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CHAPTER 2, STRUCTURE OF ATOM, • Atom is the smallest indivisible particle of the matter. Atom is made of electron,, proton and neutrons., PARTICLE, ELECTRON, PROTON, NEUTRON, Sir. J. J. Thomson, Goldstein (1886), Chadwick (1932), Discovery, (1869), Negative, Positive, Neutral, Nature of charge, -19, -19, Amount of charge 1.6 x 10 Coloumb 1.6 x 10 Coloumb 0, 9.11 x 10-31kg, 1.672614 x 10-27kg 1.67492 x10-27kg, Mass, Electrons were discovered using cathode ray discharge tube experiment., Nucleus was discovered by Rutherford in 1911., Cathode ray discharge tube experiment: A cathode ray discharge tube madeof, glass is taken with two electrodes. At very low pressure and high, voltage,current starts flowing through a stream of particles moving in the tube, fromcathode to anode. These rays were called cathode rays. When a, perforatedanode was taken, the cathode rays struck the other end of the glass, tube atthe fluorescent coating and a bright spot on the coating was developed, Results:, a. Cathode rays consist of negatively charged electrons., b. Cathode rays themselves are not visible but their behavior can be, observed with, help of fluorescent or phosphorescent materials., c. In absence of electrical or magnetic field cathode rays travel in, straight lines, d. In presence of electrical or magnetic field, behaviour of cathode rays, is similar to that shown by electrons, e. The characteristics of the cathode rays do not depend upon the, material of the electrodes and the nature of the gas present in the cathode ray, tube., Charge to mass ratio of an electron was determined by Thomson. The chargeto, mass ratio of an electron as 1.758820 x 1011 C kg-1, Charge on an electron was determined by R A Millikan by using an oil, dropexperiment. The value of the charge on an electron is -1.6 x 10-19C., The mass on an electron was determined by combining the results, ofThomson’s experiment and Millikan’s oil drop experiment. The mass of, anelectron was determined to be 9.1094 x 10-31kg., Discovery of protons and canal rays: Modified cathode ray tube, experimentwas carried out which led to the discovery of protons., Characteristics of positively charged particles:, a. Charge to mass ratio of particles depends on gas from which these originate, 17

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b. The positively charged particles depend upon the nature of gas present in the, cathode ray discharge tube, c. Some of the positively charged particles carry a multiple of fundamental of, electrical charge., d. Behaviour of positively charged particles in electrical or magnetic field is, opposite to that observed for cathode rays, Neutrons were discovered by James Chadwick by bombarding a thin sheet of, beryllium by α- particles. They are electrically neutral particles having a mass, slightly greater than that of the protons., Atomic number (Z) : the number of protons present in the nucleus, (Moseley1913)., Mass Number (A) :Sum of the number of protons and neutrons present in, thenucleus., Thomson model of an atom: This model proposed that atom is considered asa, uniform positively charged sphere and electrons are embedded in it.An, important feature of Thomson model of an atom was that mass of atom, isconsidered to be evenly spread over the atom.Thomson model of atom is also, called as Plum pudding, raisin pudding orwatermelon modelThomson model, of atom was discarded because it could not explain certainexperimental results, like the scattering of α- particles by thin metal foils., Observations from α- particles scattering experiment by Rutherford:, a. Most of the α- particles passed through gold foil un deflected, b. A small fraction of α- particles got deflected through small angles, c. Very few α- particles did not pass through foil but suffered large deflection, nearly180o, Conclusions Rutherford drew from α- particles scattering experiment:, a. Since most of the α-particles passed through foil undeflected, it means most, of the space in atom is empty, b. Since some of the α-particles are deflected to certain angles, it means that, there is positively mass present in atom, c. Since only some of the α-particles suffered large deflections, the positively, charged mass must be occupying very small space, d. Strong deflections or even bouncing back of α-particles from metal foil, were due to direct collision with positively charged mass in atom, Rutherford’s model of atom: This model explained that atom consists, ofnucleus which is concentrated in a very small volume. The nucleus, comprisesof protons and neutrons. The electrons revolve around the nucleus in, fixedorbits. Electrons and nucleus are held together by electrostatic forces, ofattraction., Drawbacks of Rutherford’s model of atom:, a. According to Rutherford’s model of atom, electrons which are negatively, charged particles revolve around the nucleus in fixed orbits. Thus,, 18

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b. theelectrons undergo acceleration. According to electromagnetic theory of, Maxwell, a charged particle undergoing acceleration should, emitelectromagnetic radiation. Thus, an electron in an orbit should, emitradiation. Thus, the orbit should shrink. But this does not happen., c. The model does not give any information about how electrons, aredistributed around nucleus and what are energies of these electrons, Isotopes: These are the atoms of the same element having the same, atomicnumber but different mass number.e g 1H1,1H2,1H3, Isobars: Isobars are the atoms of different elements having the same, massnumber but different atomic number.e g 18Ar40 , 20Ca40, Isoelectronic species: These are those species which have the same numberof, electrons., Electromagnetic radiations: The radiations which are associated, withelectrical and magnetic fields are called electromagnetic radiations. When, anelectrically charged particle moves under acceleration, alternating, electricaland magnetic fields are produced and transmitted. These fields, aretransmitted in the form of waves. These waves are called, electromagneticwaves or electromagnetic radiations., Properties of electromagnetic radiations:, a. Oscillating electric and magnetic field are produced by oscillating charged, particles. These fields are perpendicular to each other and both, areperpendicular to the direction of propagation of the wave., b. They do not need a medium to travel. That means they can even travel in, vacuum., Characteristics of electromagnetic radiations:, a. Wavelength: It may be defined as the distance between two, neighbouring crests or troughs of wave as shown. It is denoted by λ., b. Frequency (ν): It may be defined as the number of waves which, passthrough a particular point in one second., c. Velocity (v): It is defined as the distance travelled by a wave in, onesecond. In vacuum all types of electromagnetic radiations travel with, thesame velocity. Its value is 3 X108m sec-1. It is denoted by v, d. Wave number: Wave number is defined as the number of wavelengths, per unit length., Velocity = frequency x wavelength c = νλ, Planck's Quantum Theoryo The radiant energy is emitted or absorbed not continuously but, discontinuously in the form of small discrete packets of energy called, ‘quantum’. In case of light , the quantum of energy is called a ‘photon’, o The energy of each quantum is directly proportional to the frequency of, the radiation, i.e. E α υ or E= hυ, where h= Planck’s constant =, -27, 6.626 x 10 Js, o Energy is always emitted or absorbed as integral multiple of this, quantum. E=nhυ Where n=1,2,3,4,....., 19

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Black body: An ideal body, which emits and absorbs all frequencies, is calleda, black body. The radiation emitted by such a body is called black, bodyradiation., Photoelectric effect: The phenomenon of ejection of electrons from thesurface, of metal when light of suitable frequency strikes it is calledphotoelectric effect., The ejected electrons are called photoelectrons., Experimental results observed for the experiment of Photoelectric effecto When beam of light falls on a metal surface electrons are, ejectedimmediately., o Number of electrons ejected is proportional to intensity or brightness of, light, o Threshold frequency (vo): For each metal there is a, characteristicminimum frequency below which photoelectric effect is, not observed. Thisis called threshold frequency., o If frequency of light is less than the threshold frequency there is, noejection of electrons no matter how long it falls on surface or how, high isits intensity., Photoelectric work function (Wo): The minimum energy required to, ejectelectrons is called photoelectric work function.Wo= hvo, Energy of the ejected electrons :, Dual behavior of electromagnetic radiation- The light possesses both particle, and wave like properties, i.e., light has dual behavior . whenever radiation, interacts with matter, it displays particle like properties.(Black body radiation, and photoelectric effect) Wave like properties are exhibited when it, propagates(interference an diffraction), When a white light is passed through a prism, it splits into a series ofcoloured, bands known as spectrum., Spectrum is of two types: continuous and line spectrum, a. The spectrum which consists of all the wavelengths is called, continuous, spectrum., b. A spectrum in which only specific wavelengths are present is known, as a line, spectrum. It has bright lines with dark spaces between them., Electromagnetic spectrum is a continuous spectrum. It consists of a range, ofelectromagnetic radiations arranged in the order of increasing wavelengths, ordecreasing frequencies. It extends from radio waves to gamma rays., Spectrum is also classified as emission and line spectrum., o Emission spectrum: The spectrum of radiationemitted by a substance, that has absorbed energy is called an emissionspectrum., o Absorption spectrum is the spectrum obtained when radiation is, passedthrough a sample of material. The sample absorbs radiation of, 20

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certainwavelengths. The wavelengths which are absorbed are missing, and comeas dark lines., The study of emission or absorption spectra is referred as spectroscopy., Spectral Lines for atomic hydrogen:, , Rydberg equation, , R = Rydberg’s constant = 109677 cm-1, Bohr’s model for hydrogen atom:, a. An electron in the hydrogen atom can move around the nucleus in a circular, path of fixed radius and energy. These paths are called orbits orenergy, levels. These orbits are arranged concentrically around thenucleus., b. As long as an electron remains in a particular orbit, it does not lose or gain, energy and its energy remains constant., c. When transition occurs between two stationary states that differ inenergy,, the frequency of the radiation absorbed or emitted can becalculated, d. An electron can move only in those orbits for which its angularmomentum, is an integral multiple of h/2π, , The radius of the nth orbit is given byrn =52.9 pm x n2, Z, energy of electron in nth orbit is :, , Limitations of Bohr’s model of atom:, a. Bohr’s model failed to account for the finer details of the hydrogen, spectrum., b. Bohr’s model was also unable to explain spectrum of atoms containing, more than one electron., Dual behavior of matter: de Broglie proposed that matter exhibits, dualbehavior i.e. matter shows both particle and wave nature. de Broglie’s, relation is, 21

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Heisenberg’s uncertainty principle: It states that it is impossible to, determine simultaneously, the exact position and exact momentum (or, velocity) of an electron.The product of their uncertainties is always equal to or, greater than h/4π., , Heisenberg’s uncertainty principle rules out the existence of definite pathsor, trajectories of electrons and other similar particles, Failure of Bohr’s model:, a. It ignores the dual behavior of matter., b. It contradicts Heisenberg’s uncertainty principle., Classical mechanics is based on Newton’s laws of motion. It, successfullydescribes the motion of macroscopic particles but fails in the case, ofmicroscopic particles., Reason: Classical mechanics ignores the concept of dual behaviour of matter, especially for sub-atomic particles and the Heisenberg’s uncertainty principle., Quantum mechanics is a theoretical science that deals with the study of, themotions of the microscopic objects that have both observable wave like, andparticle like properties., Quantum mechanics is based on a fundamental equation which is, calledSchrodinger equation., Schrodinger’s equation: For a system (such as an atom or a molecule, whoseenergy does not change with time) the Schrödinger equation is written, as:, , When Schrödinger equation is solved for hydrogen atom, the solution givesthe, possible energy levels the electron can occupy and the correspondingwave, function(s) of the electron associated with each energy level.Out of the, possible values, only certain solutions are permitted. Eachpermitted solution is, highly significant as it corresponds to a definite energystate. Thus, we can say, that energy is quantized., ψ gives us the amplitude of wave. The value of ψhas no physicalsignificance., Ψ2gives us the region in which the probability of finding an electron, ismaximum. It is called probability density., , 22

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Orbital: The region of space around the nucleus where the probability, offinding an electron is maximum is called an orbital., Quantum numbers: There are a set of four quantum numbers which specifythe, energy, size, shape and orientation of an orbital. To specify an orbital only, three quantum numbers are required while to specify an electron all four, quantum numbers are required., Principal quantum number (n):It identifies shell, determines sizes and, , energy of orbitals, Azimuthal quantum number (l): Azimuthal quantum number. ‘l’ is also, known as orbital angular momentum or subsidiary quantum number. l. It, identifies sub-shell, determines the shape of orbitals, energy of orbitals in, multi-electron atoms along with principal quantum number and orbital angular, momentum, i.e.,, The number of orbitals in a subshell = 2l + 1. For a, given value of n, it can have n values ranging from 0 to n-1. Total number of, subshells in a particular shell is equal to the value of n., Subshell, s p, d f g, notation, Value of ‘l’, 0 1, 2 3 4, Number of, 1 3, 5 7 9, orbitals, Magnetic quantum number or Magnetic orbital quantum number (ml):, Itgives information about the spatial orientation of the orbital with respect, tostandard set of co-ordinate axis.For any sub-shell (defined by ‘l’ value) 2l+1, values of ml are possible.For each value of l, ml = – l, – (l –1), – (l–2)... 0,1..., (l – 2), (l–1), l, Electron spin quantum number (ms): It refers to orientation of the spin of, theelectron. It can have two values +1/2 and -1/2. +1/2 identifies the, clockwisespin and -1/2 identifies the anti- clockwise spin., The region where this probability density function reduces to zero is, callednodal surfaces or simply nodes., Radial nodes: Radial nodes occur when the probability density of wave, functionfor the electron is zero on a spherical surface of a particular radius., Numberof radial nodes = n – l – 1, Angular nodes: Angular nodes occur when the probability density, wavefunction for the electron is zero along the directions specified by a, particularangle. Number of angular nodes = l, Total number of nodes = n – 1, 23

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Degenerate orbitals: Orbitals having the same energy are called, degenerateorbitals., Shape of p and d-orbitals, , Shielding effect or screening effect: Due to the presence of electrons in, theinner shells, the electron in the outer shell will not experience the full, positivecharge on the nucleus., So, due to the screening effect, the net positive charge experienced by, theelectron from the nucleus is lowered and is known as effective nuclear, charge. Effective nuclear charge experienced by the orbital decreases with, increase of azimuthal quantum number (l)., Aufbau Principle: In the ground state of the atoms, the orbitals are filled, inorder of their increasing energies, 24

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n+l rule-Orbitals with lower value of (n+l) have lower energy. If two orbitals, have the same value of (n+l) then orbital with lower value of nwill have lower, energy., The order in which the orbitals are filled isas follows:, 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 4f, 5d, 6p, 7s..., Pauli Exclusion Principle: No two electrons in an atom can have the same, setof four quantum numbers. Only two electrons may exist in the same, orbitaland these electrons must have opposite spin., Hund’s rule of maximum multiplicity: Pairing of electrons in the, orbitalsbelonging to the same subshell (p, d or f) does not take place until, eachorbital belonging to that subshell has got one electron each i.e., it is, singlyoccupied., Electronic configuration of atoms:Arrangement of electrons in different, orbitals of an atom. The electronic configuration of differentatoms can be, represented in two ways., a. sapbdc...... notation., b. Orbital diagram:, each orbital of the subshell is represented by a box and the, electron is represented by an arrow (↑) a positive spin or an arrow (↓) a negative, spin., Stability of completely filled and half filled subshells:, a. Symmetrical distribution of electrons- the completely filled or half filled, sub-shells have, symmetrical distribution of electrons in them and are more stable., b. Exchange energy-The two or more electrons with the same spin present in, the degenerate orbitals of a sub-shell can exchange their position and the, energy released due to this exchange is called exchange energy. The, number of exchanges is maximum when the subshell is either half filled or, completely filled. As a result the exchange energy is maximum and so is, the stability., ONE MARK QUESTIONS, 1. Neutrons can be found in all atomic nuclei except in one case. Which is this, atomic nucleus and what does it consists of?, Ans. Hydrogen atom. It consists of only one proton., 2. Calculate wave number of yellow radiations having wavelength of 5800 A0., Ans. Wave number = 1/ wavelength, Wavelength = 5800 A0= 5800 x 10-10 m, Wave number = 1/5800 x 10-10 m = 1.72 x 106 m-1, 3. What are the values of n and l for 2p orbital?, Ans. n=2 and l= 1, 4. Which of the following orbitals are not possible? 1p, 2s, 3f and 4d, Ans. 1p and 3f are not possible., 5. Write the electronic configuration of the element having atomic number 24., Ans. 1s2 2s2 2p6 3s2 3p6 3d5 4s1, 6. What atoms are indicated by the following electronic configurations?, 25

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a. 1s2 2s2 2p1, b. [Ar]4s2 3d1, Ans. a. Boron, b. Scandium, 7. What is the relationship between frequency and wavelength of light?, Ans. velocity of light = frequency x wavelength. Frequency and wavelength are, inversely proportional to each other., 8. State Pauli Exclusion Principle., Ans. No two electrons in an atom can have the same set of four quantum numbers, or an orbital can have maximum two electrons and these must have opposite spin., 9. When α- rays hit a thin foil of gold, very few α- particles is deflected back., What does it prove?, Ans. There is a very small heavy body present within the atom., 10. What is the difference between a quantum and a photon?, Ans. The smallest packet of energy of any radiation is called a quantum whereas, that of light is called photon., TWO MARKS QUESTIONS, 1. Write the complete symbol for the atom with the given atomic number (Z) and, mass number(A). (a) Z = 17, A = 35, (b) Z = 92 , A = 233, 35, 233, Ans. (a) 17Cl, (b) 92U, 2. Using s,p,d and f notation, describe the orbital with the following quantum, numbers(a) n=1,l=0 (b) n=3, l=1 (c) n=4, l=2 (d) n=4, l=3, Ans. (a) 1s, (b) 3p (c)4d (d) 4f, 3. How many electrons in an atom have the following quantum numbers?, a. n=4, ms= -1/2, b. n =3 , l=o, Ans. (a) 16 electrons, (b) 2 electrons., 4. An element with mass number 81 contains 31.7 % more neutrons as compared, to protons. Assign the atomic symbol., Ans. Mass number = 81, i.e., p + n = 81, If protons = x, then neutrons = x + 31.7 X x = 1.317 x, 100, x+1.317x = 81 or 2.317x = 81, x=35, Thus proton = 35, i.e., atomic no. = 35, Hence symbol is 8135Br, 5. (i) The energy associated with the first orbit in the hydrogen atom is -2.18 x, 10-18J/atom. What is the energy associated with the fifth orbit, (ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom., Ans. (i) En = -2.18 x 10-18/ n2, E5 = -2.18 x 10-18/ 52 = -8.72 x 10-20 J, (ii) For H atom, rn= 0.529 x n2 r5 = 0.529 x 52 = 13.225 A0= 1.3225 nm, 6. Explain , giving reasons, which of the following sets of quantum numbers are, not possible., (a) n=0, l=0; ml = 0, ms= + ½ (c)n=1, l=0; ml = 0, ms= - ½, (b) n=1, l=1; ml =- 0, ms= + ½ (d) n=2, l=1; ml = 0, ms= + ½, Ans. (a) Not possible because n≠ 0 (c) Not possible because when n=1, l≠1, 26

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(b) Possible, (d) Possible, 7. (a)What is the lowest value of n that allows g orbitals to exist?, (b)An electron is in one of the 3d orbitals, Give the possible values of n,l and, mlfor this electron., Ans.(a) minimum value of n= 5, (b)n=3, l=2, ml = -2, -1, 0, +1, +2, 8. Calculate the total number of angular nodes and radial nodes present in 30, orbitals., Ans. For 3p orbitals, n=3, l= 1, Number of angular nodes = l= 1, Number of radial nodes = n-l-1 = 3-1-1= 1, 9. Mention the draw backs of Rutherford’s atomic model., Ans. 1. It could not explain the stability of an atom., 2. It could not explain the line spectrum of H- atom., 10. State de-Broglie concept of dual nature of matter. How do dual nature of, electron verified?, Ans. Just as light has dual nature, every material particle in motion has dual nature, (particle nature and wave nature). The wave nature has been verified by Davisson, and Germer’sexperiment whereas particle nature by scintillation experiment., , THREE MARKS QUESTIONS, 1. State (a)Hund’s Rule of maximum Multiplicity (b) Aufbau Principle (c) n+l, rule, Ans.(a) Pairing of electrons in the orbitals belonging to the same subshell (p, d or, f) does not takeplace until each orbital belonging to that subshell has got one, electron each i.e., it is singly occupied., (b)In the ground state of the atoms, the orbitals are filled inorder of their, increasing energies, (c) Orbitals with lower value of (n+l) have lower energy.If two orbitals have the, same value of (n+l) then orbital with lower value of n will have lower energy., 2. Write down the quantum numbers n and l for the following orbitals, a. 2p, b. 3d, c. 5f, Ans., a. n=2, l= 1, b. n= 3, l=2 c. n= 5, l=3, 3. Write the 3 points of difference between orbit and orbital., Ans., Orbit, orbital, 1. An orbit is a well defined, circular path around the, nucleus in which the, electrons revolve, 2. It represents the planar, motion of an electron around, the nucleus, 3. All orbits are circular and, disc like, , 1. An orbital is the three dimensional, space around the nucleus within, which the probability of finding an, electron is maximum(upto 90 %), 2. It represents the three dimensional, motion of an electron around the, nucleus, 3. Different orbitals have different, shapes, i.e., s-orbitals are spherically, symmetrical, p-orbitals are dumb-bell, shaped and so on., 27

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4. State Heisenberg’s uncertainty principle.calculate the uncertainty in the, position of an electron if the uncertainty in its velocity is 5.7 x 10 5 m/s., Ans. It states that it is impossible to determine simultaneously, the exact position, and exact momentum (or velocity) of an electron. The product of their, uncertainties is always equal to or greater than h/4π., Δx x (m x Δv) = h/4ᴨ, Δx = h/4ᴨ x m x Δv =, 6.6 x 10-34, = 1.0 x 10-10 m, 4 x 3.14 x 9.1 x 10-31 x 5.7 x 105, 5. Write 3 points of differences between electromagnetic waves and matterwaves., Electromagnetic waves, Matter waves, 1. These are associated with, 1. These are not associated with, electric and magnetic, electric and magnetic field., fields, 2. They do not require any, 2. They require medium for, medium for propagation., propagation, 3. They travel with the same, 3. They travel with lower speeds, speed a that of light, not constanr for all matter, waves, 6. (i) Calculate the number of electrons which will together weigh one gram., (ii) Calculate the mass and charge of one mole of electrons, Ans. (i) Mass of one electron = 9.10939 × 10–31 kg, Number of electrons that weigh 9.10939 × 10–31 kg = 1, Number of electrons that will weigh 1 g = (1 × 10–3kg), , = 0.1098 × 10–3 + 31, = 0.1098 × 1028, = 1.098 × 1027, (ii) Mass of one electron = 9.10939 × 10–31 kg, Mass of one mole of electron = (6.022 × 1023) × (9.10939 ×10–31 kg), = 5.48 × 10–7 kg, , 28

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Charge on one electron = 1.6022 × 10–19coulomb, Charge on one mole of electron = (1.6022 × 10–19C) (6.022 × 1023), = 9.65 × 104 C, 7. Find energy of each of the photons which, (i) correspond to light of frequency 3× 1015Hz., (ii) have wavelength of 0.50 Å., Ans.(i) Energy (E) of a photon is given by the expression,, E=, Where,, h = Planck’s constant = 6.626 × 10–34Js ν = frequency of light = 3 × 1015Hz, Substituting the values in the given expression ofE:, E = (6.626 × 10–34) (3 × 1015), E = 1.988 × 10–18J, (ii) Energy (E) of a photon having wavelength (λ)is given by the expression,, , h = Planck’s constant = 6.626 × 10–34Js, c = velocity of light in vacuum = 3 × 108m/s, Substituting the values in the given expression of E:, , 8. What is the wavelength of light emitted when the electron in a hydrogen atom, undergoes transition from an energy level with n = 4 to an energy level with n, = 2?, Ans.Theni= 4 to nf= 2 transition will give rise to a spectral line of the Balmer, series. The energy involved in the transition is given by the relation,, , 29

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Substituting the values in the given expression of E:, , E =– (4.0875 × 10–19 J), The negative sign indicates the energy of emission., , Wavelength of light emitted, , Substituting the values in the given expression of λ:, , 9. An atom of an element contains 29 electrons and 35 neutrons. Deduce (i)the, number of protons and (ii) the electronic configuration of the element (iii), Identify the element ., Ans.(i)For an atom to be neutral, the number of protons is equal to the number, of electrons., ∴ Number of protons in the atom of the given element = 29, (ii) The electronic configuration of the atom is 1s22s2 2p6 3s2 3p64s2 3d10, (iii) Copper, 10. Give the number of electrons in the species , H2 and, Ans. Number of electrons present in hydrogen molecule (H2) = 1 + 1 = 2, ∴ Number of electrons in, =2–1=1, Number of electrons in H2 = 1 + 1 = 2, Number of electrons present in oxygen molecule (O 2) = 8 + 8 = 16, ∴ Number of electrons in, , = 16 – 1 = 15, , FIVE MARKS QUESTIONS WITH ANSWERS, 30

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1. What are the draw backs of Bohr’s atomic model? Show that the, circumference of the Bohr orbit for the hydrogen atom is an integral multiple, of the de Broglie wavelength associated with the electron revolving around the, orbit., Ans.1.Bohr’s model failed to account for the finer details of the hydrogen spectrum., 2. Bohr’s model was also unable to explain spectrum of atoms containing more, than one electron., 3. Bohr’s model was unable to explain Zeeman effect and Stark effect i, 4. Bohr’s model could not explain the ability of atoms to form molecules by, chemical bonds, Since a hydrogen atom has only one electron, according to Bohr’s, postulate, the angular momentum of that electron is given by:, Where, n = 1, 2, 3, …, According to de Broglie’s equation:, , Substituting the value of ‘mv’ from expression (2) in expression (1):, , Since‘2πr’represents the circumference of the Bohr orbit (r), it is proved by, equation (3) that the circumference of the Bohr orbit of the hydrogen atom is, an integral multiple of de Broglie’s wavelength associated with the electron, revolving around the orbit., 2. State photo electric effect. The work function for caesium atom is 1.9 eV., Calculate (a) the threshold wavelength and (b) the threshold frequency of the, radiation. If the caesium element is irradiated with awavelength 500 nm,, calculate the kinetic energy and the velocity of the ejected photoelectron., Ans. Photoelectric effect: The phenomenon of ejection of electrons from the, surface of metal when light of suitable frequency strikes it is called, photoelectric effect. The ejected electrons are called photoelectrons., It is given that the work function (W0) for caesium atom is 1.9 eV., (a) From the expression,, , , we get:, , Where,, λ0 = threshold wavelength, h = Planck’s constant, c = velocity of radiation, Substituting the values in the given expression of (λ0):, 31

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6.53 × 10–7 m, Hence, the threshold wavelength, (b) From the expression,, , is 653 nm., , , we get:, , Where,, ν0= threshold frequency, h = Planck’s constant, Substituting the values in the given expression ofν0:, (1 eV = 1.602 × 10–19J), ν0= 4.593 × 1014s–1, Hence, the threshold frequency of radiation (ν0) is 4.593 × 1014s–1., (c) According to the question:, Wavelength used in irradiation (λ) = 500 nm, Kinetic energy = h (ν – ν0), , = 9.3149 × 10–20 J, Kinetic energy of the ejected photoelectron = 9.3149 × 10 –20J, Since K.E, , v = 4.52 × 105ms–1, Hence, the velocity of the ejected photoelectron (v) is 4.52 × 105ms–1., 3. (a)The quantum numbers of six electrons are given below. Arrange them in, order of increasing energies. If any of these combination(s) has/have the same, energy lists:, 32

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1. n= 4, l = 2, ml= –2 , ms= –1/2, 2. n= 3, l = 2, ml= 1 , ms= +1/2, 3. n= 4, l = 1, ml= 0 , ms= +1/2, 4. n = 3, l= 2, ml= –2 , ms= –1/2, 5. n = 3, l= 1, ml= –1 ,ms= +1/2, 6. n = 4, l= 1, ml= 0 , ms= +1/2, (b)Among the following pairs of orbitals which orbital will experience the, larger effective nuclearcharge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p, Ans.(a)Forn = 4 and l = 2, the orbital occupied is 4d., For n = 3 and l = 2, the orbital occupied is 3d., For n = 4 and l = 1, the orbital occupied is 4p., Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the 4d, 3d, 4p,, 3d, 3p, and 4p orbitals respectively., Therefore, the increasing order of energies is 5(3p) < 2(3d) = 4(3d) < 3(4p) =, 6(4p) < 1 (4d)., (b)Nuclear charge is defined as the net positive charge experienced by an, electron in the orbital of a multi-electron atom. The closer the orbital, the greater, is the nuclear charge experienced by the electron (s) in it., (i) The electron(s) present in the 2s orbital will experience greater nuclear, charge (being closer to the nucleus) than the electron(s) in the 3s orbital., (ii) 4d will experience greater nuclear charge than 4fsince 4d is closer to the, nucleus., (iii) 3p will experience greater nuclear charge since it is closer to the nucleus, than 3f., 4. (i) The unpaired electrons in Al and Si are present in 3p orbital. Which, electrons will experience more effective nuclear charge from the nucleus?, (ii) Indicate the number of unpaired electrons in: (a) P, (b) Si, (c) Cr, (d) Fe, Ans. (i) the electrons in the 3p orbital of silicon will experience a more, effective nuclear charge than aluminium., (ii) (a) Phosphorus (P):, Atomic number = 15, The electronic configuration of P is:1s2 2s2 2p63s2 3p3, The orbital picture of P can be represented as:, From the orbital picture, phosphorus has three unpaired electrons., (b) Silicon (Si):, Atomic number = 14, The electronic configuration of Si is:1s2 2s2 2p6 3s23p2, The orbital picture of Si can be represented as:, From the orbital picture, silicon has two unpaired electrons., (c) Chromium (Cr):, Atomic number = 24, 33

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The electronic configuration of Cr is:1s2 2s2 2p6 3s23p6 4s1 3d5, The orbital picture of chromium is:, From the orbital picture, chromium has six unpaired electrons., (d) Iron (Fe):, Atomic number = 26, The electronic configuration is:1s2 2s2 2p6 3s23p6 4s2 3d6, The orbital picture of chromium is:, From the orbital picture, iron has four unpaired electrons., HOTS QUESTIONS WITH ANSWERS, 1. Give the name and atomic number of the inert gas atom in which the total, number of d-electrons is equal to the difference between the numbers of total p, and total s electrons., Ans. electronic configuration of Kr ( atomic no.=36) =1s2 2s2 2p6 3s23p63d104s24p6, Total no. of s-electrons = 8, total no. of p-electrons = 18. Difference = 10, No. of d- electrons = 10, 2. What is the minimum product of uncertainty in position and momentum of an, electron?, Ans.h/4π, 3. Which orbital is non-directional ?, Ans. s- orbital, 4. What is the difference between the notations l and L ?, Ans. l represents the sub-shell and L represent shell., 5. How many electrons in an atom can have n + l = 6 ?, Ans. 18, 6. An anion A3+has 18 electrons. Write the atomic number of A., Ans.15, 7. Arrange the electron (e), protons (p) and alpha particle (α) in the increasing, order for the values of e/m (charge/mass)., Ans..α<p < e, , 34