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ATOMIC STRUCTURE, , ATOMIC STRUCTURE, SESSION -1, AIM, 1) To introduce Fundamental particles, 2) To introduce Thomson’s and Rutherford Atomic model, 3) To introduce terms like atomic number, mass number and isotopes, isobars, isotones, John Dalton coined the term atom. The atom is the fundamental particle of matter and considered to, be indivisible and indestructible., In fact, the atom as the whole is electrically neutral as number of protons in it is equal to number of, electrons., The electron, proton and neutron are the main fundamental particles of an atom., , Discovery of electron – study of Cathode rays:, J.J. Thomson observed that, when a high voltage is applied between the electrodes fitted in discharge, tube,at a very low pressure,some invisible radiations are emitted from the cathode. At this stage, wall of the discharge tube near cathode starts glowing., , Gas at low, Pressure, , Discharge tube, , Cathode rays, , , , Faint green glow, , To vacuum pump, Discharge tube experiment – production of cathode rays, Glowing is due to the bombardment of glass wall by the cathode rays. It may be noted that when the, gas pressure in the tube is 1 atm, no electric current flows through the tube. This is because the, gases are poor conductor of electricity., , Origin of Cathode rays:, Cathode rays are first produced in cathode due to bombardment of the gas molecules by the highspeed electrons emitted first from the cathode., , Properties of Cathode rays, i. Cathode rays travel in straight lines with high speed., ii. Cathode rays are made up of material particles., iii. Cathode rays carry negative charge, the negatively charged material particles constituting the, cathode rays are called electrons., iv. Cathode rays produce heating effect., v. They cause ionization of the gas through which they pass., vi. They produce X-rays when they strike against the surface of hard metals like tungsten, molybdenum, etc., vii.They produce green fluorescence on the glass walls of the discharge tube exp : zinc sulphide., viii. They affect the photographic plates., ix. They possess penetrating effect (i.e., they can easily pass through thin foils of metals)., , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, x. The nature of the cathode rays does not depend upon the nature of the gas, taken in the discharge, tube and the nature of cathode material., xi. For each cathode rays, the ratio of charge (e) to mass (m) is constant, , Discovery of proton – study of Anode rays:, Goldstein discovered the presence of positive rays., He performed discharge tube experiment in which he took perforated cathode and a gas at low, pressure was kept inside a discharge tube., On applying high voltage between electrodes ,new rays were coming from the side of anode and, passing through the hole in the cathode gives fluorescence on the opposite glass wall coated with, zinc sulphide., These rays were called anode rays or canal rays or positive rays., , Perforated cathode, H2 gas at low pressure, , Anode rays, , , , ZnS coating, , H. V, ., , To vacuum pump, Production of anode rays, , Origin of anode or positive rays:, In the discharge tube the atoms of gas lose negatively charged electrons. These atoms, thus, acquire a, positive charge. The positively charged particle produced from hydrogen gas was called the, proton., H →, H+(proton) + e-, , Properties of Anode rays:, i), ii), iii), iv), v), , They travel in straight lines. However, their speed is much less than that of the cathode rays., They are made up of material particles., They are positively charged,hence they called as canal rays or anode rays.’, The nature of anode rays depends on the gas taken in the discharge tube., For different gases taken in discharge tube the charge to mass ratio (e/m) of the positive particles, constituting the positive rays is different., , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, , Fundamental particles:, 1) Electron: Electron is a universal constituent discovered by the J.J. Thomson., * Charge: It was determined by Mullikan by oil drop experiment as, -1.602x10-19coulombs or 4.803x10-10 e.s.u., * Mass:9.11x10-28g (nearly equal to 1/1837th of mass of hydrogen atom)., * Specific charge:e/m ratio is called specific charge & is equal to 1.76x108 coulombs/gm., * Mass of one mole of electrons: It is 0.55 mg., * Charge on one mole of electron is 96500 coulombs or 1 faraday., * Density: 2.17x1017 g/cc., , 2. Proton: (+1p0 or 1H1), * It was discovered by Goldstein., * Charge:It carries positive charge i.e.1.602 x 10-19coulombs or 4.803x10-10 esu., * Mass:1.672x10-24g or 1.672x10-27kg.It is 1837 times heavier than an electron., * Specific charge (e/m):9.58x104coulomb/gm., , 3. Neutron (0n1), * It was discovered by Chadwick by bombarding Be atom with high speed -particles., 4Be9 +2 He4 → 6 C12 +0 n1, * Charge: Charge less or neutral particle., * Mass:1.675x10-24 g or 1.675x10-27 kg., * Density:1.5x1014 g/cm3 and is heavier than proton by 0.18%., * Specific charge: It is zero., * Among all the elementary particles neutron is the heaviest and least stable., , Properties of Electron, Proton and Neutron, Properties, Discovery, Charge, Mass, Spin, Charge, Location, , Electron, J.J.Thomson, -1.6022x10-19C, 9.109x10-31 kg, ½, -1, Outsidethe nucleus, , Proton, Goldstein, 1.6022x10-19 C, 1.672x10-27 kg, ½, +1, Inthe nucleus, , Neutron, Chadwick, Zero, 1.675x10-27 kg, ½, 0, Inthe nucleus, , Classical Models of Atom:, 1) Thomson’s Atomic Model, According to Thomson, an atom is a sphere of positive charge having a number of embedded, electrons in it and sufficient enough to neutralize the positive charge.This model is compared with a, water melon in which seeds are embedded or pudding in which raisins are embedded. Therefore, this, model, sometime called watermelon model or raisin or plum pudding model., , Thomson’s model of an atom, , Limitation: Its failed explain the scattering experiment of Rutherford and the stability of atom., Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, , 2) Rutherford’s Atomic Model:, Rutherford, performed -ray scattering experiment in which he bombarded thin foils of metals like, gold, silver, platinum or copper with a beam of fast moving radioactive particles originated from a lead, block. The presence of 𝛼 particles at any point around the thin foil of gold after striking it was detected, with the help of a circular zinc sulphide screen. The point at which a𝛼 particle strikes this screen, a, flash of light is given out., , Observations and Conclusions,  particles, , Beam of , particles, , +, , ZnS screen, , Nucleus, , Gold foil (100 nm thickness), , i. Most of the -particles passed through the gold foil without any deflection from their original path., Bcz atom has largely empty space as most of the -particles passed through the foil, undeflected., ii. A few of the alpha particles are deflected fairly at large angles while some are deflected through, small angles., Bcz there is heavy positive charge at the centre of the atom which causes repulsions., The entire mass of the atom is concentrated in the nucleus., iii. A very few -particals are deflected back along their path., , According to Rutherford,, 1., , Atom is spherical and mostly hollow with a lot of empty space in it., , 2., , It has a small positively charged part at its centre known as nucleus., , 3., , The nucleus is surrounded by electrons. The electrons revolve round the nucleus with very high, speeds in circular paths called orbits., The number of extra nuclear electrons is equal to the number of units of positive charge in the, nucleus.Therefore the atom is electrically neutral. Electrons and the nucleus are held together by, electrostatic forces of attraction., Rutherford’s model has resemblances with solar system. Hense also known as planetary model of, the atom., , 4., , 5., , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, 6., , There is an empty space around the nucleus called extra nuclear part. In this part electrons are, present. As the nucleus of the atom is responsible for the mass of the atom, the extra nuclear part, is responsible for its volume., , Drawbacks:, 1. According to the electromagnetic theory of Maxwell, when a charged particle moves under the, influence of attractive force it loses energy continuously in the form of electromagnetic, radiation. Therefore an electron in an orbit will emit radiation., As a result of this, the electron should lose energy at every turn and move closer and closer to, the nucleus following a spiral path., The ultimate result is that it will fall into the nucleus thereby making the atom unstable., i.eRutherford’s model cannot explain the stability of the atom., 2. If the electrons lose energy continuously, the spectrum is expected to be continuous but the, actual observed spectrum consists of well-defined lines of definite frequencies. Here the loss of, energy by the electrons is not continuous in an atom., , Atomic number (Z): It denotes the number of protons or the number of electrons in the neutral, atom., Atomic number (Z) = Number of protons in the nucleus of an atom or ion = Number of electrons in a neutral atom., , Mass number (A): The mass number is the total number of protons and neutrons present in the, nucleus of an atom of an element and indicated as A., Protons and neutrons present in the nucleus of an atom are collectively known as nucleons. Therefore,, the mass number is also known as nucleon number., Mass number (A) = Number of protons (Z) + Number of neutrons (n), , The number of neutrons (n) in an atom is equal to the difference between the mass number and the, atomic number., n=A–Z, , Mass Number, Atomic Number, , A, Z, , X, , OR, , Z, , XA, , Symbol of Element, where X is the symbol for the element with superscript A and subscript Z, both on the left hand side., , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, , Isotopes, Isobars and Isotones:, Isotopes: The atoms of the same element which have the same atomic number but different mass, numbers are called isotopes., Exp- 6 C12 , 6C13 , 6C14, 8, , O16 , 8 O17 , 8 O18, , 1, , 17, , H1, 1H2 , 1H3, , Cl35 , 17 Cl37, , Isotopes of an element differ in the number of neutrons present in the nucleus. But they have the, same number of protons and electrons., Bcz of same number of electrons they show same chemical properties.They have different, number of neutrons, so they will have different masses and hence different physical properties., , Isobars: The atoms of different elements which have the same mass number but different atomic, numbers are called isobars., Exp:, , 18, , Ar 40 ,, , 19, , K 40 ,, , 20, , Ca40 40, 20𝐶𝑎, , They have same number of nucleons. But they are differ chemically because the chemical, characteristics depend upon the number of electrons which is determined by the atomic number., , Isotones: Isotones are the atoms of different elements which have the same number of neutrons., Eg: i. 6 C14 , 7 N15 , 8 O16 (n = 8), , ii. 14 Si30 , 15P31,, , 16, , S32, , (n = 16), , Isotones show different physical and chemical properties., , CLASS EXERCISE, 1. The number of neutrons present in 19K39 is:, a) 39, , b) 19, , c) 20, , d) None of these, , 2. The nucleus of the atom (Z > 1) consists of:, a) Proton and neutron, , b) Proton and electron, , c) Neutron and electron, , d) Proton, neutron and electrons, , 3. The number of electrons in a neutral atom of an element is equal to it’s:, a) Atomic weight, , b) Atomic number, , c) Equivalent weight d) Electron affinity, , 4. The specific charge of the canal rays:, a) Is not constant but changes with gas filled in discharge tube, b) Remains constant irrespective of the nature of gas in discharge tube, c) Is maximum when gas present in discharge tube is hydrogen, d) Is 9.58 x 104 coulombs/g, 5. Proton is:, a) Nucleus of deuterium, , b) Ionized hydrogen molecule, , c) Ionized hydrogen atom, , d) An α-particle, , 6. According to the Rutherford which statement is correct?, a) Electron revolves in fixed circular path around the nucleus, b) Electron revolves around the nucleus, c) Electron does not decrease its energy at the time of revolution, Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, d) Electron obeys law of conservation of momentum at the time of revolution., 7. Rutherford's scattering experiments led to the discovery of, a) Nucleus, b) Presence of neutrons in the nucleus, c) Both a and b, d) Revolving nature of electrons around the nucleus, 8. Deflection back of a few particles on hitting thin foil of gold shows that:, a) Nucleus is heavy, , b) Nucleus is small, , c) Both a and b, , d) Electrons create hindrances in the movement of α –particles, 9. α-particles are represented by, a) Lithium atoms, , b) Helium nuclei, , c) Hydrogen nucleus d) None of the above, , HOMEEXERCISE:, 1., , The species in which one of the fundamental particles is missing is, a) Helium, , 2., , b) Protium, , c) Deuterium, , d) Tritium, , The discovery of neutron is late because neutron has, a) +ve charge, , b) –ve charge, , c) neutral charge, , d)lightest particle, , 3. Which of the following statements are correct?, a) Isotopes have same number of protons, b) Isobars have same nucleon number., c) Isobars have same number of protons, d) Both a and b, 4. The charge on electron is calculated by, a) Mullikan, , b) J J Thomson, , c) Ruther ford, , d) Newton, , 5. J J Thomson Model could able to explain the following?, a) Stability of Atom, , b) electrical neutrality of atom, , c) Stability of nucleus, , d) all of these, , 6. The thickness of the gold foil used in Ruther Ford α ray scattering experiment, a) 0.0004 cm, , b) 0.0004 m, , c) 0.0004 mm, , d) 0.004 cm, , c) 10-12cm, , d) 10-8cm, , 7. What is the size of atom predicted by Ruther Ford?, a) 10-13cm, , b) 10-14cm, , 8. If Thomson Model is correct what should be the observation in α-ray scattering experiment, a) All the α-rays should pass through the gold foil, b) Only few α-rays should pass through the gold foil, c) 98% of α-rays should get reflected back, d) Both b & c, 9. Which part of atom is responsible for volume of atom?, a) Nucleus, Page number, , b) extra nuclear part, 1, , AMCF-TEMPLE OF LEARNING, , c) protons, , d) unknown particles, Contact No. :

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ATOMIC STRUCTURE, , SESSION – 2 AND 3, AIM- To understand nature of Electromagnetic Radiation, Nature of Light (Electromagnetic Radiation): Electromagnetic radiation do not need any medium for, propagation e.g visible, ultra violet, infrared, x-rays, -rays, radio waves, radiant energy etc., Two theories were proposed to explain the nature and the propagation of light, i. Corpuscular theory: This theory was proposed by Newton. According to this theory light is, propagated in the form of invisible small particles. i.e.light has particle nature., The particle nature of light explained some of the experimental facts such as reflection and, refraction of light but it failed to explain the phenomenon of interference and, diffraction.Therefore, was discarded and ignored., ii. Wave theory of light (electromagnetic wave theory): was explained by James Clark Maxwell in, 1864 to explain and understand the nature of electromagnetic radiation., , Features of this theory are:, a. The light is a form of electromagnetic radiations., b. The light radiations consist of, electric and magnetic fields, oscillating perpendicular to, each other., , Components of radiation, iii) The vertical component of wave, ‘E’ indicates the change in the strength of the electric field and the, horizontal component of the wave ‘H’ indicates the change in the strength of the magnetic field., iv) These radiation do not require any medium for propagation., v) The radiations posses wave character and travel with the velocity of light i.e. 3x108 m/sec because, of the above characteristics, the radiation are called electromagnetic radiations or waves., , Electromagnetic radiation is explained by following characteristics:, , 1. Wave length:, The distance between two successive crests, troughs or between any two consecutive identical, points in the same phase of a wave is called wave length.It is denoted by the letter (lambda)., The wave length is measured in terms of meters (m), centimeters (cm), angstrom units (A0), nanometers (nm), picometers (pm) and also in millimicrons (m)., The S.I. unit of wavelength is meter, m, 1A0 = 10–10 m or 10–8 cm, 1nm = 10–9 m or 10–7 cm = 10A0, 1pm = 10–12m or 10–10 cm =10−2 A0, , 2. Frequency:, The number of waves that pass through a given point in one second is known as frequency of, radiation. It is denoted by the ‘v’(nue)., Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, , Crest, , , , Crest, , a, a, , , , Trough, , Trough, , Wave motion of the radiation, SI unit of frequency is per second(s–1) or Hertz (Hz). A cycle is said to be completed when a wave, consisting of a crest and a trough passes through a point., , 3. Velocity:, The distance travelled by the wave in one second is called velocity or speed of the wave (C)., SI unit is meters per second (ms–1)., C of electromagnetic radiation in vaccum is a constant commonly called the speed of light and is, denoted by ‘c’.It is equal to 3 × 108ms–1., , 4. Wave number:, The number of waves that can be present at any time in unit length is called wave number., It is denoted by  (nue bar)., It is the reciprocal of wave length., Wave number =  =, , 1, , , It is expressed in per centimeter (cm–1) or per meter (m–1)., The SI unit of wave number is m–1., , Wave length, wave number𝝂̅ , frequency 𝝂 and velocity c are related as, , c = , , 5. Amplitude:, The height of the crest or the depth of the trough of the wave is called amplitude of the wave. It is, denoted by A., The amplitude determines the strength or intensity or brightness of radiation., , 6. Time period:, It is the time taken by the wave for one complete cycle or vibrations. It is denoted by T. It is expressed, in second per cycle., 1, 𝑉, , Page number, , T=, , 1, , 1, ( where  = frequency), , , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, , Electromagnetic spectrum:, The arrangement of different types of electromagnetic radiations in the order of increasing, wavelengths or decreasing frequencies is known as electromagnetic spectrum., , , 10-16, , 10-12, , 10-10, , 10-8, , 10-7, , 10-6, , 10-4, , xUltra Visible Near Far, IR, IR, rays rays violet, , Rays Cosmic, rays, , V, , I, , B, , G, , Y, , 10-2, , 101, , Micro, wave, , Radio, waves, , O, , 106, , Long, RW, ,  increases,  decreases, E decreases, , R, , Violet Indigo Blue Green Yellow Orange Red, 3800 Å, , 4300, , 4800 5300, , 5800, , 6300, , 6900, , 7600 (in Å), , Limitations of Electromagnetic Wave Theory :, Electromagnetic wave theory was successful in explaining the properties of light such as interference,, diffraction etc., But it could not explain the following:, (i), The phenomenon of black body radiation., (ii), The photoelectric effect., (iii), The change heat capacity of solids as a function of T., (iv), The line spectra of atoms with special reference to hydrogen., These phenomena could be explained only if electromagnetic waves are supposed to have particle, nature., , Black body radiation:, When a radiant energy falls on the surface of a body, a part of it is absorbed, a part of it is reflected and, the remaining energy is transmitted., An ideal body is expected to absorb completely the radiant energy falling on it is known as a black, body. A black body is not only a perfect absorber but also a perfect emitter of radiant energy., A hollow sphere coated inside with a platinum black, which has a small hole in its wall can act as a near, black body., The radiation emitted by a black body kept at high temperature is called black body radiation.A black, body radiation is the visible glow that the solid object gives off when heated., A graph is obtained by plotting the intensity of radiation against wave length gives the following, details., , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, 1., , The nature of radiation depends upon the T of the black body., , 2., , If the energy emitted is continuous the curve should be as shown by the dotted lines., , 3., , At a given temperature the intensity of radiation increases with the wave length, reaches, maximum and then decreases., , 4., , The intensity of radiation is greatest at the medium wave lengths and least at highest and lowest, wave lengths., , 5. As the temperature increases the peak of maximum intensity shifts towards the shorter wave, lengths., , Planck’s quantum theory:, In order to explain black body radiation, Max Planck proposed quantum theory of radiation., , Postulates, 1. The emission of radiation from a body is due to the vibrations of the charged particles in the body., 2., , The energy is emitted or absorbed by a body discontinuously in the form of small packets of, energy called quanta., , 3., , The energy of each quantum of light is directly proportional to the frequency of the radiation., E , or,, E = h, Where ‘h’ is known as Planck’s constant., The value of ‘h’, 6.6256 × 10–34 Jsec- or 6.6256 × 10–27ergs sec-, , 4. In case of light, the quantum of energy is called a photon., The total amount of energy emitted or absorbed by a body will be some whole number multiple, of quantum,, E = nh  , where n is an integer such as 1,2,3 . . . . ., This means that a body can emit or absorb energy equal to hv, 2hv, 3hv . . . . . Or any other, integral multiple of h. This is called quantization of energy., 5. The emitted radiant energy is propagated in the form of waves., , PhotoElectric Effect:, When radiations with certain minimum frequency (ν0 ) strike the surface of a metal, the electrons are, ejected from the surface of the metal. It is called photoelectric effect,electrons emitted are called, photoelectron., Incident light, Detector, , Metal Surface, , Evacuated glass tube, , Electrons, −, , +, −, , +, , Photoelectric effect, , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, For each metal a certain minimum frequency is needed to eject the electrons called as threshold, frequency (  ) which differs from metal to metal., , K.E. of photoelectron, , K.E. of photoelectron, , o, , o, , K. E. constant, , Intensity of Incident, radiation, , Frequency of absorbed, photon, , K.E. as a function of frequency, , K.E. as a function of intensity, , It was explained by Einstein. When light of suitable frequency falls on a metal surface, the light photon, gives its energy to the electron of metal atom and the electron is ejected from metal surface by, absorbing this energy. The minimum energy of a photon required to eject an electron from a metal is, called work function () of the metal. The remaining part of the energy (h  - ) of photon is used to, increase the kinetic energy of the ejected electron. If o is the threshold frequency and  , the, frequency of incident light then, Work function,  = h o .,  According to Einstein, E = h ,  Kinetic energy of photo electron Ek = E -  = h − ho, , CLASS EXERCISE, 1. The frequency of a radiation whose wave length is 600 nm is, a) 3 x 1014 sec-1, , b) 4 x 1014 sec-1, , c) 5 x 1014 sec-1, , d) 3 x 1015 sec-1, , 2. The wavelength of light having wave number 4000 cm-1 is, a) 2.5  m, , b) 250  m, , d) 25 nm  m, , c) 25  m, , 3. What is the energy of photons that corresponds to a wave number of 2.5 × 10-5 cm-1?, a) 2.5 × 10-20 erg, , b) 5.1 × 10-23 erg, , c) 5.0 × 10-22 erg, , d) 8.5 × 10-22 erg, , 4. The frequency of radiation having wave number 10m-1 is:, a) 10s-1, , b) 3×107s-1c) 3×1010s-1, , d) 3×109s-1, , 5. The wavelengths of two photons are 2000Å and 4000Å respectively. What is the ratio of their, energies?, a) 1/4, , b) 4, , c) 1/2, , d) 2, , 6. In photo electric effect the number of photo electrons emitted is proportional to, a) Intensity of incident beam, , b) Frequency of incident beam, , c) Velocity of incident beam, , d) Work function of photo cathode, , 7. The kinetic energy of the photo electrons does not depend upon, a) Intensity of incident radiation, , b) frequency of incident radiation, , c) Wavelength of incident radiation, , d) wave number of incident radiation, , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, 8. The work function of a metal is 3.1 x 10-19 J. Which frequency of photons will not cause the, ejection of electrons?, a) 5 x 1014 s-1, , b) 5 x 1015 kHz, , c) 6 x 1014 s-1, , d) 5 x 1012 Hz, , 9. The work function of a metal is 4.2 eV. If radiation of 2000 falls on the metal, then the kinetic, energy of the fastest photo electrons is, a) 1.6 × 10-19 J, , b) 16 × 1010 J, , c) 3.2 × 10-19 J, , d) 6.4 × 10-10 J, , 10. A photo electric emitter has a threshold frequency v0. When light of frequency 2v0 is incident, the, speed of photo electrons is V. When light of frequency 5v0 is incident, speed of photo electrons is, a) 4V, , b) 2V, , c) 2.5V, , d) 2.5V, , HOME EXERCISE, 1. Wave theory failed to explain the following properties, a) diffraction, , b) interference, , c) black body radiation d) all the above, , 2. Plank’s quantum theory is explained which of the following properties, a) quantization, 3., , b) black body radiation c) diffraction, , d) both a &b, , The electromagnetic radiation with high energy, a) radio waves, , b) X-rays, , c) Infra-red radiation d) visible light, , 4. The atomic transition gives rise to radiation of frequency 104 Hz. The change in energy per mole of, atoms taking place would be:, a) 3.99 × 10–6J, , b) 3.99J c) 6.62×10––24Jd) 6.62× 10–30J, , 5. Two electromagnetic radiations having energy ratio 3:2 is falling on metal surface and producing, metallic luster what is the ratio of wave numbers of those radiation?, a) 1:2, , b) 2:3, , c) 3:2, , d) 9:4, , 6. The energy of the photons which corresponds to light of frequency 3 1015 sec-1 is, a) 1.9876×10-15 ergs, , b) 2.9876×10-8 ergs, , c) 1.9876×10-10 ergs, , d) 1.9876×10-11 ergs, , 7. Find the frequency of light that correspond to photons energy 5.0 x 10 -5 erg, a) 2.2 x 1011 sec-1, , b) 7.5 x 1021 sec-1, , c) 4.0 x 10-5 sec-1, , d) 4.0 x 104 sec-1, , 8. Photoelectric effect shows:, a) Particle-like behavior of light, , b) Wave like behavior of light, , c) Both wave like and particle-like behavior of light behavior of light, d) none, 9. When the frequency of light incident on a metallic plate is doubled, the KE of the emitted, photoelectrons will be:, a) Doubled, , b) Halved, , c) Increased but more than doubles of previous KE, , d) Unchanged, , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, , SESSION - 4, AIM - To introduce Atomic Spectra, ATOMIC SPECTRA, Spectrum is the impression produced on a screen when radiations of a particular wavelengths, areanalyzed through a prism or diffraction grating. Spectra are broadly classified into two., (i), , Emission Spectrum., , (ii), , Absorption Spectrum., , 1., , Emission Spectrum:When the radiation emitted from some source, e.g., from the sun or by, passing electric discharge through a gas at low pressure or by heating some substance to high, temperature etc. is passed directly through the prism and then received on the photographic, plate, the spectrum obtained is called ‘Emission spectrum’., , The spectrum of a radiation emitted by a substance in its excited state is an emission spectrum., Emission Spectrum is of two types:, a) Continuous Spectrum and b) Discontinuous Spectrum, a. Continuous Spectrum:When white light from any source such as sun, a bulb or any hot glowing, body is analyzed by passing through a prism, it is observed that it splits up into seven, , Beam, , Slit, , 7-colours, •, Prism, , VIBGYOR, , White light, , differentcolours from violet to red,(like rainbow), as shown in fig ., , Photographic, plate, , These colors are so continuous that each of them merges into the next. Hence, the spectrum is called, continuous spectrum., It may be noted that on passing through the prism, red colour with the longest wavelength is dedicated, least while violet colour with shortest wavelength is deviated the most., b. Discontinuous Spectrum:When gases or vapours of a chemical substance are heated in an, electricArc or in a Bunsen flame, light is emitted. If the ray of this light is passed through a, prism, a line spectrum is produced., •, , A discontinuous spectrum consisting of distinct and well defined lines with dark areas in, between is called line spectrum. It is also called atomic spectrum., , •, , The emission spectrum consisting of a series of very closely spaced lines is called, bandspectrum., , Band spectrum is the characteristic of molecules. Hence it is also known as molecular spectrum. The, band spectrum is due to vibrations and rotations of atoms present in a molecule., , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, , Differences between line and band spectrum, Line spectrum, , Band spectrum, , 1., , The line spectrum has sharp,, distinct well defined lines., , 1., , The band spectrum has many closed, lines., , 2., , The line spectrum is the, characteristic of atoms and is also, called atomic spectrum., , 2., , The band spectrum is characteristic of, molecules, and, is, also, called, molecularspectrum., , 3., , The line spectrum is due to, transition of electrons in an atom., , 3., , The band spectrum is due to vibrations, androtations, of atoms, in a molecule, rotations, of atoms, in a, , 4., , The line spectrum is given by inert, gases, metal vapours and, atomised nonmetals., , 4., , The band spectrum is given by hot metals, and molecular nonmetals., , 2., Absorption spectra: When white light from any source is first passed through the solution or, vapours of a chemical substance and then analysed by the spectroscope, it is observed that some dark, lines are obtained.Further, it is observed that the dark lines are at the same place where coloured lines, are obtained in the emission spectra for the same substance., , Difference between emission spectra and absorption spectra, 1., , 2., 3., , EMISSION SPECTRA, Emission spectrum is obtained 1., when the radiation from the, source are directly analyses in, the spectroscope., It consists of bright coloured 2., lines separated by dark spaces., Emission spectrum can be 3., continuous spectrum (if source, emits, white, light), or, discontinuous, i.e., line spectrum, if source emits some coloured, radiation., , ABSORPTION SPECTRA, Absorption spectrum is obtained when the, white light is first passed through the, substance and the transmitted light is, analyzed in the spectroscope., It consists of dark lines in the otherwise, continuous spectrum., Absorption spectrum is always discontinuous, spectrum of dark lines., , Emission Spectrum of Hydrogen:, When hydrogen gas at low pressure is taken in the discharge tube and the light emitted on passing, electric discharge is examined with a spectroscope,the spectrum obtained is called the emission, spectrum of hydrogen which contain large number of lines which are grouped into different 5, different series,, • Lyman series,, • Balmer series, • Paschen series, • Brackett series, • Pfund series., • Humpry series, The wave numbers of all the lines in all the series can be calculated by the Rydberg equation., Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, ν̅ =, , 1, 1, 1, = RZ2 ( 2 − 2 ), λ, n1 n2, , Where n1 and n2 are whole numbers, n2> n1., For one electron species like He+, Li2+ and Be3+, the value of R is 109677 cm–1× Z2, where Z is the, atomic number of the species., , Different series of spectral lines in hydrogen emission spectrum, Name of the series, Lyman series, Balmer series, Paschen series, Brackett series, Pfund series, , n1, 1, 2, 3, 4, 5, , n2, 2,3,4,5,6,7….., 3,4,5,6,7…, 4,5,6,7……, 5,6,7…., 6,7…., , Spectral region, Ultraviolet, Visible, Near infrared, Infrared, Far infrared, , The wave number for any single electron species like He+, Li2+ and Be3+ can be calculated from the, 1, , 1, , ν̅ = Z 2 R H (n2 − n2), , equation, , 1, , 2, , CLASS EXERCISE, 1. Number of spectral lines possible when an electron falls from fifth orbit to ground state in hydrogen, atom is, a) 4, b) 15, c) 10, d) 21, 2. Which of the following electronic transitions require the largest amount of energy?, a) n = 1 to n = 2, b) n = 2 to n = 3, c) n = 3 to n = 4, d) n = 4 to n = 5, 3. Which of the following spectral line is associated with a minimum wavelength?, a) n = 5 to n = 1, b) n = 4 to n = 1, c) n = 3 to n = 1, d) n = 2 to n = 1, 4. Of the following transitions in hydrogen atom the one which gives an absorption line is lowest, frequency is, a) n = 1 to n= 2, b) n = 3 to n = 5, c) n = 2 to n = 1, d) n = 5 to n = 3, 5. The first emission line of Balmer series in H spectrum has wave number equal to, 9𝑅, 7𝑅, 3𝑅, 5𝑅, a) 400𝐻 𝑐𝑚−1, b) 144𝐻 𝑐𝑚−1, c) 4𝐻 𝑐𝑚−1, d) 36𝐻 𝑐𝑚−1, 0, , 6. If the series limit of wave length of the Lyman series for hydrogen atoms is 912 A .then the series, limit of wave length for the Balmer series of hydrogen atom is, Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, a) 912𝐴𝑜, , b) 2 × 912𝐴°, , c) 4×912A°, , d), , 912, 2, , 𝐴°, , HOMEEXERCISE, 1. There are three energy levels in an atom. How many spectral lines are possible in its emission, spectra?, a) One, b) Two, c) Three, d) Four, 2. The wave length of second line in the Balmer series of hydrogen spectrum is equal to (R=Rydberg, constant), a) 36/5R, b) 5R/36, c) 3R/16, d) 16/3R, 3. When an electron falls from higher orbit to third orbit in hydrogen atom, the spectral time, observed, a) Balmer series b) Lyman series, c) Brackett series, d) Paschen series, 4. Which of the following electronic transitions require the largest amount of energy?, a) n = 1 → n =2, b) n = 2 → n = 3, c) n = 3 → n = 4, d) n = 4 → n = 5, 5. The wave number of the series limiting line for the Lyman series for hydrogen atom is, (R = 109678 cm-1)., a) 82259 cm-1, , Page number, , 1, , b) 109678 cm-1, , AMCF-TEMPLE OF LEARNING, , c) 1.2157 x 10-5 cm, , d) 9.1176 x 10-6 cm, , Contact No. :

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ATOMIC STRUCTURE, , SESSION – 5 AND 6, AIM-To introduce Bohr’s and Sommerfeld’s Atomic models, To overcome the objections of Rutherford model and to explain the hydrogen spectrum,Bohr proposed, a quantum mechanical model., , POSTULATES OF BOHR’S THEORY, •, , The electrons revolve round the nucleus with definite velocity in certain fixed closed circular, paths called orbits (or) shells (or) stationary state. These shells are numbered as 1, 2, 3, 4 or, termed as K, L, M, N from the nucleus., , •, , Each orbit is associated with a definite amount of energy. As long as an electron is revolving in, an orbit it neither loses nor gains energy. Hence these orbits are called stationary states or, stable orbits+, The centrifugal force of the revolving electron in a stationary orbit is balanced by the, electrostatic attraction between the electron and the nucleus., Electron can revolve only in orbits whose angular momentum are an integral multiple of the, factor h/2 π., , •, •, , nh, , mvr = 2π, , Where m = mass of electron,, v = velocity of electron,, r = radius of the orbit and, ‘n’ is the integral number like, 1, 2, 3, 4 . . . , is called principal quantum number and h = Planck’s, constant, •, , The energy of an electron changes only when it moves from one orbit to another. Outer orbits, have higher energies while inner orbits have lower energies., The energy is absorbed when an electron moves from inner orbit to outer orbit. The energy is, emitted when the electron jumps from outer orbit to inner orbit., • The energy emitted or absorbed in a transition is equal to the difference between the energies, of the two orbits (E2 – E1). Energy emitted or absorbed is in the form of quanta., E=E2 – E1 = hv, Here E1 and E2 are the lower and higher allowed energ states., • Expressions for radius of orbit:, Consider an electron of mass ‘m’ and charge ‘e–’ revolving round the nucleus of charge ‘Ze’ in a circular, orbit of radius ‘r’., Let ‘v’ be the tangential velocity of the electron. As per coulomb’s law, the electrostatic force of, attraction between the moving electron andthenucleus is –Ze2/r., , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, , Explanation of Hydrogen Spectrum by Bohr’s Theory:, Bohr’s theory successfully explains the origin of lines in hydrogen emission spectrum. Hydrogen atom, has only one electron. It is present in K shell of the atom (n = 1). When hydrogen gas is subjected to, electric discharge, energy is supplied. The molecules absorb energy and split into atoms. The electrons, in different atoms absorb different amounts of energies. By the absorption of energy the electrons are, excited to different higher energy levels., Atoms in the excited state are unstable. Therefore the electrons jump back into different lower energy, states in one or several steps. In each step the energy is emitted in the form of radiation and is, indicated by a line., Each line has a definite frequency and thus the emission spectrum of hydrogen has many spectral lines., • Lyman series are obtained in UV region, when electron returns to the ground state from higher, energy levels 2, 3, 4, 5 ......... and so on., • Balmer series are obtained in visible region when electron returns to second energy level from, higher energy levels 3, 4, 5, 6 and so on., • Paschen series are obtained in near infrared region, when electron returns to third energy level, from higher energy levels 4, 5, 6.... And so on., • Brackett series are obtained in mid infrared region when electron returns to fourth energy level, from higher energy levels 5, 6, 7 . . . and so on., • Pfund series are obtained in far infrared region when electron returns to the fifth energy level, from higher energy levels 6, 7……., The maximum number of lines produced when electrons jumps from nth level to ground level is equal, to,, , 𝑛(𝑛−1), 2, , Or, , Where,, , ∑(𝑛2 − 𝑛1 ), , n2 = higher energy level., n1 = lower energy level., n = difference in the two energy levels., , Merits and demerits of Bohr’s Atomic model:, 1., , Bohr’s model explains the stability of the atom. The electron revolving in a stationary orbit does, not lose energy and hence it remains in the orbit forever., , 2., , Bohr’s theory successfully explains the atomic spectrum of hydrogen., , 3., , This theory not only explains hydrogen spectrum but also explains the spectra of one electron, 2+, , 3+, , species such as He+, Li and Be etc., 4., , The experimentally determined frequencies of spectral lines are in close agreement with those, calculated by Bohr’s theory., , 5., , The value of Rydberg constant for hydrogen calculated from Bohr’s equation tallies with the value, determined experimentally., , Limitations of Bohr’s model:, 1., , Bohr’s theory fails to explain the spectra of multielectron atoms., , 2., , It could not explain the fine structure of atomic spectrum., , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, 3., , It does not explain the splitting of spectral lines into a group of finer lines under the influence of, magnetic field (Zeeman Effect) and electric field (Stark effect)., , 4., , Bohr’s theory predicts definite orbits for revolving electron. It is against the wave nature of, electron., , 5., , Bohr’s theory is not in agreement with Heisenberg’s uncertainty principle., , Sommerfeld’s Atomic Model:, It is an extension of Bohr’s model. In this model, the electrons in an atom revolve around the nuclei, in elliptical orbit. The circular path is a special case of ellipse. Association of elliptical orbits with, circular orbits explains the fine line spectrum of atoms., Radial Velocity, , •, , Tar, , velocity, , Avg Velocity, , major axis, , •, , focus, Minor axis, , n=4,k=4, n=4,k=3, n=4,k=2, , n=4, k=1, k  0, , •, , Sommerfeld’s orbits in hydrogen atom, The main postulates are:, i) The motion of electron in closed circular orbits is influenced by its own nucleus and is set up into, closed elliptical paths of definite energy levels., ii) The nucleus is one of the foci for all these orbits., iii) The angular momentum of electron in closed elliptical paths is also quantized i.e. k (h/2), where k is, another integer except zero., iv) The ratio, , n, k, , length of major axis, , = length of, , min or axis, , suggests for the possible number of subshells in a shell. Possible, , values of k for n = 4 are 1, 2, 3, 4 respectively. For any given value of n, k cannot be zero as in that, case, the ellipse would degenerate into a straight line passing through the nucleus. When n = k,, path becomes circular., , CLASS EXERCISE, 1., , The ratio of radius of 2nd and 3rd Bohr orbit is, a) 3 : 2, , 2., , b) 9 : 4, , c) 2 : 3, , d) 4 : 9, , According to Bohr’s model, the angular momentum of an electron in 4th orbit is, a) h/3, , Page number, , b)h/2, , 1, , c) 2h/, , AMCF-TEMPLE OF LEARNING, , d) 3h/2, , Contact No. :

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ATOMIC STRUCTURE, 3., , The radius of Bohr’s first orbit in hydrogen atom is 0.053 nm. The radius of second orbit, of He+ would be, a) 0.0265 nm, , 4., , c) 0.116 nm, , d) 0.212 nm, , The minimum energy required to excite a hydrogen atom from its ground state, a) 13.6 eV, , 5., , b) 0.053 nm, b) -13.eV, , c) 3.4 eV, , d) 10.2 eV, , The ratio of kinetic energy and potential energy of an electron in a Bohr orbit of a hydrogen atom, is, a) 1:2, , 6., , b) -1:2, b)-1:2, , c) 1:1, , d) -1:2, , The ratio of kinetic energy and total energy of an electron in a Bohr orbit of a hydrogen atom is, a) 1:-1, , 8., , d) -1:1, , The ratio of potential energy and total energy of an electron in a Bohr orbit of a hydrogen atom is, a) 2:1, , 7., , c) 1:1, , b) -2:1, , c) 1:1, , d) -1:2, , In a certain Bohr orbit the total energy is - 4.9 eV for this orbit, the kinetic energy and potential, energy are respectively., a) 9.8 eV, - 4.9 eVb) 4.9 eV, - 98 eV, , c) 4.9 eV, - 4.9 eV, , d) 9.8 eV, - 9.8 eV, , 9. If speed of electron in first Bohr orbit of hydrogen be ‘x’, then speed of the electron in second orbit, of He+ is:, a) x/2, , b) 2x, , c) x, , d) 4x, , 10. The ratio of the difference in energy between the first and second Bohr orbits to that between the, second and third Bohr orbit is, a) 1/2, , b) 1/3, , c) 4/9, , d) 27/5, , HOME EXERCISE, 1., , Calculate the ratio of the radius of in 3rd energy level of Li+2ion of 2nd energy level of He+ ion, a)3:2, , 2., , b)1:2, , c)2:3, , d)1:1, , Of the following, which of the statement(s) regarding Bohr’s theory is wrong?, a) Kinetic energy of an electron is half of the magnitude of its potential energy, b) Kinetic energy of an electron is negative of total energy of electron, c) Energy of electron decreases with increase in the value of the principal quantum number, d) The ionization energy of H-atom in the first excited state is negative of one fourth of the, energy of an electron in the ground state., , 3. If first ionization energy of hydrogen is E, then the ionization energy of He + would be:, a) E, , b) 2E, , c) 0.5E, , d) 4E, , 4. The ratio of radii of first orbits of H, He+ and Li2 is:, a) 1:2:3, 5., , b) 6:3:2, , c) 1:4:9, , d) 9:4:1, , The angular momentum of an electron in the M shell of hydrogen atom is, , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, b) h/ 2 , , a) 3h /2 , 6., , d) 2h / , , If ionization potential of H-atom is 13.6 eV, the ionization potential of He+ is, a) 54.4 eV, , 7., , c) h / , , b) 6.8eV, , c)13.6eV, , d) 27.2eV, , The ionization energy of H-atom is its ground state is 2.17 x 10-18J. The ionization energy of Li +2, in the ground state will be, a) 1.953 x 10-15 J, , 8., , b) 1.953 x 10-16 J, , d) 1.953 x 10-18 J, , If the value of E = - 78.5 K.cal /mole. The order of the orbit in hydrogen atom is, a) 4, , 9., , c) 1.953 x 10-17J, , b) 3, , c) 2, , d) 1, , The ionization potential of hydrogen atom is 13.6 eV. The energy required to remove an electron, in the n = 2 state of the hydrogen atom is, a) 3.4 eV, , b) 6.8 eV, , c) 13.6 eV, , d) 27.2 eV, , 10. The minimum energy (numerical value) required to be supplied to H-atom to push its electron, from 2nd orbit to the 3rd orbit, a) 1.9 eV b) 2.2 eV, , Page number, , 1, , c) 2.7 eV, , d) 7.0 eV, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, , SESSION- 7, AIM, 1) To introduce de Broglie’s theory, 2) To introduce Heisenberg’s Uncertainty principle., , DUAL NATURE OF MATTER(de-BROGLIE’S WAVE THEORY), Light exhibits different properties such as diffraction, interference, photoelectric effect, compton, effect, reflection and refraction. The phenomenon of diffraction and interference can be explained by, the wave nature of the light. But the phenomenon of photoelectric effect and Compton Effect can be, explained by the particle nature of the light.Thus light has dual nature.DeBroglie proposed that, matter like radiation, should also exhibit dual behaviour., Einstein’s generalization of Planck’s theory is given as, E = hν =, Einstein’s mass energy relationship is E = mc, Equating above two equations, we get, hc, λ, , 2, , = mc, , hc, λ, , 2, , h, , h, , or λ= mc or λ = mc, h, , Where ‘c’ is the velocity of light. If the velocity of micro particle is ‘v’ then, λ = mV, This is de Broglie’s equation,, Where ‘λ’ is the de Broglie’s wave length, ‘m’ is the mass of the moving particle and ‘h’ is Planck’s, constant., h, , P = mv or λ = P ., Here 𝜆 signifies wave nature and P signifies particle nature., This is applicable to microparticles like electron, proton, etc., and not applicable for macrobodies like, cricket ball, bullet etc., The electron moving with high speed possesses both the particle nature and the wave nature. The, waves associated with material particles are known as matter waves or particle waves., , The Heisenberg’s uncertainty principle:, , “It is impossible to determine simultaneously and accurately the exact position and momentum or, velocity of a sub-atomic particle like electron in an atom”., One can determine the position of a particle very accurately, and then the determination of its velocity, becomes less accurate. Similarly, one can determine the velocity of a particle very accurately, and then, the determination of its position becomes less accurate. The certainty in one factor introduces the, uncertainty in another factor., , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, If the uncertainty in the determination of the position of a small particle is given by Δx and uncertainty, in its momentum is Δp, then, ℎ, , (Δx) (Δp) ≥ 𝑛𝜋, Where n = 1,2,3,4........., For an electron revolving around the nucleus in an atom the value of n is nearly 4., Thus Heisenberg’s principle can also be stated as the product of uncertainty in position and, momentum of an electron like micro particle moving with high speed cannot be less than h/4., Heisenberg’s equation can also be written as,, ℎ, , (Δx) (Δv) ≥ 4𝜋𝑚, Where m is the mass of the particle and Δv is uncertainty in velocity., If the position of the particle is known exactly (Δx = 0), Δv becomes infinity (∞) and vice versa., Heisenberg's uncertainty principle is not applicable to those objects which cannot change their position, by themselves when a light falls on them. It is applicable for micro particles like electrons., , Significance of Heisenberg’s uncertainty principle:, Like de Broglie equation, although Heisenberg’s uncertainty principle holds good for all objects but it, is significance only for microscopic particles. The reason for this is quite obvious. The energy of the, photon is insufficient to change the position and velocity of bigger bodies when it collides with them., For example, the light from a torch falling on a running rat in a dark room, neither change the speed of, the rat nor its direction, i.e., position., This may be further illustrated with the following examples:, For a particle of mass 1 mg, we have, ℎ, , Δx.Δ𝜐 = 4𝜋𝑚 =, , 6.625×10−34 𝑘𝑔𝑚2 𝑠 −1, 4×3.1416×(10−6 𝑘𝑔), , = 10−28 𝑚2 𝑠 −1, , Thus, the product of Δx and Δ𝜐 is extremely small. For particles of mass greater than 1 mg, the product, will still smaller. Hence, these values are negligible., For a microscopic particle like an electron, we have, ℎ, , 6.625×10−34 𝑘𝑔𝑚2 𝑠 −1, , Δx.Δ𝜐 = 4𝜋𝑚 = 4×3.1416×(9×10−31 𝑘𝑔) ≈ 10−4 𝑚2 𝑠 −1, , CLASS EXERCISE, 1. A ball of 100 g mass is thrown with a velocity of 100 ms–1. The wavelength of the de Broglie wave, associated with the ball is about, a) 6.63 × 10–35 m, b) 6.63 × 10–30 m, c) 6.63 × 10–35 cm d) 6.63 × 10–33 m, 2. If kinetic energy of a proton is increased nine times the wavelength of the de-Broglie wave, associated with it would become, a) 3 times, b) 9 times, c) 1/3 times, d) 1/9 times, 3. Number of waves made by a Bohr electron in one complete revolution in the 3rd orbit, a) 1, b) 2, c) 3, d) 4, 4. The uncertainty in position and velocity of a particle are 10 -10m and 5.27 x 10-24ms-1 respectively., Calculate the mass of mass of the particle.(h=6.625 10-34 J-s), 5. Calculate the uncertainty in velocity a cricket ball of mass 150g. if the uncertainty in its positionis, 0, , the order of 1 A (h=6.6 x 10-34kg m2 s-1), , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, 6. In an atom, an electron is moving with a speed of 600 m sec-1 with an accuracy of 0.005%, certainty with the position of the electron can be located is:(h=6.6 x 10 -34kg m2 s-1, mass of, electron=9.1 x 10-31kg), a) 1.52 x 10-4 m, b) 5.1 x 10-3 m, c) 1.92 x 10-3 m, d) 3.84 x 10-3, , HOME EXERCISE, 1. The de Broglie wavelength of 1 mg grain of sand blown by a 20ms-1 wind is:, a) 3.3 x 10-29, b) 3.3 x 10-21 m, c) 3.3  10-49 M, d) 3.3  10-42 m, 2. If the kinetic energy of an electron is increased 4 times, the wavelength of the Broglie wave, associated with it would become:, 1, 1, a) 4times, b) 2times, c) times, d) times, 2, 4, 3. The momentum of the particle having the wave length of 1Å is, a) 6.6 x 10-19 gram cm/sec, b) 6.6 x 1019 gram cm/sec, c) 6.6 x 1034 gram cm/sec, , d) 6.6 x 10-34 gram cm/sec, 4. If the uncertainty in the position of an electron is 10-8cm, the uncertainty in its velocity is, a) 3×108 cm/sec, b) 5.8×107 cm/sec, c) 6.625×109 cm/sec, , d) 7.35 × 10-8 cm/sec, 5. The uncertainty in momentum of an electron is 1 x 10 -5 kg-m/s. The uncertainty in its position will, be (h = 6.6 x 10-34 Joule-sec), a) 1.05 x 10-28m, , b) 1.05 x 10-26 m, , c) 5.27 x 10-30 m, d) 5.25 x 10-28 m, 6. The uncertainty in the momentum of an electron is 10-5kg.m/sec. The uncertainty in its position will, be, a) 1.05 x 10-28 m, , Page number, , 1, , b) 1.05 x 10-26m, , AMCF-TEMPLE OF LEARNING, , c) 5.27 x 10-30m, , Contact No. :, , d) 5.25 x 10-25m

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ATOMIC STRUCTURE, , SESSION –8 AND 9, AIM, 1) To introduce wave mechanical model of atom., 2) To introduce Quantum numbers., 3) To understand shapes of orbitals and Probability distribution., , THEORY, Classical mechanics, based on Newton’s laws of motion, was successful in explaining the motion of, macroscopic bodies like falling stones or motion of planets around the sun etc. But it failed when, applied to microscopic particles like electrons, atoms, molecules etc.Hence new branch introduced, called as ‘Quantum mechanics’., , Schrodinger Wave Equation:, Quantum mechanics, as developed by Erwin Schrodinger is based on the wave motionassociated with, the particles. The Schrodinger differential wave equation is given by, ∂2 ψ, ∂x2, , +, , ∂2 ψ, ∂z2, , +, , ∂2 ψ, ∂y2, , +, , 8π2 m, h2, , (E − V)ψ, , Here x, y, z are Cartesian coordinates of the electron, m = mass of electron, h = Planck’s constant, E = total energy of the electron (KE + PE), V = potential energy of the electron (PE), ψ= wave function of the electron., Significance of ψ:ψ is the wave function. It gives the amplitude of the electron wave., , The intensity of light is proportional to the square of amplitude (ψ2).Just as 𝛙2 indicates the densityof, photons in space, 𝛙2 in case of electron wave denotes the probability of finding an electron in the, space or probability of finding the electron is also maximum., , Quantum numbers:, The behaviour of an electron in an atom is described mathematically by a wave function or orbital.They, are principal quantum number, azimuthal quantum number,magnetic quantum number and spin, quantum number., ‘Set of numbers used to describe energy,size,shape of orbitals in an atom’ called as quantum, numbers., 1.Principal quantum number(n):, • ‘n’ can be any whole number value such as 1,2,3,4, etc.The energy shells corresponding to, these numbers are K, L, M, N, etc., • Principal Quantum no. indicates the main energy level to which the electron belongs. It also, indicates the average distance of an electron from nucleus and also the speed of the atomic, electron., , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, •, •, , As the ‘n’ value increases the distance of electron from the nucleus increases and its energy also, increases., The maximum no. of electrons that can be present in an orbit is given by 2𝑛2 . The maximum no., of electron in K, L, M, and N shells are 2,8,18 and 32 respectively., , •, •, , The radius of the orbit is given by the expression: rn = 𝑍 Ao., The energy of the electron/orbit is given by the expression., , 0.529×𝑛2, , En=, , −13.6×𝑍 2, 𝑛, , cm/sec, 2.18×108 ×𝑍, , •, , The velocity of the electron is given by the expression. Vn=, , 2., , Azimuthal Quantum Number:, Azimuthal Quantum number was introduced by Sommerfeld’s to explain the fine spectrum., It is also called as secondary quantum no. or orbital angular momentum quantum number or, subsidiary quantum number., It is denoted by l., ‘l’ can have the values from 0 to (n-1), a total of ‘n’ values. ‘l’ values 0,1,2,3 indicates s,p,d,f., s,p,d and f are spectroscope terms which indicates sharp. Principle, diffuse and fundamental, respectively., Azimuthal Quantum number indicates the sub-shell to which the electron belongs. It also, determines the shapes of the orbital in which the electron is present., Each main energy shell can have ‘n’ number of sub-shells., , •, •, •, •, •, •, , n, , 𝑛, , cm /sec., , l, , 1, , 0 (1s), , 2, , 0 (2s), 1 (2p), ℎ, , •, , The orbital angular momentum, (L) of 0an(3s),, electron, is 2(3d), given by the expression:, 3, 1 (3p),, , 3., , 4, Magnetic Quantum number:, , •, , Magnetic quantum number was introduced by Lande to explain Zeeman Effect., , •, , It is denoted by m or ml., , •, , This quantum number refers to different orientations of electron could in a particular subshell., These orientations are called the orbitals., , •, , An electron due to its orbital motion around the nucleus generates an electric .This electric field, in turn produces a magnetic field which can interact with the external magnetic field. Thus,, under the influence of the external magnetic field, the electrons of a subshell can orient, themselves in certain preferred regions of space around the nucleus called orbitals. The, magnetic quantum number determines the number of preferred orientations of the electron, present in a subshell., Since each orientation corresponds to an orbital, therefore, the magnetic quantum number, determines the number of orbitals present in any subshell., , Page number, , 1, , L = √𝑙(𝑙 + 1) 2𝜋, , 0 (4s), 1(4p), 2(4d), 3(4f), , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, •, , ‘m’ can have values from – 𝑙 to +𝑙 including zero, a total (2 𝑙+1) values., Subshell, , 𝒍, , m values, , No. of orientations, (Orbitals), , s, p, , 0, 1, , 0, -1, 0, +1, , 1, 3, , d, , 2, , -2, -1, 0, +1, +2, , 5, , F, , 3, , -3, -2, -1, 0, +1, +2, +3, , 7, , •, , When l = 0, m has only one value, m = 0. The sub-level‘s’ has one orbital called s orbital., , •, , When l =1, m can have 3 values m = –1, 0, +1. The sub-level ‘p’ has three space orientations or, three orbitals. The three orbitals are designated as px, py and pz., , •, , When l = 2, m can have 5 values m = –2,–1, 0, +1, +2. The sub-level ‘d’ has five space, orientations or five orbitals. The five orbitals are designated as dxy, dyz, dzx, dx2−y2 and dz2 ., , •, , When l = 3, m can have 7 values m = –3,–2,–1,0,+1,+2,+3. The sub-level ‘f’ has seven space, orientations or seven orbitals., The magnetic quantum number gives orientation of orbitals in space. All the orbitals present in, a sublevel have same energy and shape. They are called ‘degenerate orbitals’, which differ in, their spatial orientation., , •, , Each value of ‘m’ constitutes an orbital in the sublevel., , •, , Maximum no. of electrons in subshell : 2(2𝑙+1) or (4 𝑙+2)., , 4., , Spin Quantum Number:, , •, , Spin Quantum number was proposed by Uhlenbeck and Goudsmith., , •, , It is denoted by ‘s’ or ‘ms’., , •, , It indicates the direction of spinning of electron present in any orbital., , •, , Since the electron in an orbital can spin either in the clockwise direction or in anti-clockwise, direction, hence for a given value of m, s can have only two values, i.e., +1/2 and -1/2 or these, are very often represented by two arrows pointing in the opposite direction, i.e.,↑and ↓., If an orbital contains two electrons, the two magnetic moments oppose and cancel each other., Thus, in an atom, if all the orbitals are fully filled, net magnetic moment is zero and the, substance is diamagnetic (i.e., repelled by the external magnetic field). However, if some halffilled orbitals are present, the substance has a net magnetic moment and is paramagnetic (i.e.,, attracted by the external magnetic field)., The spin angular momentum (𝜇 s) of an electron is given by, h, μs = √s(s + 1) 2π, , •, , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, , Atomic Orbital:, The three dimensional space around the nucleus where the probability of finding the electron is, maximum is called an atomic orbital., Differences between orbit and orbital:, Orbit, 1. n orbit is a well-defined circular path, around the nucleus in which the, electronrevolves., , Orbital, 1. 1.AnAn orbital isisthethe, region, region, of space, of around, space the, aroundthe around the nucleus where, theprobability of finding the electron is, maximum (95%), 2. An orbit represents the movement of 2. An orbital represents the movement of, electron in one plane., electron in three dimensional spaces., 3. An orbit means the position as well as, thevelocity of the electron can be known, with Certainty., 4. Orbits are circular or elliptical shaped., , 3. In an orbital it is not possible to find, theposition as well as velocity of the, electroncan be known with certainty., 4. They have different shapes like, spherical,dumbbell etc, , 5. Orbits do not have directional, characteristics., 6. An orbit can have a maximum number, of2n2 electrons., , Orbitals h, , 5. Except ‘s’ orbitals, all other orbitals have, directional characteristics, 6. An orbital can accommodate a, maximumof only two electrons., , Node- The three dimensional space around the nucleus where the probability of finding the electron, is minimum or zero., y, , z, Nucleus, , node, x, (2s), (1s), , Types of Nodes:, Nodes are of two types: a) Radial Node b) Angular Node, A radial node is the spherical region around then nucleus, where the probability if finding the electron, is zero (Ψ2 = 0)., Similarly,nodal plane(angular plane) have zero probability of finding electron., Calculation of no. of nodes:, No. of Radial nodes = n−𝑙 − 1, No. of angular nodes = 𝑙, Total no. of nodes = n-1, Ex: In a 3p -orbital, No. of Radial nodes = 3-1-1 = 1, No. of angular nodes = 1, Total no. of nodes = 2., Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, , Shapes of Orbitals:, •, , s –Orbitals: s- Orbital can accommodate electrons with l = 0 and these orbitals are present in, every orbit starting from 1st orbit., , Orbital in which e-s with n=1 , l = 0 are present is called 1s - orbital., All s-orbitals are spherical in shape and the size of sphere increases with ‘n’ value. s - Orbitals are, spherically symmetrical because the probability of finding the electron around the nucleus is same in, all directions., • p – Orbitals:, p- Sublevel begins from 2nd orbit. For p - sublevel l = 1, indicates that each p - sub level contains three, orbitals with ‘m’ values –1, 0, +1. These are designated as px, py and pz, depending on the axis in which, electron density is present., , In px-orbital, electron density is concentrated along the x-axis. p-Orbitals have dumb-bell shape. Each p, -orbital has two lobes separated by one nodal plane. The probability density function is zero on the, plane where the two lobes touch each other. The nodal planes for px, py and pz - orbitals are YZ, ZX, and XY - planes respectively., The three orbitals present in a given p - sublevel will have same shape, size and energy but different, orientations (differ in m value). These three orbitals are perpendicular to each other and the angle, between any two p - orbitals is 90o., •, , d - Orbitals:begins from 3rd orbit (n = 3). For d- sub level l= 2, indicates that each d - sublevel, contains five orbitals with ‘m’ values –2, –1, 0, +1, +2. These are designated as dxy,dyz,dzx,, 𝑑𝑥 2 −𝑦 2 and d𝑧 2 ., , All the d-orbitals (except d𝑧 2 ) have double dumb-bell shape. Each d-orbital has four lobes separated by, two nodal planes., In case of dxy, dyz and dzxorbitals, lobes are present in between the corresponding axes. i.e.,between x, and yaxis in case of dxy orbital. Whereasin d𝑥 2 −𝑦 2 and d𝑧 2 orbitals lobes are present along the axes. dxy, Orbital contains yz and zx as nodal planes. dyz and dzx contain (xy,zx) and (xy,yz) planes respectively., d 2 2 orbitalcontains two nodal planes perpendicular to each other and which make an angle of 45 o, 𝑥 −𝑦, , with respect to x and y axes. 𝑑𝑧 2 orbital does not contain nodal planes., 5 dorbitals present in a given d- sublevel will have same energy in the ground state., , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, , CLASS EXERCISE, 1. If the above radial probability curve indicates ‘2s’ orbital, the distance between the peak points X.Y, is:, , a) 2.07Å, b) 1.59Å, c) 0.53Å, d) 1.1Å, 2. The wave function curve which crosses ‘x’ axis maximum number of times in the graph drawn, between distance from nucleus r(on x axis) and radial wave function R(ψ r)., a)4d, b)4p, c)4s, d)4f, 3. The number of nodal planes is greatest for the orbital:, a) 4s, b) 2p, c) 3d, 4., , The radial distribution curve of the orbital with double dumbbell shape in the 4 th principle shell, consists of ‘n’ nodes, n is, a) 2, , 5., , d) 2s, , b) 0, , c)1, , d) 3, , Which one of the following sets of quantum numbers represents as impossible arrangements?, n, , 1, , m, , s, , a) 3, , 2, , –2, , ½, , b) 4, , 0, , 0, , ½, , c) 3, , 2, , –3, , ½, , d) 5, , 3, , 0, , –1/2., , 6., , Correct set of four quantum numbers for the valence (outermost) electron of rubidium (Z = 37) is, 1, 1, 1, 1, a) 5, 0, 0, +, b) 5, 1, 0, +, c) 5, 1, 1, +, d) 6, 0, 0, +, 2, 2, 2, 2, 7. The maximum number of electrons in an orbital having same spin quantum number will be:, a) l + 2, b) 2l + 1, c) l(l + 1), d) l (l + 1), 1, 8. The four quantum number of last electron of an atom are 4, 0, 0, +, then atomic number of that, 2, element could be, a) 19, b) 55, c) 36, d) 37, 9. The number of atomic orbitals with quantum numbers n = 3, l = 1, m = 0, a) 1, b) 6, c) 3, d) 5, 10. The number of electrons that can have n = 4 andl = 3 is, a) 10, b) 14, c) 6, d) 5, , HOME EXERCISE, 1. Which of the following can be negative?, a) 4πr2ψ2, b) 4πr2ψ2dr, c) ψ, d) ψ2, 2. The quantum number not obtained from the Schrodinger’s wave equation is, a) n, b) l, c) m, d) s, 3., , Maxima’s in Radial probability distribution curve of 2s is, a) One, , Page number, , b) Two, 1, , AMCF-TEMPLE OF LEARNING, , c) Three, , d) Four, Contact No. :

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ATOMIC STRUCTURE, 4., , 5., , 6., , In which of following case would the probability of finding an electron in dxyorbital be zero?, a) Xy and yz plane, , b) xy and planes, , c) xz and yz planes, , d) z-direction, yz and xz planes, , The principal quantum number of an atom is related to the, a) Size of the orbital, , b) spin angular momentum, , c) Orbital angular momentum, , d) orientation of the orbital in space, , The orbital angular momentum of an electron in 2s orbital is:, a), , 7., , 1 h, ., 2 2, , b) zero, , c), , h, 2, , d), , 2., , h, 2, , What will be all 4-Sets of Quantum Number for last electron of sodium?, a) n = 3, , 1=0, , m=0, , s = +1/2, , b) n = 3, , 1=1, , m=1, , s = +1/2, , c) n = 2, , 1=0, , m=0, , s = +1/2, , d) n = 2, 1=1, m=1, s = +1/2, 8. p-orbitals of an atom in presence of magnetic field are:, a) Three fold degenerate b) Two fold degenerate c) Non-degenerated) none of these, 9. The quantum number that is no way related to an orbital, a) principal, b) azimuthal, c) magnetic, d) spin, 10. Which one of the following set of quantum number is not possible for a 4pelectron?, 1, 1, a) n = 4 l = 1, m = +1 s = +, b) n = 4 l = 1 m = 0 s = +, 2, 2, 1, 1, c) n= 4, l = 0, m = 2, s = +, d) n = 4, l = 1, m = 1, ms = 2, 2, , SESSION – 10, Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, , AIM- To introduce Electronic Configuration, ENERGY OF ORBITALS, The energy of an electron in a hydrogen atom is determined only by the principal quantum number., Within a shell, all hydrogen orbitals havethe same energy, independent of the other quantum numbers., 1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f, Although the shapes of 2s and 2p orbitals are different, an electron has the same energy when it is in 2s, orbital or 2p orbital. The energy of an electron in a multielectron atom depends, not only on its, principal quantum number, but also on its azimuthal quantum number. The s, p, d and f orbitals within, a given shell have slightly different energies in a multi electron atom., , Electronic configuration of multi electron atoms:, The distribution and arrangement of electrons in the main shells, subshells and orbitals of an atom is, called electronic configuration of the element., • Aufbau Principle:, “In the ground state of the atoms, the orbitals are filled in order of their increasing energies”., In other words electrons first occupy the lowest energy orbital available to them and enter into higher, energy orbitals only after the lower energy orbitals are filled., The relative energy of an orbital is given by, (n +l )rule. As(n+l) value increases, the energy of orbital increases., •, •, , The orbital with the lowest (n + l) value is filled first., When two or more orbitals have the same (n +l) value, the one with the lowest ‘n’ value (or), highest ‘l ’ value is preferred in filling., Exp-Consider two orbitals 3d and 4s., n+l value of 3d = 3 + 2 = 5 and of 4s = 4 + 0 = 4. Since 4s has lowest(n +l) value, it is filled first before, filling taking place in 3d., Consider the orbitals 3d, 4p and 5s, The (n + l) value of 3d = 3 + 2 = 5, The (n +l) value of 4p = 4 + 1 = 5, The (n +l) value of 5s = 5 + 0 = 5, These three values are same. Since the ‘n’ value is lower to 3d orbitals, the electrons prefer to enter in, 3d, then 4p and 5s., The order of increasing energy of atomic orbitals is:, 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d <, 6p < 7s and so on., The sequence in which the electrons occupy various orbitals can, be easily remembered with the help of Moeller’s diagram as, shown in Fig, , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, 2s, , 2s, , 2p, , a), , 2p, , 2s, , 2s, , 2p, , 2p, , b), , c), d), 5. Electronic configuration of Ni is [Ar] 3d 8 , 4s 2 . The electronic configuration of next, element is:, a) [Ar] 3d 10 , 4s 1, b) [Ar] 3d 9 , 4s 2, c) [Ar] 3d 8 , 4s 2 , 4p 1, , d) none of these, , ATOMIC STRUCTURE EXCERSISE PROBLEMS, LEVEL – 1, 1. According to Moseley the relationship between the frequency (v) of lines in X-rays spectra and, atomic number (Z) of an element is given by the expression, a) v = az - b, b) z = a (v - b), c) z = a (v - b), d) v = a (z - b), 2. The increasing order of specific charge of electron (e) proton (P), alpha particle (  ) and neutron, (n) is, a) e, P, n, , b) n, P, e, , c) n,  , P, e, d) n, P,  , e, 3. The experimental evidence for the existence of atomic nucleus comes from, a) Millikan’s oil drop method, b) Atomic absorption spectroscopy, c) The magnetic bending of cathode rays, d) Alpha scattering by a thin metal foil., 4. The pair having identical value of e/m:, a) A proton and a neutron, b) A proton and a deuterium ion, c) An  -particle and a deuterium ion, d) an electron and  -rays, 5., , The radius of an atomic nucleus is of the order of, a) 10-10m b) 10-13m, , 6., , 77, 33, , As, , c), , 77, 34, , Se, , d), , 78, 34, , Se, , When alpha particles are sent through a thin metal foil, most of them go straight through the foil, because:, a) alpha particles are positively charged, c) alpha particle move with high velocity, , 8., , d) 10-8 m, , 76, An isotope of is: 32, Ge, 76, a) 32, Geb), , 7., , c) 10-15m, , b)most part of the atom is empty space, d) none of these, , Rutherford’s experiment on scattering of α-particles showed for the first time that the atom has, , a) electrons, b) protons, c) nucleus, d) neutrons, 9. All type of electromagnetic radiation possess same, a) Wave length, b) frequency, c) energy, d) Velocity when they passed through vaccum, 10. The radiation having maximum wave length is, a) ultra violet rays, b) radio waves, c) X-rays, d) Infrared rays, 11. The wavelength of two different radiations is 3000Å and 5000Å. The ratio of their frequency., a) 3 : 5, b) 5 : 3, c) 9 : 25, d) 25 : 9, 12. Which of the following relates to photon both as wave motion and as a stream of particles?, a) Interference, b) E = mc2, c) Diffraction, d) E = hv, 13. The source of ultra violet radiation is, Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, a) hydrogen vapour lamp, b) tungsten filament lamp, c) radioactive decay, d) rare earth oxides are heated at high temperature, 14. What is the wave number for light with a wavelength of 4000 Å?, a) 2.5 x 104 cm, b) 5 x 1016 cm, c) 7.5 x 104 cm, d) 8.5 x 104 cm, 15. The frequency of a wave of light is 1.0 x 106 sec-1. The wave length for this wave is, a) 3 x 104 cm, b) 3 x 10-4 cm, c) 6 x 104 cm, d) 6 x106 cm, , 16. If λv,λx and m represent the wavelength of visible light, X-ray and microwave respectively, then:, a) m >λv>  x, b) m  x  v, c) v  m  x d) v  x  m, 17., Which of the following does not characterize X-rays?, a) The radiation can ionize gases, b) It causes ZnS to show fluorescence, c) Defected by electric or magnetic field, d) Have wavelength shorter than the UV rays, 18. Transition of an electron from n = 3 level to n = 1 level results in, a) X -ray spectrum b) Emission spectrum c) Band spectrumd) IR spectrum, 19. Which of the following transition corresponds to 3rd line in visible region?, a) n = 3 to n = 2, b) n = 4 to n = 2, c) n = 5 to n = 2, d) n = 6 to n =3, 20. What will be the number of total no. of spectral lines obtained when a electron of H-atom jumps, from its fifth excited state to its ground state?, a) 10, , b) 15, , c) 21, , d) 6, , 21. The spectrum of He is expected to be similar to that of, a) H, , c) He+, , b) Na, , d) Li+, , 22. The wavelength of a spectral line for an electronic transition is inversely related to, a) The number of electrons undergoing the transition, b) The nuclear charge of the atom, c) The difference of the energy of the energy levels involved in the transition, d) The velocity of the electron undergoing the transition., 23. If the wave number of the first line in the Balmer series of hydrogen atom is 15000 cm -1, the wave, number of the first line of the Balmer series of Li2+ is, a) 1.43×104 cm-1, b) 1.66×109 cm-1 c) 3.5×105 cm-1, d) 1.30×5105 cm-1, 24. If the wavelength of the first line of the Balmer series of hydrogen atom is 656. 1 nm, the, wavelength of the second line of this series would be, a) 218.7 nm, b) 328.0 nm, c) 486.0 nm, d) 640.0nm, 25. The transition from the state n=4 to n=3 in a H-; like atom results in UV radiation. The infra-red, radiation will be obtained in the transition:, a) 2→ 1, b) 3→ 2, c) 4→ 2, d) 5 → 4, 26. The number of spectral lines produced according to Bohr’s concept when one electron jumps from, 5th to 2nd shell are:, a) 6, b) 8, c) 10, d) 12, 27. The first emission line in the atomic spectrum of hydrogen in the Balmer Series appears at, 9𝑅, , 7𝑅, , 3𝑅, , 5𝑅, , a) 400𝐻 𝑐𝑚−1b) 144𝐻 𝑐𝑚−1, c) 4𝐻 𝑐𝑚−1, d) 36𝐻 𝑐𝑚−1, 28. The energy level which allows the hydrogen to absorb photon but not to emit it, a) 2s, b) 3d, c) 2p, d) 1s, , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, 29. The difference in angular momentum associated with the electron in two successive orbits of, hydrogen atom is, a) h / 2 , b) h/ , c) h/2 , d) (n -1 ) h/2 , 30. According to Bohr’s model, angular momentum of an electron in the 3 rd orbit is, a), , 3ℎ, , 1.5ℎ, , b), , 𝜋, , 𝜋, , 3𝜋, , 9ℎ, , c) ℎ, , d) 𝜋, , 31. The energy of an electron in the first Bohr orbit of hydrogen atom is __ 21.7610-19 J. The energy, required to remove an electron from the third Bohr orbit of He + ion will be:, a) - 9.671×10-19J, , b) +9.671×10-19J, , c) -2.42×10-19J, , d) +2.42×10-19J, , 32. The velocity of an electron in the first Bohr orbit of hydrogen atom is 2.19 x 10 6 ms-1. Its velocity in, the second orbit would be, a) 1.10 x 106 ms-1, b) 4.38 x 106 ms-1 c) 5.5 x 105 ms-1, d) 8.76 x 106 ms-1, 33. If the speed of electron in the Bohr’s first orbit of hydrogen atom is x, the speed of the electron in, the third Bohr’s orbit is, a) x /9, b) x /3, c) 3x, d) 9x, 34. Bohr model can explain, a) The spectrum of hydrogen atom only, b) Spectrum of an atom or ion containing one electron only, c) The spectrum of hydrogen molecule, , d) The solar spectrum, 35. If an electron is moving with a velocity of 1.1 x 106 ms-1, de Broglie wave length approximately is, a) 4.1 x 10-11 m, , b) 4.9 x 10-9 m, , c) 7 x 10-10 m, , d) 3.306 x 10-8 m, , 36. The wavelength of an electron moving with a velocity of 4.4 x 10 3 cm/sec is, a) 1.65 Å, , b) 1.65nm, , c) 1.65 cm, , d) 1.67 cm-1, , 37. If the Planck’s constant h = 6.6 x 10-34 Joule -sec, the deBroglie wavelength of a particle having, momentum 3.3 x 10-24 Kg-m/sec will be, a) 0.02Å, b) 9.05Å, c) 2Å, d) 500Å, 38. If a particle of one gram is moving with a velocity100 m/sec. its wave length is, a) 6.625 x 10-29 m b) 6.625 x 10-33 m, c) 6.625 x 10-27 m, d) 6.625 x 10-30 m, 39. If wavelength of photon is 2.2 x 10-11 m, h = 6.6 x 10-34J-sec. Then momentum of a photon is, a) 3 x 10-23kg-m/s, b) 1.452 x 10-44kg-m/s, c) 3.33 x 10-22 kg-m/s, d) 6.89 x 10-43 kg-m/s, 40. Assuming the velocity to be same, which subatomic particle possesses smallest de Broglie, wavelength:, a) An electron, b) A proton, c) An  -particle, d) All have same wavelength, 41. If uncertainty of position of electron is zero, the uncertainty in its momentum would be:, a)Zero, b) h/2 , c) h/4 , d) Infinity, 42. The incorrect set of quantum numbers is, a) n = 3, l = 0, m = 0, b) n = 2, l = 1, m = + 1, c) n = 2, l = 1, m = _ 1, d) n = 2, l = 0, m = + 1, 43. For a d-electron, the orbital angular momentum is, a) 6 (h/2  ), b) 2 (h/2  ), c) (h / 2  ), d) 2 (h /2  ), 44. The four quantum numbers of the valence electron of potassium atom are, Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, , 1, 1, 1, 1, b) 4, 1, 0,, c) 4, 0, 0,, d) 4, 1, 1,, 2, 2, 2, 2, 45. The orbital notation of electron having the quantum numbers n = 4, l = 1, m = 0, s = is, a) 4dxy, b) 4d, c) 4s, d) 4pz, a) 4.0, 1., , 46. Maximum number of orbitals presentin n = 4 energy level of an atom, a) 16, b) 25, c) 8, d) 4, 47. The maximum number of electrons present in n = 5 energy level is, a) 50, b) 18, c) 56, d) 32, 48. Which orbital has two angular nodal planes?, a) s, b) p, c) d, d) f, , 49. Degenerate orbitals means:, a) Orbitals having same energy, b) Orbitals having same wave function, c) Orbitals having different energy but different wave function, d) Orbitals having different energy and same wave function, 50. Which of the following sets of the four quantum numbers, n, l, m and describes one of the, outermost electrons in a ground state magnesium atom?, a) 3, 1, 1,, , 1, 2, , b) 3, 0, 0, -, , 1, 2, , c) 3, 0, 1,, , 1, 2, , d) 3, 1, 0, ., , 1, 2, , 51. Any p-orbital can accommodate up to, a) four electrons, , b) six electrons, , c) two electrons with parallel spins, d) two electrons with opposite spins, 52. What is the correct orbital designation for the electron with the quantum numbers?, n = 5, l = 1, m = – 1, s = 1/2, a) 3s, b) 4f, c) 5p, d) 6s, 53. What is the correct orbital designation for the electron with the quantum numbers,, n = 4, l = 3,, m = – 2, s = 1/2, a) 3s, b) 4f, c) 5p, d) 6s, 54. Non-directional orbital is:, a) 3s, b) 4f, c) 4d, d) 4p, 55. Which of the following has the maximum number of unpaired electrons?, a) Mn+2, b) Ti+3, c) V+3, d) Fe+2, 56. Isoelectronic structures among the following structures are:, I) CH3+, II)H3O+, III) NH3,, IV) CH3−, a) I and III, b) III & IV, c) I &II, d) II, III&IV, 57. Which element is represented by the, , Following electronic configuration:, a)Nitrogen, b) Oxygen, , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , c) Fluorine, , d) None of these, , Contact No. :

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ATOMIC STRUCTURE, LEVEL –II, 1. The ratio between the neutrons present in carbon and silicon with respect to mass number of 12, and 28 respectively is:, a) 3:7, b) 7:3, c) 3:4, d) 6:2, 2. In photo electric effect, the photo current, a) Increase with increase of frequency of incident photon, b) Decreases with increase of frequency of incident photon, c) Does not depend on the frequency of photon but depends only on the intensity of incident light, d) Depends both on intensity and frequency of the incident photon, 3. Choose the incorrect statement, a) Every object emits radiation whose predominant frequency depends on temperature, b) Quantum energy of a wave is proportional to its frequency., c) Photons are quanta of light, d) The value of Plank’s constant is energy dependent., 4. Einstein’s PhotoElectric states that: Ek = ℎ𝜐 − 𝑤, Here, Ek refers to, a) Kinetic energy of all ejected electrons, b) Mean kinetic energy of emitted electrons, c) Minimum kinetic energy of emitted electrons, d) Maximum kinetic energy of emitted electrons, 5. The line spectra of two elements are not identical because, a) The elements do not have the same number of neutrons, b) They have different mass numbers, c) Their outermost electrons are at different energy levels, d) All of the above, 6. The maximum kinetic energy of the photoelectrons is found to be 6.63×10−19J. When the metal is, irradiated with a radiation of frequency 2×1015 Hz, the threshold frequency of the metal is about, a) 2×1015 𝑠 −1, b) 1×1015 𝑠 −1, c) 2.5×1015 𝑠 −1 d) 4×1015 𝑠 −1, 7. An electron jumps from an outer orbit to an inner orbit with an energy difference of 3.0 eV. What, will be the wavelength of the line emitted?, a) 3660 Ao, b) 3620 Ao, c)4140 Ao, d)4560 Ao, 8. What is the minimum energy that photon must possess in order to produce photoelectric effect, with platinum metal? The threshold frequency for platinum is 1.3×1015 𝑠 −1, a) 3.6×10−13 𝑒𝑟𝑔b) 8.2×10−12 𝑒𝑟𝑔 c) 8.2×10−14 𝑒𝑟𝑔 d) 8.6×10−12 𝑒𝑟𝑔, 9. The work function for a metal is 4 eV. To emit a photo-electron of zero velocity from the surface of, the metal, the wavelength of incident light should be:, a) 2700Å, b) 1700Å c) 5900Å, d)3100Å, 10. Among the first lines of Lyman, Balmer, Paschen and Brackett series in hydrogen spectra which has, higher energy, a) Lyman, b) Balmer, c) Paschen, d) Brackett, 11. The𝜆for 𝐻𝛼 line of Balmer series is 6500Å. Thus𝜆for 𝐻𝛽 line of Balmer series is, a) 4814Å, b) 4914Å, c) 5814Å, d) 4714Å, 12. The wavelength of a spectral line for an electronic transition is inversely related to:, a) No. of electrons undergoing transition, b)The nuclear charge of the atom, c) The velocity of an electron undergoing transition, d) The difference in the electron levels involved in the transition, 13. The longest𝜆 for the Lyman series is (Given RH=109678cm-1), a) 1215, b) 1315, c) 1415, d) 1515, Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, 14. When greater number of excited hydrogen atoms reach the ground state:, a) More number of lines in is found in Lyman series., b) The intensity of lines in Lyman series increases., c) Both the intensity of and number of lines in Lyman series increases., d) Two electrons A and B in an atom have the following set of quantum numbers., 15. The transition from the state n=4 to n=3 in a He+ ion results in ultraviolet radiation. Infrared, radiation will be obtained in the transition fro, a) n=2 → n=1, b) n=3 → n = 2, c) n= 4 → n = 2 d) n = 5 → n =4, 16. Which element has hydrogen like spectrum whose lines have Wavelength one fourth of atomic, hydrogen?, a) He+, b) Li2+, c) Be3+, d) B4+, 17. The wavelength of the electromagnetic radiation K  line emitted by a hydrogen atom like, Ion is 0.32Å.Find the wavelength of K  line emitted by the same element, a) 0.32Å, b) 0.27Å, c) 0.12Å, d) 0.10Å, 18. Electrons with energy 12 eV are shot into a gas of hydrogen atom in the ground state. How, many different wavelength will be emitted out from this gas sample in Blamer series?, a) One, b) Two, c) Three, d) Zero, 19. Of the following transition in hydrogen atom the one which given an absorption line of, lowest frequencies, a) n = 1 to n = 2, b) n = 3 to n = 8, c) n = 2 to n = 1 d) n= 8 to n = 3, 20. The ratio of the wave numbers of the radiations corresponding to the 3rd line of Balmer series and, the 2nd line of the Pachen series of hydrogen spectrum is, 21, 9, 25, 9, 21, 9, 16, 9, a) ×, b) ×, c) ×, d) ×, 16, 4, 16, 4, 25, 4, 25, 4, 21. The amount of energy emitted when an electron jumps from n = 4 to n =2 in H-atom, a) 4.08 x 10-18 J, b) 0.408 x 10-18 J, c) 40.8 x 10-18 J d) 2.02 x 10-18 J, 22. The ionization potential of hydrogen atom is 13.6eV. The wave length of the energy radiation, required for the ionization of H-atom, a) 1911 nm, b) 912nm, c) 68 nm, d) 91.2 nm, 23. The energy of the electron in first orbit of He+is 871.6 x 10-20 J. The energy of the electron in the, first orbit of hydrogen would be, a) -871.6x10-26J b) -43.5x10-20J, , c) -217.9x10-20J d) -108.9x10-20 J, 24. Of the following, which of the statement(s) regarding Bohr theory is wrong:, a) Kinetic energy of an electron is half of the magnitude of its potential energy, b) Kinetic energy of an electron is negative of total energy of electron, c)Energy of electron decreases with increase in the value of the principal quantum number, d)The ionization energy of H-atom in the first excited state is negative of one fourth of the energy, of an electron in the ground state., 25. The ratio of energy of the electron in ground state of hydrogen to the electron in first excited, state of Be3+ is:, a) 1: 4, b) 1 : 8, c) 1 : 16, d) 16 :1, th, th, 26. The difference between n and (n + 1) Bohr’s radius of H atom is equal to its (n - 1)th, Bohr’s radius. The value of n is:, a)1, b) 2, c) 3, d) 4, 27. Excitation potential (in eV) for IInd and Ist excited state in hydrogn atom is, a) 12.1 eV & 10.2 eV b) 10.2 eV and 12.1 eV, c) 3.4 eV & 13.6 eV, Page number, , 1, , d) 13.6 eV and 3.4 eV, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, 28. The potential energy of the electron present in the ground state of Li2+ ion in represented by, a)+, , 3𝑒 3, , b)−, , 4𝜋𝜖0 𝑟, , 3𝑒, , c)−, , 4𝜋𝜖0 𝑟, , 3𝑒 2, , d)−, , 4𝜋𝜖0 𝑟 2, , 3𝑒 2, 4𝜋𝜖0 𝑟, , 29. The amount of energy required to remove electron from a Li +2 ion in its ground state is how many, times greater than the amount of energy needed to remove the electron from an H atom in its, ground state?, a)9, b) 6, c)4, 30. Choose the incorrect relation on the basis of Bohr’s theory., a) Velocity of electron ∝, , 1, , d) 3, , b) Frequency of revolution ∝, , 𝑛, , 2, , c) Radius of orbit ∝ 𝑛 𝑍 d) Force on electron ∝, , 1, , 1, 𝑛2, , 𝑛4, , 31. The radius of which of the following orbit is same as that of the first Bohr’s orbit of hydrogen atom?, a) Li2+(n = 2), b) Li2+(n = 3), c) Be3+ (n = 2), d) He+ (n = 2), 32. The wavelength associated with a golf ball weighing 200 g and moving at a speed of 5 m/h is, of the order, a) 10-10m, , b) 10-20m, , c) 10-30m, , d)10-40m, , 33. The velocity of electron of H-atom in its ground state is 2.2 x 10-6ms-1. The de-Broglie wavelength, of this electron would be, a) 0.33 nm, b) 23.3 nm, c)45.6 nm, d) 100 nm, 34. The de-Broglie wavelength relates to applied voltage for -particles as, a) 𝜆 =, , 12.3, √𝑉, , Å, , b) 𝜆 =, , 0.286, √𝑉, , Å, , c) 𝜆 =, , 0.101, √𝑉, , Å, , d)𝜆 =, , 0.856, √𝑉, , Å, , 35. An electron beam is accelerated through a potential difference of 10,000 volt. The de-Broglie, wavelength of the electron beam is, a) 0.123A0, b)0.356 A0, c) 0.186 A0, d) 0.258 A0, 36. A particle A moving with a certain velocity has the de-Broglie wavelength of 1 A0. For, Particle B with mass 25% of A and velocity 75% of A, calculate the de-Broglie wavelength., a)3 A0, b) 5.33 A0, c) 6.88 A0, d) 0.48 A0, 37.  2 the wave function represents the probability of finding electron. Its value depends, a) inside the nucleus b) far from the nucleus, c) near the nucleus, d) upon the type of orbital, 38. The electron density of 3dxy orbital in YZ plane is, a) 50%, b) 95%, c) 33.33%, 39. Consider the following statements, A) Electron density in the XY plane in 3d 2x - 2y orbital is zero, , d) zero, , B) Electron density in the XY plane in 3d 2z orbital is not zero, C) 2S orbital has one nodal surface, D) For 2Pz orbital YZ is the nodal plane, Which are the correct statements?, a) A and C, b) B and C, c) only B, d) A, B, C, D, 40. For the differentiating electron of the atom (at. no = 37) the value of azimuthal quantum number, is, a) 0, b) 1, c) 2, d) 3, 41. Which orbital is represented by the complete wave function  420 :, a) 4d, b) 3d, c) 4p, d) 4s, 42. Two electrons A and Bin an atom have the following set of quantum numbers.What is true for A, and B: A: 3, 2, -2, +1/2 B: 3, 0, 0, +1/2, Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, a) A and B have same energy, c)B has more energy than A, , b)A has more energy than B, d)A and B represent same electron, , 43. Which of the following statements concerning Px orbitals is false?, a)The electron density has two regions of high magnitude pointed in the +x and –x directions, b)These two high probability regions are separated by a node passing through the nucleus, c)The electrons stay half the time in one of these regions and the remaining time in the, other region, d) None of these., 44. The electrons, identified by quantum numbers n and 1, (i) n = 4, l = 1 (ii) n = 4, l = 0, (iii) n = 3, l = 2 (iv) n = 3, l = 1 can be placed in order of increasing energy, from the lowest to, highest, as, a)iv < ii < iii < I, b)ii < iv < i < iii, 45., The graph representing node is, a), b), Ψ, , c) i < iii < ii < iv, , d) iii < i < iv < ii., , c), , d), , Ψ, , Ψ, , 𝑎0, , Ψ, , 𝑎0, , 𝑎0, , 𝑎0, , 46. Choose the incorrect statement., a) The shape of an atomic orbital depends upon the azimuthal quantum number., b) The orientation of an atomic orbitals depends upon the magnetic quantum number, c) The energy of an electron in an atomic orbital of multi-electron atom depends on principle, quantum number., d) The number of degenerate atomic orbitals of one type depends on the value of azimuthal, and magnetic quantum number., 47. Which of the following radial distribution graphs correspond to n = 3, 𝑙=2 for an atom?, a) b) c) d), Rrr, , Rrr, , Rrr, 2, , 𝑟 Ψ, , 2, , 2, , 𝑟 Ψ, , Rrr, 𝑟 2 Ψ2, , 2, , 𝑎0, , 𝑎0, , 𝑟 2 Ψ2, , 𝑎0, , 𝑎0, , 48. For a particular value of azimuthal quantum number, the total number of magnetic quantum, number values are given:, a) 𝑙 =, , 𝑚+1, 2, , b) 𝑙 =, , 𝑚−1, 2, , c) 𝑙 =, , 2𝑚+1, 2, , d) 𝑚 =, , 2𝑙+1, , 49. Consider the following statements:, 1) Electron density in xy plane in 3𝑑𝑥 2 −𝑦 2 orbital is zero., 2) Electron density in xy plane in 3𝑑𝑥 2 orbital is zero, 3) 2s orbital has only one spherical node., 4) For 2pz orbital yz is the nodal plane., The correct statements are, Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :, , 2

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ATOMIC STRUCTURE, a) 2 and 3, b) 1, 2, 3, 4, c) Only 2, d) 1 and 3, 50. The presence of five unpaired electrons in 3d orbitals of manganese atom is according to, a) Pauli’s principle b) Hund’s rule, c) Aufbau principle d) deBroglie’s theory, 51. If the electronic configuration of Li is written as 1s3, then it is against to, a) Pauli’s principle b) Hund’s rule, c) Aufbau principle, d) deBroglie’s theory, 52. The atomic number of the element having maximum number of unpaired 3p electron is, a) 16, b) 17, c) 15, d) 20, 53. For an atom with atomic number 14 the number of orbits and orbitals in which electrons are, present are respectively:, a)3,6, b) 6,3, c) 7,3, d) 3,8, 54. The orbital diagram in which Aufbau principle is violated is:, a), , b), , c), d), +, 55. Which will be the most stable among Cu ,Fe+, Fe2+, and Fe3+:, a) Fe2+, b) Fe+, c) Cu+, d) Fe3+, 56. Total number of orbital containing electron(s) in Ni (28) atom in its ground state is, a) 15, b) 13, c) 12, d) 10, 57. Which of the following will violates Aufbau principle as well as Pauli’s exclusion principle?, 1s, 2s, 2p 1s, 2s, 2p, , 1s, , ↿⇂, , ↿⇂, , a), 2s, c), , ↿⇂, , ↿, , ↿⇂, , ↿, , b), , ↿⇂, , ↿⇂, , ↿⇂, , ↿⇂, , ↿, , 2p, , ↿, , ↿⇂, , ↿, , d) None of the above, , 58. Select the correct statements:, 1) The total intensity of radiations (Power emitted/surface area) over all wavelength is directly, proportional to fourth power of absolute temperature during black body radiations., 2) The degeneracy of orbitals of a shell in absence of a shell in absence of magnetic field exists only, in 1 electron systems., 3) Absorption can be defined by four quantum numbers., 4) Absorption spectra are used by astronomers top identify elements in the outer layer of stars., a) 1,2,3, b) 1,2,4, c) 2,3,4, d) 1,3,4, 59. In a hydrogen atom, if energy of and electron in ground state is 13. eV, then that in the 2 nd excited, state is, [AIEEE 2002], a) 1.51 eV, b) 3.4 eV, c) 6.04 eV, d) 13.6 eV, 60. Uncertainty in position of a minute particle of mass 25 g in space is 10 -5.m. What is the uncertainty, in its velocity? (in ms-1)? (h = 6.6  10-34Js), [AIEEE 2002], -34, -34, -28, a) 2.1  10, b) 0.5  10, c) 2.1  10, d) 0.5  10-23, 61. In Bohr series of lines of hydrogen spectrum, the third line from the res end corresponds to which, one of the following inter-obit jumps of the electron for Bohr orbits ion an atom of hydrogen?, [AIEEE 2003], a) 3 → 2, b) 5 → 2, c) 4 → 1, d) 2 → 5, 62. The de Broglie wavelength of a tennis ball of mass 60 g moving with a velocity of 10 metres per, second is approximately (Planck’s constant, h=6.63  10-34Js), [AIEEE 2003], , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, a) 10-33metres, , b) 10-31metres, , c) 10-16metres, , d) 10-25metres., , 63. The orbital angular momentum for an electron revolving in an orbit is given by l (l + 1)., momentum of an s-electron will be given by, 1 h, a) + ., 2 2, , h, c), 2, , b) zero, , h, . This, 2, , [AIEEE 2003], d), , h, 2., 2, , 64. Which of the following sets of quantum numbers is correct for an electron in 4f orbital?, [AIEEE 2004], 1, 2, 1, c) n = 4, l = 3, m = +1, s = +, 2, , 1, 2, 1, d) n = 3, l = 2, m = -2, s = +, 2, , a) n = 4, l = 3, m = +4, s= +, , b) n = 4, l = 4, m = -4, s= -, , 65. Consider the ground state of Cr atom (Z =24). The numbers of electrons with the azimuthal, quantum numbers, l = 1 and 2 are, respectively, [AIEEE 2004], a) 12 and 4, b) 12 and 5, c) 16 and 4, d) 16 and 5., 66. The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to, stationary stat 1, would be (Rydberg constant = 1.097  107 m-1), [AIEEE 2004], a) 91 nm, b) 192 nm, c) 406 nm, d) 9.1  10-8 nm., 67. In a multi-electron atom, which of the following orbitals described by the three quantum numbers, will have the same energy in the absence of magnetic and electric fields?, [AIEEE 2004], i) n = 1, l = 0, m = 0, ii) n = 2, l = 0, m = 0, iii) n = 2, l = 1, m = 1 iv) n = 3, l = 2, m = 1 v) n = 3, l = 2, m = 0, a) (i) and (ii), b) (ii) and (iii), c) (iii) and (iv), d) (iv) and (v), 68. Which of the following statements in relation to the hydrogen atom is correct?, a) 3s orbital is lower in energy than 3p orbital, [AIEEE 2005], b) 3p orbital is lower in energy than 3d orbital, c) 3s and 3p orbitals are of lower energy than 3d orbital, d) 3s, 3p and 3d orbitals all have the same energy, 69. Uncertainty in the position of an electron (mass = 9.1  10-31 kg) moving with a velocity 300 ms-1,, accurate up to 0.001% will be (h= 6.6  10-34Js), [AIEEE 2006], a) 19.2  10-2n, b) 5.76  10-2 m, c) 1.92  10-2 m d) 3.84  10-2 m., 70. According to Bohr’s theory, the angular momentum of an electron in 5th orbit is, [AIEEE 2006], ℎ, , ℎ, , 𝜋, , 𝜋, , ℎ, , ℎ, , a) 25 b) 1.0 c)10 𝜋 d) 2.5 𝜋, 71. Which one of the following sets of ions represents a collection of isoelectronic species?, [AIEEE 2006], −, 2−, +, 2+, 3+, 2+, 2+, +, a) K , C𝑙 , Ca , Sc b)Ba , Sr , K , 𝑆, c) 𝑁 3− , 𝑂2− , 𝐹 − , 𝑆 2− d)Li+,Na+,Mg2+, Ca2+, 72. Which of the following sets of quantum numbers represents the highest energy of an atom?, [AIEEE 2007], 1, 1, a) 𝑛 = 3, 𝑙 = 0, 𝑚 = 0, 𝑠 = + 2, b) 𝑛 = 3, 𝑙 = 1, 𝑚 = 1, 𝑠 = + 2, 1, , 1, , c) 𝑛 = 3, 𝑙 = 2, 𝑚 = 1, 𝑠 = + 2, d) 𝑛 = 4, 𝑙 = 0, 𝑚 = 0, 𝑠 = + 2, 73. The ionization enthalpy of hydrogen atom is 1.312  106 J mol-1. The energy required to excite the, electron in the atom from n = 1 to n = 2 is, [AIEEE 2008], 5, -1, 5, -1, a) 9.84  10 J mol, b) 8.51  10 J mol, 5, -1, c) 6.56  10 J mol, d) 7.56  105 mol-1, 74. If the uncertainty in position and momentum arte equal, then uncertainty in velocity is, Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, [AIPMT 2008], a), , 1, , 2𝑚, , √, , ℎ, 𝜋, , b) √, , ℎ, , 1, , ℎ, , ℎ, , 𝑚, , 𝜋, , 𝜋, , c) √ d)√, , 2𝜋, , 75. The energy absorbed by each molecule (A2) of a substance is 4.4×10−19J and bond energy per, molecule is 4.0×10−19 J. The kinetic energy of the molecule per atom will be, [AIEEE 2009], a) 2.2×10−19 J, b) 2.0×10−19J, c) 4.0×10−20 J, d) 2.0×10−20J, 76. Maximum number of electron in a subshell of an atom is determined by the following, [AIEEE 2009], 2, a) 2𝑙+1, b) 4 𝑙=2, c) 2𝑛, d) 4 𝑙+2, 77. Which of the following is not permissible arrangement of electrons in an atom?, [AIEEE 2009], 1, 1, a) n= 5, 𝑙 = 3, 𝑚 = 0, 𝑠 = + 2, b) n=3, 𝑙 = 2, 𝑚 = −3, 𝑠 = − 2, 1, , 1, , c) n= 3, 𝑙 = 2, 𝑚 = −2, 𝑠 = − 2, d) n=4, 𝑙 = 0, 𝑚 = 0, 𝑠 = − 2, 78. In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.05%. Certainty, with which the position of the electron can be located is (h= 6.6×10−34kg𝑚2 𝑠 −1 ,mass of electron,, me = 9.9×1−31 kg), [AIEEE 2009], −4, −4, −4, −4, a) 1.52×10 m, b) 5.10×10 m, c) 1.92×10 m, d) 3.84×10 m, 79. Calculate the wavelength (on manometer) associated with a proton moving at 1.0×103 𝑚𝑠 −1, [AIEEE 2009], a) 0.032nm, b) 0.40nm, c) 2.5nm, d) 14.0nm, 80. Which one of the following statement is wrong about photon?, [AFMC 2009], a) Photon’s energy is h𝜈, b) Photon’s rest mass is zero., c) Momentum of photon is, , h𝜈, 𝑐, , d) Photon exerts no pressure, , 81. For principal quantum number n=4, the total number of orbitals having 𝑙 = 3 is[AFMC09], a) 3, b) 7, c) 5, d) 9, +, -18, -1, 82. Ionization energy of He is 19.6×10 J atom .The energy of the first stationary state (n=1) of Li2+ is, [AIEEE 2010], a) −2.2×10-15J atom-1, b) 8.82×10-17J atom-1, c) 4.41×10-16J atom-1 d) −4.41×10-17J atom-1, 83. A 0.66 Kg ball is moving with a speed of 100m/s. The associated wavelength will be, (h= 6.6×10–34Js), [AIPMT Mains 2010], –32, –34, –35, a) 6.6×10 m, b) 6.6×10 m, c) 1.0×10 m, d) 1.0×10–32 m, 84. Deuterium nucleus contains, [AFMC 2010], a) 1p + 1n, b) 2p + 0n, c) 1p + 1e–, d) 2p + 2n, 85. The total number of atomic orbitals in fourth energy level of an atom is [AIPMT 2011], a) 4, b) 8, c) 16, d) 32, 86. The energies E1 and E2 of two radiations are 25 eV and 50 eV respectively. The relation between, their wavelengths i.e., 𝜆1 and 𝜆2 will be, [AIPMT 2011], 1, a) 𝜆1 = 2 𝜆2, b) 𝜆1 = 𝜆2, c) 𝜆1 = 2𝜆2, d) 𝜆1 = 4𝜆2, 87. If n=6, the correct sequence of filling of electrons will be, [AIPMT 2011], a) 𝑛𝑠 → 𝑛𝑝(𝑛 − 1)𝑑 → (𝑛 − 2)𝑓, b)𝑛𝑠 → 𝑛(𝑛 − 2)𝑓 → (𝑛 − 1)𝑑 → 𝑛𝑝, c)𝑛𝑠 → (𝑛 − 1)𝑑 → (𝑛 − 2)𝑓 → 𝑛𝑝, d) 𝑛𝑠 → (𝑛 − 2)𝑓 → 𝑛𝑝 → (𝑛 − 1)𝑑, 88. According to the Bohr Theory, which of the following transitions in the hydrogen atom will give, rise to the least energetic photon, [AIPMT Mains 2011], a) n=6 to n=1, b) n=5 to n=4, c) n=6 to n=5, d) n=5 to n=3, , Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, , KEY ANSWER, SESSION – 1, CLASS EXERCISE:, 1)c, , 2)d, , 3)b, , 4) a&c, , 5)c, , 6)b, , 7)c, , 8)b, , 6)a, , 7)d, , 8)d, , 9)b, , 9)b, , HOME EXERCISE:, 1)b, , 2)c, , 3)d, , 4)a, , 5)b, , SESSION – 2 AND 3, CLASS EXERCISE:, 1)c, , 2)a, , 3)c4)d, , 5)d6)a 7)a 8)d9)c 10)b, , HOME EXERCISE:, 1)c, 2)d, 3)b 4)b, , 5)c, , 6)d, , 7)b, , 8)a, , 9)c, , SESSION – 4, CLASS EXERCISE:, 1) c 2)a, 3)a, HOME EXERCISE:, 1)c, 2)d, 3)d, , 4)b, , 4)a, , 5)d, , 6)c, , 5)b, SESSION – 5 AND 6, , CLASS EXERCISE:, 1)d, 2)c, 3)c, , 4) d, , 5) b6) a 7) a 8) b 9)c 10)d, , HOME EXERCISE:, 1)a, 2)c, 3)d, , 4)b, , 5)a 6)a 7)c 8)c 9)a 10)a, SESSION – 7, , CLASS EXERCISE:, 1)a, 2)c, 3)c, , 4)0.099kg, , 5)3.499×10–24 m/s–1 6)c, , HOME EXERCISE:, 1)a, 2)c, 3)a, , 4)b, , 6)c, , 5)c, , SESSION – 8 AND 9, CLASS EXERCISE:, Page number, , 1, , AMCF-TEMPLE OF LEARNING, , Contact No. :

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ATOMIC STRUCTURE, 1)a, , 2)c, , 3)c, , 4)c, , 5)c, , HOME EXERCISE:, 1)c 2)d, 3)b, , 4)c, , 5)a, , 6)a, , 6)b, , 7)b, , 8)a, , 9)a, , 10)b, , 7)a 8)a 9)d 10)c, SESSION – 10, , CLASS EXERCISE:, 1)c 2)b, 3)a, , 4)b, , HOME EXERCISE:, 1)b 2)c 3)a, 4)a, , 5)a, , 5)c, , MAIN, LEVEL –I, 1)d 2) c, 13) a, 25) d, 37) c, 49) a, , 3)d, 14) a, 26) a, 38) b, 50) b, , 4)b, 15) a, 27) d, 39) a, 51) d, , 5) b, 16) a, 28) d, 40) c, 52) c, , 6)a, 17) c, 29) a, 41) d, 53) b, , 7)b, 18)b, 30) b, 42) d, 54) a, , 8) c, 19) c, 31) b, 43) a, 55) a, , 9) d, 20) b, 32) a, 44) c, 56) d, , 10) b, 21) d, 33) b, 45) b, 57) c, , 11) b, 22) c, 34) b, 46) a, , 12)d, 23) d 24) c, 35) c 36) a, 47) a 48) c, , 8) d, 20) a, 32) c, 44) a, 56) a, 68) d, 80) d, , 9) d, 21) b, 33) a, 45) c, 57) c, 69) c, 81) b, , 10) a, 22) d, 34) c, 46) c, 58) b, 70) d, 82) d, , 11) a, 23) c, 35) a, 47) b, 59) a, 71) a, 83) c, , LEVEL –II, 1) a, 13) a, 25) a, 37) d, 49) d, 61) b, 73) a, 85) c, , Page number, , 2) a, 14) c, 26) d, 38) d, 50) b, 62) a, 74) a, 86) c, , 1, , 3) d, 15) d, 27) a, 39) b, 51) a, 63) b, 75) d, 87) b, , 4) d, 16) b, 28) d, 40) a, 52) c, 64) c, 76) d, 88) c, , 5) c, 17) b, 29) a, 41) a, 53) d, 65) b, 77) b, , 6) b, 18) d, 30) c, 42) b, 54) b, 66) a, 78) c, , AMCF-TEMPLE OF LEARNING, , 7) c, 19) b, 31) c, 43) d, 55) c, 67) d, 79) b, , Contact No. :, , 12) d, 24) c, 36) b, 48) b, 60) c, 72) c, 84) a