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Atoms, Recap Notes, , Rutherford’s a-scattering experiment, X Rutherford and his two associates, Geiger, and Marsden, studies the scattering of the, a-particles from a thin gold foil in order to, investigate the structure of the atom., , at various angles from 0 to p. This, shows that atom has a small positively, charged core called ‘nucleus’ at centre, of atom, which deflects the positively, charged a-particles at different angles, depending on their distance from, centre of nucleus., – Very few a-particles (1 in 8000) suffers, deflection of 180°. This shows that size, of nucleus is very small, nearly 1/8000, times the size of atom., N(), , Thomson’s model of, atom : It was proposed, by J. J. Thomson in, 1898. According to this, model, the positive charge of the atom is, uniformly distributed throughout the volume, of the atom and the negatively charged, electrons are embedded in it like seeds in a, watermelon., , 0°, ,  = 180° , ,  his graph shows deflection of number of, T, particles with angle of deflection q., , X, , Rutherford’s observations and results :, – Most of the a-particles pass through, the gold foil without any deflection., This shows that most of the space in an, atom is empty., – Few a-particles got scattered, deflecting, , Rutherford’s a-scattering formulae, X Number of a particles scattered per unit, area, N(q) at scattering angle q varies, inversely as sin4(q/2),, 1, i.e., N(θ) ∝ 4, sin (θ / 2), X Impact parameter : It is defined as, the perpendicular distance of the initial, velocity vector of the alpha particle from, the centre of the nucleus, when the, particle is far away from the nucleus of, the atom., – The scattering angle q of the a particle, and impact parameter b are related as, Ze 2 cot(θ / 2), b=, 4 πε0 K, where K is the kinetic energy of the, a-particle and Z is the atomic number of, the nucleus.

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X, , – Smaller the impact parameter, larger, the angle of scattering q., Distance of closest approach : At the, distance of closest approach whole kinetic, energy of the alpha particles is converted, into potential energy., – Distance of closest approach, 2Ze 2, r0 =, 4 πε0 K, , Rutherford’s nuclear model of the atom :, According to this model the entire positive, charge and most of the mass of the atom is, concentrated in a small volume known as the, nucleus with electrons revolving around it, just as planets revolve around the sun., Bohr’s model : Bohr combined classical and early, quantum concepts and gave his theory of hydrogen, and hydrogen-like atoms which have only one orbital, electron. His postulates are, X An electron can revolve around the nucleus, only in certain allowed circular orbits of definite, energy and in these orbits it does not, radiate. These orbits are known as, stationary orbits., X Angular momentum of the electron in a, stationary orbit is an integral multiple of, h/2p., nh, nh, i.e., L =, or mvr =, 2π, 2π, This is known as Bohr’s quantisation condition., X The emission of radiation takes place, when an electron makes a transition from, a higher to a lower orbit. The frequency of, the radiation is given by, E − E1, υ= 2, h, where E2 and E1 are the energies of the, electron in the higher and lower orbits, respectively., , =, , 13.6Z 2, , eV, n2, The potential energy of the electron in, the nth orbit, 2,  1  4 π2 me 4 Z 2, 1 Ze 2, Un = −, = −, 4 πε0 rn,  4 πε0 , n2 h2, =, , −27.2Z 2, , eV, n2, X Total energy of electron in the nth orbit, 2, ,  1  2 π2 me 4 Z 2, En = U n + K n = − ,  4 πε0 , n2 h2, 2, 13.6Z, =−, eV, n2, Kn = –En, Un = 2En = –2Kn, X Frequency of the electron in the nth orbit, 2, ,  1  4 π2 Z 2 e 4 m 6.62 × 1015 Z 2, =, υn = ,  4 πε0 , n3h3, n3, X Wavelength of radiation in the transition, 1, 1, 1, from n2 → n1 is given by = RZ 2  2 − 2 , λ,  n1 n2 , where R is called Rydberg’s constant.,  1  2 π2 me 4, R=, = 1.097 × 107 m −1,  4 πε0  ch3, , Spectral series of hydrogen atom :, When the electron in a H-atom jumps from, higher energy level to lower energy level,, the difference of energies of the two energy, levels is emitted as radiation of particular, wavelength, known as spectral line. Spectral, lines of different wavelengths are obtained, for transition of electron between two, different energy levels, which are found to, fall in a number of spectral series given by, , Bohr’s formulae, X Radius of nth orbit, 4 πε0n2 h2, 0.53n2, ; rn =, rn =, Å, Z, 4 π2 mZe 2, X, , Velocity of the electron in the nth orbit, 1 2 πZe 2 2.2 × 106 Z, m/s, vn =, =, 4 πε0 nh, n, The kinetic energy of the electron in the, nth orbit, 2,  1  2 π2 me 4 Z 2, 1 Ze 2, Kn =, =, 4 πε0 2rn,  4 πε0 , n2 h 2, , X, , Lyman series, – Emission spectral lines corresponding, to the transition of electron from higher, energy levels (n2 = 2, 3, ...,∞) to first energy, level (n1 = 1) constitute Lyman series., 1, 1, 1, = R  2 − 2  where n2 = 2, 3, 4, ......,∞, λ, 1 n2 

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– Series limit line (shortest wavelength), of Lyman series is given by, 1 , 1, 1, 1, = R  2 − 2  = R or λ =, λ, R, 1, ∞ , – The first line (longest wavelength) of, the Lyman series is given by, 1, 1  3R, 4, 1, =R 2 − 2=, or λ =, λ, 4, 3, R, 1, 2 , – Lyman series lie in the ultraviolet, region of electromagnetic spectrum., – Lyman series is obtained in emission, as well as in absorption spectrum., X Balmer series, – Emission spectral lines corresponding, to the transition of electron from higher, energy levels (n2 = 3, 4, ....∞) to second energy, level (n1 = 2) constitute Balmer series., 1, 1, 1, = R  2 − 2  where n2 = 3, 4, 5.........,∞, λ, n2 , 2, – Series limit line (shortest wavelength), of Balmer series is given by, 1, 1  R, 4, 1, = R  2 − 2  = or λ =, λ, 4, R, 2, ∞ , , – The first line (longest wavelength) of, the Balmer series is given by, , 1, 1  5R, 36, 1, =R 2 − 2=, or λ =, λ, 5R, 3  36, 2, – Balmer series lie in the visible region of, electromagnetic spectrum., – This series is obtained only in emission, spectrum., X Paschen series, – Emission spectral lines corresponding, to the transition of electron from, higher energy levels (n2 = 4, 5, .....,∞), to third energy level (n1 = 3) constitute, Paschen series., 1, 1, 1, = R  2 − 2  where n2 = 4, 5, 6.........,∞, λ, n2 , 3, , – Series limit line (shortest wavelength), of the Paschen series is given by, 1, 1  R, 9, 1, =R 2 − 2=, or λ =, λ, R, 3, ∞  9, The first line (longest wavelength) of the, Paschen series is given by, 1, 1  7R, 144, 1, =R 2 − 2=, or λ =, λ, 7R, 3, 4  144, – Paschen series lie in the infrared region, of the electromagnetic spectrum., – This series is obtained only in the, emission spectrum., , Brackett series, – Emission spectral lines corresponding, to the transition of electron from higher, energy levels (n2 = 5, 6, 7, ..... ,∞) to, fourth energy level (n1 = 4) constitute, Brackett series., 1, 1, 1, = R  2 − 2  where n2 = 5, 6, 7..........,∞, λ, n, 4, , 2, Series limit line (shortest wavelength) of, Brackett series is given by, 1, 1  R, 16, 1, =R 2 − 2=, or λ =, λ, R, 4, ∞  16, – The first line (longest wavelength) of, Brackett series is given by, 1, 1  9R, 400, 1, =R 2 − 2=, or λ =, λ, 9R, 4, 5  400, – Brackett series lie in the infrared region, of the electromagnetic spectrum., – This series is obtained only in the, emission spectrum., X Pfund series, – Emission spectral lines corresponding, to the transition of electron from higher, energy levels (n2 = 6, 7, 8,.......,∞) to, fifth energy level (n1 = 5) constitute, Pfund series., 1, 1, 1, = R  2 − 2  where n2 = 6, 7,...........,∞, λ, n2 , 5, X, , – Series limit line (shortest wavelength), of Pfund series is given by, 1, 1  R, 25, 1, =R 2 − 2=, or λ =, λ, R, 5, ∞  25, – The first line (longest wavelength) of, the Pfund series is given by, 1, 1  11R, 900, 1, =R 2 − 2=, or λ =, λ, 11R, 5, 6  900, – Pfund series also lie in the infrared, region of electromagnetic spectrum., – This series is obtained only in the, emission spectrum., X Number of spectral lines due to transition, of electron from nth orbit to lower orbit is, n(n − 1), N=, 2, Ionization energy and ionization potential, X Ionisation : The process of knocking, an electron out of the atom is called, 13.6, ionisation. ionisation energy = 2 eV, n, X Ionisation energy : The energy required, to, knock an electron completely out of the, atom., 13.6Z 2, X Ionisation potential =, V, n2

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OBJECTIVE TYPE QUESTIONS, , Multiple Choice Questions (MCQs), 1. The first model of atom in 1898 was proposed, by, (a) Ernest Rutherford (b) Albert Einstein, (c) J. J. Thomson, (d) Niels Bohr, 2. The transition from the state n = 3 to n = 1, in a hydrogen like atom results in ultraviolet, radiation. Infrared radiation will be obtained in, the transition from, (a) 2 → 1, (b) 3 → 2, (c) 4 → 2, (d) 4 → 3, 3. In a hydrogen atom the total energy of, electron is, e2, 4πε 0 r, −e2, (c), 8πε 0 r, (a), , (b), , −e2, 4πε 0 r, , (d), , e2, 8πε 0 r, , 4. The relation between the orbit radius and, the electron velocity for a dynamically stable, orbit in a hydrogen atom is (where, all notations, have their usual meanings), (a) v =, , 4 π ε0, me2 r, , (b) r =, , e2, 4πε 0 v, , (c) v =, , e2, 4πε 0 mr, , (d) r =, , v e2, 4πε 0 m, , 5. In the Geiger-Marsden scattering experiment, the number of scattered particles detected are, maximum and minimum at the scattering angles, respectively at, (a) 0°° and 180°°, (b) 180°° and 0°°, (c) 90°° and 180°°, (d) 45°° and 90°°, , postulate applies to a satellite just as it does, to an electron in the hydrogen atom, then the, quantum number of the orbit of satellite is, (a) 5.3 × 1040, (b) 5.3 × 1045, 48, (c) 7.8 × 10, (d) 7.8 × 1050, 8. Which of the following is not correct about, Bohrs model of the hydrogen atom?, (a) An electron in an atom could revolve in, certain stable orbits without the emission, of radiant energy., (b) Electron revolves around the nucleus only in, those orbits for which angular momentum, nh, Ln =, ., 2π, (c) When electron make a transition from one of, its stable orbit to lower orbit then a photon, emitted with energy hu = Ef – Ei., (d) Bohr model is applicable to all atoms., 9. The shortest wavelength present in the, Paschen series of spectral lines is, (a) 720 nm, (b) 790 nm, (c) 800 nm, (d) 820 nm, 10. In a Geiger-Marsden experiment. Find the, distance of closest approach to the nucleus of a, 7.7 MeV a-particle before it comes momentarily, to rest and reverses its direction., (Z for gold nucleus = 79), (a) 10 fm, (b) 20 fm, (c) 30 fm, (d) 40 fm, , 6. Rutherford’s experiments suggested that the, size of the nucleus is about, (a) 10–14 m to 10–12 m (b) 10–15 m to 10–13 m, (c) 10–15 m to 10–14 m (d) 10–15 m to 10–12 m, , 11. If an electron in hydrogen atom is revolving, in a circular track of radius 5.3 × 10–11 m with, a velocity of 2.2 × 106 m s–1 around the proton, then the frequency of electron moving around, the proton is, (a) 6.6 × 1012 Hz, (b) 3.3 × 1015 Hz, (c) 3.3 × 1012 Hz, (d) 6.6 × 1015 Hz, , 7. A 10 kg satellite circles earth once every, 2 h in an orbit having a radius of 8000 km., Assuming that Bohr ’s angular momentum, , 12. The Rydberg formula, for the spectrum of, the hydrogen atom where all terms have their, usual meaning is

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4, (a) hυif = me, 2 2, 8ε 0 h, ,  1 1, n −n ,  f, i, , 4, , , (b) hυif = me  1 − 1 , 2 2, 2, 2, 8ε 0 h  nf, ni , , (c) hυif =, , 8 ε 0 2 h2  1 1 , −, me4  nf ni , , (d) hυif =, , 8 ε 0 2 h2, me4, ,  1 1,  2 − 2,  nf ni , , 13. In which of the following systems will the, radius of the first orbit (n = 1) be minimum ?, (a) Doubly ionized lithium., (b) Singly ionized helium., (c) Deuterium atom., (d) Hydrogen atom., 14. An electron in a hydrogen atom makes a, transition from n = n1 to n = n2. The time period, of the electron in the initial state is eight times, that in the final state. The possible values of n1, and n2 are, (a) n1 = 4, n2 = 2, (b) n1 = 8, n2 = 2, (c) n1 = 8, n2 = 1, (d) n1 = 6, n2 = 2, 15. Energy is absorbed in the hydrogen atom, giving absorption spectra when transition takes, place from, (a) n = 1 → n′ where n′ > 1, (b) n = 2 → 1, (c) n′ → n, (d) n → n′ = ∞, 16. The value of ionisation energy of the, hydrogen atom is, (a) 3.4 eV, (b) 10.4 eV, (c) 12.09 eV, (d) 13.6 eV, 17. The moment of momentum for an electron, in second orbit of hydrogen atom as per Bohr’s, model is, 2h, π, h, (a), (b) 2h, (c), (d), π, h, π, 18. In the Geiger-Marsden scattering experiment,, in case of head-on collision the impact parameter, should be, (a) maximum, (b) minimum, (c) infinite, (d) zero, 19. In an atom the ratio of radius of orbit of, electron to the radius of nucleus is, , (a) 103, (c) 105, , (b) 104, (d) 106, , 20. The radius of nth orbit rn in terms of Bohr, radius (a0) for a hydrogen atom is given by the, relation, (a) na0, (b) n a0, (c) n2a0, (d) n3a0, 21. In the Bohr model of the hydrogen atom, the, lowest orbit corresponds to, (a) infinite energy, (b) maximum energy, (c) minimum energy, (d) zero energy., 22. In an experiment on a-particle scattering,, a-particles are directed towards a gold foil and, detectors are placed in position P,Q and R. What, is the distribution of a-particles as recorded at, P, Q and R ?, , (a), (b), (c), (d), , P, all, none, a few, most, , Q, none, none, some, some, , R, none, all, most, a few, , 23. From quantisation of angular momentum,, one gets for hydrogen atom, the radius of the, 2,  n2   h   4 π 2 ε 0 , nth orbit as rn = , , ,  2 ,  me   2π   e , For a hydrogen like atom of atomic number Z,, (a) the radius of the first orbit will be the same, (b) rn will be greater for larger Z values, (c) rn will be smaller for larger Z values, (d) none of these., 24. Bohr’s basic idea of discrete energy levels, in atoms and the process of emission of photons, from the higher levels to lower levels was, experimentally confirmed by experiments, performed by, (a) Michelson–Morley (b) Millikan, (c) Joule, (d) Franck and Hertz, 25. The binding energy of an electron in the, ground state of He is equal to 24.6 eV. The, energy required to remove both the electrons is, (a) 49.2 eV, (b) 54.4 eV, (c) 79 eV, (d) 108.8 eV

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26. Out of the following which one is not a, possible energy for a photon to be emitted, by hydrogen atom according to Bohr’s atomic, model?, (a) 0.65 eV, (b) 1.9 eV, (c) 11.1 eV, (d) 13.6 eV, 27. The ratio of the speed of the electron in the, ground state of hydrogen atom to the speed of, light in vacuum is, 1, 2, 1, 1, (b), (c), (d), 137, 237, 2, 237, 28. The diagram shows the energy levels for an, electron in a certain atom., Which transition shown represents the emission, of a photon with the most energy ?, (a), , (a) I, , (b) II, , (c) III, , (d) IV, , 29. If u1 is the frequency of the series limit of, Lyman series, u2 is the frequency of the first line, of Lyman series and u3 is the frequency of the, series limit of the Balmer series, then, (a) u1 – u2 = u3, (b) u1 = u2 – u3, (c), , 1, 1, 1, =, +, υ2 υ1 υ3, , (d), , 1, 1, 1, =, +, υ1 υ2 υ3, , 30. Suppose an electron is attracted towards the, origin by a force k/r, where k is a constant and r, is the distance of the electron from the origin. By, applying Bohr model to this system, the radius, of nth orbit of the electron is found to be rn and, the kinetic energy of the electron is found to be, Tn. Then which of the following is true?, 1, (a) Tn ∝ 2, n, (b) Tn is independent of n; rn ∝ n, 1, (c) Tn ∝ ; rn ∝ n, n, 1, (d) Tn ∝ and rn ∝ n2, n, 31. The first line of the Lyman series in a, hydrogen spectrum has a wavelength of 1210 Å., The corresponding line of a hydrogen-like atom, of Z = 11 is equal to, (a) 4000 Å (b) 100 Å (c) 40 Å, (d) 10 Å, , 32. The de-Broglie wavelength of an electron in, the first Bohr orbit is, (a) equal to one-fourth the circumference of the, first orbit, (b) equal to half the circumference of first orbit, (c) equal to twice the circumference of first, orbit, (d) equal to the circumference of the first orbit., 33. The excitation energy of Lyman last line is, (a) the same as ionisation energy, (b) the same as the last absorption line in, Lyman series, (c) both (a) and (b), (c) different from (a) and (b), 34. If the wavelength of the first line of the Balmer, series of hydrogen is 6561 Å, the wavelength of the, second line of the series should be, (a) 13122 Å, (b) 3280 Å, (c) 4860 Å, (d) 2187 Å, 35. The electric current I created by the electron, in the ground state of H atom using Bohr model, in terms of Bohr radius (a 0 ) and velocity of, electron in first orbit v0 is, v0, 2πa, ev0, 2πa, (a), (b), (c), (d), ev0, 2πa, v0, 2πa0, 36. If the radius of inner most electronic orbit, of a hydrogen atom is 5.3 × 10–11 m, then the, radii of n = 2 orbit is, (a) 1.12 Å, (b) 2.12 Å, (c) 3.22 Å, (d) 4.54 Å, 37. The wavelength limit present in the Pfund, series is (R = 1.097 × 107 m–1), (a) 1572 nm, (b) 1898 nm, (c) 2278 nm, (d) 2535 nm, 38. An electron is revolving in the nth orbit of, radius 4.2 Å, then the value of n is (r1 = 0.529 Å), (a) 4, (b) 5, (c) 6, (d) 3, 39. A hydrogen atom initially in the ground, level absorbs a photon and is excited to n = 4, level then the wavelength of photon is, (a) 790 Å, (b) 870 Å, (c) 970 Å, (d) 1070 Å, 40. If muonic hydrogen atom is an atom in which, a negatively charged muon (m) of mass about, 207 me revolves around a proton, then first Bohr, radius of this atom is (re = 0.53 × 10–10 m), (a) 2.56 × 10–10 m, (b) 2.56 × 10–11 m, –12, (c) 2.56 × 10, m, (d) 2.56 × 10–13 m

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Case Based MCQs, Case I : Read the passage given below and answer, the following questions from 41 to 43., Electron Transitions for the Hydrogen Atom, Read the Bohr’s model explains the spectral lines, of hydrogen atomic emission spectrum. While the, electron of the atom remains in the ground state,, its energy is unchanged. When the atom absorbs, one or more quanta of energy, the electrons, moves from the ground state orbit to an excited, state orbit that is farther away., , the orbit and n = 1, 2, 3, .... When transition takes, place from Kth orbit to Jth orbit, energy photon is, emitted. If the wavelength of the emitted photon, 1, 1 ,  1, = R  2 − 2  , where R is, is l, we find that, λ, K , J, Rydberg’s constant., On a different planet, the hydrogen atom’s, structure was somewhat different from ours. The, angular momentum of electron was P = 2n(h/2p),, i.e., an even multiple of (h/2p)., , e–, m, Ze, , The given figure shows an energy level diagram, of the hydrogen atom. Several transitions are, marked as I, II, III and so on. The diagram is, only indicative and not to scale., 41. In which transition is a Balmer series photon, absorbed?, (a) II, (b) III, (c) IV, (d) VI, 42. The wavelength of the radiation involved in, transition II is, (a) 291 nm, (b) 364 nm, (c) 487 nm, (d) 652 nm, 43. Which transition will occur when a hydrogen, atom is irradiated with radiation of wavelength, 1030 nm?, (a) I, (b) II, (c) IV, (d) V, Case II : Read the passage given below and, answer the following questions from 44 to 48., Second Postulate of Bohr’s Theory, Hydrogen is the simplest atom of nature. There, is one proton in its nucleus and an electron moves, around the nucleus in a circular orbit. According, to Niels Bohr, this electron moves in a stationary, orbit. When this electron is in the stationary, orbit, it emits no electromagnetic radiation. The, angular momentum of the electron is quantized,, i.e., mvr = (nh/2p), where m = mass of the electron,, v = velocity of the electron in the orbit, r = radius of, , F, , vn, rn, , 44. The minimum permissible radius of the, orbit will be, ε 0 h2, 4 ε 0 h2, ε 0 h2, 2ε 0 h 2, (a), (b), (c), (d), mπe2, mπe2, 2mπe2, mπe2, 45. In our world, the velocity of electron is v0, when the hydrogen atom is in the ground state., The velocity of electron in this state on the other, planet should be, (a) v0, (b) v0/2, (c) v0/4, (d) v0/8, 46. In our world, the ionization potential energy, of a hydrogen atom is 13.6 eV. On the other, planet, this ionization potential energy will be, (a) 13.6 eV, (b) 3.4 eV, (c) 1.5 eV, (d) 0.85 eV, 47. Check the correctness of the following, statements about the Bohr model of hydrogen, atom., (i) The acceleration of the electron in n = 2 orbit, is more than that in n = 1 orbit., (ii) The angular momentum of the electron in, n = 2 orbit is more than that in n = 1 orbit., (iii) The kinetic energy of the electron in n = 2, orbit is less than that in n = 1 orbit.

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(a), (b), (c), (d), , Only (iii) and (i) are correct., Only (i) and (ii) are correct., Only (ii) and (iii) are correct., All the statements are correct., , 48. In Bohr’s model of hydrogen atom, let PE, represent potential energy and TE the total, energy. In going to a higher orbit, (a) PE increases, TE decreases, (b) PE decreases, TE increases, (c) PE increases, TE increases, (d) PE decreases, TE decreases, Case III : Read the passage given below and, answer the following questions from 49 to 52., Hydrogen Emission Spectrum, Hydrogen spectrum consists of discrete bright, lines in a dark background and it is specifically, known as hydrogen emission spectrum. There, is one more type of hydrogen spectrum that, exists where we get dark lines on the bright, background, it is known as absorption spectrum., Balmer found an empirical formula by the, observation of a small part of this spectrum and, it is represented by, , 1, 1,  1, = R  2 − 2  , where n = 3, 4, 5, ..... ., , λ, n , 2, For Lyman series, the emission is from first state, to nth state, for Paschen series, it is from third, state to nth state, for Brackett series, it is from, fourth state to nth state and for Pfund series, it is, from fifth state to nth state., 49. Number of spectral lines in hydrogen atom, is, (a) 8, (b) 6, (c) 15, (d) ∞, 50. W h i c h se ri e s o f h y dr o ge n s p ec t rum, corresponds to ultraviolet region?, (a) Balmer series, (b) Brackett series, (c) Paschen series, (d) Lyman series, 51. Which of the following lines of the H-atom, spectrum belongs to the Balmer series?, (a) 1025 Å, (b) 1218 Å, (c) 4861 Å, (d) 18751 Å, 52., (a), (b), (c), (d), , Rydberg constant is, a universal constant, same for same elements, different for different elements, none of these., , Assertion & Reasoning Based MCQs, For question numbers 53-60, two statements are given-one labelled Assertion (A) and the other labelled Reason (R)., Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below., (a) Both A and R are true and R is the correct explanation of A, (b) Both A and R are true but R is NOT the correct explanation of A, (c) A is true but R is false, (d) A is false and R is also false, 53. Assertion (A) : The force of repulsion, between atomic nucleus and a-particle varies, with distance according to inverse square law., Reason (R) : Rutherford did a-particle scattering, experiment., 54. Assertion (A) : According to classical theory,, the proposed path of an electron in Rutherford, atom model will be circular., Reason (R) : According to electromagnetic theory, an accelerated particle continuously emits, radiation., 55. Assertion (A) : Between any two given energy, levels, the number of absorption transitions, is always less than the number of emission, transitions., Reason (R) : Absorption transitions start from, , the lowest energy level only and may end at any, higher energy level. But emission transitions, may start from any higher energy level and end, at any energy level below it., , 56. Assertion (A) : Electrons in the atom are held, due to coulomb forces., Reason (R) : The atom is stable only because, the centripetal force due to Coulomb’s law is, balanced by the centrifugal force., 57. Assertion (A) : Total energy of revolving, electron in any stationary orbit is negative., Reason (R) : Energy is a scalar quantity. It can, have positive or negative value., 58. Assertion (A) : Balmer series lies in the, visible region of electromagnetic spectrum.

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1, 1 ,  1, Reason (R) :, = R  2 − 2  , where K = 3, 4,, , λ, 2, K , 5, ..., 59. Assertion (A) : For the scattering of a-particles, at large angles, only the nucleus of the atom is, responsible., , Reason (R) : Nucleus is very heavy in comparison, to electrons., 60. Assertion (A) : Fraunhofer lines are observed, in the spectrum of the sun., Reason (R) : The different elements have different, spectra., , SUBJECTIVE TYPE QUESTIONS, , Very Short Answer Type Questions (VSA), 1. Why is the classical (Rutherford) model for, an atom of electron orbiting around the nucleus, not able to explain the atomic structure?, 2. When is Hα line of the Balmer series in the, emission spectrum of hydrogen atom obtained?, 3. What is the maximum number of spectral, lines emitted by a hydrogen atom when it is in, the third excited state?, 4. Find the radius and energy of a He+ ion in, the states n = 2., 5. State Bohr ’s quantization condition of, angular momentum., , 6., , Find the wavelength of the IInd line of, , Balmer series., 7., , What is the wavelength of the electronic de, , Broglie wave in the 3rd orbit of hydrogen?, 8., , Find the radius of the ground state orbit of, , hydrogen atom., 9., , Find the speed in the ground state, , 10. Calculate the shortest wavelength of the, Brackett series and state to which part of the, electromagnetic spectrum does it belong., , Short Answer Type Questions (SA-I), 11. Find the number of unique radiations that, can be emitted for a sample of hydrogen atoms, excited to the nth level., 12. An electron in the hydrogen atom makes, a transition from n = 2 energy state to the, ground state (corresponding to n = 1). Find the, wavelength and frequency of the emitted photon., 13. Use the Bohr ’s model to estimate the, wavelength of the Kα line in the X-ray spectrum, of platinum (Z = 78)., 14. The radius of the innermost electron orbit, of a hydrogen atom is 5.3 × 10–11 m. What are, the radii of the n = 2 and n = 3 orbits?, 15. If Bohr’s quantisation postulate (angular, momentum = nh/2π) is a basic law of nature,, it should be equally valid for the case of planetary, motion also. Why then do we never speak of, quantisation of orbits of planets around the sun?, 16. Would the Bohr formula for the H-atom, remain unchanged if proton had a charge, , (+4/3)e and electron a charge (–3/4)e, where, e = 1.6 × 10–19C. Give reasons for your answer., 17. Consider a gas consisting Li ++ (which is, hydrogen like ion)., (i) Find the wavelength of radiation required, to excite the electron in Li++ from n = 1 and, n = 3., (ii) How many spectral lines are observed in, the emission spectrum of the above excited, system?, 18. Using Bohr model, calculate the electric, current created by the electron when the H-atom, is in the ground state., 19. The kinetic energy of the electron orbiting in, the first excited state of hydrogen atom is 3.4 eV., Determine the de Broglie wavelength associated, with it., 20. Calculate the shortest wavelength in the, Balmer series of hydrogen atom. In which, region (infrared), visible, ultraviolet of hydrogen, spectrum does this wavelength lie?

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Short Answer Type Questions (SA-II), 21. Assume that their is no repulsive force, between the electrons in an atom but the, force between positive and negative charges is, given by Coulomb’s law as usual. Under such, circumstances, calculate the ground state energy, of a He-atom., 22. Show that the first few frequencies of light, that is emitted when electrons fall to the n th, level from levels higher than n, are approximate, harmonics (i.e. in the ratio 1 : 2 : 3...) when, n > >1., 23. What is the minimum energy that must be, given to a H atom in ground state so that it can, emit an Hγ line in Balmer series. If the angular, momentum of the system is conserved, what, would be the angular momentum of such H γ, photon?, 24. The electron in a given Bohr orbit has a total, energy of –1.5 eV. Calculate its, (i) kinetic energy. (ii) potential energy., (iii) wavelength of radiation emitted, when this, electron makes a transition to the ground, state., [Given : Energy in the ground state = –13.6 eV, and Rydberg’s constant = 1.09 × 107 m–1], 25. A 12.5 eV electron beam is used to bombard, gaseous hydrogen at room temperature. Upto, which energy level the hydrogen atoms would, be excited?, Calculate the wavelengths of the first member of, Lyman and first member of Balmer series., 26. The value of ground state energy of hydrogen, atom is –13.6 eV., (i) Find the energy required to move an electron, from the ground state to the first excited, state of the atom., (ii) Determine (a) the kinetic energy and, (b) orbital radius in the first excited state of, the atom., (Given the value of Bohr radius = 0.53 Å)., 27. (i) , How does de-Broglie hypothesis explain, Bohr ’s quantization condition for, stationary orbits ?, (ii) , Find the relation between the three, wavelengths l1, l2 and l3 from the energy, level diagram shown in the figure., , 28. Show that the radius of the orbit in hydrogen, atom varies as n 2 , where n is the principal, quantum number of the atom., 29. Direction : Read the following passage and, answer the questions given below., In 1911, Rutherford, along with his assistants,, H. Geiger and E. Marsden, performed the alpha, particle scattering experiment. H. Geiger and, E. Marsden took radioactive source (214, 83Bi) for, a-particles. A collimated beam of a-particles, of energy 5.5 MeV was allowed to fall on, 2.1 × 10–7 m thick gold foil. Observations of this, experiment are as follows, (I) Most of the a-particles passed through the, foil without deflection., (II) Only about 0.14% of the incident a-particles, scattered by more than 1°., (III) Only about one a-particle in every 8000, a-particles deflected by more than 90°., Most -particles pass, straight through, -particles, About 1 in 8000 -particles, is repelled back, , (i) Gold foil used in Geiger-Marsden experiment, is about 10–8 m thick. What does it ensures?, (ii) On which factor, the trajectory traced by an, a-particle depends?, (iii) In Rutherford scattering experiment the, fact that only a small fraction of the number, of incident particles rebound back. What it, indicates?, 30. Positronium is just like a H-atom with, the proton replaced by the positively charged, antiparticle of the electron (called the positron, which is as massive as the electron). What would, be the ground state energy of positronium?

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31. The photons from Balmer series in hydrogen, spectrum having wavelength between 450 nm to, 700 nm are incident on a metal surface of work, function 2 eV. Find the maximum kinetic energy, of one photoelectron., 32. A s m a l l p a r t i c l e o f m a s s m m o v e s, in such a way that the potential energy, U = ar2 where a is a constant and r is the distance, of the particle from the origin. Assuming Bohr’s, model of quantisation of angular momentum for, circular orbits, find the radius of nth allowed, orbit., 33. A particle known as µ-meson, has a charge, equal to that of an electron and mass 208 times, the mass of the electron. It moves in a circular, orbit around a nucleus of charge +3e. Take the, , mass of the nucleus to be infinite. Assuming that, the Bohr’s model is applicable to this system,, (i) derive an expression for the radius of the, nth Bohr orbit,, (ii) find the value of n for which the radius of, the orbit is approximately the same as that, of the first Bohr orbit for a hydrogen atom., 34. A gas of hydrogen like atoms can absorb, radiations of 68 eV. Consequently, the atoms emit, radiations of only three different wavelength. All, the wavelengths are equal or smaller than that, of the absorbed photon., (i) Determine the initial state of the gas atoms., (ii) Identify the gas atoms., (iii) Find the minimum wavelength of the, emitted radiations., , Long Answer Type Questions (LA), 35. In the Auger process an atom makes a, transition to a lower state without emitting a, photon. The excess energy is transferred to an, outer electron which may be ejected by the atom., (This is called an Auger electron). Assuming the, nucleus to be massive, calculate the kinetic, energy of an n = 4 Auger electron emitted by, Chromium by absorbing the energy from a, n = 2 to n = 1 transition., , 37. Using Bohr ’s postulates, derive the, expression for the frequency of radiation emitted, when electron in hydrogen atom undergoes, transition from higher energy state (quantum, number ni) to the lower state, (nf). When electron, in hydrogen atom jumps from energy state, ni = 4 to nf = 3, 2, 1. Identify the spectral series, to which the emission lines belong., , 36. (a) Write two important limitations of, Rutherford model which could not explain the, observed features of atomic spectra., (b) How were these explained in Bohr’s model, of hydrogen atom?, , 38. The first four spectral lines in the Lyman, series of a H-atom are λ = 1218 Å, 1028 Å, 974.3 Å, and 951.4 Å. If instead of Hydrogen, we consider, Deuterium, calculate the shift in the wavelength, of these lines., , OBJECTIVE TYPE QUESTIONS, 1. (c) : Sir J. J. Thomson proposed the first model of, atom called plum pudding model of atom. According to, this model, the positive charge of the atom is uniformly, distributed throughout the volume of the atom and the, negatively charged electrons are embedded in it like seeds, in a watermelon., 2., , (d), , 3., , (c) : The kinetic energy of the electron in hydrogen atom are, , e2, 1, K = mv 2 =, 2, 8 πε 0 r, , , e2 , 2, ∵ v =, , 4 πε 0 mr , , , Electrostatic potential energy,, − e2, and U =, 4 π ε0 r, ⇒, , The total energy E of the electron in a hydrogen atom is, E=K+U,  −e 2 , e2, e2, =, −, E =, +, 8 π ε 0 r  4 π ε 0 r , 8 π ε0 r, Here negative sign shows that electron is bound to the, nucleus., 4. (c) : In hydrogen atom electrostatic force of attraction, (Fe) is acting between the revolving electrons and the nucleus

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provides the requisite centripetal force (Fc) to keep them in, their orbits. Thus,, Fe = Fc, mv 2, 1 e2, ∴, =, r, 4 π ε0 r 2, 2, , 2, , e, e, ⇒ v =, 4 π ε0 m r, 4 π ε0 m r, 5. (a) : The number of scattered particles detected will be, maximum at the angle of scattering q = 0° and minimum at, q = 180°., 6. (c), 7. (b) : Here, m = 10 kg, r n = 8 × 106 m, T = 2 × 60 × 60 = 7200 s, nh, 2π rn, Velocity of nth orbit, v n =, and from m v n rn =, 2π, T, 2π, 2π rn, n=, ×m×, × rn, h, T, m, (2π × 8 × 106 )2 × 10, 2, = (2 π rn ) ×, =, = 5.3 × 1045., T × h 7200 × 6.64 × 10 −34, or v 2 =, , 8. (d) : The first three options (a), (b) and (c) are Bohr’s, postulates of atomic model whereas option (d) is not correct, as Bohr’s model is applicable to hydrogen atom only., 1, 1 1, 9. (d) : The wavelength for Paschen series, = R  2 − 2 , λ, 3 n , For shortest wavelength n = ∞, 1, 1 R, 1, = R − 2 =, \, λ, 9 ∞  9, 9, 9, λ= =, R 1.097 × 107, = 8.20 × 10–7 m = 820 nm., 10. (c) : Let d be the distance of closest approach then by, the conservation of energy,, Initial kinetic energy of incoming a-particle, K, = Final electric potential energy U of the system, (2e )( Ze ), 1, ×, As K =, 4π ε0, d, 1 2 Z e2, ...(i), 4 π ε 0 K , 1, = 9 × 109 N m2 C−2 , Z = 79, e = 1.6 × 10 −19 C., Here,, 4 π ε0, ∴, , d =, , K = 7.7 MeV = 7.7 × 106 × 1.6 × 10 –19 J = 1.2 × 10–12 J, Substituting these values in (i), 2 × 9 × 109 × (1.6 × 10 −19 )2 × 79, d=, 1.2 × 10 −12, –14, d = 3 × 10 m, = 30 fm, (∵ 1 fm = 10 −15 m), , 11. (d) : Frequency of the electron moving around the, proton is, velocity of electron (v ), υ=, circumference (2 π r ), Here, v = 2.2 × 106 m s–1 and r = 5.30 × 10–11 m, ∴, , υ=, , 2.2 × 106, 2 × 3.14 × 5.3 × 10 −11, , u = 6.6 × 1015 Hz., , 12. (b), 1, ,, Z, For doubly ionized lithium, Z (= 3) will be maximum, hence, for doubly ionized lithium, r will be minimum., 14. (a) : In the nth orbit, let rn = radius and vn = speed of, electron., 2πr n rn, Time period, Tn =, ∝ ., vn, vn, 13. (a) : Radius of first orbit, r ∝, , Now, rn ∝ n 2 and v n ∝, , 1, n, , rn, ∝ n 3 or Tn ∝ n 3 ., vn, 3, n , Here, 8 =  1 ,  n2 , , ∴, , or, , n1, = 2 or n1 = 2n2, n2, , 15. (a) : Absorption is from the ground state n = 1 to n′, where n′ > 1., 16. (d) : The minimum energy required to free the electron, from the ground state of the hydrogen atom is called the, ionisation energy of hydrogen atom and its value is 13.6 eV., 17. (a) : According to Bohr’s second postulate, nh, Angular momentum, L =, 2π, Angular momentum is also called a moment of momentum., For second orbit, n = 2, 2h h, L=, =, 2π π, 18. (b) : At minimum impact parameter a particles rebound, back (q ≈ p) and suffers large scattering., 19. (c) : The radius of orbit of electrons = 10–10 m, radius of nucleus = 10–15 m, \, , Ratio =, , 10 −10, = 105, 10 −15, , Hence the radius of electron orbit is 105 times larger than, the radius of nucleus.

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20. (c) : The radius of nth orbit, 2 4 π ε 0, rn = n 2, me 2,  4 π ε0, = a0 (Bohr radius), m e2, 2, , where, , Hence, rn = n 2 a0 ., 21. (c) : In hydrogen atom, the lowest orbit corresponds to, minimum energy., 22. (c), 23. (c) : For an atom with a single electron, Bohr atom, model in applicable., As the value of attraction force between a proton and electron, e2, is proportional to e2, for an ion with a single electron,, 4πε 0, Ze 2, is replaced by, 4πε 0, n2, i.e. rn ∝ ., Z, 24. (d) : Franck and Hertz showed that exciting mercury, vapour by electrons of energy 4.9 eV and more, mercury lines, of energy 4.9 eV were obtained., , First they were excited to level B and then thec atoms emitted, spectral lines of 2530 Å i.e. 4.9 eV, Bohr’s concepts were, verified., 25. (c) : The energy needed to remove one electron from the, ground state of He = 24.6 eV., As the He + is now hydrogen-like, ionisation energy, 22, = | −13.6 | 2 eV ⇒ E = 54.4 eV, 1, \ To remove both the electrons, energy needed, = (54.4 + 24.6) eV = 79 eV, 26. (c) : The energy of nth orbit of hydrogen atom is given, as, 13.6, E n = − 2 eV, n, \ E1 = –13.6 eV, E2 = −, , 13.6, = −3.4 eV, 22, , \, , E3 = −, , 13.6, = −1.5 eV, 32, , E4 = −, , 13.6, = −0.85 eV, 42, , E3 – E2 = –1.5 – (–3.4) = 1.9 eV, E4 – E3 = –0.85 – (–1.5) = 0.65 eV, , 27. (c) : Speed of the electron in the ground state of, hydrogen atom is, 2πe 2, c, =, = cα, 4 πε 0 h 137, where, c = speed of light in vacuum,, v=, , e2, is the fine structure constant. It is a pure number, 2ε 0 hc, 1, ., whose value is, 137, v, 1, ∴ =, c 137, α=, , 28. (c) : Ist transition is showing absorption of a photon., From rest of three transitions, III is having maximum energy, from level n = 2 to n = 1,  1 1, ∆E ∝  2 − 2 ,  n1 n2 , 29. (a) : For Lyman series, 1 1 , υ = Rc  2 − 2 , 1 n , where n = 2, 3, 4,......., For the series limit of Lyman series, n = ∞, 1, 1, ∴ υ1 = Rc  2 − 2  = Rc, 1 ∞ , For the first line of Lyman series, n = 2, 1 1 3, ∴ υ2 = Rc  2 − 2  = Rc, 1 2  4 , For Balmer series, 1 1, υ = Rc  2 − 2 , 2 n , where n = 3, 4, 5...., For the series limit of Balmer series, n = ∞, 1  Rc, 1, ∴ υ3 = Rc  2 − 2  =,  2 ∞  4 , From equations (i), (ii) and (iii), we get, u1 = u2 + u3 or u1 – u2 = u3, , ...(i), , ...(ii), , ...(iii), , 30. (b) : Applying Bohr model to the given system,, mv 2 k, =, rn, rn , , ...(i)

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1, 40. (d) : According to Bohr’s atomic model, r ∝, m, rµ me, ⇒, =, , re, mµ, Here, re = 0.53 × 10–10 m, mµ = 207 me, me, ∴ rµ =, × 0.53 × 10 −10, 207 me, , ...(i), , 49. (d) : Number of spectral lines in hydrogen atom is ∞., 50. (d) : Lyman series lies in the ultraviolet region., (using (i)), , = 2.56 × 10–13 m., , 41. (d) : For Balmer series, n1 = 2; n2 = 3, 4,…,    (lower), , (higher), , Therefore, in transition (VI), photon of Balmer series is, absorbed., 42. (c) : In transition II,, E2 = –3.4 eV, E4 = –0.85 eV,, hc, hc, DE = 2.55 eV ⇒ DE =, ⇒l=, = 487 nm, λ, ∆E, 43. (d) : Wavelength of radiation = 1030 Å, 12400, DE =, = 12.0 eV, 1030 Å, So, difference of energy should be 12.0 eV (approx.), Hence for n1 = 1 to n2 = 3, E n3 − E n1 = −1.51 eV − ( −13.6 eV) ≈ 12 eV, , Therefore, transition V will occur., 44. (b) : On other planet : mvr = 2n, , h, nh, ⇒v =, 2π, πmr, , mv 2, 1 e2, mn 2h 2, 1 e2, =, ⇒, =, r, 4 πε0 r 2, n 2m 2r 3 4 πε0 r 2, , Putting n = 1, we get r =, , 4h 2 ε 0, mπe 2, , e2, 45. (b) : On our planet : v 0 =, 2ε 0 nh, v, e2, On other planet : v =, = 0, 2ε 0 (2n )h 2, , 46. (b) : On our planet : E n = −, On other planet : E n′ = −, , 13.6, (2n )2, , 13.6, n2, , En, = −3.4 eV, 4, 47. (c) : Centripetal acceleration = mv2/r, Further, as n increases, r also increases. Therefore, centripetal, acceleration for n = 2 is less than that for n = 1. So,, statement (i) is wrong. Statement (ii) and (iii) are correct., ⇒, , E n′ =, , 48. (c) : Potential energy = –C/r2 and, total energy = –Rhc/n2. With higher orbit, both r and, n increase. So, both become less negative; hence both, increase., , 51. (c) : The shortest Balmer line has energy, = |(3.4 – 1.51)| eV = 1.89 eV, and the highest energy = |(0 – 3.4)| = 3.4 eV, The corresponding wavelengths are, 12400 eV Å, 12400 eV Å, = 6561 Å and, = 3647 Å, 1.89 eV, 3.4 eV, Only 4861 Å is between the first and last line of the Balmer, series., , 52. (a), 53. (b) : In Rutherford’s α-particle scattering experiment,, some of α-particles were found to be scattered at very large, angles inspite of having very high kinetic energy. This shows, that there are α-particles which will be passing very close, to nucleus. Rutherford confirmed the repulsive force on, α-particles due to nucleus varies with distance according, to inverse square law and that the positive charges are, concentrated at the centre and not distributed throughout, the atom. This is the nuclear model of Rutherford., 54. (b) : According to classical electromagnetic theory, an, accelerated charge continuously emits radiation. As electrons, revolving in circular paths are constantly experiencing, centripetal acceleration, hence they will be losing their energy, continuously and the orbital radius will go on decreasing and, form spiral and finally the electron will fall into the nucleus., 55. (a) : Absorption transition, C, B, , Absorb, radiation, , A, , Two possibilities in absorption transition., Emission transition, C, B, A, , Three possibilities in emission transition. Therefore number of, absorption transition < number of emission transition., For any two states A and B such that EA < EB we have, absorption spectrum for A → B transition and emission B → A., But most of the time atoms are in ground state, absorption is, only from the ground state.

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56. (c) : According to postulates of Bohr’s atomic model,, the electron revolve around the nucleus in fixed orbit of, definite radii. As long as the electron is in a certain orbit it, does not radiate any energy. Not only the centripetal force, has to be the centrifugal force, even the stable orbits are, fixed by Bohr’s theory., , These lines correspond to Lyman series., 4., , r = 0.529, , n2, Z, , 22, = 1.058 Å, 2, 2, 2, E (n = 2) = −13.6 × 2 = −13.6 eV, 2, r (n = 2) = 0.529 ×, , 57. (b) : The reason is correct, but does not explain the, assertion properly. Negative energy of revolving electron, indicates that it is bound to the nucleus. The electron is not, free to leave the nucleus., 58. (b) : When we put R = 107 m–1 and K = 3, 4, 5 in, the given formula, values of l calculated lie between 4000 Å, and 8000 Å, which is the visible region. The reason is true,, but does not explain the assertion properly., 59. (a) : We know that an electron is very light particle as, compared to an α-particle. Hence electron cannot scatter the, α-particle at large angles, according to law of conservation of, momentum. On the other hand, mass of nucleus is comparable, with the mass of α-particle, hence only the nucleus of atom, is responsible for scattering of α-particles., 60. (b) : When white light from the photosphere (central, portion of the sun) passes through vapours of various, elements present in the outer chromosphere, then these, elements absorb those wavelengths which they themselves, emit to bring incandescent. Hence dark lines (absence of light), appear in the continuous solar spectrum, due to absorption, of these lines. Absorption is possible in the sun, not only, from the ground state but also in higher states because of, the high temperature of the sun., , SUBJECTIVE TYPE QUESTIONS, 1. According to electromagnetic theory, electron revolving, around the nucleus are continuously accelerated. Since an, accelerated charge emits energy, the radius of the circular path, of a revolving electron should go on decreasing and ultimately, it should fall into the nucleus. So, it could not explain the, structure of the atom. As matter is stable, we cannot expect, the atoms to collapse., 2. Hα line of the Balmer series in the emission spectrum, of hydrogen atoms obtained when the transition occurs from, n = 3 to n = 2 state., 3. Number of spectral lines obtained due to transition, of electron from n = 4 (3rd excited state) to n = 1 (ground, state) is, (4)(4 − 1), N=, =6, 2, , 5. Bohr’s quantization condition : The electron can revolve, round the nucleus only in those circular orbits in which, angular momentum of an electron is an integral multiple, h, of 2π, nh, i.e., mvr = , n = 1, 2, 3,...., 2π, 1, 1, 1 ,  1 1, = RZ 2  2 −, = RZ 2  − , 6., 2, 4 9, λB, (, ), +, 2, 2, 1, , , 2, , ⇒ λB 2 =, 7., , 36, 5RZ 2, , 2πr = nλ, , n2, Here, n = 3 ⇒ 2πr = 3λ and r = 0.53 Å, Z, 9, or r3 = 0.53   Å, 1, ⇒λ=, , 2πr3 2π, =, (0.53 × 9) Å ≈ 9.99 Å., 3, 3, , 8., , rn =, , 9., , v=, , n 2h 2, , =, , n2, (0.053 nm), Z, , 4π2mekZe 2, For our case, n = 1 and Z = 2, and the result is, r1 = 0.027 nm., , Ze 2, c Z m, =, 2 ε 0nh 137  n  s, , 2c, m/s., 137, 10. The shortest wavelength of Brackett series is given as, Here n = 1, Z = 2, v1 =, , 1, 1  1.097 × 107, 1, = 1.097 × 107  2 − 2  =, 16, λ, 4, ∞ , –6, ⇒ λ = 1.4585 × 10 m, This wavelength lies in the infrared region of electromagnetic, spectrum., 11. The first excited level is 2nd line., From the 2nd level electron can go to level 1 ⇒ one radiation

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3rd level electron can go to levels 1, 2 ⇒ three radiations, 4th level electron can go to levels 1, 2, 3 ⇒ six radiations, nth level electron can go to levels 1, 2, 3, ...(n – 1), ∴ Total number of radiations, (n − 1) ⋅ n, = 1 + 2 + ...... + (n – 1) =, ., 2, hc, 12. λ =, (E 2 − E1), , 1242 eV - nm, λ=, = 122 nm., (−3.4 eV) − (−13.6 eV), This wavelength lies in the ultraviolet region., Since, c = υ λ, the frequency of the photon is, υ=, , 13., , 8, , c 3.00 × 10 m / s, =, = 2.46 × 1015 Hz, λ 1.22 × 10 −7 m, , 1 1, 1, = R (Z )2  2 − 2 , λ, n n , 1, , 2, , For characteristic X-rays, we replace Z by Z – a or (Z – 1), here. For K α , the electron jumps to the K-shell, hence,, n1 = 1 and n2 = 2., 1, 1 1, ⇒ = R (Z − 1)2  2 − 2 , λ, 1 2 , , Force ∝ (–e) (e) = –e2, ,  4 , If charge on proton is  + e  and charge on electron is,  3 , 3, , , 2, 4  3 ,  − 4 e  , then their product  e   − e  = –e . Thus, 3  4 , Bohr Formula remains the same., , 1 1 , eV = 13.6 × 8 eV,  1 32 , , 17. (i) As, ∆E = 13.6 × 32 ×  −, ⇒, , λ=, , hc, 12400, =, Å = 113.7 Å, ∆E 13.6 × 8, , (ii) Number of spectral lines =, , n(n − 1) 3(3 − 1), =, =3, 2, 2, , 18. Let a0 = Bohr radius., v0 = velocity of electron in first orbit., ∴ Time taken by electron to complete one revolution, 2πa 0, T=, ., v0, e, ∴ Current created by electron, I =, T, e, ev, =, = 0 ., 2, πa 0, 2, π, a, , 0,  v , 0, , 1, 1 1, = (1.097 × 107 m–1) (78 – 1)2  2 − 2 , λ, 1 2 , 1, = 4.9 × 1010 m–1 or λ = 2 × 10–11 m = 0.2 Å., λ, , 19. Kinetic energy in the first excited state of hydrogen atom, EK = 3.4 eV = 3.4 × 1.6 × 10–19 J, h, de-Broglie wavelength, λ =, 2m E K, , 14. Radius of innermost electron, , =, , r=, , n 2h 2ε 0, πme 2, , 2, , h ε0, , = 5.3 × 10–11 m, πme 2, For n = 2, r2 = (2)2 r1 = 2.12 × 10–10 m, For n = 3, r3 = (3)2 r1 = 4.77 × 10–10 m., For n = 1, r1 =, , h, associated with, 2π, planetary motion are incomparably large relative to h., For example angular momentum of earth in its orbital motion, h, is of the order of 1070 ., 2π, For such large value of n, the difference in successive energies, and angular momenta of the quantised levels of the Bohr, model are so small that one can predict the energy level, continuous., 16. In Bohr’s formula,, 15. Angular momentum mvr = n, , mv 2, 1, =, (e)(−e), r, 4π ε 0, , 6.63 × 10 −34, 2 × 9.1 × 10 −31 × 3.4 × 1.6 × 10 −19, , = 0.67 nm, , 20. Wavelength (l) of Balmer series is given by, 1 1, 1, = R H  2 − 2  where ni = 3, 4,5,..., λ,  2 ni , For shortest wavelength, when transition of electrons take, place from ni = ∞ to nf = 2 orbit, wavelength of emitted, photon is shortest., 7,  1 1  1.097 × 10, = RH  2 −  =, 4, λ min, 2 ∞ , –7, \ lmin = 3.646 × 10 m = 3646 Å, This wavelength lies in visible region of electromagnetic, spectrum., 21. There are two protons and two neutrons in helium atom., Two electrons are revolving around the nucleus in first orbit., It is assumed that there is no interaction between the two, electrons to He-atom., so, replacing Z = 1 by Z = 2, , 1, ,  −13.6 Z 2 , En = ,  eV,, 2,  n, 

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For ground state (n = 1) of helium atom, energy,  −13.6 , E1 =  2 (2)2  eV = −54.4 eV,  1, , , Helium has two electrons in ground state, so total energy, E = 2E1 = –108.8 eV., 1, 1, 2 1, 22. As λ = RZ  2 − 2  ,,  nf n i , , 1, 1 , υ = cRZ 2  2 −, 2,  n (n + p ) , [as c = ul, nf = n, ni = (n + p), where p = 1, 2, 3...],  (n + p )2 − n 2 , or υ = cRZ 2  2, 2 ,  n (n + p ) ,  (n 2 + p 2 + 2pn) − n 2 , = cRZ 2 , , n 2(n + p )2, , , 2,  2pn   2cRZ , u ≈ cRZ2  4  ≈  3  p (∴ n >> 1, p << n), n,  n , , or, , obviously u ∝ p, i.e., the values of υ are approximately in, the ratio 1 : 2 : 3., 23. In Balmer series, Hγ line corresponds to transition from, state ni = 5 to state nf = 2., Energy required, E = E5 – E1,  −13.6   −13.6 , =  2  −  2  = 13.06 eV.,  5   1 , If angular momentum of system is conserved,, change in angular momentum of electron = change in angular, momentum of photon,  h   h  3h, = 5  −2  =,  2π   2π  2π, =, , 3 × 6.6 × 10 −34, = 3.17 × 10 −34 J s, 2 × 3.14, , 24. (i) The kinetic energy (Ek) of the electron in an orbit is, equal to negative of its total energy (E)., Ek = –E = – (–1.5) = 1.5 eV, (ii) The potential energy (Ep) of the electron in an orbit is, equal to twice of its total energy (E)., Ep = 2E = –1.5 × 2 = –3.0 eV, (iii) Here, ground state energy of the H-atom = –13.6 eV, When the electron goes from the excited state to the ground, state, energy emitted is given by, E = –1.5 – (–13.6) = 12.1 eV = 12.1 × 1.6 × 10–19 J, hc, Now,, E=, λ, λ=, , hc 6.62 × 10 −34 × 3 × 108, =, E, 12.1 × 1.6 × 10 −19, , λ = 1.025 × 10–7 = 1025 Å, 25. Here, ∆E = 12.5 eV, Energy of an electron in nth orbit of hydrogen atom is,, 13.6, E n = − 2 eV, n, In ground state, n = 1, E1 = –13.6 eV, Energy of an electron in the excited state after absorbing a, photon of 12.5 eV energy will be, En = –13.6 + 12.5 = –1.1 eV, −13.6 −13.6, ∴ n2 =, =, = 12.36 ⇒ n = 3.5, En, −1.1, Here, state of electron cannot be in fraction., So, n = 3 (2nd excited state)., The wavelength λ of the first member of Lyman series is, given by, 1, 1 1 3, =R  2 − 2 = R, λ, 1 2  4, 4, 4, ⇒ λ=, =, 3R 3 × 1.097 × 107, ⇒ λ = 1.215 × 10–7 m, ⇒ λ = 121 × 10–9 m ⇒ λ = 121 nm, The wavelength λ′ of the first member of the Balmer series, is given by, 1, 1 1 5, =R  2 − 2 = R, λ′,  2 3  36, ⇒ λ′ =, , 36, 36, =, 5R 5 × (1.097 × 107 ), , = 6.56 × 10–7 m = 656 × 10–9 m = 656 nm, −13.6, 26. (i) ∵ E n = 2 eV, n, Energy of the photon emitted during a transition of the, electron from the first excited state to its ground state is,, ∆E = E2 – E1, −13.6  −13.6  −13.6 13.6, = 2 − 2 =, +, = −3.40 + 13.6, 4, 1,  1 , 2, = 10.2 eV, This transition lies in the region of Lyman series., (ii) (a) The energy levels of H-atom are given by, Rhc, 13.6, E n = − 2 = − 2 eV, n, n, For first excited state n = 2, 13.6, E 2 = − 2 eV = −3.4 eV, (2), Kinetic energy of electron in (n = 2) state is, K2 = – E2 = + 3.4 eV

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(b) Radius in the first excited state, r1 = (2)2 (0.53) Å, r1 = 2.12 Å, , From Bohr’s quantization condition, , 27. (i) de-Broglie hypothesis may be used to derive Bohr’s, formula by considering the electron to be a wave spread over, the entire orbit, rather than as a particle which at any instant, is located at a point in its orbit. The stable orbits in an atom, are those which are standing waves. Formation of standing, waves require that the circumference of the orbit is equal in, length to an integral multiple of the wavelength. Thus, if r is, the radius of the orbit, nh, h, , 2πr = nλ =, s, ∵ λ = p , p, which gives the angular momentum quantization., h, L = pr = n, 2π, , 1 e2, m ⋅ n 2h 2, 1 e2,  nh , =, =, m⋅, or, , 2, 2, 2, 4πε 0 r,  2πmr  4πε 0 r, 4π m r, , mvr =, , nh, nh, ...(ii), or v =, 2π, 2πmr, , Using equation (ii) in (i), we get, 2, , or, , n 2h 2ε 0, , ...(iii), πme 2, where n = 1, 2, 3, ... is principal quantum number., Equation (iii), gives the radius of nth orbit of H-atom. So, the radii of the orbits increase proportionally with n2 i.e.,, [r ∝ n2]. Radius of first orbit of H-atom is called Bohr radius, a0 and is given by, a0 =, , r=, , h 2ε 0, , πme 2, , for n = 1 or a0 = 0.529 Å, , So, radius of nth orbit of H-atom then becomes, r = n2 × 0.529 Å, , (ii), , C, , EC, , 1, , EB, , 3, , B, , 2, , A, , EA, , Clearly, from energy level diagram,, EC – EA = (EC – EB) + (EB – EA), (On the basis of energy of emitted photon)., hc hc hc, = +, \, λ 3 λ1 λ2, , ⇒, , 1, 1, 1, λλ, = +, ⇒ λ3 = 1 2, λ1 + λ2, λ 3 λ1 λ2, , which is the required relation between the three given, wavelengths., 28. Radius of nth orbit of hydrogen atom : In H-atom, an, electron having charge –e revolves around the nucleus of, charge +e in a circular orbit of radius r, such that necessary, centripetal force is provided by the electrostatic force of, attraction between the electron and nucleus., 2, 1 e2, i.e., mv = 1 e.e or mv 2 =, ...(i), 2, 4πε r, , r, , 4πε 0 r, , 0, , +e r, , F, –e, , mv2, r, , 29. (i) As the gold foil is very thin, it can be assumed that, a-particles will suffer not more than one scattering during, their passage through it. Therefore, computation of the, trajectory of an a-particle scattered by a single nucleus is, enough., (ii) Trajectory of a-particles depends on impact parameter, which is the perpendicular distance of the initial velocity, vector of the a particles from the centre of the nucleus. For, small impact parameter, a particle close to the nucleus suffers, larger scattering., (iii) In case of head-on-collision, the impact parameter is, minimum and the a -particle rebounds back. So, the fact, that only a small fraction of the number of incident particles, rebound back indicates that the number of a -particles, undergoing head-on collision is small. This in turn implies, that the mass of the atom is concentrated in a small volume., 30. According to Bohr’s formula,, , En =, , −me 4, 8ε20n 2h 2, , where m is called reduced mass., In case of hydrogen, m = me = mass of electron., For positronium,, m × me me, m= e, =, me + me, 2, Since for H-atom, E1 =, , m ee 4, , 8ε20n 2h 2, , So, for positronium E 1′ =, , = −13.6 eV, , −13.6, = −6.8 eV., 2

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31. Wavelengths corresponding to minimum wavelength, (λmin) or maximum energy will emit photoelectrons having, maximum kinetic energy., Wavelengths belonging to Balmer series and lying in the given, range (450 nm to 750 nm) corresponds to transition from, (n = 4 to n = 2). Here,, , E4 =, , 13.6, (4)2, , = −0.85 eV and E 2 = −, , 13.6, (2)2, , = −3.4 eV, , ∴ ∆E = E4 – E2 = 2.55 eV, Kmax = Energy of photon – Work function, = 2.55 – 2.0 = 0.55 eV, 32. The force at a distance r is, , F =−, , dU, = −2ar, dr, , th, , mv 2, = 2ar ...(i), r, Further, the quantisation of angular momentum gives,, nh, ...(ii), mvr =, 2π, 1/ 4,  n 2h 2 , Solving, equations (i) and (ii) for r, we get r = ,  8amπ2 , 33. (i) We have, , mv 2, Ze 2, Ze 2, =, or v 2r =, , 2, 4πε 0m, r, 4πε 0r, nh, The quantisation rule is vr =, 2πm, , (vr )2, , =, , 2, , rh =, , 0, 2, , 2 2, , 0, 2, , πme e 2, , or n 25, , n 2h 2ε 0, , 624πme e 2, , =, , h 2ε 0, , πme e 2, , or n2 = 624, , n(n − 1), =3 ∴ n = 3, 2, i.e. after excitation atom jumps to second excited state., Hence, nf = 3. So ni can be 1 or 2., If ni = 1 then energy emitted is either equal to, greater than or, less than the energy absorbed. Hence, the emitted wavelength, 34. (i), , RhcZ 2, , where R is Rydberg constant., n2, In the transition from n = 2 to n = 1, energy released,, 1  3, ∆E = −RhcZ 2  − 1 = RhcZ 2, 4  4, , En = −, , The energy required to eject a n = 4 electron, , ∴, , h 2ε 0, , For rµ = rh ,, , 35. As the nucleus is massive, recoil momentum of the atom, can be ignored. We can assume that the entire energy of, transition is transferred to the Auger electron., As there is a single valence electron in chromium (Z = 24),, the energy states may be thought of as given by Bohr model., The energy of the nth state is, , 2, , , v r, For the given system, Z = 3 and m = 208 me, n 2h 2ε 0, Thus, rµ =, 624πme e2, (ii) The radius of the first Bohr orbit for a hydrogen atom is, 2, , ( ), , 2,  1  RhcZ, = RhcZ 2   =, 16, 4, , ( ) 4πεZe m = Znπhmeε, nh, 2πm, , (ii) E3 – E2 = 68 eV, 1 1, ∴ (13.6)(Z 2 ) − = 68, 4 9, ∴ Z=6, The gas atoms correspond to carbon., 12400, 12400, (iii) λ min =, =, E 3 − E1, 1, (13.6)(6)2 1 −, 9, 12400, =, = 28.49 Å, 435.2, , ( ), , Suppose r be the radius of n orbit. The necessary centripetal, force is provided by the above force. Thus,, , The radius is r =, , is either equal to, less than or greater than the absorbed, wavelength., Hence, ni ≠ 1, If ni = 2, then Ee ≥ Ea. Hence λe ≤ λ0, , KE of Auger electron =, , 3RhcZ 2 RhcZ 2, −, 4, 16, ,  3 1  11, KE = RhcZ 2  −  = RhcZ 2,  4 16  16, 11, = (13.6 eV) × 24 × 24 = 5385.6 eV [ Rhc = 13.6 eV], 16, 36. (a) (i) Limitation of Rutherford’s model :, e–, Rutherford’s atomic model is inconsistent, with classical physics. According to, +, electromagnetic theory, an electron is a, charged particle moving in the circular orbit, around the nucleus and is accelerated, so, it should emit radiation continuously and thereby loose, energy. Due to this, radius of the electron would decrease, continuously and also the atom should then produce, continuous spectrum, and ultimately electron will fall into, the nucleus and atom will collapse in 10–8 s. But the atom is, fairly stable and it emits line spectrum., (ii) Rutherford’s model is not able to explain the spectrum, of even most simplest H-spectrum., (b) Bohr’s postulates to resolve observed features of atomic, spectrum :