans. (i) 4.34 ´ 10 -6 eV and (ii) 236, . ´ 10 - 9 eV], , 14. (i) The total energy of electron in the ground state of, , Short Answers Type Questions, , 5. Explain, why scattering of a-particles in Rutherford’s, experiment is not affected by the mass of the, nucleons., , 6. Calculate the radius of the first orbit of H-atom., Show that the velocity of electron in the first orbit is, 1/137 times the velocity of light., (Ans. 0.53 Å), , 7. What does an emperical formula mean? Hence,, , hydrogen atom is - 13.6 eV. Find the kinetic energy of, an electron in the first excited state., (ii) Find the kinetic energy, potential energy and total, energy of electron in first and second orbit of, hydrogen atom, if potential energy in first orbit is, taken to be zero., [Ans. (i) 3.4 eV,, (ii) First orbit K = 1360, . eV, E = 1360, . eV, Second orbit K = 3. 40 eV, U = 20. 40 eV,, E = 2380, . eV], , explain that how Balmer proposed this formula., , Answers, Multiple Choice Questions, 1. (d), , 2. (a), , 3. (d), , 4. (d), , For Detailed Solutions, Scan the code

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104, , CBSE Term II Physics XII, , EXPLANATIONS, PART 1, 1. (b) A positively charged nucleus in an atom was first, suggested by E Rutherford through its gold foil experiment., 2. (a) In the a-particle scattering experiment, the shape of the, trajectory of the scattered a-particles depends upon only on, impact parameter., 3. (a) In Rutherford’s nuclear model of the atom, the entire, positive charge and most of the mass of the atom are, concentrated in the nucleus with the electrons some, distance away., 4. (a) The electrostatic force of attraction Fe between the, revolving electrons and the nucleus provides the requisite, centripetal force Fc to keep them in their orbits., Thus, for a dynamically stable orbit in a hydrogen atom, Fe = Fc, 5. (b) For the distance of closest approach, we can write, 1, K ´ ( Z e ) ( Ze ), 1, mv2 =, Þ r0 µ, 2, r0, m, So, the distance of closest approach for the alpha nucleus, 1, will be proportional to ., m, 6. (c) It has been seen that, an a-particle close to the nucleus, suffers large scattering angle., In case of head-on-collision, the impact parameter is, minimum and the a-particle rebounds back (q @ p)., For large impact parameter, the a-particle goes nearly, undeviated and has a small deflection (q @ 0)., 7. (b) In Rutherford’s experiment, number of particles, scattered at large angles is very less and most of the, particles are scattered at small angles., Hence, graph of Y ( = number of a- particles) and, q ( = scattering angle) is shown correctly in option (b)., 8. (a) We know that, the angular momentum of the electron of, hydrogen atom,, nh, …(i), Ln =, 2p, But according to question, the angular momentum in 2nd, orbit is L., 2h h, So,, …(ii), L=, =, 2p p, Hence, the angular momentum in 4th orbit is, 4h 2h, L 4¢ =, =, = 2L, 2p, p, 2, n, 1, 9. (d) Radius of nth orbit, rn µ, or rn µ, Z, Z, So, here lithium has least atomic number, thus option (d) is, correct., 10. (b) From the formula of angular momentum and before, æ hö, assumption, mvr = n ç ÷, è 2p ø, Here, n = 1, h, Þ mvr =, 2p, , 2pZe 2 k, nh, For hydrogen atom in ground state, n = 1, Z = 1, 2pe 2 k, v 2pe 2 k, v=, Þ, =, h, c, hc, 12. (a) Given, radius of orbit, r = 1.5 ´ 1011 m,, , 11. (d) Speed of an electron in nth orbit, vn =, , Orbital speed, v = 3 ´ 104 ms -1 and mass of earth,, m = 6.0 ´ 1024 kg, Angular momentum, L = mvr =, , nh, 2p, , 2pvrm, h, where, n is the quantum number of the orbit., 2 ´ 3.14 ´ 3 ´ 104 ´ 1.5 ´ 1011 ´ 6.0 ´ 1024, Þn=, 6.63 ´ 10-34, 74, = 2.57 ´ 10 or n = 2.6 ´ 1074, n=, , or, , 13. (c) The atomic number of lithium is 3, therefore the radius, of Li + + ion in its ground state, on the basis of Bohr’s model, 1, will be about times to that of Bohr radius. Therefore, the, 3, 53, radius of lithium ion is near, » 18 pm., 3, 14. (a) Kinetic energy, e2, ( 9 ´ 109 Nm 2 / C 2 ) ( 1.6 ´ 10-19 C ) 2, K=, =, 8pe0 r, ( 2) ( 4.7 ´ 10-11 m), = 2 .45 ´ 10-18 J = 15.3 eV, 15. (a) A set of atoms in an excited state decays in general to any, of the states with lower energy., 16. (d) Infrared radiation found in Paschen, Brackett and Pfund, series and it is obtained when electron transition is from, higher energy level to minimum third level., Therefore, infrared radiation will be obtained in the, transition from 4 ® 3., 17. (d) In Pfund series,, 1, 1 ö, æ1, = R ç 2 - 2 ÷ , where n = 6, 7, L, l, n ø, è5, Maximum wavelength is given by, 1, l max, , 1 ö, 11 R, æ1, =R ç 2 - 2÷=, 6 ø 36 ´ 25, è5, , (in transition 6 ® 5), , Minimum wavelength is given by, R, 1, 1ö, æ1, =Rç 2 - ÷ =, 25, l min, ¥, è5, ø, So, ratio of, , (in transition ¥ ® 5), , l max 36, = ., l min 11, , 18. (c) First member is obtained in transition n =`3 to n = 2,, 1, 1 ö, æ1, = Rç 2 - 2 ÷, l1, 3 ø, è2, , ...(i)

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105, , CBSE Term II Physics XII, , Second member is obtained in transition n = 4 to n = 2,, 1, 1 ö, æ1, ...(ii), = Rç 2 - 2 ÷, l2, 4 ø, è2, From Eqs, (i) and (ii), we have, 1, 1 ö, æ1, ç 2 - 2÷, l1, 2, 3, ø, =è, 1, 1 ö, æ1, - ÷, ç, l 2 è 22 42 ø, æ1 1 ö, ç - ÷, l2, 5 64, 4 9ø, = è, =, ´, l1 æ 1 1 ö 36 12, ç - ÷, è 4 16 ø, l 2 5 4 20, Þ, = ´ =, l1 9 3 27, 20, Þ, l 2 = l1 ´, 27, 20, = 6561 ´, = 4860 Å = 486 nm, 27, 19. (c) Rutherford’s experiment suggested that, the size of the, nucleus to be about 10-15 m to 10-14 m. From kinetic theory,, the size of an atom was known to be 10-10 m, about 10,000 to, 100,000 times larger than the size of the nucleus., Thus, the electrons would seem to be at a distance from the, nucleus about 10,000 to 1,00,000 times the size of the, nucleus itself., Hence, most of space in an atom is empty., Therefore, A is true but R is false., 20. (a) In an atom, electron revolves around nucleus, for this, required centripetal force is provided by electrostatic force, of attraction between negatively charged electron and, positive nucleus., If the electrons were stationary, then the electrostatic force, will remain unbalanced, which leads to the electron to fall, into the nucleus., Therefore, both A and R are true and R is the correct, explanation of A., 21. (d) Total energy E of an electron in a hydrogen atom,, -e 2, E=, 8pe0 r, So, E is negative., This implies the fact that, the electron is bound to the, nucleus., Bohr model is valid for only one electron atom., Therefore, A is false and R is also false., 22. (a) For an atom, number of electrons is equal to number of, protons and neutrons are electrically neutral. So, net charge, on an atom is zero., Therefore, both A and R are true and R is the correct, explanation of A., 23. (b) All atoms have different number of electrons. These, electrons correspond to different energy level. Whenever an, atom is excited, we get some energy in the form of radiation, and the obtained radiation is called line spectra of that, particular atom. Therefore, atoms of each element are stable, and emit characteristic spectrum., , Different atoms produce different characteristic spectrum, and hence we can differentiate between atoms by their line, spectrum., Hence, it provides useful information about the atomic, structure., Therefore, both A and R are true but R is not the correct, explanation of A., 24. (c) According to electromagnetic theory, an accelerated, charge particle continuously emits radiation., But Rutherford could not explain the revolution of electrons, around nucleus. The path of revolution is circle, not parabola., Therefore, A is true but R is false., 25. (d) The energy of an atom is the least, when its electron is, revolving in an orbit closest to the nucleus, i.e. the one for, which n = 1. However, the highest energy state corresponds, to n = ¥ and has an energy of 0 eV., 1, Since,, En µ 2, n, So, as the value of n increases, energies of excited states, comes closer and closer., Therefore, A is false and R is also false., æ hö, 26. (b) mv2 r2 = 2ç ÷, è 2p ø, æ h ö, ÷÷ = 2l 2, \ 2pr2 = 2çç, è mv2 ø, h, Further, l =, p, Speed of momentum is maximum in ground state, hence l is, minimum., Therefore, both A and R are true but R is not the correct, explanation of A., 27. (i) (d) Rutherford’s atom had a positively charged centre and, electrons were revolving outside it. It is also called the, planetary model of the atom as in option (d) ., (ii) (d) As the gold foil is very thin, it can be assumed that, a-particles will not suffer more than 1° scattering, during their passage through it. Therefore,, computation of the trajectory of an a-particle scattered, by a single nucleus is enough., (iii) (d) The impact parameter is defined as the perpendicular, distance between the path of a projectile and the centre, of potential field created by an object that the projectile, is approaching., (iv) (b) Given, E = 15.0 MeV = 15 ´ 1.6 ´ 106 ´ 10-19 J, = 15 ´ 1.6 ´ 10-13 J, E=, , 1, Ze 2, ×, 4pe0 r0, 2, , 9 ´ 109 ´ 79 ´ ( 1.6 ´ 10-19 ), = 7.58 fm, 15 ´ 1.6 ´ 10-13, (v) (d) In case of head-on-collision, the impact parameter is, minimum and the a-particle rebounds back. So, the fact, that only a small fraction of the number of incident, particles rebound back indicates that the number of, a-particles undergoing head-on collision is small., This in turn implies that the mass of the atom is, concentrated in a small volume., Hence, options (a) and (b) are correct., Þ, , r0 =

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106, , CBSE Term II Physics XII, , 28. (i) (d) Number of spectral lines in hydrogen atom is ¥., (ii) (d) Lyman series lies in the ultraviolet region., (iii) (a) Absorption lines are obtained when an e - absorbs a, photon in ground state and is excited to a higher state., Initially e - is in ground state, so transition states are 1, 2,, and 3 and wevelengths are l1 , l 2 and l 3., æ 1, 1, 1 ö, 1, (iv) (c) = RZ 2 çç 2 - 2 ÷÷ Þ l µ 2, l, n2 ø, Z, è n1, We have, ZLi = 3,, ZHe = 1, and, ZH = 1., So, l Li + + : l He+ : l H = 4 : 9 : 36, , Þ, Þ, Þ, Þ, , 1 ´ (1.6 ´ 10-19 ) 2, 2 ´ 2 ´ (6.62 ´ 10-34 ) ´ (8.85 ´ 10-12 ), , (2) 2 ´ (6.62 ´ 10-34 ) 2 ´ (8.85 ´ 10-12 ), 3.14 ´ (9.1 ´ 10-31 ) ´ (1.6 ´ 10-19 ) 2, = 212, . ´ 10-10 m, Time period or orbital period,, 2pr2 2 ´ 314, . ´ 212, . ´ 10-10, T=, =, = 1.22 ´ 10-15 s, v2, 1.09 ´ 106, , Þ, , r2 =, , 5. The total energy of the electron in the stationary states of, the hydrogen atom is given by, me 4, En = - 2 2 2, 8n e0 h, , PART 2, 1. According to Rutherford’s experiment, following, observations were made, (i) Most of the a-particles passed through the gold foil, without any appreciable deflection., (ii) Only 014, . % of incident a-particles scattered by more than, 1°. But about 1 a-particle in every 8000 particles, deflected by more than 90°., Thus, all these leads to the conclusion that atom has a lot of, empty space and practically the entire mass of the atom is, confined to an extremely small centered core called nucleus,, whose size is of the order from 10-15 m to 10-14 m., 2. Kinetic energy of a-particle is given as, 1 2 e × Ze, KE =, 4pe0 d 2, where, d is the distance of closest approach., 2 Ze 2, Þ d=, 4pe0 KE, , h = 6.62 ´ 10-34 J-s and n = 2 (in 1st excited state), , = 1.09 ´ 106 m/s, n 2 h 2 e0, Radius of orbit, r2 =, pme 2, -31, Here, m = 91, . ´ 10 kg, , é 1, 1, 1 ù, =Rê 2 l, ( 4) 2 úû, ë ( 2), 1, æ1 1 ö, =Rç - ÷, l, è 4 16 ø, 1, æ4 -1ö, =Rç, ÷, l, è 16 ø, 1 3R, =, l 16, 16, l=, 3R, , d2 =, , Here, Z = 1, e = 1.6 ´ 10-19 C, e0 = 8.85 ´ 10-12 NC 2 m –2 ,, , Þ v2 =, , (v) (a) Wavelength is given by, æ 1, 1, 1 ö, = R çç 2 - 2 ÷÷, l, n, n, 2ø, è 1, Þ, , where, n is frequency of radiation emitted, Ei and E f are the, energies associated with stationary orbits of principal, quantum numbers n i and n f , respectively (where, n i > n f )., 1 Ze 2, 4. The velocity of electron, vn =, n 2he0, , 2 Ze 2, 4pe0 KE, , This is the required expression for the distance of closest, approach d in terms of kinetic energy (KE)., 3. An atom can emit or absorb radiation in the form of discrete, energy photons only when an electron jumps from a higher, to a lower orbit or from a lower to a higher orbit,, respectively. The frequency of radiated wave is given by, æ E f - Ei ö, ÷÷, hn = E f - Ei Þ n = çç, è h ø, , where, signs are as usual and the m that occurs in the Bohr, formula is the reduced mass of electron and proton. Also, the, total energy of the electron in the ground state of the, hydrogen atom is - 13.6 eV.For H-atom reduced mass m e,, whereas for positronium, the reduced mass is, me, m», 2, Hence, the total energy of the electron in the ground state of, the positronium atom is, -13.6 eV, = - 6.8 eV, 2, 6. Given, total energy, E = - 3.36 eV, (i) Kinetic energy in the given state, K = - E = 3.36 eV, (ii) Potential energy in the given state, U = 2E = - 6.72 eV, (iii) The potential energy is zero at infinity. If the reference of, potential energy is changed, the kinetic energy will, remain constant but total energy will change., 7. The ionisation energy is qualitatively defined as the amount, of energy required to remove the most loosely bound, electron, the valence electron of an isolated gaseous atom to, form a cation. Since, total energy is directly proportional to, the mass of electron., m e4, E0 = 2 2 , i. e. E0 µ m, 8 e0 h, So, the ionisation energy becomes 200 times on replacing an, electron by a particle of mass 200 times of the electron and, of same charge., 8. From the nth state, the atom may go to ( n – 1 ) th state, … ,, 2nd state or 1st state. So, there are ( n – 1 ) possible, transitions starting from the nth state.

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107, , CBSE Term II Physics XII, , The atoms reaching ( n – 1 )th state may make n – 2 different, transitions. Similarly, for other lower states, the total, number of possible transitions is, n ( n – 1), ( n - 1 ) + ( n – 2) + ( n – 3) + L + 2 + 1 =, 2, 9. (i) Given, l=496 nm=496´10-9 m, hc 6.62 ´ 10-34 ´ 3 ´ 108, Q, E=, =, J, l, 496 ´ 10-9, 6.6 ´ 10-34 ´ 3 ´ 108, eV = 2.5 eV, =, 496 ´ 10-9 ´ 1.6 ´ 10-19, – 0.85 eV, – 1.51 eV, – 3.40 eV, , A B C, D E, F, , – 13.6 eV, , n=4, n=3, n=2, , n=1, , This energy corresponds to the transition, B(n = 4 to n = 2) for which the energy change = 2 eV., (ii) Energy of emitted photon is given by, hc, E=, l, 1, \ l max µ, Emin, Transition A, for which the energy emission is minimum,, corresponds to the emission of radiation of maximum, wavelength., 10. For Paschen series, n1 = 3, æ 1, 1, 1 ö, Wave number, n = = R çç 2 - 2 ÷÷, l, n2 ø, è n1, where, n 2 = 4, 5, 6, .... ¥, For shortest wavelength, n 2 = ¥, 1, æ1 1 ö, = Rç - ÷, \, l¥, è9 ¥ø, 1, R, Þ, =, l¥, 9, 9, 9, Þ l¥ = =, R 1.097 ´ 107, = 82041, . ´ 10-10 m = 82041, . Å, 11. Lyman series, n = 2, 3, 4,… to n = 1, For short wavelength, n = ¥ to n = 1, 12375 12375, eV = 13.54 eV, =, l(Å), 913.4, Also, energy of nth orbit, E = 13.54 / n 2, So, energy of n = 1 , energy level = 13.54 eV, Energy of n = 2 , energy level = 13.54 / 22 = 3. 387 eV, 12375, So, short wavelength of Balmer series =, = 3653 Å, 3. 387, Energy, E =, , 12. The energy level of each atom are definite but different, from the energy levels of other elements., Therefore, the spectrum of the emission lines by atom of an, element has lines of certain different frequencies which are, different from those of all other elements., And this property helps us to distinguish different elements, from each other., , 1, 1 ö, æ1, =R ç 2 - 2÷ r, lB, n ø, è2, 4, For highest energy n ® ¥ Þ l B =, R, 1, R R, Þ, =, =, l B 22, 4, , 13. For Balmer series,, , For Paschen series,, , 1, 1 ö, æ1, = Rç 2 - 2 ÷, lP, n ø, è3, , For highest energy,, n ® ¥ Þ lP =, Þ, , lB : lP =, , 9, R, , 4 9, : =4:9, R R, , 14. We know that, de-Broglie wavelength, l =, , h, h, =, p mv, , h, l, hr nh, Þ, mvr =, =, l 2l, 2p, 2pr, Þ, l=, ´ hr =, nh, n, As,, r µ n2, 1, Þ, l µ ( n2 ) = n, n, l, l, 3, Thus, we can say that, 3 = Þ l1 = 3, 3, l1 1, Þ, , mv =, , Thus, wavelength decreases 3 times as an electron jumps, from third excited state to the ground state., 15. According to Bohr’s quantisation condition electrons are, permitted to revolve in only those orbits in which the, angular momentum of electron is an integral multiple of, , h, ., 2p, , nh, , where n = 1, 2, 3, ... , m, v, r are mass, speed, 2p, and radius of electron and h being Planck’s constant., 1, 1 ö, æ1, For Brackett-series, = R ç 2 - 2 ÷ , where n = 5, 6, 7, K, l, n ø, è4, For shortest wavelength, n = 5, 1, 1 ö, æ1, Þ, = 1.097 ´ 107 ç ÷, l, 16, 25, è, ø, 9, 7, = 1.097 ´ 10 ´, = 0.0246 ´ 107, 16 ´ 25, Þ, l = 40.514 ´ 10-7 ~, - 4051 nm, i.e. mvr =, , It lies in infrared region of electromagnetic spectrum., 16. Bohr’s second postulate (quantum condition) states that, the, electron revolves around the nucleus in certain privileged, orbit which satisfy certain quantum condition that angular, momentum of an electron is an integral multiple of h/ 2p,, where h is Planck’s constant., i.e., L = mvr = nh / 2p, where, m = mass of electron, v = speed of electron and, r = radius of orbit of electron., Þ, 2pr = n ( h / mv)

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108, , CBSE Term II Physics XII, , Since, de-Broglie wavelength associated with an electron is, given as, h, h, l= =, p mv, \Circumference of electron in nth orbit, = n ´ de-Broglie wavelength associated with electron, 17. The angular momentum in planetary motion is very large as, compared to the angular momentum of electrons. So, it, corresponds to a very large value of n., At such large quantum numbers, the quantum physics is, prevailed by classical physics and motion of planets around, the sun can be governed by Newton’s law of motion., 18. Given, atomic number, Z = 80,, KE = K = 8 MeV = 8 ´ 106 ´ 1.6 ´10- 19 J, ( Ze ) ( 2e ), Q Energy, K =, 4p e0 r0, where, r0 = distance of closest approach., Þ, r0 = 2 Ze 2 / 4p e0 ( K ), 2 ´ 9 ´ 109 ´ 80 ´ (1.6 ´ 10- 19 ) 2, 8 ´ 106 ´ 1.6 ´ 10-19, = 2.88 ´ 10- 14 m, As,, r0 µ ( 1 / K ), If KE gets doubled, distance of closest approach reduces to, half., 19. The Rutherford nuclear model of the atom describes the, atom as an electrically neutral sphere consisting of a very, small, massive and positively charged nucleus at the centre, surrounded by the revolving electrons in their respective, dynamically stable orbits. The electrostatic force of, attraction Fe between the revolving electrons and the, nucleus provides the requisite centripetal force Fc to keep, them in their orbits. Thus, for a dynamically stable orbit in a, hydrogen atom, Fc = Fe, 1, mv2, e2, =, × 2, (Q Z = 1 ), 4pe0 r, r, =, , Thus, the relation between the orbit radius and the electron, velocity is, e2, r=, 4pe0 mv2, The kinetic energy K and electrostatic potential energy U of, the electron in hydrogen atom are, 1, e2, K = mv2 =, 2, 8pe0 r, and, , U =-, , e2, 4pe0 r, , (The negative sign in U signifies that the electrostatic force is, attractive in nature.), Thus, the total mechanical energy of the electron in a, hydrogen atom,, e2, e2, e2, E = K+U =, =8pe0 r 4pe0 r, 8pe0 r, , The total energy of the electron is negative. This implies the, fact that, the electron is bound to the nucleus. If E were, positive, an electron will not follow a closed orbit around the, nucleus and it would leave the atom., 20. When an a-particle is bombarded over a gold nucleus, it is, repelled by electrostatic repulsion. As a result, KE of, a-particle is converted into electrostatic PE., At a certain distance from the nucleus, whole of the KE of, a-particle converts into electrostatic potential energy and, a-particles cannot go further close to nucleus, this distance, is called distance of closest approach., In this process, all the kinetic energies of moving particle is, converted into potential energies., From the given data,, 1, 2 e ´ Ze, …(i), Initially,, ×, =K, 4pe0, r, Let r0 be the new distance of closest approach for a twice, energetic a-particle, then we have, 1, 2 e ´ Ze, …(ii), ´, = 2K, 4pe0, r0, On dividing Eq. (i) by Eq. (ii), we get, r0 1, r, = Þ r0 =, r, 2, 2, Limitations of Rutherford Nuclear Model, Limitations of Rutherford nuclear model are as given below, (i) Rutherford’s model of an atom could not explain the, stability of an atom., According to him, charged electrons revolve around atom, in circular paths, so it should experience acceleration due, to which it should lose energy continuously in the form of, electromagnetic radiations. It then eventually fall into, the nucleus thereby making the atom unstable., (ii) Rutherford’s model of an atom could not explain as how, the electrons are arranged in the orbits around the, nucleus., 21. Given, work function, f = 2 eV, æ 1, 1, 1 ö, = R çç 2 - 2 ÷÷, l, n, n, 2ø, è 1, æ 1, hc, 1 ö, E =, = hcR çç 2 - 2 ÷÷ = f + KE, l, n2 ø, è n1, Also, KE = eV0, n1 = 2, n 2 = n, æ1 1 ö, hcR ç - 2 ÷ = 2 ´ 1.6 ´ 10-19 + 1.6 ´ 10-19 ´ 055, ., è4 n ø, 1 ö, æ1, Þ 6.62 ´ 10-34 ´ 3 ´108 ´ 1.097 ´ 107 ç - 2 ÷, è4 n ø, = ( 3. 2 + 0.88) ´ 10-19, 1 ö, æ1, Þ 21.786 ´ 10-19 ç - 2 ÷ = 4.08 ´ 10-19, 4, n ø, è, 1, 1, - 2 = 0187, ., Þ n~, -4, 4 n, 22. Given, DE =12.5 eV, Energy of an electron in nth orbit of hydrogen atom,, 13.6, En = - 2 eV, n

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109, , CBSE Term II Physics XII, , In ground state, n = 1 Þ E1 = -13.6 eV, Energy of an electron in the excited state after absorbing a, photon of 12.5 eV energy will be, En = -13.6 +12.5 = -1.1 eV, -13.6 -13.6, \, n2 =, =, = 12.36, En, -1.1, Þ, n = 3.5, Here, state of electron cannot be fraction., So, n = 3 (2 nd excited state)., 1, 1 ö, æ 1, Lyman Series = R ç 2 - 2 ÷, l, 1, n, è, ø, For first member, n = 2, 1, 1 ö, æ1, æ4 - 1ö, \, = R ç 2 - 2 ÷ = 1 . 097 ´ 107 ç, ÷, l1, 2 ø, è 4 ø, è1, Þ, , l1 = 1.215 ´ 10, , -7, , m, , 1, 1 ö, æ1, Balmer Series = R ç 2 - 2 ÷, l, n ø, è2, For first member, n = 3, 1, 1 ö, æ1, \, =R ç 2 - 2÷, l1, 3 ø, è2, æ1 1 ö, = 1 . 097 ´ 107 ç - ÷, è 4 9ø, Þ, l1 = 6.56 ´ 10-7 m, , So, the wavelength of emitted spectral line,, 1242 eV - nm 1242 eV - nm, l=, =, = 1. 88 ´ 10-6 m, E (in eV), 0.66 eV, As here, l = 1.88 ´ 10-6 m » 18751 ´ 10-10 m, The wavelength belongs to Paschen series of hydrogen, spectrum., 24. (i) An electron undergoes transition from second excited, state to the first excited state which corresponds to, Balmer series and then to the ground state which, corresponds to Lyman series., (ii) The wavelength of the emitted radiations in the two, cases., – 1.5 eV, , n=2, , Balmer series, , – 3.4 eV, , Lyman series, n=1, , hc, We know that, l =, DE, From n 3 ® n 2 ,, hc, l1 =, E3 - E2, hc, hc, =, =, ( -1.5) - ( -3.4) 1.9, , 25. The frequency of any line in a series in the spectrum of, hydrogen like atoms corresponding to the transition of, electrons from ( n + p) level to nth level can be expressed as, a difference of two terms, é, 1, 1 ù, vmn = cRZ 2 ê, - 2ú, 2, n û, ë ( n + p), where, m = n + p, (p = 1, 2, 3, K) and R is Rydberg constant., For, p << n,, -2, é, 1 æ, pö, 1 ù, vmn = cRZ 2 ê 2 ç1 + ÷ - 2 ú, nø, n úû, êë n è, 2p 1 ö, æ 1, vmn = cRZ 2 ç 2 - 3 - 2 ÷, n, n ø, èn, [by binomial theorem ( 1 + x ) n = 1 + nx, if|x| < 1], æ 2cRZ 2 ö, ç, ÷, ç n3 ÷ p, è, ø, Thus, the first few frequencies of light that is emitted when, electrons fall to the nth level from levels higher than n , are, approximate harmonic (i.e. in the ratio 1 : 2 : 3 …) when, n >> 1., 26. Bohr combined classical and early quantum concepts and, gave his theory in the form of three postulates, These three postulates are as follows, (i) Bohr’s first postulate states that, an electron in an atom, could revolve in certain stable orbits without the, emission of radiant energy, contrary to the predictions of, electromagnetic theory. According to this postulate, each, atom has certain definite stable states in which it can, exist and each possible state has definite total energy., These are called the stationary states of the atom., (ii) Bohr’s second postulate states that, the electron revolves, around the nucleus only in those orbits for which the, angular momentum is some integral multiple of h/ 2p,, where h is the Planck’s constant ( = 6.63 ´ 10-34 J- s)., vmn = cRZ 2, , 23. According to Bohr’s theory of hydrogen atom, energy of, photon released, E2 - E1 = hn, Given,, E1 = - 1 .15 eV, E2 = - 0.85 eV, E2 - E1 = - 0.85 - ( - 1.51) = 1 .51 - 0.85, E2 - E1 = 0.66 eV, , n=3, , From n 2 ® n1,, hc, l2 =, E2 - E1, hc, hc, =, =, ( - 3.4) -( - 13.6) 10.20, l1 10.20, =, = 5.3, \, l2, 1.9, , –13.6 eV, , 2p, ~, n3, , Thus, the angular momentum L of the orbiting, electron is quantised., nh, i.e., L=, 2p, As, angular momentum of electron = mvr, \ For any permitted (stationary) orbit,, nh, mvr =, 2p, where, n = any positive integer (1, 2, 3, ....)., It is also called principal quantum number., In second excited state, i.e. n = 3, three spectral lines, namely Lyman series and Balmer series can be obtained, corresponding to transition of electron from n = 3 to n = 1, and n = 3 to n = 2, respectively and n = 2 to n = 1.

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114, , CBSE Term II Physics XII, , CHAPTER 06, , Nuclei, In this Chapter..., l, , Nucleus and its Composition, , l, , Nuclear Force, , l, , Mass-Energy and Nuclear, Binding Energy, , l, , Nuclear Energy, , Nucleus and Its Composition, In every atom, the positive charge and mass are densely, concentrated at the centre of the atom forming its nucleus., The radius of the nucleus is smaller than the radius of an, atom by a factor of 104 . This means the volume of a nucleus, is about 10 - 12 times the volume of the atom., , Composition of Nucleus, The nucleus was first discovered in 1911 by Lord Rutherford, and his associates by experiments on scattering of a-particles, by atoms. The result of scattering experiment concluded that, atoms consists of a small, central, massive and positive core, surrounded by orbiting electrons., The positive charge in the nucleus is due to the protons. A, proton carries one unit of fundamental charge., , Atomic Mass Unit, The mass of an atom is very small. Kilogram cannot be used, to measure such small quantity of mass. It is measured by a, unit called atomic mass unit (amu, i.e. u)., It is defined as, mass of one 12 C atom, 1u =, = 1.660539 ´ 10 27 kg, 12, Atomic masses are measured by an instrument called mass, spectrometer., , Atomic Number (Z ), Atomic number of an element is the number of protons, present inside the nucleus of an atom of the element., Atomic number, Z = Number of protons, = Number of electrons (in a neutral atom), , Mass Number ( A), Mass number of an element is the total number of protons, and neutrons inside the nucleus of an element., Mass number, A = Number of protons, Z, + Number of neutrons, N, The term nucleon is also used for neutron and proton., Nuclear species or nuclides are shown by the notation, where X is the chemical symbol of the species., , A, Z, , X,, , Size of Nucleus, The volume of the nucleus is directly proportional to the, number of nucleons (mass number)., If R is the radius of the nucleus having mass number A, then, 4, pR 3 µ A, 3, Þ, R µ A1 / 3, R = R 0 A1 / 3, where, R 0 = 1. 2 ´ 10 -15 m is the range of nuclear size., It is also known as nuclear unit radius., , Nuclear Density, Density of nuclear matter is the ratio of mass of nucleus and, its volume., 3m, \, r=, 4 p R 03, Thus, the density of nucleus is a constant, independent of A,, for all nuclei. The density of nuclear matter is approximately, 2.3 ´ 10 17 kg/m 3 .

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115, , CBSE Term II Physics XII, , Mass-Energy and Nuclear, Binding Energy, Mass-Energy, Einstein’s mass-energy equivalence equation is E = mc 2 ,, where E is the energy, m is the mass and c is the velocity of, light in vacuum (approximately equal to 3 ´ 10 8 m/s)., , Nuclear Binding Energy, The sum of the masses of neutrons and protons forming a, nucleus is more than the actual mass of the nucleus. This, difference of masses is known as mass defect., \, E = Dmc 2, ( Dm = mass defect), E b = [ Zm p + ( A - Z )m n - M ] c 2, where, M is mass of nucleus, m p is the mass of proton and, m n is the mass of neutron., ‘‘The binding energy E b of a nucleus is defined as the, minimum energy required to separate its nucleons and place, them at rest and infinite distance apart’’., , Nuclear Force, , Potential energy (MeV), , Nuclear forces are the strongest attractive forces between, nucleons. They are non-conservative forces and do not obey, inverse square law. They are also non-central force., , 100, 0, , –100, r0 1, , 2, r (fm), , 3, , Potential energy versus distance between a pair of nucleon, , Some of the important characteristics of these forces are as, given below, (i) Nuclear forces among a pair of neutrons, a pair of, protons and also between a neutron-proton pair, is, approximately the same. This shows that, nuclear, forces are independent of charge., (ii) The nuclear forces are very short range forces. They, are operative upto distances of the order of a few fermi., (iii) The nuclear force is much stronger than the coulomb, force acting between charges or gravitational forces, acting between masses., (iv) Nuclear force between two nucleons falls rapidly to, zero as their distance is more than a few femtometres, (fm). This leads to saturation of forces in a medium or, large sized nucleus, i.e. each nucleon interacts with its, immediate neighbours only, rather than with all the, other nucleons in the nucleus., (v) The nuclear forces are dependent on spin or angular, momentum of nuclei., , Nuclear Energy, Nuclear energy is the energy released during the, transformation of nuclei with less total binding energy to, nuclei with greater binding energy., Two distinct ways of obtaining energy from nucleus are as, follows, (i) Nuclear fission, (ii) Nuclear fusion, , Nuclear Fission, Nuclear fission is the phenomenon of splitting of a heavy, nucleus (usually A > 230) into two or more lighter nuclei by, the bombardment of proton, neutron, a-particle, etc., 1, 141, 92, 1, e.g. 235, 92 U + 0 n ¾® 56 Ba + 36 Kr + 3 0 n + Q, (Energy released), The Q-value is equal to the difference of mass of products, and reactants multiplied by square of velocity of light., Energy released per fission of 235, 92 U is approximately, 200.4 MeV., Nuclear Chain Reaction, In the nuclear fission reaction, there is a release of extra, neutrons. The extra neutrons in turn initiate fission process,, producing still more neutrons and so on. This is called, nuclear chain reaction. It is of two types, Uncontrolled Chain Reaction During fission reaction,, neutrons released are again absorbed by the fissile isotopes,, this cycle repeats to give a chain reaction, i.e. self-sustaining, and produces energy at a rate that increases rapidly with time, leading to large amount of radiation. This is called, uncontrolled chain reaction., Controlled Chain Reaction The reaction is controlled in, such a way that, only one of the neutrons emitted in a fission, causes another fission, then the fission rate remains constant, and the energy is released steadily. It is used in a nuclear, reactor., , Nuclear Reactor, It is a device that can initiate a self-sustaining controlled, chain reaction of a fissionable material. They are used at, nuclear power plants for generating electricity and in, propulsion of ships., Coolant, , Control rods, , Superheated steam, , Shielding, Steam, turbine, , Electric, generator, , Heat, exchanger, Water, condenser, Fission Fuel, chamber rods, , Used, steam, , Moderator, Pump, , Water, , Nuclear reactor, , Cold water

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116, , CBSE Term II Physics XII, , Construction, The key components of nuclear reactor are as follows, (i) Nuclear Fuel It is a material that can be used in, nuclear fission to derive nuclear energy. The common, fuels used in nuclear reactor are 233 U, 235 U, 239 Pu, etc., (ii) Nuclear Reactor Core It is the portion of a nuclear, reactor containing the nuclear fuel components,, where the nuclear reaction takes place., (iii) Moderator It is a medium to slow down the fast, moving secondary neutrons produced during the, fission. Heavy water, graphite, deuterium, paraffins,, etc., acts as moderator., (iv) Control Rods It is used in nuclear reactors to control, the rate of fission of uranium and plutonium. These, are made of chemical elements capable of absorbing, many neutrons without fissioning themselves such as, silver, indium, boron and cadmium., (v) Coolant It is a liquid used to remove heat from, nuclear reactor core and transfer it to electrical, generator and environment. Ordinary water under, high pressure can be used as coolant., (vi) Shielding It is the protective covering made of, concrete wall to protect from harmful radiations., , Nuclear Fusion, Nuclear fusion is the phenomenon of fusing two or more, lighter nuclei forming a single heavy nucleus., 1, 1, 2, +, e.g., 1 H + 1 H ¾® 1 H + 1 e + n + 0.42 MeV, 1H, , 2, , + 1 H 2 ¾® 2 He 3 + 0 n1 + 3.27 MeV, , 2, , + 1 H 2 ¾® 1 H 3 + 1 H1 + 4.03 MeV, , 1H, , Fusion of hydrogen nuclei into helium nuclei is the source of, energy of most of the stars including the sun., , The fusion reaction in the sun is a multi-step process in, which the hydrogen is fused into helium. (The proton-proton, (p, p) cycle) is given as, 1, 1, 2, +, ...(i), 1 H + 1 H ¾® 1 H + e + n + 0.42 MeV, e + + e - ¾® g + g + 1.02 MeV, 2, 1, 1H + 1H, 3, 3, 2 He + 2 He, , ¾®, , 3, 2, , ¾®, , 4, 2, , He + g + 5 .49 MeV, He +, , 1, 1, 1 H +1, , ...(ii), ...(iii), , H + 12 .86 MeV ...(iv), , For the fourth reaction to occur, the first three reactions, must occur twice, in this case two light helium nuclei unite to, form ordinary helium nucleus., Thus, four hydrogen atoms combine to form 42 He atom, releasing 26.7 MeV energy., , Distinction between Nuclear, Fission and Nuclear Fusion, Distinction between nuclear fission and nuclear fusion are as, given below, (i) Fission is the splitting of large nucleus into two or, more smaller ones. On the other hand, fusion is the, combining of two or more lighter nuclei to form larger, one., (ii) Fission does not normally occur in nature but fusion, occurs in stars such as the sun., (iii) Fission requires critical mass of the substance and, high speed neutrons but in fusion, high density and, high temperature environment are required., (iv) In fission, energy released is million times greater, than in chemical reactions, but lower than energy, released by nuclear fusion., (v) Uranium is the primary fuel for fission reaction and, hydrogen isotopes are the primary fuel in nuclear, fusion reaction.

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117, , CBSE Term II Physics XII, , Solved Examples, Example 1. Two stable isotopes of lithium 63 Li and 73 Li, have respective abundance of 7.5% and 92.5%., These isotopes have masses 6.01512 u and, 7.01600 u, respectively. Find the atomic mass of, lithium., [NCERT], Sol. Given, abundance per cent of 6 Li = 7.5%, Abundance per cent of 7 Li = 92.5%, Atomic mass of 6 Li = 6.01512 u, Atomic mass of 7 Li = 7.01600 u, Atomic mass = Weighed average of the isotopes, 6.01512 ´ 75, . + 7.01600 ´ 925, ., =, 75, . + 925, ., 45 .1134 + 648.98, =, = 6.941 u, 100, , Sol. Here in order to remove a neutron, energy has to be, supplied., 1, 40, Mass defect, Dm = M( 39, 20 Ca ) + M( 0 n ) - M( 20Ca ), = 38.970691 + 1.008665 - 39.962589, = 0.016767 amu, Equivalent energy = 0.06767 ´ 931, (Q 1 amu = 931 MeV ), = 15.6 MeV, , Example 5. A neutron breaks into a proton and, electron. Calculate the energy produced in this, reaction. (Take, m e = 9 ´ 10 -31 kg,, m p = 1.6725 ´ 10 -27 kg, m n = 1.6747 ´ 10 -27 kg and, c = 3 ´ 10 8 m/s), , Example 2. Boron has two stable isotopes 10, 5 B, and 11, 5 B. Their respective masses are 10.01294 u, and 11.00931 u and the atomic mass of boron is, 11, 10.811 u. Find the abundances of 10, 5 B and 5 B., , Sol. We have 10 n ® 11 p + -10 e, Mass defect, Dm = m n - m p - m e, = [1.6747 ´ 10-27 - 1.6725 ´ 10-27 - 0.0009 ´ 10-27 ], = 1.3 ´ 10-30 kg, Energy produced = Dmc 2 = 1.3 ´ 10-30 ´ ( 3 ´ 108 ) 2, = 11.7 ´ 10-14 J =, , [NCERT], 10, , Sol. Given, mass of B = 10.01294 u, Mass of 11B = 11.00931 u, Atomic mass of boron = 10.811 u, Let the abundance of 10 B be x%, so the abundance of 11B be, ( 100 - x )%., Atomic mass = Weighted average of the isotopes, x ´ 10.01294 + ( 100 - x ) ´ 11.00931, 10.811 =, ( x + 100 - x ), Abundance of 10 B, x = 19.9%, Abundance of 11B,( 100 - x ) = 100 - 19.9 = 80.1%, Thus, the abundance of 10 B is 19.9% and the abundance of, B is 80.1%., , 11, , Example 3. Select the pairs of isotones from the, following nuclei, , 12 Mg, , 24, , , 1 H 3 , 2 He 4 , 11 Na 23 ., , Sol. Isotones have same number of neutrons, N = A - Z, (i) 1H3 and 2 He4, Number of neutrons = 3 - 1 or 4 - 2 = 2, (i) 12 Mg24 and 11Na 23, Number of neutrons = 24 - 12 or 23 - 11 = 12, , Example 4. Calculate the energy required to remove, , 11.7 ´ 10-14, = 0.73 MeV, 1.6 ´ 10-13, (Q 1 MeV =1.6 ´ 10-13 J), , Example 6. Obtain the binding energy (in MeV) of a, , nitrogen nucleus ( 147 N), given m( 147 N) = 14.00307 u., (Take, m p = 1.00783 u and m n = 1.00867 u), , [NCERT], Sol. Given, mass of proton, m p = 1.00783,, Mass of neutron, m n = 1.00867 u, 14, 7 N nucleus contains 7 protons and 7 neutrons., Mass defect ( Dm ) = Mass of nucleons – Mass of nucleus, = 7m p + 7m n - m N, = 7 ´ 1.00783 + 7 ´ 1.00867 - 14.00307, = 7.05481 + 7.06069 - 14.00307, = 0.11243 u, Binding energy of nitrogen nucleus = Dm ´ 931 MeV, = 0.11243 ´ 931 MeV = 104.67 MeV, Thus, the binding energy is 104.67 MeV., , Example 7. Consider the fission of, , 238, 92 U by, , fast, , neutrons. In one fission event, no neutrons are, emitted and the final end products, after the beta, decay of the primary fragments, are 140, 58 Ce and, 99, 44 Ru., , the least tightly bound neutron in 20 Ca 40 ., Given that,, Mass of 20 Ca 40 = 39.962589 amu,, Mass of 20 Ca 39 = 38.970691 amu, , Calculate Q for this fission process. The, relevant atomic and particle masses are, m( 238, 92 U) = 238.05079 u,, , and mass of neutron = 1.008665., , and, , m( 140, 58 Ce ) = 139.90543 u, m( 99, 44 Ru ) = 98.90594 u., , [NCERT]

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118, , CBSE Term II Physics XII, , Sol. The fission reaction is given by, 238, 1, 140, 99, 92 U + 0 n ¾® 58 Ce + 44Ru + Q, Mass defect,, Dm = m(, , 238, 92 U ), , 99, + m( 0 n1 ) - m( 140, 58 Ce) - m( 44Ru ), , = 238.05079 + 1.00867 - 139.90543 - 98.90594, = 0.24809 u, Q-value for the given decay process,, Q = Dm ´ 931.5, = 0.24809 ´ 931.5 = 231.1 MeV, , Example 8. Consider the D-T reaction, (deuterium-tritium fusion), 2, 3, 4, 1 H + 1 H ¾® 2 He + n, (a) Calcualte the energy released in MeV in this, reaction from the data, m( 12 H) = 2.014120 u, and m( 13 H) = 3.016049 u., (b) Consider the radius of both deuterium and, tritium to be approximately 2.0 fm. What is the, kinetic energy needed to overcome the coulomb, repulsion between two nuclei? To what, temperature must the gas be heated to initiate, the reaction?, [NCERT], Sol. (a) The D-T reaction is given by, 2, 3, 4, 1, 1 H + 1 H ¾® 2 He + 0 n + Q, Mass defect, Dm = m( 12H) + m( 13H) - m( 42He) - m( 0 n1 ), = 2.014102 + 3.016049 – 4.002603 – 1.00867, = 0.018878 u, Q-value for the given decay process, = Dm ´ 931 = 0.018878 ´ 931 = 17.58 MeV, (b) Repulsive potential energy of two nuclei, when they, almost touch each other, 1 q 2 9 ´ 109 ( 1.6 ´ 10-19 ) 2, U=, ×, =, = 5.76 ´ 10-14 J, 4pe0 2r, 2 ´ 2 ´ 10-15, Also we know that, kinetic energy required for one fusion, event = average thermal kinetic energy available with, the interacting particles, 3, Kinetic energy = ´ kT ´ 2 (two nuclei), 2, = 3kT, Kinetic energy, T=, 3k, 5.76 ´ 10-14, =, 3 ´ 1.38 ´ 10-23, , 3.017 and 1.009, respectively. If 1kg of deuterium, undergoes complete fusion, then find the amount of, total energy released., (Take, 1 amu = 931.5 MeV/c 2 ), Sol. Given, fusion reaction, 1 H2 + 1 H2 ® 2 He3 + 0 n1, Mass defect, Dm = 2 ´ m(1 H2 ) - [ m( 2 He4 ) + Mn ], = [(2.015 ´ 2) - (3.017 + 1.009)], = 4 ´ 10-3 amu, Equivalent energy, E = Dmc 2 = Dm ´ 931.5 MeV / c 2, = 4 ´ 10-3 ´ 931.5 = 3.726 MeV, Number of atoms in 1 kg of 1 H2, Given mass, =, ´ Avogadro number, Atomic mass, 1000, =, ´ 6.023 ´ 1023, 2, In one reaction, two atoms of 12 H are used., So, total energy released, 1 æ 1000, ö, = ç, ´ 6.023 ´ 1023 ÷ ´ 3.726 ´ 1.6 ´ 10-13, 2è 2, ø, ~, - 9 ´ 1013 J, , Example 10. In a nuclear reactor, U 235 undergoes, fission releasing energy of 100 MeV. The reactor, has 20% efficiency and the power produces is, 2000 MW. If the reactor is to function for 5 yr, find, the total mass of uranium required., Sol. Given, output power, Po = 2000MW = 2 ´ 109 W, Output power, Po, Efficiency, h =, Input power, Pi, Þ, , = m ´ 0.041 ´ 1012 J, , = 1 .39 ´ 10 K, This temperature cannot be achieved in actual, behaviour., , 1, , H 2 + 1 H 2 ® 2 He 3 + 0 n 1 , the masses of deuteron,, , helium and neutron expressed in amu are 2.015,, , Þ Pi = 1010 W, , Consider the mass of uranium be m., Number of fissions = Number of atoms, Given mass, =, ´ Avogadro number, Atomic mass, m, m, =, ´ NA =, ´ 6.023 ´ 1023, A, 235, Energy produced, = Number of fissions ´ Energy per fission, m, =, ´ 6.023 ´ 1023 ´ 200 MeV, 235, m, =, ´ 6.023 ´ 1023 ´ 100 ´ 1.6 ´ 10-13 J, 235, (Q 1 MeV = 1.6 ´ 10-13 J ), , 9, , Example 9. In the fusion reaction,, , 20 2 ´ 109, =, 100, Pi, , Þ, , = m ´ 4.1 ´ 1010 J, Energy, Power =, Time, m ´ 4.1 ´ 1010, 1010 =, 5 ´ 365 ´ 24 ´ 3600, m = 3.84 ´ 107 g = 38.4 ´ 103 kg

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119, , CBSE Term II Physics XII, , Chapter, Practice, PART 1, Objective Questions, , 7. The gravitational force between a H-atom and, another particle of mass m will be given by, M×m, Newton’s law F = G, , where r is in km., r2, What is M in this formula?, , 1. As compared to 12 C atom, 14 C atom has, (a), (b), (c), (d), , two extra protons and two extra electrons, two extra protons but no extra electrons, two extra neutrons but no extra electrons, two extra neutrons and two extra electrons, , (a) Gravitational mass of H-atom, (b) Effective mass of H-atom, (c) Nuclear mass of H-atom, (d) Mass of electrons in H-atom, , 8. Nuclear force is, , 2. Density of a nucleus is, (a), (b), (c), (d), , more for lighter elements and less for heavier elements, more for heavier elements and less for lighter elements, very less compared to ordinary matter, a constant, , 3. The nuclear radius of a certain nucleus is 7.2 fm, -17, , and it has charge of 1.28 ´ 10, C. The number of, neutrons inside the nucleus is, (a) 136, (c) 140, , (b) 142, (d) 132, , 4. The ratio of mass densities of nuclei of, 16, , 40, , Ca and, , O is close to, , (a) 5, (c) 0.1, , (a) strong, short range and charge independent force, (b) charge independent, attractive and long range force, (c) strong, charge dependent and short range attractive force, (d) long range, charge dependent and attractive force, , 9. Fpp, Fnn and Fnp are the nuclear forces between, proton-proton, neutron-neutron and, neutron-proton, respectively. Then, relation, between them is, (a) Fpp = Fnn ¹ Fnp, (c) Fpp = Fnn = Fnp, , 10. Which amongst the following is a correct graph of, potential energy U of a pair of nucleons as a, function of their separation r ?, , (b) 2, (d) 1, , m p = 1.0072676 u, mass of neutrons, m n = 1.008665 u, and mass of 2 He 4 = 4.001506 u), [All India 2020], , (a), , r, , m p and m n denotes the mass of proton and neutron, respectively and BE the binding energy (in MeV),, then, 2, , (b) BE =[ Zm p + ( A - Z ) m n - m( A, Z )] c 2, , r0, , r, , 0, , U, , 6. A nucleus Z X A has mass represented by m ( A, Z ). If, , (d) BE = m( A, Z ) - Zm p - ( A - Z ) m n, , (b), , r0, , 0, , (a) 0.016767 u, (b) 1.00726 u, (c) 2.00686 u, (d) 0.0303592 u, , (c) BE = [ Zm p + Am p - m( A, Z )] c 2, , U, , U, , 5. Mass defect of helium ( 2 He 4 ) is (Take, mass of proton,, , (a) BE = [ m( A, Z ) - Zm p - ( A - Z ) m p ] c, , (b) Fpp ¹ Fnn = Fnp, (d) Fpp ¹ Fnn ¹ Fnp, , U, , (c), , (d), 0, , r0, , r, , 0 r0, , r, , 11. Heavy stable nuclei have more neutrons than, protons. This is because of the fact that,, [NCERT Exemplar], (a) neutrons are heavier than protons, (b) electrostatic force between protons is repulsive, (c) neutrons decay into protons through beta decay, (d) nuclear forces between neutrons are weaker than that, between protons

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120, , CBSE Term II Physics XII, , 12. In any fission process, the ratio of, , 17. In a nuclear reactor , moderators slow down the, neutrons which come out in a fission process. The, moderator used have light nuclei. Heavy nuclei will, not serve the purpose, because, [NCERT Exemplar], , mass of fission products, is, mass of parent nucleus, (a) less than 1, (b) greater than 1, (c) equal to 1, (d) depends on the mass of parent nucleus, , (a) they will break up, (b) elastic collision of neutrons with heavy nuclei will not, slow them down, (c) the net weight of the reactor would be unbearably high, (d) substances with heavy nuclei do not occur in liquid or, gaseous state at room temperature, , 13. On bombarding U 235 by slow neutron, 200 MeV, energy is released. If the power output of atomic, reactor is 1.6 MW, then the rate of fission will be, , 18. Nuclear fusion is common to the pair, , (a) 5 ´ 10 22 s– 1, , I. uranium based reactor, II. hydrogen bomb, III. energy production in sun, IV. atom bomb, , 16 – 1, , (b) 5 ´ 10 s, , (c) 8 ´ 1016 s– 1, (d) 20 ´ 1016 s– 1, , (a) Both I and II, (c) Both III and IV, , 14. Which of the following fusion reactions will not, result in the net release of energy?, I. 6 Li + 6 Li, 4, , 19. Which of the following is fusion process?, (a) 12 H +12 H ¾® 42He + 10 n, , 4, , II. He + He, , 141, (b) 10 n + 235, 92 U ¾® 56 Ba +, , III. 12 C + 12C, IV., , 35, , Cl +, , 35, , Cl, l, , (b) III, (d) II, , 15. Schematic diagram of a nuclear reactor based on, thermal neutron fission is as shown below, C=Coolant, Steam to, turbine, , B=Control, rods, , Heat exchanger, (Steam generator), Water from, condenser, , A=core, , Here, the correct purpose of the parts specified by A, B, C, and D is given in, (a) A ® reduces leakage, (b) B ® can shut down the reactor, (c) C ® site of nuclear fission and contains 235, 92 U, (d) D ® transfers heat to a working fluid which in turn may, produce steam, , 16. For sustaining the chain reaction in a sample, (of small size) of, fast neutrons by, (a), (b), (c), (d), , 92, 36, , Kr + 3 ( 10 n ), , (c) Uranium decay, (d) None of the above, , (a) IV, (c) I, , D=Reflector, , (b) Both II and III, (d) Both II and IV, , 235, 92, , U, it is desirable to slow down, , friction, elastic damping/scattering, absorption, None of the above, , Assertion-Reasoning MCQs, Direction (Q. Nos. 20-25) Each of these questions, contains two statements Assertion (A) and Reason (R)., Each of these questions also has four alternative, choices, any one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given, below., (a) Both A and R are true and R is the correct, explanation of A, (b) Both A and R are true, but R is not the correct, explanation of A, (c) A is true, but R is false, (d) A is false and R is also false, , 20. Assertion The heavier nuclei tend to have larger, N/Z ratio because neutron does not exert electric, force., Reason Coulomb forces have longer range, compared to nuclear force., , 21. Assertion It is not possible to use, for fusion energy., Reason The binding energy of, , 35, , 35, , Cl as the fuel, , Cl is too small., , 22. Assertion Nuclei having number about 60 are most, stable., Reason When two or more light nuclei are, combined, the some energy is lost during the, process.

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121, , CBSE Term II Physics XII, , (ii) For sustaining the nuclear fission chain reason in a, sample (of small size) of 235, 92 U, a process called, elastic damping uses, , 23. Assertion Initially, it was believed that mass and, energy are conserved separately in a nuclear, reaction., Reason According to Einstein, one can convert, mass-energy into other forms of energy., , 24. Assertion Fission of, , 235, 92 U, , is brought about by a, thermal neutron, whereas that of 238, 92 U is brought, about by a fast neutron., 238, Reason 235, 92 U is an even-odd nucleus, whereas 92 U, is an even-even nucleus., , (a) electrons, (c) heavy nuclei, , (iii) Which of the following is/are fission reaction(s)?, 236, 133, 99, 1, I. 10 n + 235, 92 U ® 92 U ® 51 Sb ® 41Nb + 4 0 n, II. 10 n +, III. 12 H +, , nuclear reactor., , l, , Case Based MCQs, , (a) 216 MeV, (b) 200 MeV, (c) 100 MeV, (d) Cannot be estimated from given data, , 26. Nuclear Fission, , Unstable nucleus, , + 3 0 n1 + Q, (i) Fusion processes, like combining two deuterons to, form a He nucleus are impossible at ordinary, temperature and pressure.The reasons for this can, be traced to the fact, (a) nuclear forces have short range, (b) nuclei are negatively charged, (c) the original nuclei must be completely ionised before, fusion can take place, (d) the original nuclei must first break up before combining, with each other, , + 210 n, , (iv) If a nucleus with mass number A = 240 with, E bn = 7.6 MeV breaks into two fragments of A = 120, and E bn = 8. 5 MeV, then released energy is around, , Direction Read the following passage and answer the, questions that follows, In the year 1939, German scientist Otto Hahn and, Strassmann discovered that when a uranium, isotope was bombarded with a neutron, it breaks, into two intermediate mass fragments. It was, observed that, the sum of the masses of new, fragments formed were less than the mass of the, original nuclei. This difference in the mass, appeared as the energy released in the process., Thus, the phenomenon of splitting of a heavy, nucleus (usually A> 230) into two or more lighter, nuclei by the bombardment of proton, neutron,, a-particle, etc., with liberation of energy is called, nuclear fission., Fission reaction resulting from the absorption of, neutron is known as induced fission., 235, + 0 n 1 ® 92 U 236 ® 56 Ba 144 + 36 Kr 89, 92 U, , 235, 140, 94, 92 U ® 54 Xe + 38 Sr, 2, 3, 1, 1 H ® 2He + 0 n, , (a) Both II and III, (b) Both I and III, (c) Only II, (d) Both I and II, , 25. Assertion Heavy water is used to slow neutron in, Reason It does not react with slow neutron and, mass of deuterium is comparable to the neutron., , (b) lighter nuclei, (d) None of these, , (v) Assuming that about 20 MeV of energy is released, in a fusion reaction 1 H 2 + 1 H 3 ® 0 n 1 + 2 He 4 , then, the mass of 1 H 2 consumed per day in a fusion, reactor of power 1 MW will approximately be, (a) 0.001 g, (c) 10.0 g, , (b) 0.1 g, (d) 1000 g, , PART 2, Subjective Questions, l, , Short Answer (SA) Type Questions, 1. How the size of a nucleus is experimentally, determined? Show that the density of nucleus is, independent of its mass number. [Delhi 2012, 2011C], , 2. Calculate the surface area of a nucleus (assuming it, to be a perfect sphere)., , 3. The mass of a nucleus is less than the sum of the, masses of constituent neutrons and protons., Comment., , 4. Obtain approximate ratio of the nuclear radii of the, gold isotope, , 197, 79 Au, , and the silver isotope, , 107, 47 Ag., , [NCERT], , 5. Given the mass of iron nucleus as 55.85u and, A = 56. Find the nuclear density., , [NCERT]

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122, , CBSE Term II Physics XII, , 6. Supposing that, protons and neutrons have equal, masses. Calculate how many times nuclear matter, is denser than water. (Take, mass of a nucleon, = 1.67 ´ 10 -27 kg and R 0 = 1.2 ´ 10 -15 m), , 7. The neutron separation energy is defined as the, energy required to remove a neutron from the, nucleus. Obtain the neutron separation energies of, 27, the nuclei 41, 20 Ca and 13 Al from the following data, [NCERT], , m( 40, 20 Ca ) = 39.962591 u, m( 41, 20 Ca ) = 40.962278 u, 26, m( 13, Al ) = 25.986895 u, 27, m( 13, Al ), , 23, 11 Na?, , (ii) Which nuclide out of the two mirror isobars, have greater binding energy and why?, 20, 21, 9. The three stable isotopes of neon 10, Ne, 10, Ne and, 22, 10 Ne have, , respective abundances of 90.51%, 0.27%, and 9.22%. The atomic masses of three isotopes are, 19.99 u, 20.99 u and 21.99 u, respectively. Obtain, the average atomic mass of neon., [NCERT], , 10. Nuclei with magic number of protons, Z = 2, 8, 20, 28, 50, 52 and magic number of, neutrons N = 2, 8, 20, 28, 50, 82 and 126 are found, to be very stable., Verify this by calculating the proton separation, energy Sp for 120 Sn (Z = 50) and 121 Sb (Z = 51). The, proton separation energy for a nuclide is the, minimum energy required to separate the least, tightly bound proton from a nucleus of that nuclide., It is given by, Sp = ( M Z - 1,N + M H - M Z, N ) c 2, , and, , In = 118 .9058 u,, Sn = 119.902199 u,, H = 1.0078252 u., , are very similar to, those of, The average energy released per, fission is 180 MeV. How much energy in MeV is, released, if all the atoms in 1 kg of pure 239, 94 Pu, undergo fission?, [NCERT], 235, 92 U., , + 12H ® 32He + 10 n + 3.27 MeV, although number, of nucleons is conserved, yet energy is released., Explain, how., , 16. A given coin has a mass of 3.0 g. Assume that, the, coin is entirely made of 63, 29 Cu atoms (of mass, 62.92960 u). Calculate the nuclear energy that, would be required to separate all the neutrons and, protons from each other., , 17. (i) Write three characteristic properties of nuclear, force., (ii) Draw a plot of potential energy of a pair of, nucleons as a function of their separation. Write, two important conclusions that can be drawn, from the graph., [Delhi 2015], , 18. The Q-value of a nuclear reaction A + b ¾® C + d, is defined by Q = [ m A + m b - m C - m d ] c 2 , where, the masses refer to the respective nuclei., Determine from the given data, the Q-value of the, following reactions and state whether the reactions, are exothermic or endothermic., (i) 11 H + 31 H ¾® 21 H + 21 H, 12, 20, 4, (ii) 12, 6 C + 6 C ¾® 10 Ne + 2 He, , m ( 12 H) = 2.014102 u, , Sb = 120.903824 u, , 1, , 239, 94 Pu, , Atomic masses are given to be, m( 1 H1 ) = 1.007825 u, , 119, , 121, , 14. The fission properties of, , 2, 1H, , = 26.981541 u, , nuclide 2, if Z1 = N 2 and Z 2 = N 1 ., (i) What nuclide is a mirror isobar of, , 120, , 28, two equal fragments, 13, Al. Is the fission, energetically possible? Argue by working out Q of, the process. Given, m ( 56, 26 Fe) = 55.93494 u and, 28, m( 13 Al) = 27.98191 u., [NCERT], , 15. In a typical nuclear reaction,, , 8. A nuclide 1 is said to be the mirror isobar of, , Given,, , 13. Suppose we think of fission of a 56, 26 Fe nucleus into, , [NCERT Exemplar], , 11. Draw a plot of potential energy between a pair of, nucleons as a function of their separation. Mark the, regions, where potential energy is, [Delhi 2013], (i) positive and (ii) negative., , 12. Complete the following fission reaction and, calculate the amount of energy it releases., 1, 235, 88, 136, 0 n + 92 U ¾® 38 Sr + 54 Xe + (?), , m ( 13 H) = 3.016049 u, m ( 12, 6 C) = 12.000000 u, 20, m ( 10, Ne) = 19.992439 u, , [NCERT], , 19. A 1000 MW fission reactor consumes half of its fuel, in 5 yr. How much 235, 92 U did it contain initially?, Assume that, the reactor operates 80% of the time, that all the energy generated arises from the fission

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123, , CBSE Term II Physics XII, , of 235, 92 U and that this nuclide is consumed only by, the fission process., , fissionable uranium would our country need per year, by 2020? (Take, the heat energy per fission of 235 U to, be about 200 MeV), [NCERT], , 20. How long can an electric lamp of 100 W be kept, glowing by fusion of 2 kg of deuterium? Take, the, fusion reaction as, 2, 1H, , + 12H ¾® 32H + n + 3.27 MeV, , [NCERT], , 21. Calculate and compare the energy released by, (i) fusion of 1 kg of hydrogen deep within sun and, (ii) the fission of 1 kg of, , 235, , U in a fission reactor., [Delhi 2020], , l, , Long Answer Questions (5 Marks), 22. Explain giving necessary reactions, how energy is, released during, (i) fission and, , (ii) fusion., , 23. Binding energy of deuteron is 2.2 MeV. It is bound, by electrostatic forces, i.e. F =, , 1 e¢ 2, ., 4pe 0 r, , e¢, Calculate the ratio of , where e¢ is the effective, e, charge. A g-ray with energy E is aimed at this, deuteron nucleus to break it into a (neutron +, proton) pair, such that n and p move in direction of, the g-ray. If E = B, show that it cannot happen and, hence calculate how much E must be greater than B, for this process to happen., , 24. Suppose India had a target of producing, 200000 MW of electric power by 2020 AD, 10% of, which was to be obtained from nuclear power plants., Assume that on an average, the efficiency of utilisation, (i.e. conversion to electric energy) of thermal energy, produced in a reactor was 25%. How much amount of, , l, , Case Based Questions, Direction Read the following passage and answer the, questions that follows, , 25. Nuclear Fission, Nuclear fission is the phenomenon of splitting of a, heavy nucleus (usually A > 230) into two or more, lighter nuclei by the bombardment of proton,, neutron, a-particle, etc. Energies associated with, nuclear processes are about a million times larger, than chemical process., In fission, a heavy nucleus like 235, 92 U breaks into, two smaller fragments by the bombardment of, thermal neutron (low energy or slow moving)., 1, 141, 92, 1, e.g. 235, 92 U + 0 n ¾® 56 Ba + 36Kr + 3 0 n + Q, (Energy released), , Q-value here refer to the energy released in the, nuclear process, which can be determined using, Einstein’s mass - energy relation, E = mc 2 . The, Q-value is equal to the difference of mass of products, and reactants multiplied by square of velocity of light., Energy released per fission of 235, 92 U is 200.4 MeV. The, fragment nuclei produced in fission are highly, unstable. They are highly radioactive and emit, b-particles in succession until each reaches to a stable, end product., (i) What are moderators? Give few examples., (ii) What is multiplicity in nuclear fission? Explain., (iii) Define the Q-value of a nuclear process. When can, a nuclear process not proceed spontaneously?

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Chapter Test, Multiple Choice Questions, , 1. If the nuclear radius of, , 27, , Al is 3.6 fm, the aproximate, nuclear radius of 64 Cu (in fm) is, (a) 2.4, (c) 4.8, , (b) 1.2, (d) 3.6, , 2. Two protons are attracting each other, then separation, between them is, (a) 10 -10 m, (c) 10 -8 m, , (b) 10 -2 m, (d) 10 -15 m, , The binding energy per nucleon for helium nucleus will, be, (b) 7 MeV, (d) 1 MeV, , produce electric power of 200 MW for one hour?, (c) 1 ´ 10, , -6, , (b) 8 ´ 10 -6 kg, (d) 3 ´ 10 -6 kg, , kg, , (a) hydrogen is converted into carbon, (b) hydrogen and helium are converted into carbon and other, heavier metals/elements, (c) helium is converted into hydrogen, (d) hydrogen is converted into helium, , Short Answer Type Questions, , 6. If both the numbers of protons and neutrons are, conserved in a nuclear reaction like, 12, , + 6C 12 ¾®, , 10 N, , 20, , consist of still smaller parts? One way to find out is to, probe a nucleon just as Rutherford probed an atom., What should be the kinetic energy of an electron for it, to be able to probe a nucleon? Assume, the diameter, of a nucleon to be approximately 10 -15 m. (Ans. 10 9 eV), how in both these processes energy is released., Calculate the energy release in MeV in the, deuterium-tritium fusion reaction., 2, 3, 1H+ 1H, , ¾® 42 He + n, , Using the data, , 5. In fusion reaction occurring in the sun,, , 6C, , produced through the fission of 235, 92 U nuclei (by, neutrons) to sustain a chain reaction? What type of, nuclei are (preferably) needed for slowing down fast, neutrons?, , 13. Distinguish between nuclear fission and fusion. Show, , 4. How much mass has to converted into energy to, (a) 2 ´ 10 -6 kg, , 11. Why is it necessary to slow down the neutrons,, , 12. Are the nucleons fundamental particles or do they, , 3. The mass defect of helium nucleus is 0.0303 amu., , (a) 28 MeV, (c) 14 MeV, , (ii) Which nuclide out of the two mirror isobars have, greater binding energy and why?, , + 2He 4, , In what way, is the mass converted into the energy?, Explain., , 7. Explain, why elements like helium are very stable., 8. He 32 and He 31 nuclei have the same mass number. Do, they have the same binding energy?, , m( 21 H) = 2. 014102 u,, m( 31 H) = 3 . 016049 u,, m( 42 He) = 4. 002603 u,, mn = 1 . 008665 u, MeV, 1 u = 931.5 2, c, , and helium. Which species will have greater kinetic, energy?, , 15. Stars are giant ball of hydrogen which fuses to form, helium and energy. What happens, when all the, hydrogen is used up?, , Long Answer Type Questions, , 16. (i) What is the source of stellar energy? Also, explain, , 9. Why do stable nuclei never have more protons than, , the thermonuclear reactions of sun., (ii) Discuss radiation hazards., , neutrons?, , 10. A nuclide 1 is said to be the mirror isobar of nuclide 2,, if Z1 = N2 and Z2 = N1 ., (i) What nuclide is a mirror isobar of, , 23, 11, , Na?, , 17. Describe uncontrolled and controlled nuclear chain, reactions and therefore explain nuclear reactor & its, components., , Answers, Multiple Choice Questions, 1. (c), , 2. (d), , (Ans. 17.589 eV), , 14. In a nuclear reaction, uranium breaks into thorium, , 3. (b), , 4. (b), , 5. (d), , For Detailed Solutions, Scan the code

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125, , CBSE Term II Physics XII, , EXPLANATIONS, Hence, the BE of nucleus is, BE = [ Zm p + Nm n - m( A, Z )] c 2, , PART 1, 1. (c) For 612 C, A = 12 = N + Z, Z = 6, , BE = [ Zm p + ( A - Z ) m n - m ( A, Z )] c 2, , Þ, N =6, For 614 C, A = 14 = N + Z, Z = 6, Þ, N =8, Also, number of electrons in both atoms, = number of protons = Z = 6, Mass, mA, 3m, 2. (d) Density =, ,, =, =, Volume 4 pR 3 A 4pR 03, 0, 3, m = m p = Mn, = 2.3 ´ 1017 kg m -3, which is a constant., , 8., , 3. (a) Nuclear radius, R = R 0 A1/ 3, where, R = 7.2 ´ 10-15 m, R 0 = 1.2 ´ 10-15 m, 3, , GMm, r2, where, M = effective mass of hydrogen atom,, G = gravitational constant, and, r = distance between H-atom & particle of mass m., (a) Nuclear force has the following properties, I. Nuclear force is a short range force, whose range is of, the order of 2 to 3 femtometre., II. Nuclear force is the strongest force in nature., III. Nuclear force is an attractive force acting between, nucleons, which is charge independent., Therefore, nuclear force is strong, short range and charge, independent force., (c) Nuclear force between two particles is independent of, charges of particle., Fpp = Fnn = Fnp, (b) Potential energy of a pair of nucleons as a function of, their separation is correctly depicted in option (b). For a, separation greater than r0 , the force is negative and, attractive and potential energy decreases upon increasing, the distance and for a separation less than r0 , the force is, strongly repulsive, so potential energy increases for r < r 0 ., (b) Stable heavy nuclei have more neutrons than protons., This is because, electrostatic force between protons is, repulsive , which may reduce stability., , 7. (b) Given, F =, , 3, , æ 7.2 ´ 10-15 ö, æ Rö, ÷ = ( 6) 3 = 216, A = çç ÷÷ = çç, -15 ÷, è R0 ø, è 1.2 ´ 10 ø, q 1.28 ´ 10-17, Also, atomic number, Z = =, = 80, e 1.6 ´ 10-19, , 9., , \, , Therefore, number of neutrons,, N = A - Z = 216 - 80 = 136, 4. (d) Mass density of nuclear matter is a constant quantity for, all elements. It does not depend on element’s mass number, or atomic radius., \The ratio of mass densities of 40 Ca and 16 O is 1 : 1., 5. (d) Mass defect, Dm, = Mass of nucleons - Mass of nucleus, = [ Zm p + ( A - Z ) m n ] - Mn, Here, mass number, A = 4,, Atomic number, Z = 2,, Number of protons = 2,, Number of neutrons = A - Z = 4 - 2 = 2,, m p = 1.0072676 u,, m n = 1.008665 u, and, Mn = 4.001506 u., Mass of nucleon in 2 He4 =, Mass of 2 protons + Mass of 2 neutrons, So, mass of nucleons, = 2 ´ 1.0072676 + 2 ´ 1.008665, \, Dm = 2 ´ 1.0072676 + 2 ´ 1.008665 - 4.001506, = 0.0303592 u, 6. (b) According to mass defect, if the quantity of mass, disappearing is Dm, then the binding energy,, BE = Dmc 2, From the above discussion, the Dm is given by, Dm = Zm p - Nm n - m( A, Z ), where, m ( A, Z ) is the mass of the atom of mass number A, and atomic number Z ., , 10., , 11., , 12. (a) In fission process, when a parent nucleus breaks into, daughter products, then some mass is lost in the form of, energy. Thus,, Mass of fission products < Mass of parent nucleus, Mass of fission products, Þ, <1, Mass of parent nucleus, 13. (b) Energy released on bombarding U235 by neutron, = 200 MeV, Power output of atomic reactor = 1.6 MW, Power out of reactor, \ Rate of fission =, Energy released per fission, 1.6 ´ 106, =, 200 ´ 106 ´ 1.6 ´ 10-19, (Q 1 eV = 1.6 ´ 10-19 J), = 5 ´ 1016 s -1, 14. (a) The reaction 35 Cl + 35 Cl will not result in the net, release of energy because heavy nucleus are not used in, fusion reaction., 15. (b) Here,, A ® is the core of the reactor, which is the site of nuclear, fission. It contains elements in suitably fabricated form. The, fuel may be enrinched uranium say 235, 92 U., D ® is a reflector surrounding the core which is used to, reduce the leakage.

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126, , 16., 17., , 18., , 19., , CBSE Term II Physics XII, , C ® is coolant which helps in removing the energy (heat), released in fission. It transfers heat to working fluid which in, turn may produce steam., B ® is control rods that shut down the reactor as they have, high absorption of neutrons., Hence, only option (b) shows correct purpose of given part., (b) Fast neutrons are slowed down by elastic scattering with, light nuclei. Each collision takes away nearly 50% of energy., (b) According to the question, the moderator used have light, nuclei (like proton). When protons undergo perfectly elastic, collision with neutron emitted, their velocities are, exchanged, i.e. neutrons come to rest and protons move with, the velocity of neutrons., Heavy nuclei will not serve the purpose because elastic, collisions of neutrons with heavy nuclei will not slow them, down., (b) Energy produced in sun and hydrogen bomb is due to, nuclear fusion., However, energy released in uranium based nuclear reactor, and atom bomb is due to nuclear fission., (a) When two lighter nuclei combine to form a heavier, nucleus, the process is called nuclear fusion., 2, 2, 4, 1, e.g., 1 H +1 H ¾® 2 He + 0 n, , 20. (a) The heavier nuclei have greater number of neutrons than, N, protons. Thus, heavier nuclei have larger ratio. Neutrons, Z, are chargeless particle, hence they are not affected by, electric force. The electric forces have longer range as, compared to nuclear forces, which have a very short range., As, Coulomb force is effective upto several metres whereas, nuclear force are only effective within the vicinity of nucleus., Therefore, both A and R are true and R is the correct, explanation of A., 21. (c) The fusion of two lighter nuclei produce sufficient energy., Since, 35 Cl is a relatively heavier nuclei, it does not produce, significant energy. The fusion of two lighter nuclei produces, sufficient energy. The binding of 35 Cl is 298 MeV, hence it is, not suitable for fusion energy., Therefore, A is true but R is false., 22. (c) Nuclei having mass number A = 60, have highest binding, energy. Thus, those are most stable. When two or more light, nuclei fuse together, some energy is gained., Therefore, A is true but R is false., 23. (b) Before the advent of special theory of relativity, it was, believed that mass and energy were conserved separately in, a reaction., However, Einstein showed that mass is an another form of, energy and one can convert mass-energy into other forms of, energy, say kinetic energy and vice-versa., Therefore, both A and R are true but R is not the correct, explanation of A., 24. (b) Fission of U235 occurs by slow neutrons only (of energy, about 1 eV) or even by thermal neutrons (of energy about, 0.025 eV). Fission of 238, 92 U is brought about by a fast, neutrons. 235, has, odd, mass number and even atomic, U, 92, number., , Hence it is an even-odd nucleus whereas 238, 92 U has even, mass number and even atomic number, hence it is an, even-even nucleus., Therefore, both A and R are true but R is not the correct, explanation of A., 25. (a) Heavy water (D2O) is used to slow down fast moving, neutrons in nuclear reactor, hence it is called moderator. It, does remove neutrons from the system by absorbing them., When fast moving neutrons are passed through heavy water, (D2O), then they make elastic collisions with protons or, deuterium having mass comparable to neutrons, which have, comparatively much smaller than velocities. In few, interactions, the velocities of neutrons get interchanged, with those of protons., Heavy water does not react with slow moving neutrons, i.e. it, slows down-fast moving neutrons., Therefore, both A and R are correct and R is the correct, explanation of A., 26. (i) (a) Fusion processes are impossible at ordinary, temperatures and pressures. Because nuclei are, positively charged and nuclear forces are short range, strongest forces., (ii) (b) Fast neutrons are slowed down by elastic scattering, with lighter nuclei as each collision takes away nearly, 50% of energy., (iii) (d) Reactions I and II represent fission of uranium, isotope 235, 92 U, when bombarded with neutrons that, breaks it into two intermediate mass nuclear fragments., However, reaction III represents two deuterons fuses, together to form the light isotope of helium., (iv) (a) The energy released (i.e. Q-value) in the fission, reaction of nuclei like uranium is of the order of 200 MeV, per fissioning nucleus. This is estimated as follows, Let us take a nucleus with A = 240 breaking into two, fragments each of A = 240, then, Ebn for A = 240 nucleus is about 7.6 MeV, Ebn for the two A = 120 fragment nuclei is about, 8.5 MeV, So, gain in binding energy for nucleon is about 0.9 MeV., Hence, the total gain in binding energy is, 240 ´ 0.9 or 216 MeV., (v) (b) Energy produced, U = Pt, = 106 ´ 24 ´ 36 ´ 102 = 24 ´ 36 ´ 108 J, Energy released per fusion reaction, = 20 MeV = 20 ´ 106 ´ 1.6 ´ 10-19, = 32 ´ 10-13 J, Energy released per atom of 1 H2, = 32 ´ 10-13 J, Number of 1 H2 atoms used, 24 ´ 36 ´ 108, = 27 ´ 1021, =, 32 ´ 10-13, Mass of 6 ´ 1023 atoms = 2 g, \ Mass of 27 ´ 1021 atoms, 2, =, ´ 27 ´ 1021 = 01, . g, 6 ´ 1023

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127, , CBSE Term II Physics XII, , Density of water, r¢= 103 kg/ m 3, , PART 2, 1. The size of the nucleus is experimentally determined using, Rutherford’s a-scattering experiment and the distance of, closest approach and impact parameter., The relation between radius and mass number of nucleus is, R = R 0 A1/ 3, where, R 0 = 1.2 fm, A = mass number and R = radius of, nucleus., Mass of nucleus, mA, Nuclear density, r =, =, Volume of nucleus 4 p( R A1/ 3 ) 3, 0, 3, where, m = mass of each nucleon., mA, m, r=, Þ r=, 4 3, 4 3, pR 0 A, pR 0, 3, 3, From the above formula, it is clear that r does not depend on, the mass number., 2. Surface area of nucleus,, SA = 4pR 2 = 4p( R 0 A1/ 3 ) 2, = 4pR 02 × A 2 / 3, Substituting constant ( R 0 ) value = 1.2 ´ 10, SA = 4( 314, . )( 1.2 ´ 10, , -15, , )A, , -15, , m, we get, , 2/ 3, , = ( 1.8 ´ 10-29 ) A 2 / 3, 3. When nucleons approach each other to form a nucleus, they, strongly attract each other. Their potential energy decreases, and becomes negative. It is the potential energy which holds, the nucleons together in the nucleus. The decrease in PE, results in the mass of nucleons inside the nucleons., 4. Radius of nuclei, R = R 0 A1/ 3, where, A is the mass number of nucleus and R 0 is an, empirical constant., \, R µ A1/ 3, 1/ 3, , 1/ 3, æ Agold ö, 197 ö, ÷÷ = æç, = çç, ÷ = 1.225 = 1.23, Rsilver è Asilver ø, è 107 ø, 5. Given, mass, m = 55 .85 u, = 55 .85 ´ 1 .67 ´ 10-27 kg, 4, 4, 4, Volume, V = pR 3 = p ( R 0 A1/ 3 )3 = pR 03 ´ A, 3, 3, 3, m, \ Nuclear density, r =, V, 3 ´ 55.85 ´ 1.67 ´ 10-27, =, 22, 4 ´ ´ (1.2 ´ 10-15 ) 3 ´ 56, 7, = 2.29 ´ 1017 kg/m 3, , Þ, , R gold, , 6. Density of nucleus of water,, 3m, 3 ´ 1.67 ´ 10-27, r=, =, 3, 4pR 0 4 ´ 22 ´ (1.2 ´ 10-15 ) 3, 7, 7 ´ 3 ´ 1.67 ´ 1018, =, 88 ´ 1.2 ´ 1.2 ´ 1.2, = 2.307 ´ 1017 kg/ m 3, , \, , r 2.307 ´ 1017, =, r¢, 103, = 2.307 ´ 1014, , 7. When a neutron is separated from, 40, 20 Ca and the reaction becomes, 41, 20 Ca, , ¾®, , 41, 20 Ca,, , 40, 20 Ca, , Mass defect,, 1, Dm = m ( 40, 20 Ca) + m( 0 n ) - m (, , we are left with, , + 0 n1, 41, 20, , Ca), , = 39.962591 + 1 .008665 - 40.962278, = 0.008978 u, Energy for separation of neutron = Dm ´ 931, = 0.008978 ´ 931, = 8.358 MeV, 27, When a neutron is separated from 13, Al, we are left with, 26, 13 Al. Thus, the reaction becomes, 27, 13 Al, , ¾®, , 26, 13 Al, , + 10 n, , Mass defect,, 26, 27, Dm = m (13, Al ) + m ( 0 n1 ) - m( 13, Al ), = 25.986895 + 1.00865 - 26.981541, = 0.014019, \ Energy for separation of neutron, = Dm ´ 931 = 0.014019 ´ 931, = 13.06 MeV, 8. (i) According to the question, a nuclide 1 is said to be, mirror isobar of nuclide 2, if Z1 = N 2 and Z2 = N1., Now, in 1123 Na, Z1 = 11, N1 = 23 - 11 = 12, \ Mirror isobar of 1123 Na is 1223 Mg, for which, Z2 = 12 = N1 and N 2 = 23 - 12 = 11 = Z1, (ii) As, 1223 Mg contains even number of protons (12), against 1123 Na which has odd number of protons ( 11 ),, therefore 1123 Mg has greater binding energy than 1123 Na., 9. Given, abundance per cent of Ne20 = 90.51%, Abundance per cent of Ne21 = 0.27%, Abundance per cent of Ne22 = 9.22%, Mass of Ne20 = 19.99 u, Mass of Ne21 = 20.99 u, Mass of Ne22 = 21.99 u, Average atomic mass, m = Weighted average of all isotopes, 90.51 ´ 19.99 + 0.27 ´ 20.99 + 9.22 ´ 21.99, =, 90.51 + 0.27 + 9.22, 1809.29 + 5.67 + 202.75 2017.71, =, =, = 20.18, 100, 100, 10. The proton separation energy is given by, Sp (Sn) = ( M119 , 70 + MH - M120 , 70 ) c 2, = (118.9058 + 1.0078252 - 119.902199) c 2, = 0.0114262 c 2

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128, , CBSE Term II Physics XII, , = (119.902199 +1.0078252 -120.9038224) c, , 2, , = 0.006202 c 2, Since, Sp (Sn) > Sp (Sb) , Sn nucleus is more stable than, Sb nucleus., 11. The graph between the potential energy of a pair of, nucleons as a function of their separation is given below, , PE (MeV), , 100, , Positive region, 10 fm, x, r (fm), , 0.8 fm, , Negative, region, , 100, , (i) For distance less than 0.8 fm, negative PE decreases to, zero and then becomes positive., (ii) For distance greater than 0.8 fm, negative PE goes on, decreasing., 12. By conservation of charge and mass, given equation can be, written as, 1, 235, 88, 136, 1, 0 n + 92 U ¾® 38Sr + 54 Xe + 120 n + Q (Energy), For amount of energy released, use, Q = Dm ´ 931 MeV, 13. The given reaction for decay process,, 56, 28, 26 Fe ¾® 213 Al, 28, Mass defect, Dm = m ( 56, 26 Fe) - 2m (13 Al ), , = 55.93494 - 2(27.98191), = - 0.02888 u, Þ, Q = Dm ´ 931, = - 0.02888 ´ 931, = -26.88728 MeV, Because the energy is negative, so the fission is not possible, energetically., 14. According to the concept of Avogadro number,, 23, the number of atoms in 239 g of 239, 94 Pu = 6.023 ´ 10, Number of atoms in 1 kg of, , 239, 94 Pu, 23, , 6.023 ´ 10 ´ 1000, = 2.52 ´ 1024, 239, The average energy released in one fission, = 180 MeV, So, total energy released in fission of 1 kg of, 239, . ´ 1024, 94 Pu = 180 ´ 252, =, , = 4.53 ´ 1026 MeV, 15. In a nuclear reaction, the sum of the masses of the target, nucleus (12 H) and the bombarding particle (12 H) may be, greater than the product nucleus ( 32 He) and the outgoing, neutron 10 n. So, from the law of conservation of mass-energy,, some energy (3.27 MeV) is evolved due to mass defect in the, nuclear reaction. This energy is called Q-value of the, nuclear reaction., , 16. Given, mass of coin = 3g, Atomic mass of Cu = 63, Mass of 63, 29 Cu, M = 62.92960 u, Avogadro’s number = 6.023 ´ 1023, Mass of proton, m p = 1.007825 u, Mass of neutron, m n = 1.008665 u, Nuclear energy required to separate neutrons and protons,, Eb = ?, Since, each atom of copper contains 29 protons and 34, neutrons. Therefore, mass defect of each atom using the, relation,, Dm = [ Z m p + ( A - Z ) m n ] - M, D m = [29 ´ 1.007825 + (63 - 29) ´ 1.008665] - 62.92960, = 0.591935 u, 6.023 ´ 1023 ´ 3, Number of atoms in 3 g coin =, = 2.868 ´ 1022, 63, Total mass defect of all atoms,, ( D m ) total = 0.591935 ´ 2.868 ´ 1022 = 1.6977 ´ 1022, The nuclear energy required ( Eb ) to separate all the neutrons, and protons from each other and can be calculated by using, the relation,, Eb = ( D m ) ´ c 2 = ( D m ) c 2 ´ 931 MeV/c 2 (Q 1 u = 931 MeV ), = 1.6977 ´ 1022 ´ 931 MeV = 1.58 ´ 1025 MeV, 17. (i) Characteristics properties of nuclear force, (a) Nuclear forces act between a pair of neutrons, a pair of, protons and also between a neutron-proton pair, with, the same strength. This, shows that, nuclear forces are, independent of charge., (b) The nuclear forces are dependent on spin or angular, momentum of nuclei., (c) Nuclear forces are non-central forces. This shows that,, the distribution of nucleons in a nucleus is not, spherically symmetric., (ii), Potential energy (MeV), , Similarly, Sp(Sb) = ( M120 , 70 + MH - M121, 70 ) c 2, , 100, 0, , –100, r01, , 2, r (fm), , 3, , (Potential energy versus distance), , From the plot, it is concluded that, (a) The potential energy is minimum at a distance, r0 ( » 0.8 fm ) which means that, the force is attractive, for distances greater than 0.8 fm and repulsive, for the, distance less than 0.8 fm between the nucleons., (b) Nuclear forces are negligible, when the distances, between the nucleons is more than 10 fm., 18. (i) The given reaction, 11H + 13H ¾® 12 H + 12 H, Mass defect, Dm = m (11H) + m (13H) -2 m (12 H), = 1.007825 + 3.016049 - 2(2.014102), = - 0.00433 u

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129, , CBSE Term II Physics XII, , Q-value of the reaction,, Q = Dm ´ 931 = - 0.00433 ´931, Q = - 4.031 MeV, As, the energy is negative, so the reaction is, endothermic., (ii) The given reaction,, 12, 12, 20, 4, 6 C + 6 C ¾® 10 Ne + 2 He, 20, 4, Mass defect, Dm = 2 m (12, 6 C) - m (10 Ne) - m ( 2 He), , = 2 ´ 12 - 19.992439 - 4.002603, Dm = 0.00495 u, Q -value of the reaction,, Q = Dm ´ 931, = 0.00495 ´ 931, = 4.62 MeV, Since, the energy is positive, thus the reaction is, exothermic., 19. Given, power of reactor, P = 1000 MW, 235, The energy generated in one fission of 92, U is 200 MeV., 1, 235, Number of 92, ´ 6.023 ´ 1023, U atoms in 1 g =, 235, \ Energy generated per gram of 235, 92 U, , æ 1, ö, =ç, ´ 6.023 ´ 1023 ´ 200 ´ 1.6 ´ 10-13 ÷, è 235, ø, Total energy generated in 5 yr with 80% of the time, 80, = 1000 ´ 106 ´ 5 ´ 365 ´ 24 ´ 60 ´ 60 ´, 100, ( as, E = Pt ), consumed, in, 5, yr,, \ Mass of 235, U, 92, Total energy, m=, Energy consumed per gram, 1000 ´ 106 ´ 5 ´ 365 ´ 24 ´ 60 ´ 60 ´ 0.8, æ 1 ö, 23, -13, ç, ÷ ´ 6.023 ´ 10 ´ 200 ´ 1.6 ´ 10, è 235 ø, = 1.538 ´ 106 g = 1538 kg, =, , \ Initial amount of, , 235, 92 U, , = (1544 ´ 2) kg = 3076 kg, , 20. Let t be the time., According to the Avogadro number concept,, Number of atoms in 2 g of deuterium = 6.023 ´ 1023, Number of atoms in 2 kg of deuterium, 6.023 ´ 1023 ´ 2 ´ 103, =, 2, +26, = 6.023 ´ 10 nuclei, Energy released during fusion of two deuterium, = 3.27 MeV, \ Energy released per deuterium = 1 .635 MeV, Energy released in 6.023 ´ 1026 deuterium atoms, = 1.635 ´ 6.023 ´ 1026, = 9.848 ´ 1026 MeV, = 9.848 ´ 1026 ´ 1.6 ´ 10-13, = 15.75 ´ 1013 J, , 100 J energy used by lamp in time = 1 s, 1 ´ 15.75 ´ 1013, s, 15 .75 ´ 1013 J energy used in time =, 100, = 15.75 ´ 1011s, 15 .75 ´ 1011, yr, =, 60 ´ 24 ´ 60 ´ 365, [Q 1 yr = (60 ´ 24 ´ 60 ´ 365) s], = 4.99 ´ 104 yr, Thus, the bulb glows for 4.99 ´ 104 yr., 21. (i) In sun, four hydrogen nuclei fuse to form a helium, nucleus and release 26MeV energy., Q 1 g of hydrogen contains = 6.023 ´ 1023 nuclei, \ Energy released by fusion of 1 kg ( = 1000 g) of, 6.023 ´ 1023 ´ 26 ´ 103, hydrogen, E1 =, = 39 ´ 1026 MeV, 4, 235, (ii) Energy released in one fission of 92, U nucleus, = 200 MeV, Mass of uranium = 1 kg = 1000 g, We know that, 235 g of 235 U has 6.023 ´ 1023 atoms or, nuclei., \ Energy released in fission of 1 kg of U 235 ,, 6.023 ´ 1023 ´ 1000 ´ 200, E2 =, 235, 26, = 5.1 ´ 10 MeV, E1, 39 ´ 1026, \, =, = 7.65 » 8, E2, 5 .1 ´ 1026, Thus, the energy released in fusion is 8 times the, energy released in fission., 22. (i) Nuclear Fission The phenomenon of splitting of heavy, nuclei (mass number > 120) into smaller nuclei of nearly, equal masses is known as nuclear fission. In nuclear, fission, the sum of the masses of the product is less than, the sum of masses of the reactants. This difference of, mass gets converted into energy E = mc 2 and hence, sample amount of energy is released in a nuclear fission., 1, 141, 92, 1, e.g. 235, 92 U + 0 n ® 56 Ba + 36Kr + 3 0 n + Q, Masses of reactant, = 235.0439 amu + 1.0087 amu, = 236.0526 amu, Masses of product, = 140.9139 + 91.8973 + 3.0261, = 235.8373 amu, Mass defect = 236.0526 - 235.8373 = 0.2153 amu, Q 1 amu º 931 MeV, Energy released = 0.2153 ´ 931, Þ, = 200 MeV nearly, Thus, energy is liberated in nuclear fission, if, undergoes nuclear fission., , 235, 92 U, , (ii) Nuclear Fusion The phenomenon of conversion of two, lighter nuclei into a single heavy nucleus is called nuclear, fusion., Since, the mass of the heavier product nucleus is less, than the sum of masses of reactant nuclei and therefore

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130, , CBSE Term II Physics XII, , certain mass defect occurs which converts into energy as, per Einstein’s mass - energy relation. Thus, energy is, released during nuclear fusion., e.g., , 1, 1H, , + 1H1 ¾® 1H2 + e + + n + 0.42 MeV, , Also, 1H2 + 1H2 ¾® 1H3 + 1H1 + 4.03 MeV, 23. The binding energy of H-atom,, me 4, E = 2 2 = 13. 6 eV, pe0 h, , …(i), , If proton and neutron had charge e each and were governed, by the same electrostatic force, then in the above equation, we would need to replace electronic mass m by the reduced, mass m ¢ of proton-neutron and the electronic charge e by e ¢., M ´N, M 1836 m, m ¢=, = =, = 918 m, M+N, 2, 2, Here, M represents mass of a neutron/proton., 918m ( e ¢ ) 4, \ Binding energy =, = 2. 2 MeV, 8e20 h 2, , …(ii), , Dividing Eq. (ii) by Eq. (i), we get, 4, 2.2 MeV 2.2 ´ 106, æ e¢ ö, 918 ç ÷ =, =, 13.6 eV, 13.6, èeø, 4, , 6, , 2.2 ´ 10, æ e¢ ö, = 176.21, ç ÷ =, 13.6 ´ 918, èeø, e¢, = ( 176.21 )1/ 4 = 3.64, e, Given binding energy, B = 2.2 MeV, From the energy conservation law,, pp2, p2, E - B = Kn + Kp = n +, 2m 2m, From conservation of momentum,, E, pn + pp =, C, As, E = B,, From Eq. (i), pn2 + pp2 = 0, , …(iii), , …(iv), , It can only happen, if pn = pp = 0., So, the Eq. (iii) cannot be satisfied and the process cannot, take place., Let E = B + X, where X <<B for the process to take place., Put value of pn from Eq. (iv) in Eq. (iii), we get, æE, ö, ç - pp ÷, 2, c, è, ø + pp, X=, 2m, 2m, 2Epp E2, 2, or 2pp + 2 - 2mX = 0, c, c, Using the formula of quadratic equation, we get, , pp =, , 2E, ±, c, , æ E2, ö, 4E 2, - 8 çç 2 - 2mX ÷÷, 2, c, èc, ø, , 4, For real value pp , the discriminate is positive, æ E2, ö, 4E 2, \, = 8çç 2 - 2mX ÷÷, 2, c, c, è, ø, E2 ~ B 2, Þ, X=, 4mc 2 4mc 2, , 24. Total target power = 200000 = 2 ´ 105 MW, Total nuclear power = 10% of total target power, 10, =, ´ 2 ´ 105 = 2 ´ 104 MW, 100, Energy produced/fission = 200 MeV, Efficiency of power plant = 25%, Energy converted into electrical energy per fission, 25, =, ´ 200 = 50 MeV, 100, = 50 ´ 1.6 ´ 10-13 J, Total electrical energy to be produced per year, = 2 ´ 104 MW = 2 ´ 104 ´ 106 W, = 2 ´ 1010 W = 2 ´ 1010 J/s, = 2 ´ 1010 ´ 60 ´ 60 ´ 24 ´ 365 J/yr, Number of fission in one year,, 2 ´ 1010 ´ 60 ´ 60 ´ 24 ´ 365, n=, 50 ´ 1.6 ´ 10-13, 2 ´ 36 ´ 24 ´ 365, n=, ´ 1024, 8, Mass of 6.023 ´ 1023atoms of 235 U = 235 g = 235 ´ 10-3kg, Mass of 235, 92 U required to produce, 2 ´ 36 ´ 24 ´ 365, =, ´1024 atoms, 8, 235 ´ 10-3 ´ 2 ´ 36 ´ 24 ´ 365 ´ 1024, =, 6.023 ´ 1023 ´ 8, = 3.08 ´ 104 kg, Thus, the mass of uranium needed per year is, 3.08 ´ 104 kg., 25. (i) In nuclear reaction, to sustain a chain reaction, fast, moving electrons are slowed by elastic collision with, light nuclei., Therefore, in reactor moderators are provided with, fissionable nuclei for slowing down fast neutrons., Some examples of moderators are heavy water (D 2O),, boron (B) and cadmium (Cd)., (ii) The sustained fissibility of nuclear chain reaction, depends on the multiplication factor or simply called, multiplicity ( K )., Rate of production of neutrons, K=, Rate of loss of neutrons, If K = 1, the operation of reactor is said to be critical. It is, what we wish to be, steady power operation., If K > 1, the reaction rate and reactor power increases, exponentially. It is called supercritical and reactor can, explode., If K < 1, the reaction gradually stops and the condition, is called sub-critical., (iii) The Q-value of a nuclear process refers to the energy, released in the nuclear process which can be determined, using Einstein’s mass-energy relation, E = mc 2 . The, Q-value is equal to the difference of mass of products and, reactant nuclei multiplied by square of velocity of light., The nuclear process does not proceed spontaneously,, when Q - value of a process is negative or sum of masses, of product is greater than sum of masses of reactant.

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131, , CBSE Term II Physics XII, , CHAPTER 07, , Semiconductor, Electronics, (Materials, Devices and Simple Circuits), In this Chapter..., l, , Semiconductors, , l, , Diode as a Rectifier, , l, , p-n Junction Diode, , l, , Special Purpose p-n, Junction Diodes, , Classification of Conductors,, Semiconductors and Insulators, On the basis of the relative values of electrical conductivity, (s ) or resistivity (r = 1 / s ), the solids are broadly classified as, Conductors They possess very low resistivity (or high, conductivity)., r ~10 - 2 -10 - 8 W-m, s ~10 2 -10 8 Sm - 1, l, , l, , l, , Semiconductors They have resistivity or conductivity, intermediate to metals and insulators., r ~10 - 5 - 10 6 W-m, s ~10 + 5 - 10 - 6 Sm - 1, Insulators They have high resistivity (or low conductivity)., r ~10 11 - 10 19 W-m, s ~10 - 11 - 10 - 19 Sm - 1, , Energy Bands in Solids, According to Bohr’s atomic model and concept of electronic, configuration in an isolated atom, the electrons have certain, definite discrete amounts of energy corresponding to, different shells and sub-shells, i.e. there are well-defined, energy levels of electrons in an isolated atom., But in a crystal due to interatomic interaction, valence, electrons are shared by more than one atom. Due to this,, splitting of energy level takes place. The collection of these, closely spaced energy levels is called an energy band. These, bands are formed due to the continuous energy variation in, different energy levels., , These different energy levels in different electrons are, formed because inside the crystal, each electron has a unique, position and no two electrons are exactly at the same pattern of, surrounding charges., , Valence Band, The energy band, which includes the energy levels of the, valence electrons is called valence band. This band may be, partially or completely filled with electrons but is never, empty., , Conduction Band, The energy band above the valence band is called conduction, band. At room temperature, this band is either empty or, partially filled with electrons. Electrons can jump from, valence band to conduction band and contribute to the, electric current., , Energy Band Gap, The minimum energy required for shifting electrons from, valence band to conduction band is called energy band gap, ( Eg ). It can be zero, small or large depending upon the, material., If l is the wavelength of radiation used in shifting the, electron from valence band to conduction band, then energy, band gap,, E g = hn = hc / l, where, h is called Planck’s constant and c is the velocity of, light.

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132, , CBSE Term II Physics XII, , Difference between Conductor, Insulator and Semiconductor on the basis of Energy Bands, Insulator, , Semiconductor, , No energy gap between the conduction band, and valence band or the conduction band and, valence band overlap each other., So, electrons from below the fermi level can, shift to higher energy levels above the fermi, level in the conduction band and behave as, free electrons., , The valence band is completely filled and, the conduction band is completely empty., The energy gap is quite large., Electrons are bound to valence band and are, not free to move. Hence, electric conduction, is not possible in this type of material., , The valence band is totally filled and the, conduction band is empty but the energy gap, between conduction band and valence band,, unlike insulators is very small., At room temperature, some electrons in the, valence band acquire thermal energy greater, than energy band gap and jump over to the, conduction band, where they are free to, move under the influence of even a small, electric field and acquire small conductivity., , EV, Valence, band, , Valence, band, For metals, , Empty conduction band, , Electron energy, , Electron, energy E, C, , Conduction, band, Eg » 0, , Overlapped, area, , Conduction, band, , EC, Eg > 3 eV, EV, , Fermi Energy, , Valence, band, , l, , It is the maximum possible energy possessed by free, electrons of a material at absolute zero temperature (i. e. 0 K)., The value of fermi energy is different for different materials., , Electron energy, , The materials whose conductivity lie between metals and, insulators are known as semiconductors., At absolute zero temperature, all states in valence band are, filled and all states in conduction band are empty. At low, temperature, pure semiconductors are insulators., On the basis of purity, semiconductors are of two types, , l, , l, , l, , l, , Under the action of an electric field, holes move towards, negative potential giving hole current I h . The total, current I is the sum of the electron current I e and the, hole current I h . i.e. I = I e + I h ., At equilibrium, the rate of generation is equal to rate of, recombination of charge carriers. The recombination, occurs due to an electron colliding with a hole., An intrinsic semiconductor behaves like an insulator at, T = 0 K., , Conduction, band, , Eg < 3 eV, EV, , Valence band, , Electrons, , EC, Eg, , EV, , Eg, EV, , (a), , Holes, , (b), , Fig. (a) an intrinsic semiconductor at T = 0K behaves like, insulator and Fig. (b) is representing four thermally, generated electron-hole pairs at T > 0K, , Intrinsic Semiconductors, This type of semiconductor is also called an undoped, semiconductor or i-type semiconductor. It is a pure, semiconductor without any significant presence of dopant, species., Some characteristics of these semiconductors are as given, below, In intrinsic semiconductors, the number of excited, electrons is equal to number of holes, i.e. n h = n i , where, n i is called intrinsic carrier concentration., , EC, , Thermally excited electrons at T > 0 K, partially occupy the, conduction band. They have come from the valence band, leaving equal number of holes there., EC, , Semiconductors, , Electron energy, , Conductor (Metal), , Extrinsic Semiconductors, Those semiconductors in which some impurity atoms are, embedded are known as extrinsic or impurity, semiconductors., When some desirable impurity is added to intrinsic, semiconductors deliberately, then this process is called doping, and the impurities are called dopants., There are two types of dopants used in doping, Trivalent (valency 3) atoms: e.g. Indium (In), Boron (B),, aluminium (Al), etc., Pentavalent (valency 5) atoms: e.g. Arsenic (As), Antimony, (Sb), Phosphorous (P), etc., Extrinsic semiconductors are basically of two types, n-type semiconductors, p-type semiconductors, l, , l, , l, , l

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133, , CBSE Term II Physics XII, , n-Type Semiconductors, This type of semiconductor is obtained, when pentavalent, impurity is added to Si or Ge., Therefore, major conduction in n-type semiconductors is due, to electrons. So, electrons are known as majority carriers and, the holes are known as the minority carriers., This means, n e >> n h ; I e >> I h ., , p-Type Semiconductors, This type of semiconductor is obtained, when a trivalent, impurity is added to Si or Ge., In p-type semiconductor, the holes movement results in the, formation of the current. In this type of semiconductor,, majority charge carriers are holes, i.e. positively charged and, minority charge carriers are electrons., i.e. n h >> n e ; I h >> I e ., Hence, these conductors are known as p-type, semiconductors or acceptor type semiconductors., The electron and hole concentration in a semiconductor in, thermal equilibrium is given by n e n h = n i2 ., , p-n Junction Diode, , l, , l, , l, , Forward Biasing and Reverse, Biasing of Junction Diode, Biasing is the method of connecting external battery or emf, source to a p-n junction diode., , Forward Biasing, , It is an arrangement made by a close contact of n-type, semiconductor and p-type semiconductor., It is basically a p-n junction and it is a two terminal device., p, n, It is represented by the symbol, The direction of arrow indicates the conventional direction of, current., , Formation of Depletion, Region in p-n Junction, During the formation of p-n junction and due to the, concentration gradient across p and n-sides, holes diffuse, from p-side to n-side ( p ® n) and electrons diffuse from, n-side to p-side ( n ® p ). The diffused charge carriers, combine with their counterparts in the immediate vicinity of, the junction and neutralise each other., Electron drift, , The potential difference developed across the depletion, region is called the potential barrier (V0 )., It depends on dopant concentration in the semiconductor, and temperature of the junction., Due to this potential barrier, following currents arises in, semiconductor, Due to the diffusion of holes from p-side to n-side and, electrons from n-side to p-side at the junction, a current, rises from p-side to n-side, which is called diffusion, current., If an electron-hole pair is created on the depletion region, due to thermal collision, the electrons are pushed by the, electric field towards the n-side and the holes towards the, p-side, which gives rise to a current from n-side to p-side, known as drift current., In steady state, diffusion current = drift current., , Electron diffusion, , A junction diode is said to be forward biased, when the, positive terminal of the external battery is connected to the, p-side and negative terminal to the n-side of the diode., When an external voltage V is applied in forward biasing,, then, (i) The effective barrier height under forward bias is, (V0 - V )., (ii) Due to external electric field, electron-hole, recombination occur and a covalent bond breaks near, p-region producing more electron and hole that moves, to the external battery terminals. Thus, the total, forward current is the sum of hole diffusion current, and conventional current due to diffusion of electrons., Junction, Electron, Hole, p-region, n-region, Ei, , +, , –, , E, p, , n, , +, Depletion region, Hole diffusion, , Hole drift, , p-n junction formation process, , This sets up potential difference across the junction and an, internal electric field E i directed from n-side to p-side. The, region on either side of the junction which becomes depleted, (free) from the mobile charge carriers is called depletion, region or depletion layer. The width of depletion region is of, the order of 10 -6 m., , V, , –, , Battery, , Forward biasing of junction diode, , Reverse Biasing, A junction diode is said to be reverse biased, when the, positive terminal of the external battery is connected to the, n-side and negative terminal to the p-side of the diode., When an external voltage V is applied across the ends of, diode in reverse biasing, then

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134, , CBSE Term II Physics XII, , Reverse voltage (V), –10 –8 –6 –4 –2, –, +, , p-region, , Ei, , 4, Breakdown, voltage, –, , (b), (a), Reverse biased characteristic of a diode, , +, , +, , l, , Reverse biasing of junction diode, , I-V (Current-Voltage) Characteristics of, p-n Junction Diode, The graphical relations between voltage applied across p-n, junction and current flowing through the junction are called, I-V characteristics of junction diode., , Forward Biased Characteristics, The circuit diagram and graph plotted between voltage and, current for diode are shown in Figs. (a) and (b)., n, , +, , mA, , V, , +, , +, , –, , Battery, (a), l, , l, , –, , Forward current (mA), , –, , 7, , B, , 6, , In reverse biased, the applied voltage supports the flow, of minority charge carriers across the junction. So, a very, small current flows across the junction due to minority, charge carriers., The reverse current is voltage independent upto certain, voltage known as breakdown voltage and this voltage, independent current is called reverse saturation current., , Note If the reverse bias is equal to the breakdown voltage, then the, reverse current through the junction increases very rapidly (CD, portion of the graph), this situation is called avalanche, breakdown and the junction may get damaged due to excessive, heating, if this current exceeds the rated value of, p-n junction., , In diodes, a resistance is offered by the junction which, depends on the applied voltage, which is called dynamic, resistance. It is the ratio of small change in voltage to the, small change in current produced., DV, Dynamic resistance, rd =, ., DI, , Diode as a Rectifier, The process of converting alternating voltage/current into, direct voltage/current is called rectification., , 5, Ge, , 4, , 8, D, , Battery, , p, , 6, , +, , Electron, , E, , V, , 2, , C, , –, , l, , –, , 0, O, , +, , Battery, , n-region, , –, , V, , mA, , Junction, Hole, , n, , p, , Reverse current (mA), , The direction of applied voltage is same as the direction of, barrier potential, so effective barrier height will be V0 + V., There is almost no flow of current due to majority charge, carriers, a very small current due to minority charge carriers, flows across the junction. This current is called reverse, current., , Principle, , 3, 2, A, , 1, , O, , 0, (b), , 0.1 0.2 0.3 0.4 0.5, Forward voltage (V), , At the start when applied voltage is low, the current through, the diode is almost zero. It is because of the potential barrier,, which opposes the applied voltage., With further increase in applied voltage, the current, increases very rapidly and diode behaves like a conductor., The forward voltage beyond which the current through the, junction starts increasing rapidly with voltage is called knee, voltage or threshold voltage., , Diode allows current to pass only, when it is forward, biased. So, if an alternating voltage is applied across a, diode, the current flows only in that part of the cycle when, the diode is forward biased., , Diode as a Half-Wave Rectifier, In the half-wave rectifier, the AC voltage to be rectified is, connected to the primary coil of a step-down transformer, and secondary coil is connected to the diode through, resistor R L across, which output is obtained., Transformer, , Primary, , A, , X, , Secondary, , RL, , Reverse Biased Characteristics, The circuit diagram and graph plotted between voltage and, current for reverse biased diode are shown in Figs. (a) and (b)., , B, , Circuit diagram of half-wave rectifier, , Y

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135, , CBSE Term II Physics XII, , Role of Filters, , During positive half cycle of the input AC, the p-n junction, is forward biased. Thus, the resistance in p-n junction, becomes low and current flows. Hence, we get output in the, load. During negative half cycle of the input AC, the p-n, junction is reverse biased. Hence, no output is in the load., , In order to get the steady DC output from the pulsating, voltage normally, a capacitor is connected across the output, terminals (parallel to load R L ). They are called filters., , Input AC, , +, C, –, , +, –, , Voltage across R L, , AC, , t, , (a), , Output AC, , +, , +, t, , Input and output waveforms, , Diode as a Full Wave Rectifier, In the full wave rectifier, two p-n junction diodes D1 and D 2 are, used. This arrangement is as shown in the diagram below, Centre tap, transformer, Centre, tap, , t, , (b), , A full wave rectifier with capacitor filter Fig. (a) and, input and output voltage of rectifier in Fig. (b)., , Special Purpose p-n, Junction Diodes, , D1, , A, , RL DC, , Y, –, , Output with, capacitor, input filter AC input, , +, , DC component, , X, , Rectifier, , Voltage at A, , Working, , Photodiode, , X, , B, RL Output, , D2, Y, , It is a special type of junction diode used for detecting, optical signals. It is a reverse biased p-n junction made from, a photosensitive material. Its symbol is, , Circuit diagram of full wave rectifier, +, , Working, In full wave rectifier, we get output in the load resistance, In positive half cycle by D1 ., In negative half cycle by D 2 ., l, , Input, waveform at A, , l, , p, , n, , Construction, A photodiode fabricated with a transparent cover to allow, light to fall on the diode and operates under reverse bias., (hn > Eg ), , O, , t, , mA, , Input, waveform at B, , p-side n-side, O, , –, , Output waveform, (across RL), , t, , Input and output waveforms, , R, , V, A reverse biased photodiode illuminated with light, , Due to Due to Due to Due to, D1, D2, D1, D2, O, , +, , Working, t, , When the photodiode is illuminated with light (photons),, with energy greater than the energy gap of the, semiconductor, then electron-hole pairs are generated due to, the absorption of photons. These charge carriers contribute, to the reverse current.

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136, , CBSE Term II Physics XII, , V-I Characteristics, We observe from the figure that, current in photodiode changes, with the change in light intensity I, when reverse bias is applied., , I-V Characteristics, , mA, , Reverse, current, , I, , I-V characteristics is drawn, in the fourth quadrant of the, coordinate axes because a, solar cell does not draw, current but supplies the, same to the load., , Reverse bias, I1, I2, I3, I4, I 4 > I3 > I2 > I 1, , metal contact acts as negative electrode. When an external, load is connected across metal electrodes, a photocurrent, flows., , volt, , mA, , V-I characteristics of photodiode at different intensities, , Solar Cell, It is a p-n junction diode, which converts solar energy into, electrical energy., hn, , Open circuit, voltage (Voc), V, , O, ISC, , Short circuit current, , I-V characteristics of a solar cell, , Light Emitting Diode (LED), It is a heavily doped p-n junction diode which, converts electrical energy into light energy., This diode emits spontaneous radiation,, under forward biasing. Its symbol is, , hn, +, , –, p, , n, , Working, Metallised, finger electrode, , Top, surface, , On recombination of electron and hole, the energy is given, out in the form of heat and light., , n, p, , p, , n, p, , n, p, , p, n, , R, B, , Back contact, , Forward biased LED, , p-n junction of solar cell, , Its symbol is, , V-I Characteristics, , +, , hn, , The colour of light emitted by a given LED, depends on its, band gap energy. The photon emitted by an LED is of, energy equal to or slightly less than the band gap energy., Forward current conducted by the junction determines the, intensity of light emitted by LED., , p, n, , –, , I (mA), 30, , Construction, It consists of a silicon or gallium-arsenide p-n junction diode, packed in a can with glass window on the top., , Silicon, , 20, 15, 10, , RL, IL, , –10V, , hn, , 0.5 0.8, 0, I (mA), p, , V (volt), , n, , V-I characteristics of LED, Depletion region, , A typical illuminated p-n junction solar cell, , Working, When photons of light (of energy hn > Eg ) falls at the junction,, electron-hole pairs are generated near the junction and they, move in opposite directions due to junction field. They will be, collected at the two sides of the junction, giving rise to a, photovoltage between the top and bottom metal electrodes., The top metal contact acts as positive electrode and bottom, , LEDs Advantages over Incandescent, Low Power Lamps, LED’s advantages over incandescent low power lamps, which are as given below, Fast action and no warm up time required., The bandwidth of emitted light is from 100 Å to, 500 Å, so it is nearly (not exactly) monochromatic., Long life and ruggedness., Low operational voltage and less power consumed., l, , l, , l, , l

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137, , CBSE Term II Physics XII, , Solved Examples, Example 1. The maximum wavelength at which solid, begin to absorb energy is 10000 Å. Calculate the, energy gap of a solid (in eV)., hc, l, where, h = Planck’s constant, c = speed of light, and, l = wavelength at which solid absorbs energy., On putting the values of h, c and l, we get, (6.626 ´ 10-34 J- s)(3 ´ 108 m / s), Eg =, (10000 ´ 10-10 m), , Sol. The energy band gap is given by, Eg = hn =, , = 1.98 ´ 10, , -19, , n final, n final, , 21, , Factor =, », , (Q n e >> n h), 15, , n final - n initial 25, . ´ 10 - 14 ´ 10, =, n initial, 14 ´ 1015, 25, . ´ 1021, = 1.8 ´ 105, 14 ´ 1015, , Example 4. (a) Calculate the value of V 0 and i, if the, Ge, Si, , Example 2. In an intrinsic (pure) semiconductor, the, number of conduction electrons is 7 ´ 1019 per, cubic metre. Find the total number of current, carriers (electrons and holes) in the same, semiconductor of size 1 cm ´ 1 cm ´ 1mm., , Sol. In an intrinsic semiconductor, n e = n h, where, n e = number of conduction electrons, and, n h = number of holes per unit volume., Given, n e = 7 ´ 1019 per m 3, n h = n e = 7 ´ 1019 per m 3, , So, total current carrier density, n e + n h = 7 ´ 1019 + 7 ´ 1019, = 14 ´ 1019 per m 3, Now, total number of current carrier, = number density ´ volume, = ( 14 ´ 1019 per m 3 ) ´ ( 10-2 m ´ 10-2 m ´ 10-3 m, = 1.4 ´ 1013, , Example 3. The concentration of hole-electron pairs in, pure germanium at T = 300 K is 7 ´ 1015 per cubic, metre. Antimony is doped into germanium a, proportion of 1 atom 10 7 Ge atoms. Assuming that, half of the impurity atoms contribute electron in, the conduction band, calculate the factor by which, the number of charge carriers increases due to, doping the number of germanium atoms per cubic, metre is 5 ´ 10 28 ., , Sol. In pure semiconductor electron-hole pair = 7 ´ 1015 m -3, Total charge carrier, n total initial = n h + n e = 14 ´ 1015, After doping donor impurity, N D =, , So,, Þ, , silicon and germanium diode start conducting at, 0.7 V and 0.3 V, respectively., , J, , 1.98 ´ 10-19, =, eV, 1.6 ´ 10-19, = 1.24 eV, , \, , ND, = 25, . ´ 1021, 2, = nh + ne, » n e » 25, . ´ 1021, , ne =, , and, , 5 ´ 1028, = 5 ´ 1021, 107, , i, , V0, , RL 5 kW, , 12 V, , (b) If the Ge diode connection is now reversed, what, will be the new values of V 0 and i?, Sol. (a) Ge diode will start conducting before the silicon diode, does so. The effective forward voltage across Ge diode is, (12 - 0.3) V =11.7 V. This will appear as the output, voltage across the load, i.e.,, Vo = 11.7 V, The current through R L,, 11.7, i=, A = 2.34 mA, 5 ´ 103, (b) On reversing the connection of Ge diode, it will be, reverse biased and conduct no current. Only Si diode will, conduct. Therefore,, Vo = ( 12 - 0.7) V = 11.3 V, 11.3, and current, i =, A = 2.26 mA, 5 ´ 103, , Example 5. A p-n photodiode is made of a material, with a band gap of 1.5 eV. What is the minimum, wavelength of radiation that can be absorbed by the, material?, hc, l, The minimum wavelength of radiation, hc (6.4 ´ 10-34 J - s) ´ (3 ´ 108 ms -1 ), l= =, E, 1.5 ´ 1.6 ´ 10-19 J, , Sol. Energy, E = hn =, , = 8.3 ´ 10-7 m = 830 nm, 1.5 ´ 1.6 ´ 10-19 J, =, » 1.2 ´ 106 Hz, 6.6 ´ 10-34 J - s ´ 3 ´ 108 ms -2

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138, , CBSE Term II Physics XII, , Chapter, Practice, PART 1, Objective Questions, , Ec, , (c), , Ev, l, , Ec, Eg, , (d), , Eg, , Ev, , Multiple Choice Questions, Electrons, , 1. The conductivity of a semiconductor increases with, increase in temperature, because [NCERT Exemplar], (a) number density of free current carriers increases, (b) relaxation time increases, (c) Both number density of carriers and relaxation time, increase, (d) number density of carriers increases, relaxation time, decreases but effect of decrease in relaxation time is, much less than increase in number density, , 2. Correct order of relative values of electrical, conductivity s for different types of solid is, (a) s semiconductor > s insulator > s metal, (b) s metal > s semiconductor > s insulator, (c) s semiconductor > s metal > s insulator, (d) s insulator > s semiconductor > s metal, , hole in semiconductor?, (a) An anti-particle of electron, (b) A vacancy created when an electron leaves a covalent, bond, (c) Absence of free electrons, (d) An artificially created particle, , 4. Energy band gap Eg diagram for an intrinsic, , semiconductor at temperature T > 0 K is, (Here, E C is energy for conduction band and E V is, energy for valence band.), Electrons, , Holes, Ec, Eg, , Electrons, Ev, , Ec, , (b), , semiconductor to make p-type semiconductor is, (a) phosphorus, (c) aluminium, , (b) antimony, (d) arsenic, , 6. In n-type semiconductor, electrons are majority, charge carriers but it does not show any negative, charge. The reason is, (a) electrons are stationary, (b) electrons neutralise with holes, (c) mobility of electrons is extremely small, (d) atom is electrically neutral, , 7. Pure silicon at 300 K has equal electron ( n e ) and, , 3. Which of the following correctly represents the, , (a), , 5. The substance which is doped in an intrinsic, , Eg, , Electrons, , Ev, , hole ( n h ) concentration of 1.5 ´ 1016 m - 3 . Doping, by indium increases n h to 4.5 ´ 10 22 m - 3 . The n e, in doped silicon is (in m -3 ), (a) 9 ´ 105, (c) 2.25 ´ 1011, , (b) 5 ´ 109, (d) 3 ´ 1019, , 8. When an electric field is applied across a, semiconductor,, (a) electrons move from lower energy level to higher energy, level in the conduction band, (b) electrons move from higher energy level to lower energy, level in the conduction band, (c) holes in the valence band move from lower energy level, to higher energy level, (d) None of the above, , 9. The barrier potential of a p-n junction depends on, [CBSE 2014], , (i) type of semiconductor material, (ii) amount of doping, (iii) temperature

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139, , CBSE Term II Physics XII, , (c) would be like a half-wave rectifier with negative cycles in, output, (d) would be like that of a full wave rectifier, , Which one of the following is correct?, (a) Both (i) and (ii), (c) Both (ii) and (iii), , (b) Only (ii), (d) (i), (ii) and (iii), , 14. The wavelength and intensity of light emitted by a, , 10. In figure given below, V 0 is the potential barrier, , LED depend upon, , across a p-n junction, when no battery is connected, across the junction, then, [NCERT Exemplar], , (a) forward bias and energy gap of the semiconductor, (b) energy gap of the semiconductor and reverse bias, (c) energy gap only, (d) forward bias only, , 1, 2, 3, , V0, , 15. The I-V characteristic of an LED is, R YG B, , (a) 1 and 3 both correspond to forward bias of junction, (b) 3 corresponds to forward bias of junction and 1, corresponds to reverse bias of junction, (c) 1 corresponds to forward bias and 3 corresponds to, reverse bias of junction, (d) 3 and 1 both correspond to reverse bias of junction, , (a), , O, , (c), , 11. A 220 V AC supply is connected between points A, and B (figure). What will be the potential, difference across the capacitor? [NCERT Exemplar], , (b), , I, , R V, Y, G, B, , O, , V, , V, , O, , O, , V, , (d), , I, , R, G, Y, R, , R, Y, G, B, I, , Red, Yellow, Green, Blue, , A, l, , 200 AC, , C, , V, , B, , (a) 220 V, (c) 0 V, , (b) 110 V, (d) 220 2 V, , 12. V-I characteristics of a silicon diode is as shown., I (mA), 30, Silicon, 20, 15, 10, –10 V, 0, 1 mA, , (a) 10, , (b) 10, , -4, , (c) 10, , -5, , (d) 10, , band and conduction band is greater in silicon than, in germanium., , -6, , 13. The output of the given circuit in figure given, below, , Direction (Q. Nos. 16-20) Each of these questions, contains two statements Assertion (A) and Reason (R)., Each of these questions also has four alternative, choices, any one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given, below., (a) Both A and R are true and R is the correct, explanation of A, (b) Both A and R are true, but R is not the correct, explanation of A, (c) A is true, but R is false, (d) A is false and R is also false, , 16. Assertion The energy gap between the valence, , 0.8, 0.5 0.7, V (volts), , The ratio of resistance of diode at I D = 15 mA and, V D = - 10 V, is, -3, , Assertion-Reasoning MCQs, , [NCERT Exemplar], , Vm sin wt, , Reason Thermal energy produces fewer minority, carriers in silicon than in germanium., , 17. Assertion The total current I in a semiconductor is, the sum of electron current and hole current., Reason In a semiconductor, I h arises due to the, motion of holes towards positive potential and free, electrons under an applied electric field., , 18. Assertion The resistivity of a semiconductor, decreases with temperature., , Reason The atoms of a semiconductor vibrate, (a) would be zero at all times, (b) would be like a half-wave rectifier with positive cycles in, output, , with larger amplitudes at higher temperature, thereby increasing its resistivity.

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140, , CBSE Term II Physics XII, , (ii) In figure, the input is across the terminals A and C, and the output is across B and D. Then, the output, is, , 19. Assertion A hole on p-side of a p - n junction moves, to n-side just an instant after drifting of charge, carriers occurs across junction., , B, , C, , Reason Drifting of charge carriers reduces the, concentration gradient across junction., , 20. Assertion Light Emitting Diode (LED) emits, , (a) zero, (c) half wave rectified, , Reason LED are forward biased p-n junctions., l, , (a) It can convert DC to AC. (b) It can convert AC to DC., (c) It can shift voltage level. (d) None of these, , 21. Direction Read the following passage and answer, the questions that follows, Full Wave Rectifier, The process of converting alternating, voltage/current into direct voltage/current is called, rectification. Diode is used as a rectifier for, converting alternating current/voltage into direct, current/voltage., Diode allows current to pass only, when it is, forward biased. So, if an alternating voltage is, applied across a diode, the current flows only in, that part of the cycle when the diode is forward, biased. This property is used to rectify the, current/voltage., , (iv) In the given circuit,, , AC, , X, , B, RL Output, , D2, Y, , (i) If in a p-n junction, a square input signal of 10 V is, applied as shown, , (b) 2 : 1, (d) 1 : 4, , PART 2, Subjective Questions, Short Answer (SA) Type Questions, 1. Carbon and silicon both have four valence, electrons each, then how are they distinguished?, [Delhi 2011C], , 2. Draw energy band diagram of n-type and p-type, , +5 V, RL, –5 V, , Then, the output across R L will be, (b), , 10 V, , (d), , semiconductors at temperature T > 0K. Mark the, donor and acceptor energy level with their, energies., [Foreign 2014], , 3. Distinguish between intrinsic and extrinsic, semiconductors., , [All India 2015], , 4. The number of conduction electrons (in per cubic, , –10 V, , (c) –5 V, , RL DC, , (v) The ratio of output frequencies of half-wave, rectifier and a full wave rectifier, when an input of, frequency 200 Hz is fed at input, is, , l, , Circuit diagram of full wave rectifier, , C, , (a) for storing potential energy, (b) as a bypass to DC component to get AC in R L, (c) to remove sparking, (d) as a bypass to AC component to get DC in R L, , (a) 1 : 2, (c) 4 : 1, , Centre, tap, , (a), , Rectifier, , Capacitor C is used, , D1, , A, , (b) same as the input, (d) full wave rectified, , (iii) Which of the following is not true about a rectifier, circuit?, , Case Based MCQs, , Centre tap, transformer, , D, , A, , spontaneous radiation., , 5V, , metre) present in a pure semiconductor is, 6.5 ´ 1019 . Calculate the number of holes in a, sample having dimensions 1 cm ´ 1 cm ´ 2 mm ?

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141, , CBSE Term II Physics XII, , 5. Why are elemental dopants for silicon or, , 12. In half-wave rectification, what is the output, , germanium usually chosen from group XIII or, group XV?, [NCERT Exemplar], , frequency. If the input frequency is 50 Hz, what is, the output frequency of a full wave rectifier for the, same input frequency?, [NCERT], , 6. As we know that, a p-type semiconductor has large, , 13. Explain briefly how a photodiode operates., , number of holes but it is still electrically neutral., Why?, , [CBSE 2018C], , 14. Three photodiodes D1 , D 2 and D 3 are made of, , 7. The impurity levels of doped semiconductor are, , semiconductors having band gaps of 2.5 eV, 2 eV, and 3 eV, respectively. Which one will be able to, detect light of wavelength 6000 Å?, , 30 eV below the conduction band., Determine whether the semiconductor is n-type or, p-type., At the room temperature, thermal collisions occur, as a result of which, the extra electron loosely, bound to the impurity ion gets an amount of energy, kT and hence this electron can jump into, conduction band. What is the value of T? (Take, k is, Boltzmann constant = 8.62 ´ 10 -5 eV/K), , [NCERT Exemplar], , 15. Mention the important considerations required, while fabricating a p-n junction diode to be used as, a Light Emitting Diode (LED). What should be the, order of band gap of an LED, if it is required to, emit light in the visible range?, [Delhi 2013], , 16. (i) Explain with the help of a diagram the formation, , 8. Write differences between forward bias and reverse, bias., , of depletion region and barrier potential in a p - n, junction., (ii) Draw the circuit diagram of a half-wave rectifier, and explain its working., [All India 2016], , [All India 2020], , 9. There are two semiconductor materials, A and B which are made by doping germanium, crystal with indium and arsenic, respectively. As, shown in the figure, the junction of two is biased, with a battery. Will the junction be forward bias or, reverse bias?, A, , 17. Draw the circuit diagram of a full wave rectifier, and explain its working. Also, give the input and, output waveforms., [Delhi 2019], , B, , 18. (i) In the following diagram, which bulb out of B1, and B 2 will glow and why?, D1, , 10. The V-I characteristic of a silicon diode is as shown, in the figure. Calculate the resistance of the diode, at (i) I = 5 mA and (ii) V = -20V., [Delhi 2020], , D2, +9 V, , B1, , I (mA), , B2, , 30, , (ii) Draw a diagram of an illuminated, p-n junction solar cell., (iii) Explain briefly the three processes due to which, generation of emf takes place in a solar cell., , Silicon, , 20, 15, 10, –20 V, , 0, I (mA), , 0.5, 0.7 0.8 V, , [All India 2020, 17], , 19. What is photodiode ? Breifly explain its working, , 11. Assuming that, the two diodes D1 and D 2 are used, in the electric circuit shown in the figure are ideal., Find out the value of the current flowing through, 1W resistor., D1, , 2W, , D2, , 2W, , + –, , 1W, , 6V, , and draw its V-I characteristics., l, , [Delhi 2020], , Long Answer (LA) Type Questions, 20. The number of silicon atoms per m 3 is, 5 ´ 10 28 . This is doped simultaneously with, 5 ´ 10 22 atoms per m 3 of arsenic and 5 ´ 10 20 atoms, per m 3 of indium. Calculate the number of, electrons and holes. Given that, n i = 1.5 ´ 1016 m - 3 ., Is the material n-type or p-type?, [NCERT]

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142, , CBSE Term II Physics XII, , (ii) Which semiconductors are preferred to make, LEDs and why?, (iii) Give two advantages of using LEDs over, conventional incandescent low power lamps., , 21. Predict the effect on the electrical properties of a, silicon crystal at room temperature, if every, millionth silicon atom is replaced by an atom of, indium. Given, concentration of silicon atoms, = 5 ´ 10 28 m -3 , intrinsic carrier concentration, = 1.5 ´ 1016 m -3 , m e = 0.135 m 3 / V - s and, m h = 0.048m 3 / V-s., , [All India 2011], , 27. (i) What is the advantage of using GaAs for, synthesis of solar cells?, (ii) Draw V-I characteristics of solar cell and, mention its significance., , 22. A potential barrier of 0.4V exists across p-n junction., (i) If the depletion region is 4.0 ´ 10 -7 m wide, what, is the intensity of the electric field in this region?, (ii) If an electron with speed 4 ´ 105 m/s approaches, the p-n junction from the n-side, find the speed, with which it will be p- side., , 23. Assuming an ideal diode, draw the output, waveform for the circuit given in the figure, explain, the waveform., [NCERT Exemplar], R, , l, , Case Based Questions, Direction Read the following passage and answer the, questions that follows, , 28. Photodiode, Photodiode is a special type of junction diode used, for detecting optical signals. It is a reverse biased, p-n junction made from a photosensitive material., Its symbol is, , 5V, , 24. Draw V-I characteristics of a p-n junction diode., Answer the following questions giving reasons., (i) Why is the current under reverse bias almost, independent of the applied potential upto a, critical voltage?, (ii) Why does the reverse current show a sudden, increase at the critical voltage?, , 25. If each diode in figure has a forward bias resistance, of 25 W and infinite resistance in reverse bias, what, will be the values of the currents I1 , I 2 , I 3 and I 4 ?, A, C, E, I1, , I4, , 125 W, , I3, , 125 W, , I2, , 125 W, , B, , D, F, 25 W, H, , G, 5V, , [NCERT Exemplar], , 26. (i) Describe the working of Light Emitting Diodes, (LEDs)., , p, , –, , +, , 20 sin wt, , n, , A photodiode fabricated with a transparent cover to, allows light to fall on the diode and operated under, reverse bias. A photodiode is used in sensor, circuits., (hn > Eg ), , mA, p-side n-side, , –, , +, , R, , V, A reverse biased photodiode, illuminated with light, , (i) The current in the forward bias is known to be, more (~mA ) than the current in the reverse bias, (~mA ). What is the reason to operate the, photodiode in reverse bias?, [Delhi 2012], (ii) Three photodiodes D1 , D 2 and D 3 are made of, semiconductors having band gaps of 15, . eV ,, 2.5 eV and 35, . eV, respectively. Which of them, will not be able to detect light of wavelength, 450 nm?, (iii) Write main use of photodiode.

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Chapter Test, Short Answer Type Questions, , Multiple Choice Questions, , 1. In an unbiased p-n junction, holes diffuse from the, , 6. Draw the circuit diagram showing how a p - n, , p-region to n-region because, , junction diode is, , (a) free electrons in the n - region attract them, (b) they moves across the junction by the potential difference, (c) hole concentration in p-region is more as compared to hole, concentration in n-region, (d) All of the above, , (i) forward biased and (ii) reverse biased., , 2. Which of these graphs shows potential difference between, p- side and n-side of a p-n junction in equilibrium?, , How is the width of depletion layer affected in the, two cases?, , 7. Assuming that the resistances of the meters are, negligible, what will be the readings of the, ammeters A1 and A2 in the circuit shown in figure?, 20 W, A1, , (a), , p-side, , n-side, , (b), , p-side, n-side, , Junction, plane, , Junction, plane, , 4V, A2, , 20 W, , 8. How do you obtain steady DC output from the, (c), , p-side, , (d), , pulsating voltage?, , p-side, , n-side, , n-side, , Junction, plane, , Junction, plane, , 3. Which is reverse biased diode?, , 5V, , (d) 20 V, –5 V, , 4. The diode shown in the circuit is a silicon diode. The, potential difference between the points A and B will be, 2W, , S, , A, , B, , (c) 0.7 V, , (d) 0 V, , 5. The current through an ideal p-n junction shown in the, following circuit diagram will be, , 1V, , forward biased p-n junction diode which emits, spontaneous radiation. State the least band gap, energy of this diode to have emission in visible, region., , 13. Write two characteristics features to distinguish, between n-type and p- type semiconductors., , 14. A photodiode is operated under reverse bias, , 100 W, , n, , crystalline solids., (ii) Draw the energy band diagrams of (a) a metal, and (b) a semiconductor., , 12. Explain with help of circuit diagram, the action of a, , 6V, , (b) 0.6 V, , p, , (ii) In a p-n junction diode, the forward bias, resistance is low as compared to the reverse bias, resistance. Give reason., , 11. (i) Explain the formation of energy bands in, , 10 V, , (a) 6 V, , Long Answer Type Questions, joined to another n-type semiconductor slab to, form p-n junction? Justify your answer., , –10 V, , (c) 15 V, , incandescent lamps., , 10. (i) Can a slab of p-type semiconductor be physically, , (b) – 20 V, , (a), , 9. Give two advantages of LED’s over the conventional, , although in the forward bias, the current is known to, be more than the current in the reverse bias. Explain,, giving reason., , 2V, , 15. Describe the working principle of a solar cell., (a) zero, , (b) 1 mA, , (c) 10 mA, , (d) 30 mA, , Mention three basic processes involved in the, generation of emf., , Answers, Multiple Choice Questions, 1. (c), , 2. (c), , 3. (b), , 4. (a), , 5. (a), , For Detailed Solutions, Scan the code

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144, , CBSE Term II Physics XII, , EXPLANATIONS, PART 1, 1. (d) The conductivity of a semiconductor increases with, increase in temperature because the number density of, current carriers increases, relaxation time decreases but, effect of decrease in relaxation time is much less than, increase in number density., 2. (b) The values of conductivity and resistivity for different, types of solids are as follows, (i) Metal r ~10-2 -10-8 W-m, s ~102 -108 Sm -1, (ii) Semiconductor, r ~10-5 -106 W-m, s ~105 - 10-6 Sm -1, (iii) Insulator, r ~1011 - 1019 W-m, s ~10-11 - 10-19 S m -1, As 108 > 10-6 > 10-19 , so s metal > s semiconductor > sinsulator ., 3. (b) The concept of hole describes the lack of an electron at a, position, where one could exist in an atom or atomic lattice., If an electron is excited into a higher state, it leaves a hole in, its old state., Thus, hole can be defined as a vacancy created when an, electron leaves a covalent bond., 4. (d) In an intrinsic semiconductor, at T > 0K, due to thermal, energy some electrons from the valence band excites to, conduction band and partially occupy it. Some electrons are, in the conduction band, these have come from valence band, leaving equal number of holes there., 5. (c) In an intrinsic semiconductor, when an impurity of, trivalent group such as aluminium, boron, etc., mixed in, very small quantity, then the resultant crystal will be p-type, semiconductor., 6. (d) The n-type semiconductor, region has (negative), electrons as majority charge carriers and an equal number of, fixed positively charged donor ions. Again, the material as a, whole is neutral. That is a reason, atom is electrically, neutral., 7. (b) In an extrinsic semiconductor,, n e n h = n i2, Þ n e ´ 4.5 ´ 1022 = (1.5 ´ 1016 ) 2, ne =, Þ, , 2.25 ´ 1032, 4.5 ´ 1022, , n e = 5 ´ 109 m -3, , 8. (a) When an electric field is applied across a semiconductor,, the electrons in the conduction band get accelerated and, acquire energy. They move from lower energy level to higher, energy level. While the holes in valence band move from, higher energy level to lower energy level, where they will, be having more energy., 9. (d) Barrier potential depends on the material used to make, p-n junction diode (whether it is Si or Ge). It is also depends, on amount of doping due to which the number of majority, , carriers will change. It also depends on temperature due to, which the number of minority carriers will change., 10. (b) When p-n junction is forward biased, it opposes the, potential barrier junction, when p-n junction is reverse, biased, it supports the potential barrier junction, resulting in, increase in potential barrier across the junction., So, 3 corresponds to forward bias of junction and 1, corresponds to reverse bias of junction., 11. (d) As p-n junction conducts during positive half cycle only,, so the diode connected here will work in positive half cycle., Potential difference across C = peak voltage of the given AC, voltage, V0 = Vrms 2 = 220 2 V, 12. (d) From the graph, given in question at I = 20 mA, V is, 0.8 V and at I = 10 mA, V = 0.7 V, 0.1 V, DV 0.8 - 0.7, \, rf b =, =, =, DI, 20 - 10 10 mA, 0.1 V, =, = 10 W, 10 ´ 10-3 A, Also, at V = - 10 V, I = - 1 mA, 10 V, 10 V, rrb =, =, = 1 ´ 107 W, 1 mA 1 ´ 10-6 A, r, 10, \ Ratio = f b = 7 = 10-6, rrb 10, 13. (c) Due to forward biased during positive half cycle of input, AC voltage, the resistance of p-n junction is low. The current, in the circuit is maximum. In this situation, a maximum, potential difference will appear across resistance connected, in series of circuit. This result into zero output voltage, across p-n junction., Due to reverse biased during negative half cycle of AC, voltage, the p-n junction is reverse biased. The resistance of, p-n junction becomes high which will be more than, resistance in series. That is why, there will be voltage across, p-n junction with negative cycle in output., 14. (a) LED is a heavily doped p-n junction diode which emits, spontaneous radiation, under forward biasing. The colour, (wavelength and intensity) of the emitted radiation depend, on its band gap energy., 15. (a) An LED operates in forward bias mode, so its I-V, characteristics is similar to a forward biased p-n junction, diode. Secondly, the threshold voltage increases with, decrease in wavelength. Thus, option (a) is correct., 16. (b) The energy gap between valence band and conduction, band in germanium is 0.76 eV and the energy gap between, valence band and conduction band in silicon is 1.1 eV., Also, it is true that thermal energy produces fewer minority, carriers in silicon than in germanium., Therefore, both A and R are true but R is not the correct, explanation of A., 17. (c) In a semiconductor at T > 0K, electrons originally set, free are not involved in the process of hole motion. These, free electrons moves completely as conduction electron and

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145, , CBSE Term II Physics XII, , give rise to an electron current Ie , under an applied electric, field. However, motion of hole is only a convenient way of, describing the actual motion of bound electrons wherever, there is an empty bond anywhere in the crystal. Thus, under, the action of an electric field these holes move towards, negative potential giving the hole current Ih., The total current I is thus the sum of the electron current Ie, and hole current Ih,, I = Ie + Ih, Therefore, A is true but R is false., 18. (c) Resistivity of a semiconductor decreases with rise in the, temperature. The atoms of a semiconductor vibrate with, larger amplitudes at higher temperature, thereby increasing, its conductivity not resistivity., Therefore, A is true but R is false., 19. (d) In a p-n junction, due to diffusion of electrons, a positive, space-charge region on n-side of the junction and a negative, space charge region on p-side of the junction is formed. Due, to this, an electric field directed from positive charge, towards negative charge develops (electric field is from, n-side to p-side). Due to this field, an electron on p-side of, the junction moves to n-side and a hole on n-side of the, junction moves to p-side. This motion of charge carriers due, to the electric field is called drift. Thus, a drift current,, which is opposite in direction to the diffusion current starts., However, concentration gradient is due to the doping of, sides. It is not affected by drift of charge carriers., Therefore, A is false and R is also false., 20. (a) When a junction diode is forward biased as shown in, figure, energy is released at the junction due to, recombination of electrons and holes. In the junction diode, made of gallium arsenide or indium phosphide, the energy, is released in visible region., Light, p, RL, n, –, , +, , Such a junction diode is called light emitting diode or LED., It is a heavily doped p-n junction diode, so the radiations, emitted are spontaneous in forward biasing., Therefore, both A and R are true and R is the correct, explanation of A., 21. (i) (d) As it is forward biased, so it takes positive value., Hence, option (d) is correct., (ii) (d), , AC input is applied across A and C and output is taken, across BD., V0, , D1D4, , D2D3, , D1D4, , When positive cycle is fed to AC, D1 and D4 conduct and, when negative cycle is fed to AC, D3 and D2 conduct in, the same direction. Output across BD is thus full wave, rectified., (iii) (a) A rectifier can convert AC to DC. It can also shift or add, voltage level in the output but it cannot convert DC to AC., (iv) (d) In the given circuit, for the first half-cycle of rectified, output when the voltage across C will rise, it gets charged., Then, in the presence of R L, it gets discharged through it, and the voltage begins to fall., In the next half-cycle of rectified output, it again gets, charged to peak value and will similarly gets discharged, through R L. Thus, a steady DC output from pulsating, voltage is obtained. In other words, capacitor helps to filter, out the AC ripple and give a pure DC voltage or bypass AC, component to get DC., (v) (a) Output frequency of full wave rectifier is twice the, output frequency of half-wave rectifier., f half -wave 1, \, =, f full wave, 2, , PART 2, 1. The four valence electrons of carbon are present in second, orbit while that of silicon in third orbit. So, energy required, to extricate an electron from silicon is much smaller than, carbon., Therefore, the number of free electrons for conduction in, silicon is significant on contrary to the carbon. This makes, silicon’s conductivity much higher than carbon. This is the, main distinguishable property., 2. The required energy band diagram is as shown below, Conduction band, Acceptor energy level, , B, , D1, , D2D3, , 10.04 eV, Valence band, , D2, , (a) p-type, A, , C, , RL, , Conduction band, D3, , 10.045 eV, , D4, D, , Donor energy level, Valence band, (b) n-type

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146, , CBSE Term II Physics XII, , 3. Difference between intrinsic and extrinsic semiconductors, is as given below, Intrinsic semiconductor, , Extrinsic semiconductor, , It is a pure, It is prepared by doping a small, semiconductor material, quantity of impurity atoms to the, with no impurity atoms in pure semiconductor., it., The number of free, electrons in the, conduction band and the, number of holes in, valence band is exactly, equal., ne = nh = ni, , The number of free electrons, and holes is never equal. There, is an excess of electrons n e > n i, in n-type semiconductors and, excess of holes in p-type, semiconductors., nh > ni, , 4. Given, number of conduction electrons, n e = 6.5 ´ 1019 m -3, Volume of the sample = 1 cm ´ 1 cm ´ 2 mm = 2 ´ 10-7 m 3, Number of holes in the sample, n h = Number of electrons in, the sample, n e, = n e ´ V = 65, . ´ 1019 ´ 2 ´ 10-7, = 1 .3 ´ 1013, 5. The size of the dopant atom should be such that their, presence in the pure semiconductor does not distort the, semiconductor but easily contribute the charge carriers on, forming covalent bonds with Si or Ge atoms, which are, provided by group XIII or group XV elements., 6. p-type semiconductor is formed by doping it with trivalent, impurities. These impurities or dopant takes the atoms in, the crystal and its three electrons take part in chemical, bonding with three electrons of intrinsic semiconductor or, pure semiconductor, whereas the last bond are left free., Since, as whole atom is electrically neutral, so p -type, semiconductor is also neutral., 7. The separation of impurity energy level from conduction, band is less in case of n-type semiconductor and more in, case of p-type semiconductor. As, energy separation of, impurity is 30 ´ 10-3 eV is much smaller than energy gap of, pure semiconductor, i.e. E » 1 eV. Therefore, the doped, semiconductor is n-type., Eg = 30 ´ 10-3 eV = kT, Þ, , T=, , Eg, k, , =, , 9. As, semiconductor A is doped with indium, so it behaves as, p-type semiconductor and B is doped with arsenic, so it, behaves as n-type semiconductor. Thus, the figure shows, that, it is forward bias condition., 10. (i) From the given curve, we have, Voltage, V = 0.5 V for current of 10 mA, 0 .5, So, for 5 mA, V =, ´ 5 = 0.25 V, 10, V, 0. 25, \ Resistance, R = =, = 50W, I 5 ´ 10-3, (ii) For V = -20 V, we have, I = -1 mA = -1 ´ 10-6 A, 20, Þ R=, = 2.0 ´ 107 W, 1 ´ 10-6, 11. According to the question, D2 is reverse biased, so do not, conduct., D1, , 2W, , A, , B, D2, , D, , E, , 2W, C, , + –, , 1W, F, , 6V, , R EF = 2 + 1 = 3 W, V, 6, IEF =, = = 2A, R EF 3, 12. A half-wave rectifier rectifies only the half of AC input, i.e., it conducts once during an AC input cycle while a full-wave, rectifier rectifies both the half cycles of the AC input, i.e. it, conducts twice during a cycle., \ The output frequency for half-wave rectifier is 50 Hz, and the output frequency of a full wave rectifier is, 2 ´ 50 = 100 Hz., 13. A junction diode made from light sensitive semiconductor is, called a photodiode., A photodiode is a p-n junction diode arranged in reverse, biasing., hn, , -3, , 30 ´ 10, = 348.02 K, 8. 62 ´ 10-5, , 8. Differences between forward and reverse biasing are as, given below, Forward bias, , Reverse bias, , Positive terminal of battery is, connected to p -type and, negative terminal to n-type, semiconductor., , Positive terminal of battery, is connected to n-type and, negative terminal to p-type, semiconductor., , Depletion layer is very thin., , Depletion layer is thick., , p-n junction offers very low, resistance., , p-n junction offers very, high resistance., , An ideal diode have zero, resistance., , An ideal diode have infinite, resistance., , mA, p-side, , n-side, R, , The number of charge carriers increases when light of, suitable frequency is made to fall on the p-n junction,, because new electron-hole pairs are created by absorbing, the photons of suitable frequency. Intensity of light controls, the number of charge carriers. Due to this property,, photodiodes are used to detect optical signals., 14. Given, wavelength of light,, l = 6000 Å = 6000 ´ 10-10 m, \Energy of the light photon,

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147, , CBSE Term II Physics XII, hc, 6.6 ´ 10-34 ´ 3 ´ 108, eV = 2.06 eV, =, l, 6000 ´ 10-10 ´ 1.6 ´ 10–19, The incident radiation which is detected by the photodiode, having energy should be greater than the band gap, so it is, only valid for diode D2 . Then, diode D2 will detect this, radiation., 15. For LEDs, the threshold voltages are much higher and, slightly different for different colours. The reverse, breakdown voltages of LEDs are low generally around 5V. It, is due to this reason, the care is taken that high reverse, voltages do not appear across LEDs. There is very little, resistance to limit the current in LED. Therefore, a, resistor must be used in series with the LED to avoid any, damage to it., The semiconductor is used for fabrication of visible LEDs, must at least have a band gap of 1.8eV (spectral range of, visible light is from about 0.4 mm to 0.7 mm i.e. from about 3, eV to 1.8 eV)., 16. (i) The small region in the vicinity of the junction which is, depleted of free charge carriers and has only immobile, ions is called depletion region., , VP, , E=, , Fictitious battery, Electron, VB, Donor ion, , ¬Electron diffusion, Electron drift®, n, , ¬Depletion region, Hole diffusion®, ¬Hole drift, V, VB, X2, , X1, , The accumulation of negative charges in the p-region and, positive charge in the n-region sets up a potential, difference across the junction. This acts as a barrier and is, called barrier potential VB., (ii) Transformer, P1, , A, RL, , ~, P2, , B, , t, , 0, , T/2, , t, , Centre tap transformer, D1, A, Centre, tap, B, , X, R2 Output, , D2, Y, , Circuit diagram of full wave rectifier, , Working During the positive half-cycle of the input AC, the, diode D1 is forward biased and the diode D2 is reverse, biased. The forward current flows through diode D1., During the negative half-cycle of the input AC, the diode D1, is reverse biased and diode D2 is forward biased. Thus,, current flows through diode D2 . Thus, we find that during, both the halves, current flows in the same direction., , O, , t, (a), , O, , t, , (b), Due to Due to Due to Due to, D1, D2, D1, D2, , O, , t, , 18. (i) D1 diode is forward biased, hence current will flow in B1, bulb and D2 is reverse biased, so there will be no current, in B2 . Hence, B1 will glow., (ii) The diagram of illuminated p-n junction solar cell is as, shown below, , Circuit diagram of half-wave rectifier, , Working, (a) During positive half-cycle of input alternating voltage,, the diode is forward biased and a current flows through, the load resistor R L and we get an output voltage., (b) During other negative half-cycle of the input, alternating voltage, the diode is reverse biased and it, does not conduct (under breakdown region)., , T, , Hence, AC voltage can be rectified in the pulsating and, unidirectional voltage., 17. A rectifier is used to convert alternating current into direct, current, whose labelled circuit is as given below, , Input, waveform at B, , p-type, n-type, Depletion layer, , p, , T, , Input, waveform at A, , Hole, , 2, , 0, , Output, waveform (across RL), , Junction, Acceptor ion, , VP, , 1, , R, hn, Depletion, layer, , p, , n

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148, , CBSE Term II Physics XII, , (iii) Processes involved due to generation of emf in a solar cell, are given below, (a) When light photon reach the junction, the excited, electrons from the valence band to conduction band, creating equal number of holes and electrons., (b) These electron-hole pairs move in opposite direction, due to junction field. Their movement in opposite, direction creates potential difference (photovoltage)., (c) When load is connected in the external circuit,, current starts flowing through it due to photovoltage., 19. Photodiode, It is a special type of junction diode used for detecting, optical signals. It is a reverse biased p-n junction made from, a photosensitive material. Its symbol is, Working, When the photodiode is illuminated with light (photons),, with energy greater than the energy gap of the, semiconductor, then electron-hole pairs are generated due, to the absorption of photons. These charge carriers, contribute to the reverse current., Circuit diagram of illuminated photodiode in reverse bias, is as shown below, , I (mA), , V, I1, I2, I3, I4, , I (mA), , 20. For each atom doped with arsenic, one free electron is, received. Similarly, for each atom doped of indium, a, vacancy is created. So, number of free electrons introduced, by pentavalent impurity,, N As = 5 ´ 1022 m -3, The number of holes introduced by trivalent impurity, added,, N I = 5 ´ 1020 m -3, So, net number of electrons added,, n e = NAs - N I, = 5 ´ 1022 - 5 ´ 1020, = 4.95 ´ 1022 m -3, We know that, n e n h = n i2, So, n h =, , = 3.33 ´ 106, New electron concentration,, n 2 (1.5 ´ 1016 ) 2, ne = i =, nh, 5 ´ 10 22, = 0.45 ´ 1010 / m 3, Electron concentration has been reduced, n, 1.5 ´ 1016, = i =, n e 0.45 ´ 1010, , s = e( n em e + n hm h ), , Reverse bias, , I 4 > I 3 > I2 > I1, , The doping of indium is 1 atom in 106 atoms of Si. But, indium has three valence electrons and each doped indium, atom creates one hole in Si crystal. Hence, it acts as an, acceptor atom., \ Concentration of acceptor atoms,, n h = 5 ´ 1028 ´ 10- 6, = 5 ´ 1022 / m 3, Intrinsic carrier concentration,, n i = 1.5 ´ 1016 / m 3, \ Hole concentration is increased,, n, 5 ´ 1022, = h=, n i 1.5 ´ 1016, , = 3.33 ´ 106 / m 3, This means that, the hole concentration has been increased, over its intrinsic concentration by the same amount with, which the electron concentration has been decreased., The conductivity of doped silicon is given by, , mA, , p-side n-side, , 21. As, concentration of Si atom = 5 ´ 1028 / m 3, , n i2 (1.5 ´ 1016 ) 2, =, ne, 4.95 ´ 1022, = 4.54 ´ 109 m -3, , As, n e > n h (number of holes). So, the material is n-type, semiconductor., , = 1.6 ´ 10-19 (0.45 ´ 1010 ´ 0.135 + 5 ´ 1022 ´ 0.048), = 384 S/m, 1, 1, Resistivity, r = =, = 0.0026 W-m, s 384, Conductivity of pure Si crystal,, s = en i( m e + m h ), = 1.6 ´ 10-19 ´ 1.5 ´ 1016 (0.135 + 0.048), = 0.4392 ´ 10-3S / m, 1, 1, =, = 2276.8 W - m, s 0.4392 ´ 10-3, Thus, we see that the conductivity of Si doped within, become much greater than its intrinsic conductivity and the, resistivity has become much smaller than the intrinsic, resistivity., 22. Given, V = 0.4 V, (i) d = 4.0 ´ 10-7 m, V, 0.4, Electric field, E = =, = 1 ´ 106 V/m, d 4 ´ 10-7, (ii) v1 = 4 ´ 105 m/s, v2 = ?, Resistivity, r =, , Suppose v1 be the speed of electron, when it enters the, depletion layer and v2 be the speed, when it comes out of, the depletion layer., According to principle of conservation of energy,, KE before entering the depletion layer = Gain in PE, + KE after crossing the depletion layer

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149, , CBSE Term II Physics XII, , 1, 1, mv12 = e ´ V + mv22, 2, 2, 1, -31, ´ 9. 1 ´ 10 ´ ( 4 ´ 105 ) 2, 2, 1, = 1.6 ´ 10-19 ´ 0.4 + ´ 9.1 ´ 10-31 ´ v22, 2, v2 = 1.39 ´ 105 m/s, , Þ, Þ, , \, , 23. When the input voltage is equal to or less than 5 V, diode, will be reverse biased. It will offer high resistance in, comparison to resistance R in series. Now, diode appears as, open circuit. The input waveform is then passed to the, output terminals. The result with sine wave input is to dip, off all positive going portion above 5 V., If input voltage is more than + 5 V, diode will be conducting, as if forward biased offering low resistance in comparison to, R. But there will be no voltage in output beyond 5 V, as the, voltage beyond + 5 V, will appear across R., When input voltage is negative, there will be opposition to, 5 V battery. In p- n junction, if input voltage becomes more, than - 5 V, the diode will be reverse biased. It will offer, high resistance in comparison to resistance R in series. Now,, junction diode appears as open circuit. The input waveform, is then passed on to the output terminals., The output waveform is as shown below, Voltage, , O, –5V, , =, Þ, , 2 T 3T, 4 4, , 4T, 4, , Time, , 24. V-I characteristics of p-n junction diode is as shown below, , Current (mA), , R¢ = 75 W, , Forward, bias, , So,, I1 = I4 + I2, Here, the resistances R1 and R 2 are same., i.e., , I4 = I2, , \, , I1 = 2I2, I 0.05, I2 = 1 =, = 0.025 A and I4 = 0.025 A, 2, 2, I1 = 0.05 A, I2 = 0.025 A, I3 = 0, , Þ, Thus,, , Reverse, bias, , 1, 1, 2, +, =, 150 150 150, , Total resistance, R = R ¢ + 25 = 75 + 25 = 100 W, V, 5, Current, I1 = =, = 0.05 A, R 100, I1 = I4 + I2 + I3 (here, I3 = 0), , +5 V, , T, 4, , electrons in the p-side are accelerated due to the reverse, bias voltage., These minority carriers acquire sufficient kinetic energy, from the electric field and collide with a valence electron., Thus, the bond is finally broken and the valence, electrons move into the conduction band resulting in, enormous flow of electrons and thus formation of, electron-hole pairs., Thus, there is a sudden increase in the current at the, critical voltage., 25. Given, forward biased resistance = 25 W, Reverse biased resistance = ¥, As the diode in branch CD is in reverse biased, which have, infinite resistance., So,, I3 = 0, Resistance in branch AB = 25 + 125 = 150 W (say R1 ), Resistance in branch EF = 25 + 125 = 150 W (say R 2 ), AB is parallel to EF., 1, 1, 1, So, resultant resistance,, =, +, R ¢ R1 R 2, , Voltage (V), , (i) Under the reverse bias condition, the holes of p-side, are attracted towards the negative terminal of the, battery and the electrons of the n-side are attracted, towards the positive terminal of the battery. This, increases the depletion layer and the induced potential, barrier is also increased. However, the minority charge, carriers are drifted across a junction producing a small, current. At any temperature, a number of minority, charge carriers is constant, so there is the small current, at any applied potential., This is the reason for the current under reverse bias, known as reverse saturation current, which is almost, independent of applied potential. At the certain level of, voltage, avalanche breakdown takes place which results, in a sudden flow of large current., (ii) At the critical voltage, (the voltage at which breakdown, takes place), the holes in the n-side and conduction, , and, I4 = 0.025 A, 26. (i) Working of LED LED is a forward biased p- n junction, which converts electrical energy into optical energy of, infrared and visible light region. Being in forward bias,, thin depletion layer and low potential barrier facilitate, diffusion of electron and hole through the junction., When high energy electron of conduction band combines, with the low energy holes in valence band, then energy is, released in the form of photon, which may be seen in the, form of light., (ii) Semiconductors with appropriate band gap Eg close to, 1.5 eV are preferred to make LED, e.g. GaAs., The other reasons to select these materials are high, optical absorption, availability of raw material and low, cost., (iii) Uses of LEDs, (a) LED can operate at very low voltage and consumes, less power in comparison to incandescent lamps., (b) Unlike the lamps, they take very less operational time, and have long life.

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150, , CBSE Term II Physics XII, , 27. (i) The energy of maximum intensity of the solar radiation is, 1.5 eV. In order to excite electrons, the energy of, radiation hn must be greater than energy band gap Eg ,, i.e. hn > Eg . Therefore, the semiconductor like GaAs is, preferred because it has a band gap lower than 1.5 eV. It, captures significantly higher number of electron than any, other material., (ii) V-I characteristics of a solar cell, I, Open circuit voltage (VOC), A, V, (VOC), B, Isc, Short circuit current, , Significance :, (a) V-I curve is drawn in the fourth quadrant, because a, solar cell does not draw current but supplies current to, the load., (b) In V-I curve , the point A indicates the maximum voltage, VOC which is being supplied by the given solar cell when, no current is being drawn from it. VOC is called the open, circuit voltage., , (c) In V-I curve, the point B indicates the maximum current, ISC which can be obtained by short-circuiting the solar, cell without any load resistance. ISC is called the, short-circuit current., 28. (i) When photodiode is illuminated with light due to, breaking of covalent bonds, equal number of additional, electrons and holes come into existence whereas, fractional change in minority charge carrier is much, higher than fractional change in majority charge, carrier., Since, the fractional change of minority carrier current, is measurable significantly in reverse bias than that of, forward bias. Therefore, photodiodes are connected in, reverse bias., (ii) Given, l = 450 nm, 1240, 1240, Energy, E =, Å=, eV = 2.66 eV, l, 450, So, diodes D2 and D3 will not detect light wave of, wavelength 450 nm., (iii) Main use of photodiode The photodiode are, extensively used in high speed reading of computer,, punched cards, light-detection systems, light-operated, switches, counting of objects interrupting a light beam,, etc.

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Physics, Class 12th ( Term II ), , Practice Paper 1, , *, , (Solved), Time : 2 Hours, Max. Marks : 35, , General Instructions, , 1. There are 9 questions in the question paper. All questions are compulsory., 2. Question no. 1 is a Case Based Question, which has five MCQs. Each question carries one mark., 3. Question no. 2-6 are Short Answer Type Questions. Each question carries 3 marks., 4. Question no. 7-9 are Long Answer Type Questions. Each question carries 5 marks., 5. There is no overall choice. However, internal choices have been provided in some questions. Students have to attempt, only one of the alternatives in such questions., , * As exact Blue-print and Pattern for CBSE Term II exams is not released yet. So the pattern of this, paper is designed by the author on the basis of trend of past CBSE Papers. Students are advised, not to consider the pattern of this paper as official, it is just for practice purpose., , 1. Direction Read the following passage and answer the questions that follows, Discovery of Nucleus, The nucleus was first discovered in 1911 by Lord Rutherford and his associates by experiments on, scattering of a-particles by atoms. He found that the scattering results could be explained, if atoms consist, of a small, central, massive and positive core surrounded by orbiting electrons.The experimental results, indicated that, the size of the nucleus is of the order of 10 -14 m and is thus 10000 times smaller than the size, of atom., (i) Ratio of mass of nucleus with mass of atom is approximately, (a) 1, , (b) 10, , (c) 10 3, , (d) 1010, , (ii) Masses of nuclei of hydrogen, deuterium and tritium are is in the ratio of, (a) 1 : 2 : 3, , (b) 1 : 1 : 1, , (c) 1 : 1 : 2, , (d) 1 : 2 : 4, , (iii) Density of a nucleus is, (a) more for lighter elements and less for heavier elements, (b) more for heavier elements and less for lighter elements, (c) very less compared to ordinary matter, (d) a constant, , (iv) If R is the radius and A is the mass number, then log R versus log A graph will be, (a) a straight line, , (b) a parabola, , (v) The ratio of the nuclear radii of the gold isotope, (a) 1.23, , (b) 0.216, , (c) an ellipse, 197, 79 Au, , and silver isotope, , (c) 2.13, , (d) None of these, 107, 47 Au, , is, , (d) 3.46

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154, , CBSE Term II Physics XII, , 2. (i) Name the EM waves which are used for the treatment of certain forms of cancer. Write their frequency range., (ii) Thin ozone layer on top of stratosphere is crucial for human survival. Why?, (iii) Why is the amount of the momentum transferred by the EM waves incident on the surface so small?, Or (i) Which segment of electromagnetic waves has highest frequency? How are these waves produced? Give one, use of these waves., (ii) Which EM waves lie near the high frequency end of visible part of EM spectrum? Give its one use. In what, way, this component of light has harmful effects on humans?, , 3. Hydrogen spectrum consists of discrete bright lines in a dark background and it is specifically known as, hydrogen emission spectrum.There is one more type of hydrogen spectrum that exists where we get dark lines, on the bright background, which is known as absorption spectrum., Line spectra of the hydrogen atom is given below, whose series limit corresponds to the wavelength for n = ¥., Series limit, n=7, n=6, n=5, n=4, n=3, n=2, , Energy level (eV), , Ionised atom, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, , Paschen, series, , Brackett, series, , Balmer, series, , Pfund, series, , n=1, , Lyman, series, , By using above spectra, write the expression for the series limit for all the series obtained., Or (i) Using Bohr’s second postulate of quantisation of orbital angular momentum, show that the circumference of, the electron in the nth orbital state in H-atom is n-times the de-Broglie wavelength associated with it., (ii) The electron in H-atom is initially in the third excited state. What is the maximum number of spectral lines, which can be emitted, when it finally moves to the ground state?, , 4. Two convex lenses A and B of an astronomical telescope having focal lengths 5 cm and 20 cm, respectively are, arranged as shown below, B, A, , 15 cm, , (i) Which one of the two lenses you will select as the objective lens and why?, (ii) What should be the change in the distance between the lenses to have the telescope in its normal adjustment, position?, (iii) Calculate the magnitude of magnifying power of the telescope in the normal adjustment position., , 5. What are extrinsic semiconductors? On the basis of valence band model, explain how can a pure semiconductor, of Ge or Si be converted into n-type semiconductor.

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155, , CBSE Term II Physics XII, , 6. Describe a photocell and mention few of its applications., Or What are the observations made from the expression of de-Broglie wavelength?, , 7. (i) Double convex lenses are to be manufactured from a glass of refractive index 1.55 with both faces of the same, radius of curvature. What is the radius of curvature required, if the focal length is 30 cm?, (ii) A Cassegrain telescope (reflecting telescope) uses two mirrors as shown in figure below. Such a telescope is, built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and of the small, mirror is 140 mm, then where will be the final image of an object at infinity?, , B, , Objective, mirror (C), , Secondary, mirror, Eyepiece, , Or (i) In a single slit diffraction experiment, a slit of width d is illuminated by red light of wavelength 650 nm. For, what value of d will the, (a) first minimum fall is at an angle of diffraction of 60° and, (b) first maximum fall is at an angle of diffraction of 60°?, (ii) In Young’s double slit experiment, the two slits 0.15 mm apart are illuminated by monochromatic light of, wavelength 450 nm. The screen is 1.0 m away from the slits., (a) Find the distance of the second, I. bright fringe, II. and dark fringe from the central maxima., (b) How will the fringe pattern change, if the screen is moved away from the slits?, , 8. (i) State briefly the processes involved in the formation of p-n junction, explaining clearly how the depletion, region is formed., (ii) Using the necessary circuit diagrams, show how the V-I characteristics of a p-n junction are obtained in (a), forward biasing and (b) reverse biasing., How are these characteristics made use of in rectification?, Or (i) Explain with the help of diagram, how a depletion layer and barrier potential are formed in a junction diode., (ii) Draw a circuit diagram of a full wave rectifier. Explain its working and draw input and output waveforms., , 9. (i) Using postulates of Bohr’s theory of hydrogen atom, show that, (a) the radii of orbits increase as n 2 and, (b) the total energy of the electron increases as 1/ n 2 , where n is the principal quantum number of the atom., (ii) Calculate the wavelength of H a -line in Balmer series of hydrogen atom., (Take, Rydberg constant, R = 1.097 ´ 10 7 m -1 ), Or Using Bohr’s postulates, derive the expression for the frequency of radiation emitted, when electron in, hydrogen atom undergoes transition from higher energy state (quantum number n i ) to the lower state, ( n f )., When electron in hydrogen atom jumps from energy state n i = 4 to n f = 3, 2, 1. Identify the spectral series to, which the emission lines belong.

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156, , CBSE Term II Physics XII, , EXPLANATIONS, 1. (i) (a) As nearly 99.9%, mass of an atom is in nucleus., Mass of nucleus 99.9, =, = 0.99 » 1, \, Mass of atom, 100, (ii) (a) Since, the nuclei of deuterium and tritium are, isotopes of hydrogen, they must contain only one, proton each. But the masses of the nuclei of hydrogen,, deuterium and tritium are in the ratio, of 1 : 2 : 3, because of presence of neutral matter in, deuterium and tritium nuclei., Mass, mA, (iii) (d) Density =, =, Volume 4 pR 3 A, 0, 3, 3m, =, 4p R 03, m = mp = mn, = 2.3 ´ 1017 kgm -3, which is a constant., 1/ 3, (iv) (a) R = R 0 A, 1, log R = log R 0 + log A, 3, On comparing the above equation of straight line, y = mx + c, so the graph between log A and log R is a, straight line also., (v) (a) Given, A1 = 197 and A2 = 107, As,, , 1/ 3, , \, , R1 æ A1 ö, =ç ÷, R 2 çè A2 ÷ø, ~, -1 . 23, , 1/ 3, , æ 197 ö, =ç, ÷, è 107 ø, , = 1 . 225, , 2. (i) g-rays are used for the treatment of certain forms of, cancer. Its frequency range is 3 ´ 1019 Hz to, 5 ´ 1022 Hz., (ii) The thin ozone layer on top of stratosphere absorbs, most of the harmful ultraviolet rays coming from the, sun towards the earth. They include UVA, UVB and, UVC radiations, which can destroy the life system on, the earth., Hence, this layer is crucial for human survival., (iii) An electromagnetic wave transports linear momentum,, as it travels through space. If an electromagnetic wave, transfers a total energy, U to a totally absorbing surface in time t, then total, linear momentum delivered to the at surface,, U, hn, p=, Þ p=, c, c, This means that, the momentum range of EM waves is, 10-19 to 10- 41. Thus, the amount of momentum, transferred by the EM waves is incident on the surface, is very small., Or, (i) Gamma rays has the highest frequency in the, electromagnetic waves. These rays are of the nuclear, origin and are produced in the disintegration of, radioactive atomic nuclei and in the decay of certain, subatomic particles. They are used in the treatment of, cancer and tumours., (ii) Ultraviolet rays lie near the high frequency end of, visible part of EM spectrum. These rays are used to, , preserve food stuff. The harmful effect from exposure to, ultraviolet (UV) radiation can be life threatening and, include premature aging of the skin, suppression of the, immune systems, damage to the eyes and skin cancer., 3. The wavelengths of spectral line in these series can be, expressed by the following formulae, (i) For Lyman series, 1, 1 ö, æ1, = R ç 2 - 2 ÷ , where n = 2, 3, 4,..., l, n ø, è1, 1, For, n = ¥, l =, R, (ii) For Balmer series, 1, 1 ö, æ1, = R ç 2 - 2 ÷ , where n = 3, 4, 5, ..., l, n ø, è2, 4, For, n = ¥, l =, R, (iii) For Paschen series, 1, 1 ö, æ1, = R ç 2 - 2 ÷ , where n = 4, 5, 6, ..., l, n ø, è3, 9, For, n = ¥, l =, R, (iv) For Brackett series, 1, 1 ö, æ1, = R ç 2 - 2 ÷ , where n = 5, 6, 7,..., l, 4, n, è, ø, 16, For, n = ¥, l =, R, (v) For Pfund series, 1, 1 ö, æ1, = R ç 2 - 2 ÷ , where n = 6, 7, 8,..., l, n ø, è5, 25, For, n = ¥, l =, R, Or, (i) Bohr’s second postulate states that, the electron, revolves around the nucleus in certain privileged orbit, which satisfy certain quantum condition that angular, h, momentum of an electron is an integral multiple of ,, 2p, where h is Planck’s constant., nh, i.e., L = mvr =, 2p, where, m = mass of electron, v = velocity of electron, and r = radius of orbit of electron., æ h ö, Þ, 2pr = n ç, ÷, è mv ø, \ Circumference of electron in n th orbit, = n ´ de-Broglie wavelength associated with electron, h ö, æ, çQ l =, ÷, mv, è, ø, (ii) Given, the electron in H-atom is initially in third, excited state., \, n=4, and the total number of spectral lines of an atom that, can exist is given by the relation

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157, , CBSE Term II Physics XII, n ( n - 1), 2, Here,, n=4, So, number of spectral lines, 4 ( 4 - 1) 4 ´ 3, =, =, =6, 2, 2, Hence, when a H-atom moves from third excited state, to ground state, it emits six spectral lines., 4. (i) In telescope objective lens have large focal length than, eyepiece. So, lens B is choosen as the objective lens., (ii) In normal adjustment, the distance between objective, and eyepiece is given by, L = f o + f e = 20 + 5 = 25, So, the distance required to be increased between the, two lenses,, L ¢ = L - 15 = 10 cm, (iii) Magnifying power of telescope in normal adjustment,, f, 20, m= o =, =4, fe, 5, =, , 5. Semiconductors in which some impurity atoms are, embedded are known as extrinsic or impure, semiconductors., Extrinsic semiconductors are basically of two types, (i) n-type semiconductors, (ii) p-type semiconductors, n-type Semiconductors, This type of semiconductor is obtained when pentavalent, impurity such as phosphorus (P), arsenic (As), etc is added to, Si or Ge., During doping, four electrons of pentavalent atom bond, with the four silicon neighbours, while fifth remains very, weakly bound to its parent atom. Also, the ionisation energy, required to set this electron free is very small., Hence, these electrons are almost free to move. In other, words, we can say that these electrons are donated by the, impurity atoms., So, these are also known as donor atoms and the, conduction inside the semiconductor will take place with, the help of the negatively charged electrons. Due to this, negative charge, these semiconductors are known as n-type, semiconductors., Therefore, major conduction in n-type semiconductors is, due to electrons. So, electrons are known as majority, carriers and the holes are known as the minority carriers., This means, n e >> n h; Ie >> Ih, , +4, , +4, , +4, , +4, , +5, , +4, , +4, , +4, , +4, , e– unbonded free, electron donated, by pentavalent, (+5 valency) atom, , 6. It is a device which converts light energy into electrical, energy. It is also called an electric eye. As the photoelectric, current is set-up in the photoelectric cell corresponding to, incident light, it provides the information about the objects, as done by our eye in the presence of light., Incident, light, C, Collector (Anode), , A, , Evacuated, glass bulb, , Emitter, (Cathode), , –, , B, , +, , mA, , A photocell consists of a semi-cylindrical photosensitive, metal plate C (emitter) and a wire loop A (collector), supported in an evacuated glass or quartz bulb. When light, of suitable wavelength falls on the emitter C, photoelectrons, are emitted., Some applications of photocell are given below, (i) Used in television camera for telecasting scenes and in, photo telegraphy., (ii) Reproduction of sound in cinema film., (iii) Used in burglar alarm and fire alarm., Or, According to de-Broglie hypothesis, the wavelength of wave, associated with moving material particle is given by, h, h, l= =, p mv, which is the expression for de-Broglie wavelength., From the above expression, the following observations are, made, 1, (i) The de-Broglie wavelength l µ . So, if the particle, v, moves faster, then the wavelength will be smaller and, vice-versa., (ii) If the particle is at rest ( v = 0), then the de-Broglie, wavelength is infinite ( l = ¥ ). Such a wave cannot be, visualised., (iii) The de-Broglie waves cannot be electromagnetic in, nature because electromagnetic waves are produced by, motion of accelerated charged particles., (iv) The wavelength of a wave associated with moving, particle defines a region of uncertainty, within which, the whereabouts of the particle are unknown., 7. (i) Given f = 30 cm, m = 1.55, R1 = R and R 2 = - R, Using lens Maker’s formula,, æ1, 1, 1 ö, ÷÷, = ( m - 1 ) çç f, R, R, è 1, 2ø, Þ, Þ, , 1, 2, 1 ù, é1, . ´, = ( 1.55 - 1 ) ê ú = 055, 30, R, ë R ( -R ) û, R = 1.1 ´ 30 = 33 cm

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158, , semiconductor. There are various methods of forming, p-n junction diode., Formation of Depletion Region in p-n Junction In an, n-type semiconductor, the concentration of electrons is, more than concentration of holes. Similarly, in a p-type, semiconductor, the concentration of holes is more than, that of concentration of electrons. During formation of, p-n junction and due to the concentration gradient, across p and n-sides, holes diffuse from p-side to n-side, ( p ® n ) and electrons diffuse from n-side to p-side, ( n ® p)., Electron drift, , Electron diffusion, , p, , n, , Depletion region, Hole diffusion, , Hole drift, , The diffused charge carriers combine with their, counterparts in the immediate vicinity of the junction, and neutralise each other., Thus, near the junction, positive charge is built on, n-side and negative charge on p-side. This sets up, potential difference across the junction and an internal, electric field Ei directed from n-side to p-side. The, equilibrium is established when the field Ei becomes, strong enough to stop further diffusion of the majority, charge carriers (however it helps the minority charge, carriers to diffuse across the junction). The region on, either side of the junction which becomes depleted, (free) from the mobile charge carriers is called, depletion region or depletion layer. The width of, depletion region is of the order of 10-6 m. The potential, difference developed across the depletion region is, called the potential barrier. Potential barrier depends, on dopant concentration in the semiconductor and, temperature of the junction., (ii) (a) Forward Biased Characteristics, The circuit diagram for studying forward biased, characteristics is shown in the figure. Starting from a, low value, forward bias voltage is increased step by, step (measured by voltmeter) and forward current is, noted (by ammeter). A graph is plotted between, voltage and current. The curve so obtained is the, forward biased characteristic of the diode., 8, B, , 7, p, , –, , +, , mA, , V, , +, , n, , –, , 6, Forward current (mA), , (ii) Given, distance between objective and secondary, mirror, d = 20 mm, Radius of curvature of large mirror, R1 = 220 mm,, so its focal length, f1 = 110 mm., Radius of curvature of small mirror, R 2 = 140 mm,, so its focal length, f 2 = 70 mm, The image formed by objective mirror at infinity will, acts as a virtual object for secondary mirror., \Distance of virtual object from secondary mirror,, u = f1 - d = 110 - 20 = 90 mm, Using mirror formula for secondary mirror,, 1 1 1, 1 1, 1, + =, Þ +, =, v u f2, v 90 70, 90 ´ 70, Þ, v=, = 315 mm, 90 - 70, Hence, the final image of the object will be formed 315, mm away from the secondary mirror., Or, (i) Given, wavelength of light, l = 650 nm = 650 ´ 10-9 m, and d is slit width., (a) Angle of diffraction, q = 60°, For first minimum,, l = d sin q, Þ, 650 = dsin 60°, 2, Þ, d = 650 ´, = 75055, . nm, 3, (b) For first maximum,, l, ( 2n - 1 ) = dsin q, 2, 3l, = dsin q, 2, 3 ´ 650, Þ, = dsin 60°, 2, 3 ´ 650 2, Þ, d=, ´, = 1125.83 nm, 2, 3, (ii) Given, slit width, d = 015, . mm = 015, . ´ 10-3 m, Wavelength of light l = 450 nm = 450 ´ 10-9 m, Distance between screen and slits, D = 1 m, (a) I. For second bright fringe ( n = 2),, nlD, y=, d, 450 ´ 10-9, = 6 ´ 10-3 m, Þ, y = 2´, 015, . ´ 10-3, = 6 mm, II. For second dark fringe ( n = 2),, lD, y = ( 2n - 1 ), 2d, 450 ´ 10-9 ´ 1, = ( 2 ´ 2 - 1), 2 ´ 015, . ´ 10-3, = 45, . mm, p, (b) The fringe width, b µ D and angular separation, q = ., d, So, on moving screen away from the slits, the fringe, width will increase while separation remains constant., 8. (i) p-n Junction A p-n junction is an arrangement made by, a close contact of n-type semiconductor and p-type, , CBSE Term II Physics XII, , 5, 4, 3, 2, 1, 0, , +, , Battery, , –, , (a), , Ge, , A, O, 0.1 0.2 0.3 0.4 0.5, Forward bias (V), (b)

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159, , CBSE Term II Physics XII, , p, , V, , +, , The input and output waveforms have been given below, , O, , Time, Due to Due to Due to Due to, D1, D2, D1, D2, , 0, O, , –, , 2, , C, , 4, Breakdown, voltage, , 6, 8, , Battery, –, , Output, D2, , Output, voltage, , Reverse bias (V), –10 –8 –6 –4 –2, , 10, , Reverse current (mA), , mA, , D1, , n, , +, –, , In this way, current flows in the load in the single direction, as shown in figure., , Input, voltage, , At the beginning, when applied voltage is low, the, current through the diode is almost zero. It is because, of the potential barrier, which opposes the applied, voltage., Till the applied voltage exceeds the potential barrier,, the current increases very slowly with increase in, applied voltage (OA portion of the graph). With further, increase in applied voltage, the current increases very, rapidly (AB portion of the graph). In this situation, the, diode behaves like a conductor., The forward voltage beyond which the current through, the junction starts increasing rapidly with voltage is, called knee voltage., If line AB is extended back, it cuts the voltage axis at, potential barrier voltage., (b) Reverse Biased Characteristics, The circuit diagram for studying reverse biased, characteristics is shown in the figure, , O, , Time, , 9. (a) A hydrogen like atom consists of a tiny positively charged, nucleus and an electron revolving in a stable circular orbit, around the nucleus., v, Electron e –, , +, , (a), , r, D, , Nucleus, +Ze, , (b), , In reverse biased, the applied voltage supports the flow of, minority charge carriers across the junction. So, a very small, current flows across the junction due to minority charge, carriers., Motion of minority charge carriers is also supported by, internal potential barrier, so all the minority carriers cross, over the junction., Therefore, the small reverse current remains almost, constant over a sufficiently long range of reverse bias,, increasing very little with increasing voltage (OC portion of, the graph)., This reverse current is voltage independent upto certain, voltage known as breakdown voltage and this voltage, independent current is called reverse saturation current., Use of p-n Junction Characteristics in Rectification, From forward and reverse characteristics, it is clear that, current flows through the junction diode only in forward, bias not in reverse bias, i.e. Current flows only in one, direction., Or, (i) Refer to Q 8. (i), (ii) During the first half of input cycle, the upper end of the coil, is at positive potential and lower end at negative potential., The function diode D1 is forward biased and D2 in reverse, biased. Current flows in output load in the direction shown, in figure. During the second half of input cycle, D2 is, forward biased., , Let e, m and v be respectively the charge, mass and velocity, of the electron and r the radius of the orbit., The positive charge on the nucleus is Ze, where Z is the, atomic number (in case of hydrogen atom, Z = 1 ). As, the, centripetal force is provided by the electrostatic force of, attraction, we have, mv2, 1 ( Ze ) ´ e, =, ×, r, 4pe0, r2, or, , mv2 =, , Ze 2, 4pe0 r, , ....(i), , From the first postulate of Bohr’s atomic model, the angular, momentum of the electron is, h, ....(ii), mvr = n, 2p, where, n (= 1, 2, 3, ....) is principal quantum number., From Eqs. (i) and (ii), we get, h 2 e0, ....(iii), r = n2, pmZe 2, This is the equation for the radii of the permitted orbits., According to this equation, rn µ n 2 ., Since, n = 1, 2, 3, ... it follows that the radii of the permitted, orbits increase in the ratio 1 : 4 : 9 : 16 : ……… from the, first orbit.

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Physics, Class 12th ( Term II ), , Practice Paper 2, , *, , (Unsolved), Time : 2 Hours, Max. Marks : 35, , General Instructions, , 1. There are 9 questions in the question paper. All questions are compulsory., 2. Question no. 1 is a Case Based Question, which has five MCQs. Each question carries one mark., 3. Question no. 2-6 are Short Answer Type Questions. Each question carries 3 marks., 4. Question no. 7-9 are Long Answer Type Questions. Each question carries 5 marks., 5. There is no overall choice. However, internal choices have been provided in some questions. Students have to attempt, only one of the alternatives in such questions., , * As exact Blue-print and Pattern for CBSE Term II exams is not released yet. So the pattern of this, paper is designed by the author on the basis of trend of past CBSE Papers. Students are advised, not to consider the pattern of this paper as official, it is just for practice purpose., , 1. Direction Read the following passage and answer the questions that follows, Excited State of Atom, At room temperature, most of the H-atoms are in ground state. When an atom receives some energy (i.e. by, electron collisions), the atom may acquire sufficient energy to raise electron to higher energy state. In this, condition, the atom is said to be in excited state. From the excited state, the electron can fall back to a state, of lower energy, emitting a photon equal to the energy difference of the orbit., Total energy, E(eV), Unbound (ionised), atom, 0, –0.85, , n=5, n=4, , –0.51, , n=3 Excited, states, , –3.40, , n=2, , Ground state, –13.6, , n=1

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162, , CBSE Term II Physics XII, , In a mixture of H—He + gas (He + is single ionised He atom), H-atoms and He + ions are excited to their respective, first excited states. Subsequently, H atoms transfer their total excitation energy to He + ions (by collisions)., (i) The quantum number n of the state finally populated in He + ions is, (a) 2, , (b) 3, , (c) 4, , (d) 5, +, , (ii) The wavelength of light emitted in the visible region by He ions after collisions with H-atoms is, (a) 6.5 ´ 10 -7 m, , (b) 5 .6 ´ 10 -7 m, , (c) 4.8 ´ 10 -7 m, , (d) 4.0 ´ 10 -7 m, +, , (iii) The ratio of kinetic energy of the electrons for the H-atom to that of He ion for n = 2 is, (a), , 1, 4, , (b), , 1, 2, , (c) 1, , (d) 2, , (iv) The radius of the ground state orbit of H- atom is, (a), , e0, hpme, , 2, , (b), , h 2 e0, pme, , (c), , 2, , p m e2, h, , (d), , 2 p h e0, me 2, , (v) Angular momentum of an electron in H-atom in first excited state is, (a), , h, p, , (b), , h, 2p, , (c), , 2p, h, , (d), , p, h, , 2. For a given lens, the magnification was found to be twice as large when the object was 0.15 m distant from it, than when the distance was 0.2 m. What is the focal length of the lens?, Or An astronomical telescope has objective and eyepiece of focal lengths 40 cm and 4 cm, respectively. Find the, distance by which the lenses must be separated, so that image of an object 200 cm away from the objective can, be seen at infinity. Also, draw the ray diagram., (i), 3. Why do we need the oil drops in Millikan’s experiment to be of microscopic sizes? Why cannot we carry out, the experiment with bigger drops?, (ii) What happens to the wavelength of a photon after it collides with an electron?, (iii) Can X-rays cause photoelectric effect?, 4. (i) What is the ratio of the number of holes and the number of conduction electrons in an intrinsic, semiconductor?, (ii) Draw the energy band diagram of n-type semiconductor., (iii) Draw I versus V graph of a forward biased junction diode., Or If each diode in figure has a forward bias resistance of 25 W and infinite resistance in reverse bias, what will be, the values of the currents I1 , I 2 , I 3 and I 4 ?, A, C, E, I1, , I4, , 125 W, , I3, , 125 W, , I2, , 125 W, , B, , D, F, 25 W, H, , G, 5V, , 5. (i) Why is the core of a nuclear reactor one of its most important part?, (ii) Why is the number of neutrons in heavier nuclei more than the number of protons?, (iii) Name the element with which control rods in nuclear reactors are made up., , 6. (i) Identify the part of the electromagnetic spectrum used in (i) radar and (ii) eye surgery. Write their frequency, range., (ii) Prove that the average energy density of the oscillating electric field is equal to that of the oscillating, magnetic field.

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163, , CBSE Term II Physics XII, , Or State clearly how a microwave oven works to heat up a food item containing water molecules., Why are microwaves found useful for the raw systems in aircraft navigation?, , 7. Show that the refractive index of the material of a prism is given by, (A +dm), 2, where, symbols have their usual meanings., m=, æ Aö, sin ç ÷, è2ø, Or (i) A point object O is kept in a medium of refractive index n 1 infront of a convex spherical surface of radius of, curvature R which separates the second medium of refractive index n 2 from the first one, as shown in the, figure. Draw the ray diagram showing the image formation and deduce the relationship between the object, distance and the image distance in terms of n 1 , n 2 and R., sin, , O, , n1, , n2, , u, , R, , C, , (ii) When the image formed above acts as a virtual object for a concave spherical surface separating the medium, n 2 from n 1 ( n 2 > n 1 ), draw this ray diagram and write the similar [similar to (i)] relation. Hence, obtain the, expression for the lens Maker’s formula., , 8. (i) When the width of the slit is made double, how would this effect the size and intensity of the central, diffraction band? Justify your answer with the help of diagram., (ii) Write three characteristic features to differentiate between diffraction and interference., Or (i) Consider two coherent sources S1 and S2 producing monochromatic waves to produce interference pattern., Let the displacement of the wave produced by S1 be given by y1 = a cos wt and the displacement by S2 be, y 2 = a cos(wt + f ). Find out the expression for the amplitude of the resultant displacement at a point and show, that the intensity at that point will be, f, I = 4a 2 cos 2, 2, Hence, establish the conditions for constructive and destructive interference., (ii) What is the effect on the interference fringes in Young’s double slit experiment, when, (a) the width of the source slit is increased and, (b) the monochromatic source is replaced by a source of white light?, , 9. (i) How is a depletion region formed in p-n junction?, (ii) With the help of a labelled circuit diagram, explain how a junction diode is used as a full wave rectifier. Draw, its input and output waveforms., (iii) How do you obtain steady DC output from the pulsating voltage?, Or (i) Explain with the help of suitable diagram, the two processes which occur during the formations of a p-n, junction diode. Hence, define the terms (i) depletion region and (ii) potential barrier., (ii) Draw a circuit diagram of a p-n junction diode under forward bias and explain its working., , Answers, 1. (i) c,, , (ii) c,, , (iii) a, (iv) b,, , (v) a, , 2. 0.10 m Or 54 cm, , 4. Or 0.05 A, 0.025 A, 0A, 0.05 A

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Physics, Class 12th ( Term II ), , Practice Paper 3, , *, , (Unsolved), Time : 2 Hours, Max. Marks : 35, , General Instructions, , 1. There are 9 questions in the question paper. All questions are compulsory., 2. Question no. 1 is a Case Based Question, which has five MCQs. Each question carries one mark., 3. Question no. 2-6 are Short Answer Type Questions. Each question carries 3 marks., 4. Question no. 7-9 are Long Answer Type Questions. Each question carries 5 marks., 5. There is no overall choice. However, internal choices have been provided in some questions. Students have to attempt, only one of the alternatives in such questions., , * As exact Blue-print and Pattern for CBSE Term II exams is not released yet. So the pattern of this, paper is designed by the author on the basis of trend of past CBSE Papers. Students are advised, not to consider the pattern of this paper as official, it is just for practice purpose., , 1. Direction Read the following passage and answer the questions that follows, Total Internal Reflection, Total internal reflection is the phenomenon of reflection of light into denser medium at the interface of denser, medium with a rarer medium. Light must travel from denser to rarer and angle of incidence in denser medium, 1, must be greater than critical angle (i c ) for the pair of media in contact, we can show m =, ., sin i c, B, , Rarer, medium, (air), , r¢, , r, O1, i, Denser, medium, (water), A, , N, , O2, , O3, , i¢ N, , ic, , Water-air, D O4 interface, i > ic, N, , Totally, reflected ray, , Partially, C reflected rays, , (i) In total internal reflection, light ray is, (a) travelling through a denser medium is completely reflected back to denser medium, (b) travelling through a denser medium is completely refracted to rarer medium, (c) light ray is partially reflected back to denser medium and partially refracted to rarer medium, (d) absorbed completely by denser medium, , (ii) Total internal reflection of a light ray travelling from denser medium to rarer medium occurs only, when, angle of incidence is, (a) 45°, (c) acute, , (b) 90°, (d) more than a certain value

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165, , CBSE Term II Physics XII, , (iii) Critical angle for water-air interface is 48.6°. What is the refractive index of water?, (a) 1, , (b) 3/2, , (c) 4/3, , (d) 3/4, , (iv) Light is travelling from air to water at Ði = 50°, which is greater than critical angle for air-water interface., What fraction of light will be totally reflected?, (a) 100%, , (b) 50%, , (c) 25%, , (d) None of these, , (v) Critical angle for glass-air interface, where refractive index m of glass is 3/2 is, (a) 41.8°, , (b) 60°, , (c) 30°, , (d) 44.3°, , 2. When four hydrogen nuclei combine to form a helium nucleus, estimate the amount of energy (in MeV) released, in this process of fusion (neglect, the masses of electrons and neutrinos). (Take, mass of 11 H = 1.007825 u and mass, of helium nucleus = 4.002603 u), , 3. (i) If e 0 & m 0 are the electric permittivity & magnetic permeability of free space and e & m are the corresponding, , quantities in the medium. Find the index of refraction of the medium in terms of above parameter., (ii) An electromagnetic wave is travelling in vacuum with a speed of 3 ´ 10 8 m/s. Find the velocity in a medium, having relative electric permittivity and magnetic permeability 2 and 1, respectively., Or Answer the following questions., (i) Name the waves which are produced during radioactive decay of a nucleus. Write their frequency range., (ii) Welders wear special glass goggles while working. Explain, why., (iii) Why are infrared waves often called as heat waves? Give their one application., , 4. In the following figures, indicate which of the diodes are forward biased and which are reverse biased?, +3V, +11V, , (i), , (ii), , (iii), , +7V, , – 5V, , Or Predict the effect on the electrical properties of a silicon crystal at room temperature, if every millionth silicon, atom is replaced by an atom of indium. Given, concentration of silicon atoms = 5 ´ 1028 m -3 , intrinsic carrier, concentration = 1.5 ´ 1016 m -3 , m e = 0.135 m 3 / V- s and m h = 0.048m 3 / V-s., , 5. A neutron of mass (m) = 1.66 ´ 10 -27 kg having energy (E) = 8.28 ´ 10 -21 J at 127° C is moving in a waveform, then, its de-Broglie wavelength can be calculated as,, (given, Boltzmann constant, k = 1.38 ´ 10 -23 J mol -1 K -1 and Planck’s constant, h = 6.63 ´ 10 -34 J-s), l=, , h, 2mE, , =, , 6.63 ´ 10 -34, 2 ´ 1.66 ´ 10 -27 ´ 8.28 ´ 10 -21, , l = 1.264 ´ 10 -10 m = 1.264 Å, If the energy of neutron will not be given, then suggest an alternative method to find the wavelength., , 6. (i) A diverging lens of focal length f is cut into two identical parts, each forming a plano-convex lens. What is the, focal length of each part?, (ii) A ray of light passes through an equilateral glass prism, such that the angle of incidence is equal to angle of, convergence and each of these angles is equal to ( 3/ 4)th of angle of prism. What is the value of angle of deviation?, Or (i) Out of blue and red lights, which is more deviated by a prism? Give reason., (ii) Give one application of prism., (iii) If a prism of 5° angle gives deviation of 3.2°, then what will be the refractive index of prism?

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166, , CBSE Term II Physics XII, , 7. Define the Q-value of a nuclear process. When can a nuclear process not proceed spontaneously? If both the number, of protons and the number of neutrons are conserved in a nuclear reaction in what way is mass converted into energy, (or vice-versa) in the nuclear reaction?, Or (i) (a) Why is the mass of a nucleus always less than the sum of masses of constituents, neutrons and protons?, (b) What is obtained by fusion of two deuterons?, (c) 32 He and 13 He nuclei have the same mass number ., Do they have same bindng energy?, (ii) The bombardment of lithium with protons gives rise to the following reaction, 7, 1, 4, 4, 2 Li + 1 H ¾® 2 He + 2 He + Energy, The atomic masses of lithium, hydrogen and helium are 7.016 amu, 1.008 amu and 4.004 amu, respectively., Find the initial energy of each helium atom. (Take, 1 amu = 931 MeV/c 2 ), , 8. (i) Draw a circuit arrangement for studying V-I characteristics of a p -n junction diode in, (a) forward bias and, (b) reverse bias., Show typical V-I characteristics of a silicon diode., (ii) State the main practical application of LED. Explain, giving reason, why the semiconductor used for, fabrication of visible light LEDs must have a band gap of at least (nearly) 1.8 eV., Or (i) Draw the circuit arrangement for studying the V-I characteristics of a p-n junction diode in (a) forward and (b), reverse bias. Briefly explain how the typical V-I characteristics of a diode are obtained and draw these, characteristics., (ii) With the help of necessary circuit diagram, explain the working of a photodiode used for detecting optical, signals., , 9. What is diffraction of light? Draw a graph showing the variation of intensity with angle in a single slit diffraction, experiment. Write one feature which distinguishes the observed pattern from the double slit interference, pattern., How would the diffraction pattern of a single slit be affected, when, (i) the width of the slit is decreased and, (ii) the monochromatic source of light is replaced by a source of white light?, Or (i) A ray of light falls on a transparent sphere with centre C as shown in the figure. The ray emerges from the, sphere parallel to the line AB. Find the angle of refraction of A, if the refractive index of material of sphere is 3., Also, draw the refracted ray in the given figure., Air, , A, , C, , Air, , B, , 60°, , (ii) The image obtained with a convex lens is erect and its length is four times the length of the object. If the focal, length of the lens is 20 cm, calculate the object and image distances., , Answers, 1/ 2, , 1. (i) c,, , (ii) d,, , (iii) c,, , (iv) d,, , 6. (i) 2f , (ii) 30° Or (iii) 1.64, , (v) a, , 2. 26.72 MeV, , 7. Or (ii) 7.488 MeV, , é me ù, 3. (i) ê, ú, ë (m 0e0 ) û, , ,, , (ii), , 3, ´ 108 m/s, 2, , 9. Or (i) 30° (ii) u = 15 cm, v = 60 cm, , 5. 1.264 Å