12.

ATOMS

Single Correct Answer Type

1. The ratio of the speed of the electron in the first Bohr orbit of hydrogen and the speed of light is equal to
(where e, h, and c have their usual meanings in cgs system)
a) 2πhc/e2 b) er2h/2πc c) e2c/2πh d) 2πe2/hc

2. In Fig, E1 to E6 represent some of the energy levels of an electron in the hydrogen atom

Which one of the following transitions produces a photon of wavelength in the ultraviolet region of the
electromagnetic spectrum?
a) E - E b) E - E c) E - E d) E - E
2 1 3 2 4 3 6 4

3. If elements of quantum number greater than n were not allowed, the number of possible elements in
nature would have been

{ }
2
a) 1 n(n+1) b) n(n+1) c) 1 n(n+1)(2n + 1) d) 1 n(n+1)(2n + 1)
2 2 6 3
4. The ratio between total acceleration of the electron in singly ionized helium atom and hydrogen atom
(both in ground state) is
a) 1 b) 8 c) 4 d) 16

5. Consider a spectral line resulting from the transition n = 5 to n = 1, in the atoms and ions given below.
The shortest wavelength is produced by
a) Helium atom b) Deuterium atom

c) Singly ionized helium d) Ten times ionized sodium atom

6. A hydrogen atom in around state absorbs 10.2 eV of energy. The orbital angular momentum of the electron
is increased by
a) 1.05×10-34 Js b) 2.11×10-34 Js c) 3.16×10-34 Js d) 4.22×10-34 Js

7. In the Bohr model of a hydrogen atom, the centripetal force is furnished by the Coulomb attraction
between the proton and the electron. If a0 is the radius of the ground state orbit, m is the mass, and e is the
charge on the electron and ε0 is the permittivity of vacuum, then the speed of the electron is:
a) 0 b) e c) e d) 4πεoaom
εoaom 4πεoaom e
8. As the electron in Bohr orbit of hydrogen atom passes from state n = 2 to n = 1, the KE (K) and PE (U)
change as
a) K two-fold, U also two-fold b) K four-fold, U also four-fold

c) K four-fold, U two-fold d) K two-fold, U four-fold

9. An electron is in an excited state in a hydrogen like atom. It has a total energy = -3.4 eV. The kinetic energy
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 of electron is E and its de Broglie wavelength is λ
a) E = 6.8 eV; λ = 6.6×10-10m b) E = 3.4 eV; λ = 6.6×10-10m

c) E = 3.4 eV; λ = 6.6×10-11m d) E = 6.8 eV; λ = 6.6×10-11m

10. When the voltage applied to an X-ray tube increases from V1 = 10 kV to V2 = 20 kV, the wavelength
interval between Kα line and cut-off wavelength of continuous spectrum increase by a factor of 3. Atomic
number of the metallic target is
a) 28 b) 29 c) 65 d) 66

11. A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding
photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same
hydrogen atom inelastically with an energy of 15n eV. What will be observed by the detector?
a) 2 photon of energy 10.2 eV.

b) 2 photon of energy of 1.4 eV.

c) One photon of energy 10.2 eV and an electron of energy 1.4 eV

d) One photon of energy 10.2 eV and another photon of energy 1.4 eV

12. A hydrogen-like atom emits radiations of frequency 2.7×1015 Hz when it makes a transition from n = 2 to
n = 1. The frequency emitted in a transition from n = 3 to n = 1 will be
a) 1.8×1015 Hz b) 3.2×1015 Hz c) 4.7×105 Hz d) 6.9×1015 Hz

13. Imagine an atom made of a proton and a hypothetical particle of double the mass of the electron but
having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of
this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has
wavelength [given in terms of the Rydberg constant R for the hydrogen atom] equal to
a) 9 b) 36 c) 18 d) 4
5R 5R 5R R
14. In a hydrogen atom, the electron is in nth excited state. It comes down to first excited state by emitting 10
different wavelengths. The value of n is
a) 6 b) 7 c) 8 d) 9

15. If the series limit wavelength of the Lyman series for hydrogen atom is 912 Å, then the series limit
wavelength for the Balmer series for the hydrogen atom is
a) 912 Å/2 b) 912 Å c) 912×2 Å d) 912×4 Å

16. An electron in the ground state of hydrogen has an angular momentum L1 and an electron in the first
excited state of lithium has an angular momentum L2. Then
a) L = L b) L = 4L c) L = 2L d) L = 2L
1 2 1 2 2 1 1 2

17. According to Bohr’s theory of hydrogen atom, the product of the binding energy of the electron in the nth
th
orbit and its radius in the n orbit
a) Is proportional to n2 b) Is inversely proportional to n3

c) Has a constant value of 10.2 eV - Å d) Has a constant value of 7.2 eV - Å

18. The electron in a hydrogen atom makes a transition n1→n2, where n1 and n2 are the principal quantum
numbers of the two states. Assume Bohr model is valid in this case. The frequency of the orbital motion of
the electron in the initial state is 1/27 of that in the final state. The possible values of n1 and n2 are
a) n = 6, n = 3 b) n = 4, n = 2 c) n = 8, n = 1 d) n = 3, n = 1
1 2 1 2 1 2 1 2

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19. Mark out the correct statement regarding X-rays

a) When fast moving electrons strike the metal target, they enter the metal target and in a very short time
span come to rest, and thus an accelerated charged electron produces electromagnetic waves (X-rays)
b) Characteristic X-rays are produced due to transition of an electron from higher energy levels to vacant
lower energy levels
c) X-rays spectrum is a discrete spectra just like hydrogen spectra

d) Both (a) and (b) are correct

20. The potential difference across the Coolidge tube is 20 kV and 10 mA current flows through the voltage
supply. Only 0.5% of the energy carried by the electrons striking the target is converted into X-rays. The
power carried by the X-ray beam is P. Then
a) P = 0.1 W b) P = 1 W c) P = 2 W d) P = 10 W

21. Check the correctness of the following statements about 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 KE of the electron in n = 2 orbit is less than that in n = 1 orbit
a) All the statements are correct b) Only (i) and (ii) are correct

c) Only (ii) and (iii) are correct d) Only (iii) and (i) are correct

22. The shortest wavelength of Lyman series of hydrogen is equal to the shortest wavelength of Balmer series
of a hydrogen-like atom of atomic number Z. The value of Z is
a) 4 b) 2 c) 3 d) 6

23. Figure shows five energy levels of an atom, one being much lower than the other four. Five transitions
between the levels are indicated, each of which produces a photon of definite energy and frequency

Which one of the spectra below best corresponds to the set of transitions indicated?

a) b)

c) d)

24. The wavelength of radiation required to excite the electron from first orbit to third orbit in a doubly
ionized lithium atom will be
a) 134.25 Å b) 125.5 Å c) 113.7 Å d) 110 Å

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25. A hydrogen-like atom (atomic number Z) is in a higher excited state of quantum number n. This excited
atom can make a transition to the first excited state by successively emitting two photons of energies 10.2
eV and 17.0 eV, respectively. Alternatively, the atom from the same excited state can make transition to the
second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV, respectively.
The values of n and Z are, respectively,
a) 6 and 6 b) 3 and 3 c) 6 and 3 d) 3 and 6

26. Difference between nth and (n+1)th Bohr’s radius of hydrogen atom is equal to (n-1)th Bohr’ radius. The
value of n is
a) 1 b) 2 c) 3 d) 4

27. When the hydrogen atom emits a photon in going from n = 5 to n = 1 state, its recoil speed is nearly

a) 10-4 ms-1 b) 2×10-2 ms-1 c) 4 ms-1 d) 8×102 ms-1

28. In a hydrogen atom following the Bohr’s postulates, the product of linear momentum and angular
momentum is proportional to (n)x where ‘n’ is the orbit number. Then ‘x’ is
a) 0 b) 2 c) -2 d) 1

29. The angular momentum of an electron in a hydrogen atom is proportional to:

a) 1/ r b) 1/r c) r d) r2

30. When photons of wavelength λ1 are incident on an isolated sphere suspended by an insulated thread, the
corresponding stopping potential is found to be V. When photons of wavelength λ2 are used, the
corresponding stopping potential was thrice the above value. If light of wavelength λ3 is used, calculate the
stopping potential for this case

[
a) hc 1 + 1 - 1
e λ3 2λ2 λ1 ] [
b) hc 1 + 1 - 3
e λ3 2λ2 2λ1 ] [
c) hc 1 + 1 - 1
e λ3 λ2 λ1 ] [
d) hc 1 - 1 - 1
e λ3 λ2 λ1 ]
31. The ratio of maximum to minimum possible radiation energy in Bohr’s hypothetical hydrogen atom is
equal to
a) 2 b) 4 c) 4/3 d) 3/2

32. In order to determine the value of E0, a scientist shines photon (“light particles”) of various energies at a
cloud of atomic hydrogen. Most of the hydrogen atoms occupy the ground state. A detector records the
intensity of light transmitted through that cloud; see figure (a). Figure (b) is a graph of part of the
scientist’s data, showing the intensity of the transmitted light as a function of the photon energy. A
hydrogen atom’s electron is likely to absorb a photon only if the photon gives the electron enough energy
to knock it into a higher shell

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 According to this experiment, what is the approximate value of E0?
a) 1.6×10-18 J b) 2.1×10-18 J c) 3.2×10-18 J d) 6.4×10-18 J

33. Suppose two deuterons musts get as close as 10-14 m in order for the nuclear force to overcome the
repulsive electrostatic force. The height of the electrostatic barrier is nearest to
a) 0.14 MeV b) 2.3 MeV c) 1.8 x 10 MeV d) 0.56 MeV

34. An electron of energy 11.2 eV undergoes an inelastic collision with a hydrogen atom in its ground state.
[Neglect recoiling of atom as mH ≫ me]. Then in this case
a) The outgoing electron has energy 11.2 eV

b) The entire energy is absorbed by the H atom and the electron stops

c) 10.2 eV of the incident electron energy is absorbed by the H atom and the electron would come out with
1.0 eV energy
d) None of the above

35. The energy of an electron in an excited hydrogen atom is -3.4 eV. Then, according to Bohr’s Theory, the
angular momentum of this electron, in Js, is
a) 2.11×10-34 b) 3×10-34 c) 1.055×10-34 d) 0.5×10-34

36. If an X-ray tube operates at the voltage of 10 kV, find the ratio of the de Broglie wavelength of the incident
11
electrons to the shortest wavelength of X-rays produced. The specific charge of electron is 1.8×10 C kg
a) 1 b) 0.1 c) 1.8 d) 1.2

37. 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

38. The recoil of speed of hydrogen atom after it emits a photon in going from n = 2 state to n = 1 state is
7 -1 -34
nearly [Take R∞ = 1.1×10 m and h = 6.63×10 Js]
a) 1.5 ms-1 b) 3.3 ms-1 c) 4.5 ms-1 d) 6.6 ms-1

39. Transitions between three energy levels in a particular atom give rise to three spectral lines of wavelength,

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 in increasing magnitudes, λ1, λ2 and λ3. Which one of the following equations correctly relates λ1, λ2 and λ3?
a) λ = λ - λ b) λ = λ - λ c) 1 = 1 + 1 d) 1 = 1 - 1
1 2 3 1 3 2 λ1 λ2 λ3 λ1 λ3 λ2
40. If the Kα radiation of Mo (Z = 42) has a wavelength of 0.71 Å find the wavelength of the corresponding
radiation of Cu (Z = 29)
a) 1 Å b) 2 Å c) 1.52 Å d) 1.25 Å

41. The ratio (in S.I. units) of magnetic dipole moment to that of the angular momentum of an electron of mass
m kg and charge e coulomb in Bohr’s orbit of hydrogen atom is
a) e b) e c) 2e d) None of these
2m m m
42. An electron in a hydrogen atom makes a transition from first excited state to ground state. The equivalent
current due to circulating electron:
a) Increases 2 times b) Increases 4 times c) Increases 8 times d) Remains the same

43. If potential energy between a proton and an electron is given by |U| = ke2/2R3, where e is the charge of
electron and R is the radius of atom, then radius of Bohr’s orbit is given by (h = Planck’s constant, k =
constant)
2 2 2 2 2 2
a) ke m b) 6π ke m c) 2π ke m d) 4π ke m
2 2 2 2 2 2
h n h n h nh
44. The power of an X-ray tube is 16 W. If the potential difference applied across the tube is 5 kV, then the
number of electron striking the target per second is
a) 8.4×1016 b) 5×1017 c) 2×1016 d) 2×1019

45. An electron in H atom makes a transition from n = 3 to n = 1. The recoil momentum of H atom will be

a) 6.45×10-27N s b) 6.8×10-27N s c) 6.45×10-24N s d) 6.8×10-24N s

46. The potential energy of an electron in the fifth orbit of hydrogen atom is

a) 0.54eV b) -0.54 eV c) 1.08 eV d) -1.08 eV

47. The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest
wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is
a) 802 nm b) 823 nm c) 1882 nm d) 1648 nm

48. The radius of the Bohr orbit in the ground state of hydrogen atom is 0.5 Å. The radius of the orbit of the
+
electron in the third excited state of He will be
a) 8 Å b) 4 Å c) 0.5 Å d) 0.25 Å

49. When an electron jumps from nth

1
th
orbit to n2 orbit, the energy radiated is given by

a) hv = E /E b) hv = E /E c) hv = E - E d) hv = E - E
1 2 2 1 1 2 2 1

50. The voltage applied to an X-ray tube is 18 kV. The maximum mass of photon emitted by the X-ray tube will
be
a) 2×10-13 kg b) 3.2×10-36 kg c) 3.2×10-32 kg d) 9.1×10-31 kg

51. In order to break a chemical bond in the molecules of human skin, causing sunburn, a photon energy of
about 3.5 eV is required. This corresponds to wavelength in the
a) Infrared region b) X-ray region c) Visible region d) Ultraviolet region

52. Which of the following is true when Bohr gave his model for hydrogen atom?

a) It was not known that hydrogen lines could be explained as differences of terms like R/n2 with R being a
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 constant and n an integer
b) It was not known that positive charge is concentrated in a nucleus of small size

c) It was not known that radiant energy occurred in energy bundles defined by hv with h being a constant
and v a frequency
2
d) Bohr knew terms like R/n and in the process of choosing the allowed orbits to fit them he got “angular
momentum = ni/2π as a deduction
53. Let v1 be the frequency of series limit of Lyman series, v2 the frequency of the first line of Lyman series,
and v3 the frequency of series limit of Balmer series. Then which of the following is correct?
a) v - v = v b) v - v = v c) v = 1 (v +v ) d) v + v = v
1 2 3 2 1 3 3
2 1 2 2 1 3

54. The wavelength of the second line of Balmer series in the hydrogen spectrum is 4861 Å. The wavelength of
the first line is
a) 27 ×4861 Å b) 20 ×4861 Å c) 20×4861 Å d) 4861 Å
20 27
-
55. The orbiting speed vn of e in the nth orbit in the case of positronium is x-fold compared to that in nth orbit
in a hydrogen atom, where x has the value
a) 1 b) c) 1/ 2 d) 2
2
56. The frequency of revolution of an electron in nth orbit is fn. If the electron makes a transition from nth orbit
to (n-1)th orbit, then the relation between the frequency (v) of emitted photon and fn will be
a) v = f2 b) v = f c) v = 1 d) v = f
n n fn n

57. Electron with energy 80 keV are incident on the tungsten target of an X-ray tube. K shell electrons of
tungsten have -72.5 keV energy. X-rays emitted by the tube contain only
a) A continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of -0.155 Å

b) A continuous X-ray spectrum (Bremsstrahlung) with all wavelengths

c) A continuous X-ray spectrum of tungsten

d) A continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of 0.155 Å and the
characteristic X-ray spectrum of tungsten
58. The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength
of the second spectral line in the Balmer series of singly ionized helium atom is
a) 1215 Å b) 1640 Å c) 2430 Å d) 4687 Å

59. An X-ray tube is operating at 150 kV and 10 mA. If only 1% of the electric power is converted into X-rays,
-1
the rate at which the target is heated, in cal s , is
a) 3.57 b) 35.7 c) 4.57 d) 15

60. Kα wavelength emitted by an atom of atomic number Z = 11 is λ. The atomic number for an atom that
emits Kα radiation with wavelength 4λ is
a) 6 b) 4 c) 11 d) 44

61. Figure represents some of the lower energy levels of the hydrogen atom in simplified form
If the transition of an electron from E4 to E2 were associated with the emission of blue light, which one of
the following transitions could be associated with the emission of red light?

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 a) E to E b) E to E c) E to E d) E to E
4 1 3 1 3 2 1 3

62. Magnetic moment due to the motion of the electron in nth energy of hydrogen atom is proportional to

a) n b) n0 c) n5 d) n3

63. Consider a hypothetical annihilation of a stationary electron with a stationary positron. What is the
wavelength of the resulting radiation?
a) λ = h b) λ = 2h c) λ = h d) None of these
2 2
m0c m0c 2m0c
64. A neutron having kinetic energy 5 eV is incident on a hydrogen atom in its ground state. The collision

a) Must be elastic b) Must be completely inelastic

c) May be partially elastic d) Information is insufficient

65. When an electron accelerated by potential difference U is bombarded on a specific metal, the emitted X-
rays spectrum obtained is shown in figure. If the potential difference is reduced to U/3, the correct
spectrum is

a) b)

c) d)

66. The orbital velocity of an electron in the ground state is v. If the electron is excited to energy state -0.54 eV,
its orbital velocity will be
a) v b) v c) v d) v
3 5 7
67. If the average life time of an excited state of hydrogen is of the order of 10-8 s, then the number of
revolutions an electron will make when it is in n = 2 state before coming to ground state will be [Take
a0 = 0.53 Å and all standard data if required]
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 a) 107 b) 8×106 c) 2×105 d) None of these

68. If an electron in n = 3 orbit of hydrogen atom jumps down to n = 2 orbit, the amount of energy released
and the wavelength of radiation emitted are
a) 0.85 eV, 6566 Å b) 1.89 eV, 1240 Å c) 1.89 eV, 6566 Å d) 1.5 eV, 6566 Å

69. A proton of mass m moving with a speed v0 approaches a stationary proton that is free to move. Assume
impact parameter to be zero, i.e., head-on collision. How close will the incident proton go to other proton?
3 3 2
a) e b) e c) e 2 d) None of the above
2
πε0m v0 πε0mv0 πε0mv0
70. 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

71. In X-ray tube, when the accelerating voltage V is halved, the difference between the wavelength of Kα line
and minimum wavelength of continuous X-ray spectrum
a) Remains constant b) Becomes more than two times

c) Becomes half d) Becomes less than two times

72. When an electron jumps from a level n = 4 to n = 1, the momentum of the recoiled hydrogen atom will be

a) 6.5×10-27kg m s-1 b) 12.75×10-19kg m s-1

c) 13.6×10-27kg m s-1 d) Zero

73. Two electrons are revolving around a nucleus at distances ‘r’ and ‘4r’. The ratio of their period is

a) 1:4 b) 4:1 c) 8:1 d) 1:8

74. A stationary hydrogen atom of mass M emits a photon corresponding to the first line of Lyman series. If R
is the Rydberg’s constant, the velocity that the atom acquires is
a) 3 Rh b) Rh c) Rh d) Rh
4M 4M 2M M
75. Monochromatic radiations of wavelength λ are incident on a hydrogen sample in ground state. Hydrogen
atom absorbs the light and subsequently emits radiations of 10 different wavelengths. The value of λ is
nearly
a) 203 nm b) 95 nm c) 80 nm d) 73 nm

76. If n ≫ 1, then the dependence of frequency of photon emitted as a result of transition of an electron from
th
n orbit to (n-1)th orbit, on n will be
a) v ∝ 1 b) v ∝ 1 c) v ∝ 1 d) v ∝ 1
2 3 4
n n n n
77. The ionization potential of H atoms is 13.6 V. The energy difference between n = 2 and n = 3 levels is
nearest to
a) 1.9 eV b) 2.3 eV c) 3.4 eV d) 4.5 eV

78. In the spectrum of singly ionized helium, the wavelength of a line observed is almost the same as the first
line of Balmer series of hydrogen. It is due to transition of electron
a) From n = 6 to n = 4 b) From n = 5 to n = 3
1 2 1 2

c) From n = 4 to n = 2 d) From n = 3 to n = 2
1 2 1 2

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79. A hydrogen atom is in an excited state of principal quantum number n. It emits a photon of wavelength λ
when it returns to the ground state. The value of n is
a) λR(λR-1) b) λ(R-1) c) λR d) λ(R-1)
λR λR-1
80. A sample of hydrogen is bombarded by electrons. Through what potential difference should the electrons
be accelerated so that third line of Lyman series be emitted?
a) 2.55 V b) 10.2 V c) 12.09 V d) 12.75 V

81. A beam of 13.0 eV electrons is used to bombard gaseous hydrogen. The series obtained in emission spectra
is/are
a) Lyman series b) Balmer series c) Bracket series d) All of these

82. The ratio of minimum to maximum wavelength in Balmer series is

a) 5:9 b) 5:36 c) 1:4 d) 3:4

83. An electron with kinetic energy E eV collides with a hydrogen atom in the ground state. The collision is
observed to be elastic for
a) 0 < E < ∞ b) 0 < E < 10.2 eV c) 0 < E < 13.6 eV d) 0 < E < 3.4 eV

84. r
The electric potential between a proton and an electron is given by V = V0 In , where r0 is a constant.
r0
Assuming Bohr’s model to be applicable, write variation of rn with n, n being the principal quantum
number?
a) r ∝ n b) r ∝ 1 c) r ∝ n2 d) r ∝ 1
n n 2
n
n n
n
85. Which of the following statement is true regarding Bohr’s model of hydrogen atom?
(I) Orbiting speed of an electrons decreases as it falls to discrete orbits away from the nucleus
(II) Radii of allowed orbits of electrons are proportional to the principal quantum number
(III) Frequency with which electrons orbit around the nucleus in discrete orbits is inversely proportional
to the cube of principal quantum number
(IV) Binding force with which the electron is bound to the nucleus increase as it shifts to outer orbits
Select the correct answer using the codes given below:
a) I and II b) II and IV c) I, II and III d) II, III and IV

86. Hydrogen atoms in a sample are excited to n = 5 state and it is found that photons of all possible
wavelengths are present, in the emission spactra. The minimum number of hydrogen atoms in the sample
would be
a) 5 b) 6 c) 10 d) Infinite

87. The circumference of the second Bohr orbit of electron in hydrogen atom is 600 nm. The potential
difference that must be applied between the plates so that the electrons have the de Broglie wavelength
corresponding in this circumference is
a) 10-5 V b) 5 ×10-5 V c) 5×10-5 V d) 3×10-5 V
3
88. The wavelength of the spectral line that corresponds to a transition in hydrogen atom from n = 10 to
ground state would be [In which part of electromagnetic spectrum this line lies?]
a) 92.25 nm, ultraviolet b) 92.25 nm, infrared c) 86.95 nm, ultraviolet d) 97.65 nm, ultraviolet

89. A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding
photon is 10.2 eV. Almost instantaneously, another photon collides with same hydrogen atom inelastically
with an energy of 15 eV. What will be observed by the detector?
a) Two photons of energy 10.2 eV
P a g e | 10
 b) Two photon of energy 1.4 eV

c) One photon of energy 10.2 eV and one electron of energy 1.4 eV

d) One electron having kinetic energy nearly 11.6 eV

90. In hydrogen spectrum, the shortest wavelength in Balmer series is λ. The shortest wavelength in Brackett
series will be
a) 2λ b) 4λ c) 9λ d) 16λ

91. One of the lines in the emission spectrum of Li2+ has the same wavelength as that of the 2nd line of Balmer
series in hydrogen spectrum. The electronic transition corresponding to this line is
a) n = 4→n = 2 b) n = 8→n = 2 c) n = 8→n = 4 d) n = 12→n = 6

92. The wavelength of the first line of Balmer series is 6563 Å. The Rydberg’s constant is

a) 1.09×105 per m-1 b) 1.09×106 per m-1 c) 1.097×107 per m-1 d) 1.09×108 per m-1
++
93. A hydrogen atom and a Li ion are both in the second excited state. If lH and lLI are their respective
electronic angular momenta, and EH and ELi their respective energies, then
a) l < l and |E | > |E | b) l = l and |E | < |E |
H LI H LI H LI H LI

c) l = l and |E | > |E | d) l < l and |E | < |E |

H LI H LI H LI H LI

94. Given: mass number of gold = 197, density of gold= 19.7 g cm-3, Avogadro’s number = 6×1023. The radius
of the gold atom is approximately:
a) 1.5×10-8 m b) 1.5×10-9 m c) 1.5×10-10 m d) 1.5×10-12 m

95. Magnetic field at the center (at nucleus) of the hydrogen-like atoms (atomic number = z) due to the
th
motion of electron in n orbit is proportional to
3 4 2 3
a) n b) n c) z d) z
5 3 5
z z n n
96. The recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state is

a) 4.718 ms-1 b) 7.418 ms-1 c) 4.178 ms-1 d) 7.148 ms-1

97. The wavelength Kα of X-rays produced by the X-ray tube is 0.76 Å. The atomic number of the node material
of the tube is
a) 30 b) 40 c) 50 d) 60

98. The force acting on the electron in a hydrogen atom depends on the principal quantum number as

a) F ∝ n2 b) F ∝ 1 c) F ∝ n4 d) F ∝ 1
2 4
n n
99. As the quantum number increases, the difference of energy between conservative energy levels:

a) Decreases b) Increases

c) First decreases and then increases d) Remain the same

100. If 10,000 V are applied across an X-ray tube, find the ratio of wavelength of the incident electrons and
11 -1
shortest wavelength of X-ray coming out of the X-ray tube, given e/m, of electron = 1.8×10 C kg
a) 1:10 b) 10:1 c) 5:1 d) 1:5

101. The maximum angular speed of the electron of a hydrogen atom in a stationary orbit is

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 a) 6.2×1015 rad s-1 b) 4.1×1016 rad s-1 c) 24×1010 rad s-1 d) 9.2×106 rad s-1

102. In interpreting Rutherford’s experiments on the scattering of alpha particles by thin foils, one must
examine what were the known factors, and what did the experiment conclude. Which of the following are
true in this context?
a) The number of electrons in the target atom (i.e., Z) was settled by these experiments

b) The validity of Coulomb’s law for distance as small as 10-13 was known before these experiments

c) The experiments settled that size of the nucleus could not be larger than a certain value

d) The experiments also settled that size of the nucleus could not be smaller than a certain value

103. An electron jumps from the 4th orbit to the 2nd orbit of hydrogen atom. Given: the Rydberg’s constant
5 -1
= 10 cm . The frequency, in Hz, of the emitted radiation will be
a) 3 ×105 b) 3 ×1015 c) 9 ×105 d) 9 ×1015
16 6 16 16
++
104. The angular momentum of an electron in first orbit of Li ion is

a) 3h b) 9h c) h d) h
2π 2π 2π 6π
105. A neutron moving with a speed v makes a head-on collision with a hydrogen atom in ground state kept at
rest. The minimum kinetic energy of the neutron for which inelastic collision will take place is (assume
that mass of proton is nearly equal to the mass of neutron)
a) 10.2 eV b) 20.4 eV c) 12.1 eV d) 16.8 eV

106. An electron and a photon have same wavelength. If p is the momentum of electron and E the energy of
photon, the magnitude of p/E is S.I. unit is
a) 3.0×108 b) 3.33×10-9 c) 9.1×10-31 d) 6.64×10-34

107. An electron revolving in an orbit of radius 0.5 Å in a hydrogen atom executes 1016 revolutions per second.
The magnetic moment of electron due to its orbital motion will be
a) 1.256×10-23A m2 b) 653×10-26A m2 c) 10-3A m2 d) 256×10-26A m2

108. The frequency of emission line for any transition in positronium atom (consisting of a positron and an
electron) is x times the frequency for the corresponding line in the case of H atom, where x is
a) b) 1/2 2 c) 1/2 2 d) 1/2
2
109. If the potential difference applied across a Coolidge tube is increased, then

a) Wavelength of K will increase b) λ will increase

α min

c) Difference between wavelength of Kα and λmin d) None of these

increases
110. If R is the Rydberg constant for hydrogen, then the wave number of the first line in the Lyman series is

a) R b) 2R c) R d) 3R
2 4 4
111. How many revolutions does an electron complete in one second in the first orbit of hydrogen atom?

a) 6.62×1015 b) 100 c) 1000 d) 1

112. Angular momentum (L) and radius (r) of a hydrogen atom are related as

a) Lr = constant b) Lr2 = constant c) Lr4 = constant d) None of these

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113. The total energy of an electron in the ground state of hydrogen atom is -13.6 eV. The potential energy of an
2+
electron is the ground state of Li ion will be
a) 122.4 eV b) -122.4 eV c) 244.8 eV d) -244.8 eV

114. If elements with principal quantum number n > 4 were not allowed in nature, the number of possible
elements would have been
a) 32 b) 60 c) 64 d) 4

115. An electron collides with a hydrogen atom in its ground state and excites it to n = 3. The energy given to
hydrogen atom in this inelastic collision is [Neglect the recoiling of hydrogen atom]
a) 10.2 eV b) 12.1 eV c) 12.5 eV d) None of these

116. In terms of Rydberg constant R, the shortest wavelength in Balmer series of hydrogen atom spectrum will
have wavelength
a) 1/R b) 4/R c) 3/2R d) 9/R

117. An atom emits a spectral line of wavelength λ when an electron makes a transition between levels of
energy E1 and E2. Which expression correctly relates λ, E1 and E2?
a) λ = hc b) λ = 2hc c) λ = 2hc d) λ = hc
E1+E2 E1+E2 E1-E2 E1-E2
118. In which of the following transitions will the wavelength be minimum?

a) n = 5 to n = 4 b) n = 4 to n = 3 c) n = 3 to n = 2 d) n = 2 to n = 1
1 2 1 2 1 2 1 2

119. An electron in a Bohr orbit of hydrogen atom with the quantum number n2 has an angular momentum
-34 2 -2
4.2176×10 kg m s . If the electron drops from this level to the next lower level, the wavelength of this
line is
a) 18 nm b) 187.6 pm c) 1876 Å d) 1.876×104 Å

120. In hydrogen and hydrogen-like atoms, the ratio of E4n - E2n and E2n - En varies with atomic number z and
principal quantum number n as
2 4
a) z b) z c) z d) None of these
2 4
n n n
121. The wavelength of first line of Lyman series in hydrogen atom is 1216. The wavelength of first line of
Lyman series for 10 times ionized sodium atom will be added
a) 0.1 Å b) 1000 Å c) 100 Å d) 10 Å

122. An alpha particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of
the closest approach is of the order of
a) 10-15 cm b) 10-13 cm c) 10-12 cm d) 10-19 cm

123. Figure shows the electron energy levels, referred to the ground state (the lowest possible energy) as zero,
for five different isolated atoms. Which atom can produce radiation of the shortest wavelength when
atoms in the ground state are bombarded with electrons of energy W?

a) A b) B c) C d) D
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124. Figure shows Moseley’s plot between f and Z, where f is the frequency and Z is the atomic number. Three
lines A, B and C shown in the graph may represent

a) K , K and K lines, respectively b) K , K and K lines, respectively

α β γ γ β α

c) K , L and M lines, respectively d) Nothing

α α α

125. The ionization energy of the ionized sodium atom Na10+ is:

a) 13.6 eV b) 13.6×11 eV c) (13.6/11) eV d) 13.6×(112) eV

126. Determine the maximum wavelength that hydrogen in its ground state can absorb. What would be the next
smaller wavelength that would work?
a) 133 nm b) 13.3 nm c) 10.3 nm d) 103 nm

127. The longest wavelength that a singly ionised helium atom in its ground state will absorb is

a) 912 Å b) 304 Å c) 606 Å d) 1216 Å

128. Let the potential energy of a hydrogen atom in the ground state be zero. Then, its energy in the first excited
state will be
a) 10.2 eV b) 13.6 eV c) 23.8 eV d) 27.2 eV

129. Which of the following parameters are the same for all hydrogen-like atoms and ions in their ground
states?
a) Radius of the orbit b) Speed of the electron

c) Energy of the atom d) Orbital angular momentum of the electron

130. The minimum kinetic energy required for ionization of a hydrogen atom is E1 is case electron is collided
with hydrogen atom. It is E2 if the hydrogen ion is collided and E3 when helium ion is collided. Then,
a) E = E = E b) E > E > E c) E < E < E d) E > E > E
1 2 3 1 2 3 1 2 3 1 3 2

131. If 13.6 eV energy is required to ionize the hydrogen atom, then the energy required to remove an electron
from n = 2 is
a) 10.2 eV b) 0 eV c) 3.4 eV d) 6.8 eV

132. The energy change is greatest for a hydrogen atom when its state changes from

a) n = 2 to n = 1 b) n = 3 to n = 2 c) n = 4 to n = 3 d) n = 5 to n = 4

133. If a hydrogen atom emits a photon of energy 12.1 eV, its orbital angular momentum changes by ΔL. Then,
ΔL equals orbital angular momentum changes by ΔL. Then, ΔL equals
a) 1.05×10-34 Js b) 2.11×10-34 Js c) 3.16×10-34 Js d) 4.22×10-34

134. X-rays emitted from a copper target and a molybdenum target are found to contain a line of wavelength
22.85 nm attributed to the Kα line of an impurity element. The Kα lines of copper (Z = 29) and
molybdenum (Z = 42) have wavelengths 15.42 nm and 7.12 nm, respectively. The atomic number of the
impurity element is

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 a) 22 b) 23 c) 24 d) 25

135. In the Bohr model of a π-mesin atom, a π-menon of mass mπ and of the same charge as the electron is in a
circular orbit of radius r about the nucleus with an orbital angular momentum h/2π. If the radius of a
1
-15
nucleus of atomic number Z is given by R = 1.6×10 Z 3 m, Then the limit on Z for which
2 2
(ε0h /πme = 0.53 Å and mπ/me = 264) π-mesic atoms might exist is
a) < 105 b) > 105 c) < 37 d) > 37

136. The electron in a hydrogen atom jumps from ground state to the higher energy state where its velocity is
reduced to one-third its initial value. If the radius of the orbit in the ground state is r, the radius of new
orbit will be
a) 3r b) 9r c) r d) r
3 9
137. In which of the following transitions will the wavelength be minimum?

a) n = 5 to n = 4 b) n = 4 to n = 3 c) n = 3 to n = 2 d) n = 2 to n = 1

138. The approximate value of quantum number n for the circular orbit of hydrogen of 0.0001 mm in diameter
is
a) 1000 b) 60 c) 10000 d) 31
th rd
139. Electrons in a hydrogen-like atom (Z=3) make transitions from 4 excited state to 3 excited state and
rd nd
from 3 to 2 excited state. The resulting radiations are incident on a metal plate to eject photoelectrons.
The stopping potential for photoelectrons ejected by the shorter wavelength is 3.95 V. The stopping
potential for the photoelectrons ejected by the longer wavelength is
a) 2.0 V b) 0.75 V c) 0.6 V d) None of the above
+
140. The velocity of an electron in the first orbit of H atom is v. The velocity of an electron in the 2nd orbit of He
is
a) 2v b) v c) v d) v
2 4
141. The radius of hydrogen atom in the ground state is 5.3×10-11 m. When struck by an electron, its radius is
-11
found to be 21.2×10 m. The principal quantum number of the final state will be
a) 1 b) 2 c) 3 d) 4

142. The minimum energy to ionize an atom is the energy required to

a) Add one electron to the atom

b) Excite the atom from its ground state to its first excited

c) Remove one outermost electron from the atom

d) Remove one innermost electron from the atom

143. The shortest wavelength produced in an X-ray tube operating at 0.5 million volt is

a) Dependent on the target element

b) About 2.5×10-12 m

c) Double of the shortest wavelength produced at 1 million volt

d) Dependent only on the target material

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144. The shortest wavelength of the Brackett series of a hydrogen-like atom (atomic number = z) is the same
as the shortest wavelength of the Balmer series of hydrogen atom. The value of z is
a) 2 b) 3 c) 4 d) 6

145. A sample of hydrogen atoms is in excited state (all the atoms). The photons emitted from this sample are
made to pass through a filter through which light having wavelength greater than 800 nm can only pass.
Only one type of photons are found to pass the filter. The sample’s excited state initially is [Take
hc = 1240 eV-nm, ground state energy of hydrogen atom = -13.6 eV]
a) 5th excited state b) 4th excited state c) 3rd excited state d) 2nd excited state

146. If first excitation potential of a hydrogen-like atom is V electron volt, then the ionization energy of this
atom will be
a) V electron volt b) 3V electron volt
4
c) 4V electron volt d) Cannot be calculated by given information
3
147. In a hydrogen atom, the electron makes a transition from n = 2 to n = 1. The magnetic field produced by
the circulating electron at the nucleus
a) Decreases 16 times b) Increases 4 times c) Decreases 4 times d) Increases 32 times

148. Hydrogen atoms are excited from ground state to the state of principal quantum number 4. Then, the
number of spectral lines observed will be
a) 3 b) 6 c) 5 d) 2

149. The electric potential energy between a proton and an electron is given by U = U0 In r/r0, where r0 is a
constant. Assuming Bohr’ model to be applicable, write variation of r with n, n being the principal quantum
number
a) r ∝ n b) r ∝ 1/n c) r ∝ n2 d) r ∝ 1/n2
n n n n

150. The angular momentum of an electron in an orbit is quantized because it is a necessary condition for the
compatibility with
a) Wave nature of electron b) Particle nature of electron

c) Pauli’s exclusion behavior d) None of these

151. A beam of electron accelerated by a large potential difference V is made to strike a metal target to produce
X-rays. For which of the following values of V, will the resulting X-rays have the lowest minimum
wavelength?
a) 10 kV b) 20 kV c) 30 kV d) 40 kV

152. A Kα X-ray emitted from a sample has a energy of 7.46 keV. Of which element is the sample made?

a) Calcium (Ca, Z = 20) b) Cobalt (Co Z=27)

c) Cadmium (Cd Z=48) d) Nickel (Ni, Z = 28)

153. In the above question, if the scientist continues taking data at higher photon energies, he will find the next
major “dip” in the intensity graph at what photon energy?
a) 1 E b) 8 E c) 3E d) 9E
9 0
9 0 0 0

154. The photon radiated from hydrogen corresponding to 2nd line of Lyman series is absorbed by a hydrogen-
like atom ‘X’ in 2nd excited state. As a result then, the hydrogen-like atom ‘X’ makes a transition of n orbit
th

a) X = He+, n = 4 b) X = Li++, n = 6 c) X = He+, n = 6 d) X = Li++, n = 9

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 257 257
155. If the atom 100 Fm follows the Bohr model and the radius of 100 Fm is n times the Bohr radius, then find
n.
a) 100 b) 200 c) 4 d) 1/4

156. In a hypothetical system, a particle of mass m and charge -3q is moving around a very heavy particle
charge q. Assume that Bohr’s model is applicable to this system, then velocity of mass m in firsts orbit is
2 2
a) 3q b) 3q c) 3q d) 3q
2ε0h 4ε0h 2πε0h 4πε0h
157. The electron in a hydrogen atom makes a transition from n = n1 to n = n2 state. 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) n = 4, n = 2 b) n = 8, n = 2 c) n = 8, n = 3 d) n = 6, n = 2
1 2 1 2 1 2 1 2

158. When a hydrogen atom is raised from the ground state to fifth state:

a) Both KE and PE increase b) Both KE and PE decrease

c) PE increases and KE decreases d) PE decreases and KE increases

159. If the radius of an orbit is r and the velocity of electron in it is v, then the frequency of electron in the orbit
will be
a) 2πrv b) 2π c) vr d) v
vr 2π 2πr
160. Two hydrogen atoms are in excited state with electrons residing in n = 2. The first one is moving towards
left and emits a photon of energy E1 towards right. The second one is moving towards right with same
speed and emits a photon of energy E2 towards left. Taking recoil of nucleus into account, during the
emission process
a) E > E b) E < E
1 2 1 2

c) E = E d) Information insufficient
1 2

161. The wavelength of KαX-rays of two metals ‘A’ and ‘B’ are 4/1875R and 1/675R, respectively, where ‘R’ is
Rydberg’s constant. The number of elements lying between ‘A’ and ‘B’ according to their atomic numbers is
a) 3 b) 6 c) 5 d) 4

162. In a hydrogen atom, the transition takes place from n = 3 to n = 2. If Rydberg’s constant is 1.09×107m-1,
the wavelength of the line emitted is
a) 6606 Å b) 4861 Å c) 4340 Å d) 4101 Å

163. If λ1 and λ2 are the wavelength of the first members of the Lyman and Paschen series, respectively, then
λ1:λ2 is
a) 1:3 b) 1:30 c) 7:50 d) 7:108

164. Three photons coming from excited atomic-hydrogen sample are picked up. Their energies are 12.1 eV,
10.2 eV, and 1.9 eV. These photons must come from
a) A single atom b) Two atoms

c) Three atoms d) Either two atoms or three atoms

165. Whenever a hydrogen atom emits a photon in the Balmer series

a) It need not emit any more photon

b) It may emit another photon in the Paschen series

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 c) It must emit another photon in the Lyman series

d) It may emit another photon in the Balmer series

166. The element which has a Kα X-rays line of wavelength 1.8 Å is

7 -1
(R = 1.1×10 m , b = 1 and 5/33 = 0.39)
a) Co, Z = 27 b) Iron, Z = 26 c) Mn, Z = 25 d) Ni, Z = 28

167. The angular momentum of an electron in hydrogen atom is 4h/2π. Kinetic energy of this electron is

a) 4.35 eV b) 1.51 eV c) 0.85 eV d) 13.6 eV

168. If wavelength of photon emitted due to transition of an electron from third orbit to first orbit in a
hydrogen atom is λ, then the wavelength of photon emitted due to transition of electron from fourth orbit
to second orbit will be
a) 128 λ b) 25 λ c) 36 λ d) 125 λ
27 9 7 11
169. If the electron in an hydrogen atom jumps from an orbit with level nf = 3 to an orbit with level nf = 2, the
emitted radiation has a wavelength given by
a) λ = R b) λ = 36 c) λ = 6 d) λ = 5R
6 5R R 36
3+
170. The speed of electrons in the second orbit of Be ion will be

a) c b) 2c c) 3c d) 4c
137 137 137 137

Multiple Correct Answers Type

171. In an X-ray tube, the voltage applied is 20 kV. The energy required to remove an electron from L shell is
19.9 keV. In the X-rays emitted by the tube,
a) Minimum wavelength will be 62.1 nm

b) Energy of the characteristic x-rays will be equal to or less than 19.9 keV

c) L X-ray may be emitted

α

d) L X-ray will have energy 19.9 keV

α

172. The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the
following statements is true?
a) Its kinetic energy increases and its potential and total energies decrease

b) Its kinetic energy decreases, potential energy increases, and its total energy remains the same

c) Its kinetic and total energies decrease and its potential, energy increases

d) Its kinetic, potential and total energies decrease

173. Which of the following products, in a hydrogen atom, are independent of the principal quantum number n?
The symbols have their usual meanings
a) ω2r b) E2 c) v2r d) E
v r
174. The shortest wavelength of X-rays emitted from an X-ray tube depends on

a) The current in the tube b) The voltage applied to the tube

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 c) The nature of the gas in the tube d) The atomic number of the target material

175. In Bohr’s model of hydrogen atom:

a) The radius of nth orbit is proportional to n2

b) The total energy of electron in nth orbit proportional to n

c) The angular momentum of the electron in an orbit is an integral multiple of h/2π

d) The magnitude of the potential energy of an electron in any orbit is greater than its kinetic energy

176. An X-ray tube is operating at 50 kV and 20 mA. The target material of the tube has mass of 1 kg and
-1 -1
specific heat 495 J kg ℃ . One percent of applied electric power is converted into X-rays and the
remaining energy goes heating the target. Then,
a) A suitable target material must have high melting temperature

b) A suitable target material must have low thermal conductivity

c) The average rate of rise of temperature of the target would be 2℃ s-1

d) The minimum wavelength of X-rays emitted is about 0.25×10-10 m

177. In Bohr’ model of the hydrogen atom

a) The radius of the nth orbit is proportional of n2

b) The total energy of the electron in nth orbit is inversely proportional to n

c) The angular momentum of the electron in an orbit is an integral multiple of h/2λ

d) The magnitude of potential energy of the electron in any orbit is greater than its KE

178. Suppose the potential energy between an electron and a proton at a distance r is given by Ke2/3r3.
Application of Bohr’s theory to hydrogen atom in this cases shows that:
a) Energy in the nth orbit is proportional to n6

b) Energy is proportional to m-3 (m =mass of electron)

c) Energy in the nth orbit is proportional to n-2

d) Energy is proportional to m3(m = mass of electron)

179. If the potential energy of the electron in the first allowed orbit in hydrogen atom is E; its

a) Ionization potential is -E/2 b) Kinetic energy is –E/2

c) Total energy is E/2 d) None of these

180. The electron in a hydrogen atom makes a transition n1→n2, where n1 and n2 are the principal quantum
numbers of the two states. Assume the Bohr model to be valid. 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) n = 4, n = 2 b) n = 8, n = 2 c) n = 8, n = 1 d) n = 6, n = 3
1 2 1 2 1 2 1 2

181. A hydrogen atom having kinetic energy E collides with a stationary hydrogen atom. Assume all motions
are taking place along line of motion of the moving hydrogen atom. For this situation, mark out the correct
statement(s)
a) For E ≥ 20.4 eV only, collision would be elastic
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 b) For E ≥ 20.4 eV only, collision would be inelastic

c) For E = 2.4 eV, collision would be perfectly inelastic

d) For E = 18 eV, the KE of initially moving hydrogen atom after collision is zero

182. Whenever a hydrogen atom emits a photon in the Balmer series:

a) It may emit another photon in the Balmer series

b) It must emit another photon in the Lyman series

c) The second photon, if emitted, will have a wavelength of about 122 nm

d) It may emit a second photon, but the wavelength of this photon cannot be predicted

183. Which of the following statements are correct for an X-ray tube?

a) On increasing potential difference between the filament and target, photon flux of X-rays increases

b) On increasing potential difference between the filament and target, frequency of X-rays increases

c) On increasing filament current, cut-off wavelength increases

d) On increasing filament current, intensity of X-rays increases

184. Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but
having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of
this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has
wavelength λ (giving in terms of the Rydberg constant R for the hydrogen atom) equal to
a) 9/(5R) b) 36/(5R) c) 18/(5R) d) 4/R

185. A gas of monoatomic hydrogen is bombarded with a stream of electrons that have been accelerated from
rest through a potential difference of 12.75 V. In the emission spectrum, one can observe lines of
a) Lyman series b) Balmer series c) Paschen series d) Pfund series

186. Let An be the area enclosed by the nth orbit in a hydrogen atom. The graph of In (An/A1) against In(n)

a) Will pass through origin

b) Will be a straight line with slope 4

c) Will be monotonically increasing nonlinear curve

d) Will be a circle

187. Which of the following statements are true?

a) The shortest wavelength of X-rays emitted from an X-ray tube depends on the current in the tube

b) Characteristic X-ray spectra is sample as compared to optical spectra

c) X-rays cannot be diffracted by means of an ordinary grating

d) There exists a sharp limit on the short wavelength side for each continuous X-ray spectrum

188. Which of the following are in the ascending order of wavelength?

a) Hα, H , Hγ… lines in Balmer series of hydrogen atom

β

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 b) Lyman limit, Balmer limit, and Paschen limit in the hydrogen spectrum

c) Violet, blue, yellow, and red colors in solar spectrum

d) None of the above

189. The transition from the state n = 4 to n = 3 in a hydrogen-like atom results in ultra violet radiations.
Infrared radiation will be obtained in the transition
a) 2→1 b) 3→2 c) 4→2 d) 5→

190. As per Bohr model, the minimum energy (in eV), required to remove an electron from the ground state of
doubly ionized Li atom (Z = 3) is
a) 1.51 b) 13.6 c) 40.8 d) 122.4

191. Energy liberated in the de-excitation of hydrogen atom from 3rd level to 1st level falls on a photo-cathode.
Later when the same photo-cathode is exposed to a spectrum of some unknown hydrogen-like gas, excited
nd
to 2 energy level, it is found that the de Broglie wavelength of the fastest photoelectrons now ejected has
nd st
decreased by a factor of 3. For this new gas, difference of energies of 2 Lyman line and 1 Balmer line is
found to be 3 times the ionization potential of the hydrogen atom. Select the correct statement(s):
a) The gas is lithium b) The gas is helium

c) The work function of photo-cathode is 8.5 eV d) The work function of photo-cathode is 5.5 eV

192. Which of the following statements about hydrogen spectrum are correct?

a) All the lines of Lyman series lie in ultraviolet region

b) All the line of Balmer series lie in visible region

c) All the lines of Paschen series lie in infrared region

d) None of the above

193. The third line of the Balmer series spectrum of a hydrogen like ion of atomic number Z equals to 108.5 nm.
The binding energy of the electron in the ground state of these ions is EB. Then
a) Z = 2 b) E = 54.4 eV c) Z = 3 d) E = 122.4 eV
B B

194. The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation

a) The intensity increases b) The minimum wavelength increases

c) The intensity remains unchanged d) The minimum wavelength decreases

195. If elements with principal quantum number n > 4 were not allowed in nature, the number of possible
elements would be
a) 60 b) 32 c) 4 d) 64

196. According to Bohr’s theory of hydrogen atom, for the electron in the nth permissible orbit

a) Linear momentum ∝ 1 b) Radius of orbit ∝ n

n
c) Kinetic energy ∝ 1 d) Angular momentum ∝ n
2
n
197. According to Einstein’s theory of relativity, mass can be converted into energy and vice-versa. The lightest
elementary particle, taken to be the electron, has a mass equivalent to 0.51 MeV of energy. Then, we can
say that

P a g e | 21
 a) The minimum amount of energy available through conversion of mass into energy is 1.2 MeV

b) The least energy of a γ-ray photon that can be converted into mass is 1.02 MeV

c) Whereas the minimum energy released by conversion of mass into energy is 1.02 MeV, it is only a γ-ray
photon of energy 0.51 MeV and above that can be converted into mass
d) Whereas the minimum energy released by conversion of mass into energy is 0.51 MeV, it is only a γ-ray
photon of energy 1.02 MeV and above that can be converted into mass
198. X-ray from a tube with a target A of atomic number Z shows strong K lines for target A and weak K lines for
impurities. The wavelength of Kα lines is λz for target A and λ1 and λ2 for two impurities
λz λ 1
= 4 and z =
λ1 λ2 4
Assuming the screening constant of Kα lines to be unity, select the correct statement(s)
a) The atomic number of first impurity is 2z - 1

b) The atomic number of first impurity is 2z + 1

c) The atomic number of second impurity is (z+1)

2
d) The atomic number of second impurity is z + 1
2
199. According to Bohr’s theory of the hydrogen atom, for the electron in the nth allowed orbit:

a) The linear momentum is proportional to (1/n) b) The radius is proportional to n

c) The kinetic energy is proportional to (1/n2) d) The angular momentum is proportional to n

200. Consider the spectral line resulting from the transition n = 2→n = 1 in the atoms and ions given below.
The shortest wavelength is produced by
a) Hydrogen atom b) Deuterium atom

c) Singly ionized helium d) Doubly ionized lithium

201. When a hydrogen atom is excited from ground state to first excited state, then

a) Its kinetic energy increases by 20 eV b) Its kinetic energy decreases by 10.2 eV

c) Its potential energy increases by 20.4 eV d) Its angular momentum increases by 1.05×10-34 Js

202. The Kα X-ray emission line of tungsten occurs at λ = 0.02 nm. The energy difference between K and L
levels in this atom is about
a) 0.51 MeV b) 1.2 MeV c) 59 MeV d) 13.6 MeV

203. Electrons with energy 80 keV are incident on the tungsten target of an X-ray tube. K shell electrons of
tungsten have -72.5 keV energy. X-rays emitted by the tube contain only
a) A continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of -0.155 Å

b) Continuous X-ray spectrum (Bremsstrahlung) with all wavelengths

c) The characteristic X-ray spectrum of tungsten

d) A continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of -0.155 Å and the
characteristic X-ray spectrum of tungsten
204. Suppose frequency of emitted photon is f0 when the electron of a stationary hydrogen atom jumps from a
higher state m to a lower state n. If the atom is moving with a velocity v ( ≪ c) and emits a photon of

P a g e | 22
 frequency f during the same transition, then which of the following statements are possible?
a) f may be equal to f b) f may be greater than f
0 0

c) f may be less than f d) f cannot be equal to f

0 0

205. In Bohr’s model of the hydrogen atom, let R, V, T, and E represent the radius of the orbit, speed of the
electron, time period revolution of electron, and the total energy of the electron, respectively. The
quantities proportional to the quantum number n are
a) VR b) RE c) T d) V
R E
206. Which of the following statements are correct?

a) If angular momentum of the Earth due to its motion around the Sun were quantized according to the
74
Bohr’s relation L = nh/2π, then the quantum number n would be of the order of 10
b) If elements with principal quantum number >4 were not allowed in nature, then the number of possible
elements would be 64
c) Rydberg’s constant varies with mass number of the element

d) The ratio of the wave number of Ha line of Balmer series for hydrogen and that of Ha line of Balmer
series for singly ionized helium is exactly 4
207. The intensity of X-rays from a Coolidge tube is plotted against wavelength λ as shown in figure. The
minimum wavelength found is λC and the wavelength of the Kα line is λK. As the accelerating voltage is
increased

a) λ - λ increases b) λ - λ decreases c) λ increases d) λ decreases

K C K C K K

208. If, in a hydrogen atom, radius of nth Bohr orbit is rn, frequency of revolution of electron in nth orbit is fn, and
th
area enclosed by the n orbit is An, then which of the following graphs are correct?

a) b)

c) d)

209. Mark out the correct statement(s)

a) Line spectra contain information about atoms only

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 b) Line spectra contain information about both atoms and molecules

c) Band spectra contain information about both atoms and molecules

d) Band spectra contain information about molecules only

210. Continuous spectrum is produced by

a) Incandescent electric bulb b) Sun

c) Hydrogen molecules d) Sodium vapor lamp

211. X-rays are produced in an X-ray tube operating at a given accelerating voltage. The wavelength of the
continuous X-rays has values from
a) 0 and ∞ b) λ to ∞; where λ > 0
min min

c) 0 to λ ; where λ < ∞ d) λ to λmax; where 0 < λmin < ∞

max max min

212. For a certain metal, the K absorption edge is at 0.172 Å. The wavelength of Kα, Kβ and Kγ lines of K series
are 0.210 Å, 0.192 Å, and 0.180 Å, respectively. The energies of K, L, and M orbits are EK, EL and EM,
respectively. Then
a) E = -13.04 keV b) E = -7.52 keV c) E = -3.21 keV d) E = 13.04 keV
K L M K

213. The X-ray beam coming from an X-ray tube will be

a) Monochromatic

b) Having all wavelengths smaller than a certain maximum wavelength

c) Having all wavelengths larger than a certain minimum wavelength

d) Having all wavelengths lying between a minimum and a maximum wavelength

214. Let λα, λβ and λ'α denotes the wavelength of the X-rays of the Kα, Kβ and Lα lines in the characteristic X-rays
for a metal. Then,
a) λ' > λ > λ b) λ' > λ > λ c) 1 = 1 + 1 d) 1 = 1 + 1
' '
α α β α β α λβ λα λα λα λβ λα
215. Hydrogen atoms absorb radiation of wavelength λ0 and consequently emit radiations of 6 different
wavelength of which two wavelengths are shorter than λ0. Then,
a) The final excited state of the atoms is n = 4

b) The initial state of the atoms may be n = 2

c) The initial state of the atoms may be n = 3

d) There are three transitions belonging to Lyman series

216. An X-ray tube is operated at 6.6 kV. In the continuous spectrum of the emitted X-rays, which of the
following frequencies will be missing?
a) 1018 Hz b) 1.5×1018 Hz c) 2×1018 Hz d) 2.5×1018 Hz

217. The mass number of a nucleus is

a) Always less than its atomic number

b) Always more than its atomic number

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 c) Sometimes equal to its atomic number

d) Sometimes more than and sometimes equal to its atomic number

Assertion - Reasoning Type

This section contain(s) 0 questions numbered 218 to 217. Each question contains STATEMENT 1(Assertion)
and STATEMENT 2(Reason). Each question has the 4 choices (a), (b), (c) and (d) out of which ONLY ONE is
correct.

a) Statement 1 is True, Statement 2 is True; Statement 2 is correct explanation for Statement 1

b) Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for Statement 1

c) Statement 1 is True, Statement 2 is False

d) Statement 1 is False, Statement 2 is True

218

Statement 1: In He-Ne laser, population inversion takes place between energy levels of neon atoms.

Statement 2: Helium atoms have a meta-stable energy level.

219

Statement 1: In a hydrogen atom, energy of emitted photon corresponding to transition from n = ∞ to

n = 1 is much greater as compared to transition from n = ∞ to n = 2
Statement 2: Wavelength of photon is directly proportional to the energy of emitted photon

220
+
Statement 1: The energy of a He ion for a given n is almost exactly four times that of H atom for the
same n
+
Statement 2: Photons emitted during transition between corresponding pairs of levels in He and H
have the same energy E and the same wavelength λ = hc/E
221

Statement 1: The different lines of emission spectra (like Lyman, Balmer etc) of atomic hydrogen gas
are produced by different atoms.
Statement 2: The sample of atomic hydrogen gas consists of millions of atoms.

222

Statement 1: An alpha particle is a doubly ionized helium atom.

Statement 2: An alpha particle carries 2 units of positive charge.

223

Statement 1: It is difficult to excite nucleus to higher energy states by usual methods which we use to
excite atoms like by heating or by irradiation of light.
Statement 2: Terms like ground state or excited state for nucleus are meaningless.

224

P a g e | 25
 Statement 1: Bohr had to postulate that the electrons in stationary orbits around the nucleus do not
radiate.
Statement 2: According to classical physics all moving electrons radiate.

225

Statement 1: In an X-ray tube, if the energy with which an electron strikes the metal target increases,
then the wavelength of the characteristic X-rays also changes
Statement 2: Wavelength of characteristic X-rays depends only on the initial and final energy levels

Matrix-Match Type

This section contain(s) 0 question(s). Each question contains Statements given in 2 columns which have to be
matched. Statements (A, B, C, D) in columns I have to be matched with Statements (p, q, r, s) in columns II.

226. Match the following:

Column-I Column- II

(A) Characteristic X-ray (p) Inverse process of photoelectric effect

(B) X-ray production (q) High potential difference

(C) Cut-off wavelength (r) Moseley’s law

(D) Continuous X-ray (s) Emission of radiations

CODES :

A B C D

a) P,q r s q

b) r,s p,q p,r p

c) r p,q,r,s q, s

d) s q p,q r,s

227. Match the following:

Column-I Column- II

(A) The voltage applied to X-ray tube is increased (p) Average KE of the electrons decreases

(B) In photoelectric effect, work function of the (q) Average KE of the electrons increases
target is increased
(C) Stopping potential decrease (r) Cut-off wavelength decreased

(D) Wavelength of K∞ X-ray increased (s) Atomic number of target material decreases

CODES :

A B C D

a) P,r q,s s q

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 b) s p,q q,r p,r

c) p,r q,r p,r q,s

d) q,r p,r p,r s

228. Match the following lists.

Column-I Column- II

(A) Burning candle (p) Line spectrum

(B) Sodium vapour (q) Continuous spectrum

(C) Bunsen flame (r) Band spectrum

(D) Dark lines in solar spectrum (s) Absorption spectrum

CODES :

A B C D

a) c a b d

b) c b a d

c) b c a d

d) b a c d

229.

Column-I Column- II

(A) Kα photon of aluminium (p) Will be most energetic among the four

(B) Kβ photon of aluminium (q) Will be least energetic among the four

(C) Kα photon from sodium (r) Will be more energetic than the lithium

(D) Kβ photon of beryllium (s) Constant speed

CODES :

A B C D

a) Q,r r,s p,q p,r

b) r,s q,r,s r,s q,r,s

c) p,q q,r p,r r,s

d) q,r,s p,q r,s p,r

230. Match the entries of Column I with the entries of Column II:

Column-I Column- II

(A) Emission spectra (p) Discrete

P a g e | 27
 (B) Absorption spectra (q) Continuous

(C) X-ray spectra (r) Electronic Transition

(D) Thermal radiation spectra (s) Quantum theory of electromagnetic radiation

CODES :

A B C D

a) Q,r,s p,q,r,s q,r,s q,r,s

b) p,q r,s p,q,r,s q,r

c) r,s p,q q,r q,r,s

d) q,r,s r,s q,r,s p,q

231. In each situation of Column I, a physical quantity related to orbiting electron in hydrogen-like atom is
given. The terms ‘Z’ and ‘n’ given in Column II have usual meaning in Bohr’ theory. Match the quantities in
Column I with the terms they depend on it Column II
Column-I Column- II

(A) Frequency of orbiting electron (p) Is directly proportional to Z2

(B) Angular momentum of orbiting electron (q) Is directly proportional to n

3
(C) Magnetic moment of orbiting electron (r) Is inversely proportional to n

(D) The average current due to orbiting of (s) Is independent of Z

electron
CODES :

A B C D

a) P,r q,s q,s q,r

b) q,s p,r q,r p,q

c) p,q q,r p,r q,s

d) q,r p,q q,s p,s

232.

Column-I Column- II

(A) Radius of orbit depends on principal quantum (p) Increase

number as
(B) Due to orbital motion of electron, magnetic (q) Decrease
field arises at the center of nucleus is
proportional to principal quantum number as
(C) If electron is going from lower energy level to (r) 1
Proportional to 2
higher energy level, then velocity of electron n
will
(D) If electron is going from lower energy level to (s) Proportional to n2
higher energy level, then total energy of

P a g e | 28
 electron will
(t) 1
Is proportional to 5
n
CODES :

A B C D

a) t q p s

b) p r s t

c) s t q p

d) q p r s

233. Match the appropriate pairs from Lists I and II.

Column-I Column- II

(A) Nitrogen molecules (p) Continuous spectrum

(B) Incandescent solids (q) Absorption spectrum

(C) Fraunhoffer lines (r) Band spectrum

(D) Electric arc between iron roads (s) Emission spectrum

CODES :

A B C D

a) c a b d

b) b a d c

c) d a b c

d) a c d b

234. Take the usual meanings of the symbols to match the following:

Column-I Column- II

(A) Average kinetic energy of photoelectrons (p) Zero

(B) Minimum kinetic energy of photoelectrons (q) hc/λ - λ

(C) Maximum wavelength of continuous X-rays (r) hc/eV

(D) Minimum wavelength of continuous X-rays (s) Not predictable

CODES :

A B C D

a) p q r s

b) s p s r

c) r s p q
P a g e | 29
 d) q r p p

235.

Column-I Column- II

(A) Radius of orbit is related with atomic number (p) Is proportional to Z

(Z)
(B) Current associated due to orbital motion of (q) Is inversely proportion to Z
electron with atomic number (Z)
2
(C) Magnetic field at the center due to orbital (r) Is proportional to Z
motion of electron related with Z
(D) Velocity of an electron related with atomic (s) Is proportional to Z3
number (Z)
CODES :

A B C D

a) q r s p

b) p q r s

c) r s q p

d) s p q r

236. The spectral lines of hydrogen-like atom fall within the wavelength range from 950 Å to 1350 Å. Then,
match the following
Column-I Column- II

(A) If it is atomic hydrogen atom and energy (p) λ = 1212 Å and it corresponds to transition
E = -0.85 eV from 2 to 1
(B) If it is atomic hydrogen atom and energy (q) λ = 134 Å and it corresponds to transition
E = -3.4 eV from 2 to 1
(C) If it is doubly ionized lithium atom, then (r) λ = 303 Å and it corresponds to transition
from 2 to 1
(D) If it is singly ionized helium atom, then (s) λ = 970 Å and it corresponds to transition
from 4 to 1
CODES :

A B C D

a) p r s q

b) s p q r

c) r q p s

d) q s r p

Linked Comprehension Type

This section contain(s) 32 paragraph(s) and based upon each paragraph, multiple choice questions have to be
answered. Each question has atleast 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

P a g e | 30
Paragraph for Question Nos. 237 to -237

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/2π), where m = mass of the electron, v = velocity of the electron in the orbit, r =
th th
radius of the orbit, and n = 1, 2, 3,… When transition takes place from K orbit to J orbit, energy photon is
1
[ ]
1 1
emitted. If the wavelength of the emitted photon is λ, we find that = R 2 - 2 , where R is Rydberg’s constant
λ J K

On a different planet, the hydrogen atom’s structure was somewhat different from ours, there the angular
momentum of electron was P = 2n(h/2π), i.e., an even multiple of (h/2π)
Answer the following questions regarding the other planet based on above passage:

237. The minimum permissible radius of the orbit will be

2 2 2 2
a) 2ε0h b) 4ε0h c) ε0h d) ε0h
2 2 2 2
mπe mπe mπe 2mπe

Paragraph for Question Nos. 238 to - 238

In an ordinary atom, as a first approximation, the motion of the nucleus can be ignored. In a positronium atom,
a positron replaces the proton of hydrogen atom. The electron and positron masses are equal and, therefore,
the motion of the positron cannot be ignored. One must consider the motion of both electron and positron
about their centre of mass. A detailed analysis shows that formulae of Bohr model apply to positronium atom
provided that we replace me by what is known as the reduced mass of the electron. For positronium, the
reduced mass is me/2

238. The orbital radius of the first excited level of positronium atom is
Where a0 is the orbital radius of ground state of hydrogen atom
a) 4a b) a /2 c) 8a d) 2a
0 0 0 0

Paragraph for Question Nos. 239 to - 239

th th
The electrons in a H-atom kept at rest, jumps from the m shell to the n shell (m > n). Suppose instead of
emitting electromagnetic wave, the energy released is converted into kinetic energy of the atom. Assume Bohr
model and conservation of angular momentum are valid. Now, answer the following questions:

239. What principle is violated here?

a) Laws of motion b) Energy conservation c) Nothing is violated d) Cannot be decided

P a g e | 31
Paragraph for Question Nos. 240 to - 240

The energy levels of a hypothetical one electron atom are shown in figure
n = ∞_________________________0 eV
n = 5_________________________-0.80 eV
n = 4__________________________-1.45 eV
n = 3_________________________ -3.08 eV
n = 2_________________________-5.30 eV
n = 1_________________________-15.6 eV

240. Find the ionization potential of the atom

a) 11.2 eV b) 13.5 eV c) 15.6 eV d) 12.6 eV

Paragraph for Question Nos. 241 to - 241

The electron in a hydrogen atom at rest makes a transition from n = 2 energy state to the n = 1 ground state

241. Find the energy (eV) of the emitted photon

a) 5.8 eV b) 8.3 eV c) 10.2 eV d) 12.7 eV

Paragraph for Question Nos. 242 to - 242

Consider a hypothetical hydrogen-like atom. The wavelength, in Å, for the spherical lines for transitions from
n = p to n = 1 are given by
2
1500p
λ= 2 where p = 1, 2, 3, 4,…
p -1

242. Find the wavelength of the most energetic photons in this series

a) 1800 Å b) 1500 Å c) 1300 Å d) 1650 Å

Paragraph for Question Nos. 243 to - 243

A sample of hydrogen gas in its ground state is irradiated with photons of 10.02 eV energies. The radiation from
+ 2+
the above sample is used to irradiate two other samples of excited ionized He and excited ionized Li ,
respectively. Both the ionized samples absorb the incident radiation

243. How many spectral lines are obtained in the spectra of Li2+?

a) 10 b) 15 c) 20 d) 17

P a g e | 32
Paragraph for Question Nos. 244 to - 244

A neutron of kinetic energy 65 eV collides inelastically with a singly ionized helium atom at rest. It is scattered
at an angle 90° with respect to its original direction

244. Find the minimum allowed values of energy of the neutron

a) 0.39 eV b) 0.32 eV c) 0.25 eV d) 0.43 eV

Paragraph for Question Nos. 245 to - 245

A electron and a photon are separated by a distance r so that the potential energy between them is u = klog r,
where k is a constant

245. In such an atom, radius of nth Bohr’s orbit is

a) 2nh b) nh c) nh d) nh
π mk 2π 2mk 2π mk 4π mk

Paragraph for Question Nos. 246 to - 246

Pertain to the following statement and figure

The figure above shows as energy level diagram of the hydrogen atom. Several transitions are marked as I, II,
III,… The diagram is only indicative and not to scale

246. In which transition is a Balmer series photon absorbed?

a) II b) III c) IV d) VI

Paragraph for Question Nos. 247 to - 247

A certain species of ionized atoms produces emission line spectrum according to the Bohr model. A group of
lines in the spectrum is forming a series in which the shortest wavelength is 22.79 nm and the longest
wavelength is 41.02 nm. The atomic number of atom is Z
Based on above information, answer the following questions:

247. The value of Z is

P a g e | 33
 a) 2 b) 3 c) 4 d) 5

Paragraph for Question Nos. 248 to - 248

Simplified model of electron energy levels for a certain atom is shown in figure. The atom is bombarded with
fast moving electrons. The impact of one of these electrons can cause the removal of electron from K-level, thus
creating a vacancy in the K-level. This vacancy in K-level is filled by an electron from L-level and the energy
released in this transition can either appear as electromagnetic waves or may all be used to knock out an
electron from M-level of the atom

Based on the above information, answer the following questions:

248. The minimum potential difference through which bombarding electron beam must be accelerated from
rest to cause the ejection of electron from K-level is
a) 18750 V b) 400 kV c) 2.16 kV d) 21.6 kV

Paragraph for Question Nos. 249 to - 249

A monochromatic beam of light having photon energy 12.5 eV is incident on a sample A of atomic hydrogen gas
in which all atoms are in ground state. The emission spectra obtained from this sample is incident on another
st
sample B of atomic hydrogen gas in which all atoms are in 1 excited state
Based on the above information, answer the following questions:

249. The atoms of sample A after passing of light through it

a) May be in 1st excited state b) May be in 2nd excited state

c) May be in both 1st and 2nd excited states d) None of the above

Paragraph for Question Nos. 250 to - 250

An electron orbits a stationary nucleus of charge +ze, where z is a constant and e is the magnitude of electronic
charge. It requires 47.2 eV to excite the electron from the second Bohr orbit to third Bohr orbit

250. The value of z is

a) 5 b) 4 c) 3 d) 2

P a g e | 34
Paragraph for Question Nos. 251 to - 251

When high energetic electron beam, (i.e., cathode rays) strike the heavier metal, then X-rays are produced.
Spectrum of X-rays are classified into two categories: (i) continuous spectrum, and (ii) characteristic spectrum.
The wavelength of continuous spectrum depends only on the potential difference across the electrode. But
wavelength of characteristic spectrum depends on the atomic number (z)

251. The production of characteristic X-ray is due to the

a) Continuous acceleration of incident electrons towards the nucleus

b) Continuous retardation of incident electrons towards the nucleus

c) Electron transitions between inner shells of the target atom

d) Electron transitions between outer shells of the target atom

Paragraph for Question Nos. 252 to - 252

Light from a discharge tube containing hydrogen atom falls on the surface of a plate of sodium. The kinetic
energy of the fastest photoelectrons emitted from sodium is 0.73 eV. The work function for sodium is 1.82 eV

252. The energy of the photons causing the photoelectric emission is

a) 2.55 eV b) 0.73 eV

c) 1.82 eV d) Information insufficient

Paragraph for Question Nos. 253 to - 253

++ th
The electron in a Li ion is in the n shell, n being very large. One of the K-electrons in another metallic atom
++
has been knocked out. The second metal has four orbits. Now, we take two samples-one of Li ions and the
other of the second metallic ions. Suppose the probability of electronic transition from higher to lower energy
levels is directly proportional to the energy difference between the two shells. Take hc = 1224 eV nm, where h
is Planck’s constant and c the velocity of light in vacuum. It is found that major electromagnetic waves emitted
from the two samples are identical. Now, answer the following questions:

253. What is the X-ray having least intensity emitted by the second sample?

a) K b) L c) M d) Data insufficient
α α α

Paragraph for Question Nos. 254 to - 254

Two hydrogen-like atoms A and B are of different and each atom has ratio of neutron to proton equal to unity.
The difference in the energies between the first Lyman lines emitted by A and B is 81.6 eV. When the atoms A
and B moving with the same velocity strike separately a heavy target, they rebound back with half of the speed

P a g e | 35
before collision. However, in this process atom B imparts the target a momentum which is three times the
momentum imparted to target by atom A

254. Atom A is

a) 1H b) 2H c) 6Li d) 4Li
1 1 3 2

Paragraph for Question Nos. 255 to - 255

1.8 g of hydrogen is excited by irradiation. The study of spectra indicated that 27% of the atoms are in the first
excited state, 15% of the atoms in the second excited state, and the rest in the ground state. The ground state
-12
ionization potential energy of hydrogen atom is 21.4×10 ergs

255. The number of atoms present in the second excited state is

a) 1.61×1023 b) 0.805×1023 c) 2.92×1023 d) 1.46×1023

Paragraph for Question Nos. 256 to - 256

In a set of experiments on a hypothetical one-electron atom, the wavelengths of the photons emitted from
transition ending in the ground state (n = 1) are shown in the energy level diagram (figure)

λ5→1 = 73.86 nm
λ4→1 = 75.63 nm
λ3→1 = 79.76 nm
λ2→1 = 94.54 nm

256. The energy of the atom in level n = 1 is nearly

a) 13.14 eV b) 15.57 eV c) 17.52 eV d) 16.42 eV

Integer Answer Type

257. Find recoil speed (approximately in m/sec) when a hydrogen atom emits a photon during the transition
from n = 5 to n = 1

P a g e | 36
258. The shortest wavelength of the Brackett series of a hydrogen-like atom (atomic number Z) is the same as
the shortest wavelength of the Balmer series of hydrogen atom. Find the value of Z
259. An electron in nth excited state in a hydrogen atom comes down to first excited state by emitting ten
different wavelengths. Find value of n (an integer)
260. The average lifetime for the n = 3 excited state of a hydrogen-like atom is 4.8×10-8 sec and that for the
-8
n = 2 state is 12.8×10 sec. The ratio of the average number of revolutions made in the n = 2 state of the
average number of revolutions made in the n = 3 state before any transitions can take place from these
states is
261. Heat at the rate of 200 W is produced in an X-ray tube operating at 20 kV. Find the current in the circuit.
-2
Assume that only a small fraction of the kinetic energy of electrons is converted into X-rays (in ×10 A)
262. A Bohr hydrogen atom undergoes a transition n = 5 to n = 4 and emits a photon of frequency f.
Frequency of circular motion of electron in n = 4 orbit is f4. The ratio f/f4 is found to be 18/5m. State the
value of m
263. An electron in hydrogen atom jumps from n1 state to n2 state, where n1 and n2 represent the quantum
number of two states. The time period of revolution of electron in initial state is 8 times that in final state.
Then the ratio of n1 and n2 is
264. An atom of atomic number Z = 11 emits Kα wavelength which is λ. Find the atomic number for an atom
that emits Kα radiation with wavelength 4λ (an integer)
265. An electron in an H-atom kept at rest, jumps from the mth shell to the nth shell (m > n). Suppose instead of
emitting electromagnetic wave, the energy released is converted into the kinetic energy of the atom.
Assume Bohr model and conservation of angular momentum are valid. If I is the moment of inertia the
angular velocity of the atom about the nucleus is 4(m-n)h/kI. Calculate k
266. In the spectrum of singly ionized helium, the wavelength of a line observed is almost the same as the first
line of Balmer series of hydrogen. It is due to transition of electron from n1 = 6 to n2 = ‘*’. What is the
value of ‘*’

P a g e | 37
 12.ATOMS

: ANSWER KEY :

1) d 2) a 3) d 4) b 105) b 106) b 107) a 108) d

5) d 6) a 7) c 8) b 109) c 110) d 111) a 112) d

9) b 10) b 11) c 12) b 113) d 114) b 115) b 116) b

13) c 14) a 15) d 16) c 117) d 118) d 119) d 120) d

17) d 18) d 19) b 20) b 121) d 122) c 123) b 124) d

21) c 22) b 23) d 24) c 125) d 126) d 127) b 128) c

25) c 26) d 27) c 28) a 129) d 130) c 131) c 132) a

29) c 30) b 31) c 32) b 133) b 134) c 135) c 136) b

33) a 34) c 35) a 36) b 137) d 138) d 139) b 140) b

37) c 38) b 39) c 40) b 141) b 142) c 143) c 144) a

41) a 42) c 43) b 44) c 145) c 146) c 147) d 148) b

45) a 46) d 47) b 48) b 149) a 150) a 151) d 152) d

49) c 50) c 51) d 52) d 153) b 154) d 155) d 156) a

53) a 54) a 55) a 56) d 157) a 158) c 159) d 160) b

57) d 58) a 59) a 60) a 161) d 162) a 163) d 164) d

61) c 62) a 63) a 64) a 165) c 166) a 167) c 168) a

65) b 66) c 67) b 68) c 169) b 170) b 1) a,b,c 2) a

3) b,c 4) b
69) c 70) a 71) d 72) a
5) a,c,d 6) a,c 7) a,c,d 8)
73) d 74) a 75) b 76) c a,b

77) a 78) a 79) c 80) d 9) a,b,c 10) a,d 11) b,c,d 12)
81) d 82) a 83) b 84) a b,c

85) a 86) b 87) b 88) a 13) b,d 14) c 15) a,b,c 16)
a,b
89) d 90) b 91) d 92) c
17) b,c,d 18) b,c 19) d 20) d
93) b 94) c 95) d 96) c
21) b,c 22) a,c 23) a,b 24)
97) b 98) d 99) a 100) a c,d

101) b 102) c 103) d 104) c 25) a 26) a,c,d 27) a,b 28)
P a g e | 38
a,c

29) a,c,d 30) d 31) b,c,d 32) c

33) d 34) a,b,c 35) a,c,d 36)

a,c

37) a 38) a,b,c 39) b,d 40)

a,b

41) b 42) a,b,c 43) c 44)

a,c

45) a,b,d 46) c,d 47) c,d 1) b

2) c 3) a 4) b

5) b 6) c 7) c 8) d

1) c 2) d 3) d 4) b

5) a 6) a 7) c 8) a

9) b 10) a 11) b 1) b
2) c 3) a 4) c

5) c 6) b 7) b 8) b

9) c 10) d 11) c 12) a

13) d 14) a 15) c 16) a

17) a 18) b 19) c 20) d

1) 4 2) 2 3) 6 4) 9

5) 1 6) 5 7) 2 8) 6

9) 8 10) 4

P a g e | 39
 12.ATOMS

: HINTS AND SOLUTIONS :

3
1 (d) or a ∝ z

()
2 3
2πe a1 2
vn = k ∴ = =8
nh a2 1

We know that in cgs system k = 1 5 (d)

( )
2 2
2πe 2πe 1 1 1
∴ vn = ⇒v1 = 2
= Z R 2- 2
nh h λ 1 5
2
v1 2πe Hence, λ is minimum when Z is maximum
So =
c ch
6 (a)
2 (a)
-13.6 - (-10.2) = -3.4 eV
The wavelengths of the hydrogen spectrum could
be arranged in a formula or series named after its -13.6 2 13.6
2 = -3.4 or n = =4
discoverer. For ultraviolet spectrum the series is n 3.4
called Lyman series, for visible spectrum the
or n = 2
Balmer series, and for infrared region we have the
Paschen series 2h h h
Increase in angular momentum = - =
2π 2π 2π
The ultraviolet series is obtained when the energy
of the atom falls from higher states to the energy 6.625×10
-34
= Js
level corresponding to n = 1. Thus, ultraviolet 2×3.14
radiation can only be possible with transition
-34
from E2 to E1 out of the given transitions = 1.05×10 Js

3 (d) 7 (c)
2 2
For each principal quantum number n, number of mv 1 e
=
electrons permitted equals the number of ao 4πεo a20
elements corresponding to the quantum number
e
v=
(total number of electrons) = ∑2n 2 4πεoaom

n(n+1)(2n+1) 8 (b)
=
3
1 1
KE ∝ 2 and PE ∝
4 (b) n n
2

2
v 9 (b)
a=
r
Potential energy = -2× kinetic energy = -2E
(z)2
∴a∝ (for n = 1)
(1/z) Total energy = -2E + E = -3.4 eV = -E

P a g e | 40
 or E = 3.4 eV

p = momentum, m = mass of electron

f = cZ R
2
[ ]
1 1
2- 2
n1 n2

E=
p
2m
2 ⇒2.7×10
15 2
= cz R [ ]
1 1
2- 2
1 2

or p = 2mE
'
f = cZ R
2
[ ]
1 1
2- 2
1 3
-31 -19 -24
= 2×9.1×10 ×3.4×1.6×10 ≈ 10 Divide and solve to get: f = 3.2×10 Hz
15

de Brogile wavelength 13 (c)

h h -10

( )
λ= = = 6.6×10 m
p 2mE 1 1 1
= R 2- 2
λ n1 n2
10 (b)
For longest wavelength, n1 = 2, n2 = 3
The cut-off wavelength when V = V1 = 10 kV is

λ1 =
hc
eV1
-13
= 1243.125×10 m
∴
1
λ
1 1
2 3 [ ] 1
λ
1 1
= R 2 - 2 or = R -
4 9 [ ]
36
The cut off wavelength when V = V2 = 20 kV is, ⇒λ = for electron,
5R

hc -13 1
λ2 = = 621.56×10 m But λ ∝
eV2 m

The wavelength corresponding to Kα line is, ' 1 36 18

So λ = × =
2 5R 5R
1 3R
= (Z-1)2
λ 4 14 (a)

From given information, (λ-λ2) = 3(λ - λ1) Number of possible emission lines are n(n - 1)/2
th
when an electron jumps from n state to ground
Solving above equation, we get Z = 29 state. In this question, this value should be
(n-1)(n - 2)/2
11 (c)
(n-1)(n-2)
The first photon will excite the hydrogen atom (in Hence, 10 =
2
ground state) in first excited state (as E2 - E1 - 10.2
eV). Hence, during de-excitation a photon of 10.2 Solving this, we get n = 6
eV will be released. The second photon of energy
15 eV can ionize the atom. 15 (d)

Hence the balance energy ie, For Lyman series, the series limit wavelength is
given by
(15 - 13.6) eV = 1.4 eV is retained by the electron.

Therefore, by the second photon an electron of

1
λ
1 1
[ ]
= R 2 - 2 = R or λ =
1 ∞
1
R
energy 1.4 eV will be released.
For Balmer series, the series limit wavelength is
12 (b) given by

1
λ
'' [ ]
1 1 R ''
= R 2 - 2 = or λ =
2 ∞ 4
4
R

Clearly,

P a g e | 41
 ''
λ =4 [] 1
R
''
or λ = 4λ
P = VI

Therefore, total power drawn by Coolidge tube

16 (c) PT = VI = 200 W

h As 0.5% of the energy is carried by electron,

L1 = (1) (i)
2π
Power carried by X-ray is
(Using Bohr’ Quantization Rule)
0.5
In the first excited state of Li, 0.5% of PT = ×200 = 1 W
100

h The answer is (b)

L2 = (2) (ii)
2π
21 (c)
L
∴ 2 =2 2
L1 Centripetal acceleration = mv /r

17 (d) Further, as n increase, r also increases. Therefore,

centripetal acceleration for n = 2 is less than that
1 2 for n = 1. So, statement (i) is wrong. Statement
En ∝ 2 and rn ∝ n
n (ii) and (iii) are correct

Therefore, Enrn is independent of n 22 (b)

Hence, E1r1 = (13.6 eV)(0.53 Å)

= 7.2 eV-Å
R [ ]
1 1
2-
1 ∞
2 = RZ
2 1
2-
2 ∞[ ]
1
2 or Z = 2

23 (d)
= constant
The first three transitions from the left fall in the
18 (d) Lyman series of the hydrogen spectrum which
corresponds to ultraviolet radiation
We know that frequency of orbital motion:
The fourth transition falls in the Balmer series of
1 1
f∝ 3 and given f1 = f the spectrum which corresponds to the visible
n 27 2
light emission. The last transition falls in the

() ( )
3
n2 f 1 n2 1
1/3
1 Paschen series which corresponds to the infrared
⇒ = ⇒ = =
n1 f 2 n1 27 3 radiation

19 (b) Thus, frequencies of the last two transitions are

closer to each other on the extreme left of the
Option (a) explains the production of X-rays on frequency spectrum whereas the frequencies of
the basis of electromagnetic theory of light, which the first three transitions are closer to one
is not able to explain the characteristic X-rays and another and fall on the right corner of the
cut-off wavelength, so option (a) is wrong frequency spectrum

Option (b) correctly explains the production of The spectrum of the transitions is thus best
characteristic X-rays represented in diagram(d)

Option (c) is wrong as X-ray spectra is a 24 (c)

continuous spectra having some peaks
representing characteristic X-rays Required energy = [( ) ( )]
-13.6 -13.6
9
-
1
×9

20 (b)
[
= 13.6-
13.6
9 ]
9 = 8×13.6 eV

P a g e | 42
 12375 ke
2
Wavelength = = 113.7 Å ∴L = m
2
r⇒L = mke r
8×13.6 mr

25 (c) ⇒L ∝ r
In the first case, transition is from n state to 2nd
th
30 (b)
(1st excited) state
hc

[ ]
KEλ = - Ψ = eΔV
1 1 2 λ1
∴ (10.2+17.0) eV = 13.6×Z 2 - 2
1

2 n
hc
In 2nd case, transition is from n state to 3rd state
th KEλ = - Ψ = 2eΔV
2
λ2

∴ (4.25+5.95) eV = 13.6 Z
2
[ ]
1 1
2- 2
3 n ⇒3 ( ) hc
λ1
-Ψ =
hc
λ2
-Ψ

( )
Solving above equations, we get n = 6 and Z = 3 3 1
⇒ Ψ = hc -
2λ1 2λ2
26 (d)

We know, rn ∝ n
2
2
⇒KEλ =
3
hc
λ3
- hc
3 1
-[
2λ1 2λ2 ]
So, (n+1)2 - n = (n-1)2⇒n = 4

27 (c)
= hc [ 1 1 3
+ -
λ3 2λ2 2λ1 ]
Photon energy = h f = 13.6 1- [ ] 1
25
eV = 13 eV eΔV = hc [ 1 1 3
+ -
λ3 2λ2 2λ1 ]
Photon momentum = momentum of hydrogen
atom:
ΔV = [
hc 1 1 3
+ -
e λ3 2λ2 2λ1 ]
hf hf 31 (c)
⇒p = or mv =
c c

v=
hf
=
13×1.6×10
-27
-19

8 = 4 ms
-1
Emax = 13.6 eV;Emin = 31.6 1- ( )
1
2
2 =
3
4
×13.4 eV
mc 1.67×10 ×3×10
Emax 4
28 (a) ⇒ =
Emin 3

1 32 (b)
Linear momentum ⇒mv ∝
n
-18
From figure (b), photons of energy 1.6×10 J get
Angular momentum ⇒mvr ∝ n
absorbed in large numbers, no lower energy
Therefore, product of linear momentum and photon gets absorbed. And according to the
angular momentum ∝ n
0 passage, substantial absorption occurs only if the
photon bumps the ground state electron into a
29 (c) higher shell
2 2 -18
mv ke Therefore, 1.6×10 J photon knocks a ground
Using Bohr’s theory, = 2
r r state electron (n = 1) into the first excited state
2
(n = 2). Hence, the difference in energy between
2 ke the ground state and the first excited state must
v = ⇒L = mvr
mr -18
be 1.6×0 J

P a g e | 43
 E0 h h
Using En = As λb = =
n
2
mV 2qmV

1.6×10
-18
= E2 - E1 = -
E0
4 ( )E 3
- - 0 = E0
1 4
For cut-off wavelength of X-rays, we have
qV =
hc
λm
4(
1.6×10 J) = 2.1×10 J
-18 -18
∴ E0 =
3 hc
or λm =
qV
Therefore, the answer is (b)
qV
33 (a) λb 2m
Now, =
λm c
Barrier height
2 q 11 -1
1 e 1 e As = 1.8×10 C kg for electron,
= J= eV m
4πε0 re 4πε0 re
11 3
9 -19 λb 1.8×10 ×10×10 /2
9×10 ×1.6×10 5 We have = = 0.1
= -14 eV = 1.44×10 eV λm 3×10
8
10

34 (c) Therefore, the answer is (b)

As the collision is inelastic, it means a part of 37 (c)

kinetic energy is transformed into some other 2
Potential energy = -C/r and total energy
form due to collision. In this case, the kinetic 2
= -Rhc/n . With higher orbit, both r and n
energy of incident electron can be absorbed by H
increase. So, both become less negative; hence
atom and it can absorb only 10.2 eV out of 11.2
st both increase
eV, so that it can reach to 1 excited state and the
electron leaves with remaining, i.e., 1.0 eV 38 (b)

35 (a)

1
E = R∞ ch× 1- [ ]
1
2
2 =
3
R ×hc
4 ∞
En = -3.4 eV,En ∝ 2
n Momentum of photon emitted is,

E1 = -13.6 eV E 3R∞h
p= =
c 4
Clearly, n = 2
Recoiling speed of hydrogen atom is given by
Angular momentum
v = P/m, where m is the mass of hydrogen atom
nh 2h h -34 7 -34
= = = = 2.11×10 Js 3R∞h 3×1.1×10 ×6.63×10 -1
2π 2π π v= = -27 = 3.3 ms
4m 4×1.67×10
36 (b)
39 (c)
Kinetic energy gained by a charge q after being
Let the three energy levels be E1, E2, and E3. The
accelerated through a potential difference V volt,
wavelengths λ1, λ2, and λ3 of the spectral lines
is given by
corresponding to the three energy transitions are
1 2 depicted as shown in figure
qV = mv
2

mV = 2MqV

P a g e | 44
 2 2 2
mv mv 3ke
But, F = ⇒ = 4
R R 2R

nh
Also, mvR =
2π
2 2
6π ke m
Solve to get: R = 2 2
nh

44 (c)
h 1
E = hf = or E ∝ (given λ1 < λ2 < λ3)
λ λ Let n be the number of electrons per second
striking the target, P = VI

{
Thus, for the three wavelengths, we have
3
16W = (5×10 )×ne
h
E3-E2= (i)
λ3 16 16
∴n = 3 -19 = 2×10
h 5×10 ×1.6×10
E2-E1= (ii)
λ2
45 (a)
h
E3-E1= (iii)
λ1 The recoil momentum of atom is same as that of
photon but in opposite direction
Now, E3 - E1 = (E3-E2) + (E2-E1) Hence, recoil momentum:
h h h 1 1 1 E 12.09×1.6×10
-19
⇒ = + ⇒ = + P= =
27
Ns = 6.45×10 Ns
λ1 λ3 λ2 λ1 λ2 λ3 c 3×10
8

40 (b) Note that almost whole of the energy will be

carried away by the photon because it is very light
Use Moseley’s law
in comparison to H atom
41 (a)
46 (d)
1
As magnetic moment, μB = evr 13.6
2 E =- 2 eV
5
L
L = mvr ⇒vr = E = -0.544 eV
m
Ep = -2×0.544 eV = -1.088 eV
1 eL μB e
∴ μB = ⇒ =
2m L 2m
47 (b)
42 (c)
The series in U-V region is Lyman series. Longest
q wavelength corresponds to, minimum energy
i= which occurs in transition from n=2 to n=1.
T

2 3 6 1 1
Now T ∝ r ∝ n ⇒i ∝ 3 R
n ∴ 122 = …(i)

43 (b) ( )
1 1
2- 2
1 2
2 2
ke dU 3ke The smallest wavelength in the infrared region
U =- 3, F = - =- 4
2R dR 2R corresponds to maximum energy of Paschen
series.

P a g e | 45
 1 wavelength (maximum photon energy) and first
R line means the largest possible wavelength
∴ λ= …(ii)
( )
1 1
2-
3 ∞
(minimum photon energy) in the series

Solving Eqs.(i) and (ii) , we get

v=C [ ]1 1
2- 2
n m
(where C is a constant)

λ = 823.5 nm For series limit of Lyman series:

48 (b) n = 1, m = ∞⇒v1 = C

For third excited state, n = 4 For firsts line of Lyman series:

2
n n = 1, m = 2⇒v2 = 3C/4
rn = r0
2
For series limit of Balmer series:
4×4
or r1 = 0.5× Å = 4Å
2 n = 2, m = ∞⇒v3 = C/4

49 (c) 54 (a)
This is Bohr’ postulate

50 (c)
1
λ2 [ ]
1 1
= R 2- 2
2 4

2
Energy of photon is given by mc . Now, the
maximum energy of photon is equal to the
1
λ2
1 1
=R -
4 16[ ]
maximum energy of electrons = eV

2 eV
1
λ2
=R
3
16 [ ]
or λ2 =
16
3R
Hence, mc = eV ⇒m =

[ ]
2
c 1 1 1
Again, = R 2- 2
-19 3 λ1 2 3
1.6×10 ×18×10 -32
= = 3.2×10 kg
(3×10 )
[ ]
8 2
1 1 1
=R -
λ1 4 9
51 (d)

hc 1240 eV-nm 1 5R 36
λ= = = 354 nm = or λ1 =
E 3.5 eV λ1 36 5R

This wavelength is in the ultraviolet region λ1 36 3R 27

∴ = × =
λ2 5R 16 20
52 (d)
27
a. No, since Balmer formula was known or λ1 = ×4861 Å
20
b. No, since Rutherford scattering experiment was 55 (a)
known
2
mvn = K/rn. Since modified m is half and modified
c. No, since Einstein’s photon theory was known rn is double, vn remains the same as in H-atom
2
d. Bohr’s chose ‘allowed’ energy levels ∝ 1/n and Note: Positronium is an atom in which an electron
these led to angular momentum quantized as a - +
(e ) and a positron (e ) go around
derivation
their center of mass. Bohr’s conditions hold for it,
53 (a) as used in hydrogen atom, but the
Series limit means the shortest possible mass me of the electron is replaced by the
P a g e | 46
 modified mass μ = (me mp)/(me + mp ), where Now, as the energy of incident radiation is more
me is the than that of K-shell electrons, the characteristic X-
rays appear as peaks on the continuous spectrum
positronium mass, which is equal to m. With this,
one may treat the electron going Therefore, (d) is the answer

round the positron and apply the equations used 58 (a)

for hydrogen atom case
n= 4
n= 4
n =3 n= 3
56 (d)
n =2 n=2
2 2
2 he 2 ke
mrω = 2 or ω = 3 n=1
r mr First line of
Balmer series
n=1
Second line of
Balmer series
2 2
2 2 ke 2 ke
or 4π fn = 3 or fn = 2 3
mr 4π mr
2 2 For hydrogen or hydrogen type atoms
1 nh
But, r = × 2 2

2
k 4π me

ke (k×4π me )
2 2 2 2
1
λ
2 1 1
= RZ 2 - 2
nf ni ( )
∴f =
4π m(n h )
n 2 2 2 3
In the transition from ni→ nf

k e (4π ) m
4 8 2 2 2
2
or fn = 1

( )
∴ λ ∝
(n2h2)3 21 1
Z 2- 2
2 2 4
nf ni
4π k me
or fn =
( )
3 3
nh 2 1 1
Z1 2- 2
λ2 nf ni
[ ]
2 4

( )
2 2π me 1 1 ∴ = 1
Again, hv = k - λ1 1 1
h
2
(n-1)2 n2 2
Z2 2 - 2
nf ni
2

[ ]
2 4 2
2π me n -(n-1)2
( )
2
or, v = k 3 2 1 1
h n (n-1)2 λ1Z1
2
2- 2
nf ni

( )
λ2 = 1

[ ]
2 4
2π me (2n-1)
2 1 1
2
or v = k 3 2 Z2 2 - 2
h n (n-1)2 nf ni
2

If n is very large, then Substituting the values, we have

( )
2 4
2 2π ke 2n 1 1
v=k × 4 (6561)(1)2 2- 2
3
h n 2 3
= = 1215 Å
2 2
4π k me
4
( )(2 ) 2 1 1
2 - 2
or v = 3 3 = fn 2 4
nh
59 (a)
57 (d)
3 -3 -1
P = VI = 150×10 ×10×10 = 1500 Js
hc -11
As λ0 = = 1.55×10 m
E 1500 1 -1
Heating rate of target = × = 3.57 cal s
4.2 100
∴ λ0 = 0.155 Å
60 (a)
Which is the minimum wavelength of continuous
X-rays which carry energy equivalent of energy of From Moseley’s law, v = a(Z - 1) for Kα X-ray
incident electrons

P a g e | 47
 1 the minimum KE of striking neutron must be 20.4
i.e., = a(Z - 1)
λ eV. [This condition is derived in theory]

1 1 As the energy of the given incident neutron is less

Given, = a(11 - 1) and = a(Z - 1)
λ 4λ than 2.4 eV, the collision must be elastic
Dividing, we get Z = 6 65 (b)
61 (c) λm will increase to 3λm due to decrease in the
For emission of a photon with greater wavelength, energy of bombarding electrons. Hence, no
energy gap should be less characteristic X-rays will be visible, only
continuous X-ray will be produced
Blue light falls in the Balmer series and it is
obtained when the atom makes transition from E4 66 (c)
to E3. Red light also falls in the Balmer series and 13.6 2 13.6
En = - 2 ⇒n = -
it has a lower frequency compared to blue light. n -0.54
By quantum theory of radiation, the energy
2
change E is proportional to the frequency of or n = 25.2 or n = 5(nearly)
electromagnetic radiation f by E = hf. Thus, red
v
light is associated with a smaller energy change As v ∝ 1/n, so vn =
5
from a lower energy level (compared to E4) to the
first excited state E2. Hence, the only possible 67 (b)
transition that result in the emission of red light is th
Time period for n energy level electron is,
the E3 to E2 transition
2 2
2πrn 4π m rn
62 (a) T= = ×
vn h n
Magnetic moment e
= 2
rn = n a0
Angular momentum 2m
2
∴ Magnetic moment ∝ angular momentum 4π m 3 2
T= ×n a0
h
∝n (∵L=n
h
2π ) Required number of revolutions, N =
10
-8

T
63 (a)
-31
After substituting n = 2, m = 9.1×10 kg, and
From conservation of momentum, two identical -34
h = 6.63×10 J-s, we get N = 8×10
6

photons travel in opposite directions with equal

magnitude of momentum and energy hc/λ 68 (c)

From conservation of energy, we have

hc hc 2 2
1
λ ( )
1 1
= R 2- 2
n1 n2
+ = m0C + m0C
λ

h
λ 1
λ ( )
1 1
= R 2- 2
2 3
λ=
m0C 36 36×10
-7
-7
⇒λ = = = 6.566×10
5R 5×1.097
Therefore, the answer is (a)
λ = 6566 Å
64 (a)

For a collision of neutron with hydrogen atom in

ground state to be inelastic (partial or complete),

P a g e | 48
 ( )
-34 8
hc 6.64×10 ×3×10 -19 h 1 1
E= = -7 = 3.03×10 J = = hR 2 - 2
λ 6.566×10 λ ni nf

∴ E = 1.89 eV

69 (c)
= 6.6×10 ×10
-34 +7
( )
1 1
-
1 16
-27 -1
= 6.5×10 kg m s
The protons move toward each other till their
relative velocity becomes equal to zero. At the 73 (d)
closest distance of approach, both the proton will 3
2
be moving with the same velocity T ∝R

( ) ()
3/2 3/2
As coulombian repulsive force is internal for the TR R 1 1
= = =
system of protons, we can apply the law of T4R 4R 4 8
conservation of momentum
74 (a)
∴ mv0 = 2mv
From conservation of momentum:

()
2

( )
1 2 1 v h 1 3 hR
Change in KE = mv - 2× m 0 MV = = hR 1- ⇒V =
2 0 2 2 λ 4 4M
This change in energy is equal to the electrical 75 (b)
potential energy
In the emission spectrum 10 lines are observed,
()
2 2
mv0 v e
2
so the energy level (n) to which the sample has
-m 0 =
2 2 4πε0r been excited after absorbing the radiation is given
2 by
e
∴r = 2
πε0mv0 n(n-1)
= 10
2
70 (a)
Which gives n = 5
Radius of first orbit, r ∝ 1/Z. For doubly ionized
lithium, Z will be maximum. Hence, for doubly
ionized lithium r will be minimum
So,
hc
λ
1
= 13.6 1- 2 eV
5 ( )
71 (d) 1242 24
eV - nm = 13.6× eV
λ 25
∆λ = λK - λmin
α
∴ λ = 95 nm
When V is halved λmin becomes two times but λK
α
76 (c)
remains the same
2 2 4
4π k me
''
∴ ∆λ = λK - 2λmin v= 3 3
α nh

= 2(∆λ) - λK 1
α
v∝ 3
n
''
∴ ∆λ < 2(∆λ)
77 (a)
72 (a)

Momentum of the recoiled hydrogen atom

13.6 ( )
1 1
2 - 2 eV = 1.9 eV
2 3

= momentum of the emitted photon 78 (a)

For the first line of Balmer series of hydrogen,

P a g e | 49
 1
λ ( )
1 1
= R 2- 2 =
2 3
5R
36
⇒λ =
36
5R
atom will absorb energy from the colliding
electron only if it can go from ground state to first
excited state, i.e., from n = 1 to n = 2 state. For
For singly ionized helium (Z = 2), this, hydrogen atom must absorb energy

( )
1 1 1 E2 - E1 = -3.4 - (-13.6) = 10.2 eV
''
= 4R 2 - 2
λ n1 n2
So, if the electron possesses energy less than 10.2
'
Given λ = λ eV, it would never lose it and hence collision
would be elastic
For n1 = 6 to n2 = 4
84 (a)
1
λ'
1 1
4 6 ( )
= 4R 2 - 2 =
20 R
144
=
5R
36 U = eV = eV0ln ()
r
r0
It corresponds to transition from n1 = 6 to n2 = 4

79 (c)
∴ | |
|F| = - dU =
dr
eV0
r

hc 2 This force will provide the necessary centripetal

= Rhc(1 - 1/n )
λ force. Hence
2
λR mv eV0
or n = =
λR-1 r r

80 (d) eV0
or v= ….(i)
m
13.6 - 0.85 = 12.75 eV
Moreover
So, 12.75 V is the required potential difference
nh
81 (d) mvr = …..(ii)
2π
As the electron beam is having energy of 13 eV, it Dividing Eq. (ii) by Eq. (i), we have
can excite the atom to the states whose energy is
less than or equal to 0.6 eV (13.6-13).E5 = 0.544
eV and E4 = 0.85 eV. So, the electron beam can
mr = ( )
nh
2π
m
eV0
excite the hydrogen gas maximum to 4th energy
state, hence the transit electron can come back to Or rn ∝ n
ground state from either of three excited states,
85 (a)
thus emitting Lyman, Balmer and Paschen series
In case of Bohr’s model of hydrogen atom
82 (a)

( )
v
1 1 1 Frequency =
∝ 2- 2 2πr
λ n1 n2
1
( )
2
1 1 Here, v ∝ and r ∝ n
2- 2
n
λmin 2 3 5
= =
λmax
( )
1 1
2-
2 ∝
9
∴ Frequency ∝
1
n
3

83 (b) 86 (b)

For an elastic collision to take place, there must be The wavelengths present in emission spectra are
no loss in the energy of electron. The hydrogen shown in figure

P a g e | 50
 n1 = 2;n2 = ∞

∴
1
λ
1 1
=R -
4 ∞ [ ]
or λ =
4
R

For shortest wavelength in Brackett series,

n1 = 4;n2 = ∞

∴
1
λ
''
1 1
= R 2- 2
4 ∞ [ ]
Transitions a, e, h and j can be performed by a
single atom also. This is also true about ' 16 4
or λ = = 4× = 4λ
transitions b and i, other transitions require one R R
atom each
91 (d)
87 (b) nd
For 2 line of Balmer series in hydrogen
de Broglie wavelength of electron in hydrogen spectrum, corresponds to 4→2 transition of
atom hydrogen atom

h 2πrn It is equivalent to 4×3→2×3 i.e., 12→6 transition

= = 2+
mv n of Li

For second Bohr orbit, 92 (c)

-9

[ ]
600×10 -9
λ= = 3000×10 m 1 1 1
2 = R 2- 2
λ 2 3
150
λ= Å = 300 Å 36 36 -1
V or R = = -10 m
5λ 5×6563×10
150 5 -5
∴V = = ×10 V 36000 7 -1 7 -1
(3000)2 3 = ×10 m = 1.097×10 m
5×6563
88 (a)
93 (b)
hc
λ [
1 1
]
= 13.6 2 - 2 = 13.6×0.99
1 10
L will be same for both because it does not
depend upon Z. But for energy
1242 2
λ= nm = 92.25 nm
13.6×0.99 (En)Li = - Z ×13.6
2
13.6
and (En)H = - 2
n n
The line belongs to UV part of electromagnetic
Clearly, |EH| < |ELi|
spectrum
94 (c)
89 (d)
Volume occupied by one mole of gold
Total energy received by the atom will be 25.2 eV.
13.6 eV energy is needed to remove the electron 197 g 3
from the attraction of the nucleus. Rest of the = -3 = 10 cm
19.7 gm
energy will be almost available in the form of KE
of electron Volume of one atom

90 (b) 10 5 23 3
= -23 = ×10 cm
6×10 3
For shortest wavelength in Balmer series,
Let r be the radius of the atom. Therefore,
P a g e | 51
 4 3 5 23 -10 4
πr = ×10 or r ≅ 1.5×10 m Z=
3 3 3λR∞

95 (d) = 39.9 ≈ 40
μ0In 98 (d)
Bn =
2rn
2
mv
F=
In r
or Bn ∝
rn
1 2
But v ∝ and r ∝ n
(fn) n
∝
rn
1
⇒F ∝ 4
(vn/rn) n
∴ Bn ∝
rn
99 (a)
vn
∝ 13.6
(rn)2 th
Total energy for n level = -
n
2 eV

( )
(z/n)
∝ 1 1 13.6×3
(n2/z)2 E2 - E1 = -13.6 - =
4 1 4
3
z = 0.75×13.6 eV
∝ 5
n

96 (c) E3 - E2 = -13.6 ( )
1 1
- =
9 4
13.6×5
36

E = R∞hc 1- ( ) 1
25
= 0.14×13.6 eV

Momentum of photon emitted is E4 - E3 = -13.6 ( )

1 1
- =
16 9
13.6×7
144
eV
E
p = = R∞h
c
24
25 ( ) = 0.05×13.6 eV

Recoil momentum of H atom will also be p Obviously, the difference of energy between
consecutive energy levels decreases
mv = p
100 (a)
v=
p
=
(1.097×10 )(6.626×10 )24
7 -34

m (25)(1.67×10-27) For the incident electron,

-1
∴ v = 4.178 ms 1 2
mv = Ve
2
97 (b)
2
p = 2meV
1
λ
2 1 1
= Z R∞ 2 - 2
n1 n2 ( ) de Broglie wavelength of incident electron,

For Kα line, n1 = 1 and n2 = 2 h h

λ1 = =
p 2mVe
1
λ
2
= Z R∞
3
4 () Shortest X-ray wavelength, λ2 =
hc
Ve

P a g e | 52
 ∴
λ1
λ2
=
Ve
c 2mVe
=
1
c ()
V e
2 m
105 (b)

Let v = speed of neutron before collision,

4
10 11
v1 = speed of neutron after collision,
×1.8×10
2 1
= 8 = 0.1 =
3×10 10 v2 = speed of proton or hydrogen atom after
collision,
or λ1:λ2 = 1:10
and ΔE = energy of excitation
101 (b)
From conservation of linear momentum
Maximum angular speed will be in its ground
state. Hence, mv = mv1 + mv2 (i)
6
v1 2.2×10 From conservation of energy,
ωmax = = -10
r1 0.529×10
1 2 1 2 1 2
16 -1 mv = mv1 + mv2 + ΔE (ii)
= 4.1×10 rad s 2 2 2

102 (c) From Eq. (i),

2 2 2
a. Z was taken from X-ray scattering experiments v = v1 + v2 + 2v1v2

b. Validity not known earlier; established by From Eq. (ii),

Rutherford’s experiments
2 2 2 2ΔE
2 v = v1 + v2 +
Ze m
c. Yes, the experiments said r <
1 2
2πε0 mv
2 ( ) ∴ 2v1v2 =
2∆E
m
This sets upper limit for r
∴ (v1-v2) = (v1+v2) - 4v1v2
2 2

d. Lower limit of r not set

∆E
⇒(v1-v2) = v - 4
2 2
103 (d) m

1
λ [ ]
1 1
= R 2- 2
2 4
As v1 - v2 must be real, therefore

∆E
[ ]
2
f 1 1 v -4 ≥0
or = R - m
c 4 16
1 2
[ ]
1 1
or f = cR -
4 16
or mv ≥ 2∆E
2

The minimum energy that can be absorbed by

8 3 7
hydrogen atom in ground state to go into excited
= 3×10 ×10 ×
16 state is 10.2 eV. Therefore,
9 15
1 2
= ×10 Hz mv = 2×10.2 eV
16 2 min
104 (c) = 20.4 eV
h h 106 (b)
Angular momentum, mvr = n = (n = 1)
2π 2π
which is independent of Z h
λ= (for electron)
p
P a g e | 53
 2 2
or p = h/λ εnh
r = 02
e πm
hc
and E = (for photon)

p 1 1
λ

-9 -1
=
ε0(2πL)2
2
ε πm
L=n
h
2π (
or nh=2πL )
∴ = = 8 = 3.33×10 s m
E c 3×10 -
1
∴ Lr 2 = constant
107 (a)
113 (d)
2
M = IA = efπr
2 2
ke ke
Ep = - ,E =
= 1.6×10 ×10 ×3.14×(0.5×10 )
-19 16 -10 2 2
Am r 2r

= 1.256×10
-23
Am
2
So, Ep = 2E = 2(-13.6) eV = -27.2 eV

108 (d) Potential energy of electron in the ground state of

2+
Li ion is
Rydberg’s constant determines the frequencies.
2
We have R ∝ m. So, modified R for positronium = -3 ×27.2 eV or -244.8 eV
atom is half of H atom. Hence, frequencies are
reduced to half 114 (b)

109 (c) N= ∑2n 2

The characteristic X-ray depends on the material

N = 2(1 +2 +3 +4 ) = 60
2 2 2 2

used
115 (b)
110 (d)
The energy taken by hydrogen atom corresponds
1
λ [ ]
1 1
= R 2- 2
1 2
to its transition from n = 1 to n = 3 state

∆E (given to hydrogen atom)

or
1
[ ] 1
= R 1- or
1
=
3R
λ 4 λ 4

1 3R
= 13.6 1- ( ) 1
9
8
= 13.6× = 12.1 eV
9
Hence, wavelength =
λ 4 116 (b)
111 (a)

Frequency of electron revolution:

Frequency = R ( )
1 1
2-
2 ∞
2 c

4
mZ e
2 4
Hence, λ =
f = 2 3 3, R
4ε0n h
117 (d)
Put the various values to get
2
By quantum theory of radiation, the energy
Z 15
change ∆E between energy levels is proportional
f = 6.62×10 3
n to the frequency of electromagnetic radiation f
and is given by
Now, put Z = 1 and n = 1 to get
15 hc
f = 6.62×10 Hz ∆E = hf =
λ
112 (d)
hc hc
Hence, λ = =
∆E E1-E2

P a g e | 54
118 (d) ground state

In transition of electron from higher energy level 124 (d)

to lower energy level, the wavelength is given by
λ = hc/ΔE, where ΔE is the energy difference The K,L, and M lines have different intercepts. The
between two levels intercept of K is more than that of L, which in turn
is more than that of M
For minimum λ, ΔE should be maximum, so (d) is
the correct option 125 (d)
2
119 (d) Z
E= 2E
n 0
-34 n2h
Angular momentum, L = 4.2176×10 = 126 (d)
2π

⇒n2 = 4 When wavelength is maximum, the energy is

minimum. Hence, this is from the ground state to
For the transition from n2 = 4 to n1 = 3, the the first excited state, for which the energy is 13.6
wavelength of spectral line = λ eV - 3.4 eV = 10.2 eV

1
λ
= -( )
13.6 1 1
hc 32 42
Hence, the required wavelength is 122 nm

The text possibility is to jump from the ground

=
13.6 eV
( )7
1240 eV nm 9×16
state to the second excited state, which requires
= 13.6 - 1.5 = 12.1 eV

1240×144 Hence, it corresponds to a wavelength

λ= = 1876 nm = 18760 Å
13.6×7
λ=
c
=
hc
=
(6.63×10 )×(3×10 )
-34 8

= 1.876×10 Å
4
v E3-E1 (12.1)×(1.6×10-19)
= 103 nm
120 (d)
Therefore, (d) is the answer
E1 E1
2-
E4n-E2n 16n 4n
2
1 127 (b)
= = = constant
E2n-En E1 E1 4
2- 2 912 Å

[ ]
4n n λ=
21 1
Z 2- 2
121 (d) n1 n2

1 For singly ionized helium atom Z = 2

λ∝ 2
Z
For the wavelength to be longest, n1 = 1, n2 = 2
1216
Now, λNa = ≈ 10 Å 912 Å 912
11×11 ∴ λ= = = 304 Å

122 (c) [ ]
(22) 1- 1
4
3

1 (Ze)(e) 128 (c)

Use E =
4πε0 r0
Making potential energy zero increases the value
123 (b) of total energy by 13.6 - (-13.6) = 27.2 eV

From the graphs given, atom in graph B will Now, actual energy in second orbit = -3.4 eV
absorb most of the energy W from the electron
Hence, new value is (-3.4+27.2) eV = 23.8 eV
and re-radiate, in all directions, radiation of
shortest wavelength when the atom returns to its
P a g e | 55
129 (d) E1 = -13.6 eV

nh E2 = -3.4 eV
L=
2π
E3 = -1.50 eV
Clearly, L is constant and independent of Z
E4 = -0.85 eV
130 (c)
From above, we can see that
Assuming that ionization occurs as a result of a
completely inelastic collision, we can write E3 - E1 = 12.1 eV

mv - 0 = (m + mH)u i.e., the electron must be making a transition from

n = 3 to n = 1 level
Where m is the mass of incident particle, mH the
mass of hydrogen atom, v0 the initial velocity of h h
∆L = (3-1) =
incident particle, and u the final common velocity 2π π
of the particle after collision. Prior to collision, the -34
= 2.11×10 Js
KE of the incident particle was
2 134 (c)
mv0
E0 =
2 Using Moseley’s law, v
1/2
= a(z-b)

() ()
The total kinetic energy after collision 1/2 1/2
c c 2
= a(z1 - b) and = a(z - b)
λ1 λ2
E=
(m+mH)u 2

=
2 2
mv 0
2 2(m+mH)

()
1

( )
1
λ2 2 a(z1-b) 7.12 2 (29-b)
= ⇒ =
The decrease in kinetic energy must be equal to λ1 a(z2-b) 15.42 (42-b)
ionization energy. Therefore,
(42-b) = 1.47(29-b)⇒b = 1.44
E1 = E0 - E =
mH
E ( )
()
1
m+mH 0 λ1 2
(z-1.44)
=
λ (z1-1.44)
E1 1
i.e., =
E0 m
( )
1
1+ 15.42 2
mH (27.56) = z - 1.44⇒ = 24
22.85
i.e., the greater the mass m, the smaller the
135 (c)
fraction of initial kinetic energy that be used for
ionization nh
The angular momentum is mvr = ⇒n = 1
2π
131 (c)
2 2
mv Ze
13.6 Centripetal force, = 2
eV = 3.4 eV r 4πε0r
4

( )( )
2 2 2
εnh ε0h me 1
132 (a) r= 0 2 = 2
πmπe Z πmee mπ Z
In this case, there is the widest energy gap

[ ]
-10 -15
0.53×10 200×10 mπ
= = ∵ =264
133 (b) 264 Z Z me

13.6 Since r cannot be less than nuclear radius,

Since E = - 2 eV
n
1
-15
r > 1.6 Z 3 ×10 m
P a g e | 56
 -15 1
200×10 -15 eV02 = 2.75 - 2 = 0.75 eV
or > 1.6×10 Z3
Z
So, V02 = 0.75 V

( )
3
200 4
⇒Z < < 37 140 (b)
1.6

( )
2
cZ e
136 (b) vn = α , where α = is the fine structure
n 2hε0c

v=
1 c
137 n
or v ∝
1
n
constant α= ( ) 1
137

Since v is reduced to one-third, therefore c(2)

vHe = α
+ = αc
2
n=3
c(1)
2 and vH = α = αc = vHe
Now, r ∝ n
+
1

137 (d) 141 (b)

λmin is found for n = 2→1 since energy gap is rn ∝ n

2

maximum
'2 -11 ''2
n 21.2×10 n
138 (d) 2 = -11 or 2 = 4
n 5.3×10 n
0.001
×10 = (0.5×10 )n
'2
-3 -10 2 n '
2 or 2 = 4 or n = 2
1

{ }
th
Because radius of n orbit is equal to 142 (c)
2
rn=n r0, where r0=0.529 Å
The minimum energy to ionize an atom is the
2
∴ n = 1000 energy required to remove an outermost electron
in the atom
or n = 31
143 (c)
139 (b)
hc
λmin =
∆E1 (for 4th to 3rd excited states) eVmax

= 13.6×3
2
[ ]1 1
2 - 2 = 2.75 eV
4 5
144 (a)

Shortest wavelength of Brackett series

∆E2 (for 3rd to 2nd excited states) corresponds to the transition of electron between
n1 = 4 and n2 = ∞ and the shortest wavelength
= 13.6×3
2
[ ]1 1
2 - 2 = 5.95 eV
3 4
of Balmer series corresponds to the transition of
electron between n1 = 2 and n2 = ∞. So,
For shorter wavelength, i.e., for ∆E2, V01 = 3.95
volt (z2) ( ) ( )
13.6
16
=
13.6
4
From eV01 = hc - ϕ, 2
∴z =4
3.95 = 5.95 - ϕ or z = 2
∴ ϕ = 2 eV 145 (c)
For longer wavelength, Through filter, only those photons will pass

P a g e | 57
 through whose energy is less than nh
mvr =
hc 2π
E= = 1.55 eV
800 nm
h (2πr)
∴ =
From hydrogen’s energy level diagram, we can mv n
easily identify that when electron jumps from 3rd
h
excited state (4th energy level) to 2nd excited state = de Broglie wavelength
rd mv
(3 energy level), then photons of 1.55 eV energy
have been emitted. So, the required initial energy 151 (d)
level is the 3rd excited state
hc
146 (c) λmin =
eVmax

First excitation energy is 152 (d)

Rhc ( )
1 1
2 - 2 = Rhc
1 2
3
4
The energy of the Kα X-ray photon

EK = Ei - Ef = (Z-1)2( - 3.4 eV + 13.6 eV)

3 α

Rhc = V eV
4 = (Z-1)2(10.2 eV)
4V Given EK = 7.46 keV
∴ Rhc = eV
3 α

3
∴ 7.46×10 eV = (Z-1)2(10.2 eV)
147 (d)
3
7.46×10
μI e or (Z-1)2 = = 731.4
B = 0 and I = 10.2
2r T
or (Z-1) = 27 ⇒Z = 28
μ0e 2 3
B= [r ∝ n , T ∝ n ]
2rT 153 (b)

B∝
1
n
5 Ephoton = E3 - E1 = -
E0
3 ( )
E0
2 - - 2 =
1
8
E
9 0
148 (b) Therefore, (b) is the answer
Possible transitions are: 154 (d)
4→3, 4→2, 4→1, th
Energy of n state in hydrogen is same as energy
th ++
3→2, 3→1, and of 3n state in Li

2→1 ∴ 3→1 transition in H would give same energy as

++
the 3×3→1×3 i.e., 9→3 transition in Li
149 (a)
155 (d)

| |
2
U
|F| = -dU = mv ⇒v = 0 , which is a constant
( )
2
m
dr r m (rm) = (0.53Å) = (n× 0.3)Å
z
nh 2
So, mvnrn = ⇒r ∝ n m
2π n ∴ =n
z
150 (a)
m=5 for 100Fm257 (the outermost shell) and
z = 100

P a g e | 58
 ∴ n=
(5)2
100
=
1
4
Using
1
λ
1 1
[ ]
= R(z-1)2 2 - 2
n2 n1

156 (a) For Kα line; n1 = 2, n2 = 1

()
2 2 2
mv 3q 3q 1875R 2 3
r
=
4πε0r
2 ⇒mvr =
4πε0v
(i) For metal A; = R(z1-1)
4 4

nh ⇒z1 = 26
and = mvr (ii)
2π

Using (i) and (ii) and putting n = 1 For metal B;675R = R(z-1)2 ()
3
4
2 2
h 3q 3q ⇒z2 = 31
= ⇒v =
2π 4πε0v 2ε0h
Therefore, 4 elements lie between A and B
157 (a)
162 (a)
2 3 2 2 6 3
T ∝ r and r ∝ n ⇒T ∝ n ⇒T ∝ n
1 7 1 1
[ ]
() ()
3 3 = 1.09×10 2 - 2
T1 n n n1 λ 2 3
= 1 ⇒8 = 1 or =2
T2 n2 n2 n2 -7
⇒ λ = 6.606×10 m⇒6606 Å
Only a. satisfies the above, hence this is right
choice 163 (d)

158 (c) For Lyman series, n1 = 1 and n2 = 2 for first line

As the number of orbit increases, the velocity

decreases. The potential energy becomes less
1
λ1 [ ] [ ]
1 1 1 1
= R 2- 2 = R - =
1 2 1 4
3R
4
negative, i.e., PE increases while KE decreases
For Paschen series, n1 = 3 and n2 = 4 for firsts
159 (d) line

v = 2πrf ∴
1
λ2 [ ] [ ]
1 1 1 1
= R 2- 2 = R -
3 4 9 16
=
7R
144
v
⇒f =
2πr λ1 4/3R 7
= =
λ2 144/7R 108
160 (b)
164 (d)

If the electron jumps from n3 level to n1 level, then

photon of energy 12.1 eV is emitted. If the
electron jumps from n2 level to n1 then 10.2 eV
photon is emitted. Clearly, these transitions are
possible in minimum two atoms and maximum
three atoms

165 (c)
In the first case, KE of H atom increases due to
recoil whereas in the second case KE decreases The electron is still in the state n = 2. It has to
due to recoil reach the ground state by emitting a photon

∴ E2 > E1 166 (a)

161 (d)
P a g e | 59
 1 3R 172 (a)
= (Z-1)2
λα 4
We know that as the electron comes nearer to the
4 4 nucleus, the potential energy decreases
(Z-1) =
( )
= 7 -10 2
3Rλα 3×1.1×10 ×1.8×10 -KZe
=PE and r decreases
r
200 5 78

[ ]
= = = 26⇒Z = 27 2
3 35 3 1 1 kZe
The KE will increase ∵KE= |PE|=
2 2 r
167 (c)
[ ]
2
1 kZe
The total energy decreases TE=
As angular momentum of electron is 4h/2π, it 2 r
means electron is in the 4th orbit
173 (b,c)
th
TE of atom in 4 orbit is -0.85 eV
1 1 2
v∝ , E ∝ 2 , and r ∝ n
KE of electron = |TE| = 0.85 eV n n

168 (a) 174 (b)

1
λ
1 1
[ ]
= R 2 - 2 ⇒λ =
1 3
9
8R
Shortest wavelength or cut-off wavelength
depends only upon the voltage applied in the
Coolidge tube
Again,
1
λ [ ]
' = R
1 1 '
2 - 2 ⇒λ =
2 4
16
3R 175 (a,c,d)
' 2 2
λ 16 8R ' 128 nh 2
Now, = × or λ = λ a. rn = 2 2 , i.e., rn ∝ n
λ 3R 9 27 4π mKe

169 (b) nd nh
c. Bohr’s 2 postulate, mvr =
2π
1
λ
1 1
2 3 [ ]
= R 2- 2 =
5R
36
;λ =
36
5R
d. Kn =
KZe
2
,U =
KZe
2

2rn n rn
170 (b)
176 (a,c)
v=Z [ ]
1 c
137 n Power loss increases the temperature

1 c 178 (a,b)
⇒v = 4× ×
137 2
2

2c |F| = dU = Ke4 (i)

or v = dt r
137
2 2
Ke mv
171 (a,b,c) 4 = (ii)
r r
12400 nh
λmin = λmin = Å and mvr = (iii)
v0 2π
12400 By (ii) and (iii),
= = 62 Å
20,000
2 2
Ke 4π m m
r= 2 2 = K1 2 (iv)
h n n

1
Total energy = (potential energy)
2

P a g e | 60
 2 2 2 6
Ke -Ke -Ke n 181 (b,c,d)
3 = =
( )
3 3 3
6r Km 6K1m
6 12 Let collision between two atoms be an inelastic
n
one. From momentum conservation,
Total energy ∝ n
6
mv0 = mv1 + mv2
-3
Total energy ∝ m

∴ (a) and (b) are correct

179 (a,b,c)

U1 E
a. U1 = E, then total energy in the orbit = =
2 2

b. IE = (TE)n=∞ - (TE)n=1 = 0 - (TE)n=1 = - ()

E
2
From energy conservation,
2 2 2
mv1 mv2 mv0
c. (KE)n=1 = -(TE)n=1 = - ()
E
2
2
+
2
-
2
= -∆E

Where ∆E is the energy absorbed by the initially

180 (a,d)
stationary atom to change its state
The time period of the electron in a Bohr orbit is
Solving above equations, we get
given by T = 2πr/V
th
Since for the n Bohr orbit, mvr = n(h/2π), the (v1-v2)2 = v20 - 4∆E
m
time period becomes
For collision to be inelastic, (v1-v2) ≥ 0: a real
2

( )
2
2πr 4π m 2
T= = r quantity [equal to sign for perfect inelastic
nh/(2πmr) nh
collision]
Since the radius of the orbit r depends on n, we
The minimum value of ∆E is 10.2 eV, so for
replace r
collision to be inelastic, E ≥ 20.4 eV

For perfectly inelastic collision, v1 = v2 and hence

E = 20.4 eV

For E = 18 eV, the collision is elastic one and as

masses are the same, velocities would be
interchanged during collision

182 (b,c)

The expression of Bohr radius of a hydrogen atom Any transition causing a photon to be emitted in

( )
2
2 h ε0 the Balmer series must end at n = 2. This must be
is r = n 2
πme followed by the transition from n = 2 to n = 1,
emitting a photon of energy 10.2 eV, which

( )( ) ( )
4 4 2 3 2
4π m n h ε0
2
3 4h ε0 corresponds to a wavelength of about 122 nm.
Hence, T = 2 2 4 = n 4
nh πme me This belongs to the Lyman series

()
3
T1 n 183 (b,d)
For two orbits, = 1
T2 n2
When potential difference between filament and
It is given that T1/T2 = 8. Hence, n1/n2 = 2 target is increased, then KE of striking electrons
gets increased. Since most energetic electrons
P a g e | 61
 now strike the target, therefore more energetic the X-rays emitted depends on the energy of the
electrons are emitted. It means, frequency of X- electrons incident on the target. This, in turn,
ray photons increases. It means, penetration depends on the potential through which they have
power of X-rays gets increased fallen. In fact,

To increase the photon flux, rate of collision of hc

λmin =
electrons with the target must be increased. This eV
can be achieved only when rate of emission of
Statement (b) is true. X-ray spectra of all heavy
electrons from the filament is increased. To
elements are similar in character.
achieve this, filament current must be increased.
Therefore, options (b) and (d) are correct Statement (c) is also true. The short wavelength
of the X-rays (compared to the grating constant of
184 (c)
optical grating) makes it difficult to observe X-ray
1 diffraction with ordinary gratings.
We know that λ ∝
m
Statement (d) is also true. The sharp limit on the
For ordinary hydrogen atom, short wavelength side is dependent on the voltage
applied to the incident electrons and is give by:
1
λ
1 1
2 3 [ ]
= R 2- 2 =
5R
36
or λ =
36
5R
λmin =
hc
eV
With hypothetical particle, required wavelength
188 (b,c)
'1 36 18
λ = × =
2 5R 5R Lyman series lies in the ultraviolet region, Balmer
series in visible region, and Paschen series in
185 (a,b,c) infrared region
H H
En = E1 + ΔE = -13.6 eV + 12.75 eV = -0.85 eV c. λR > λY > λB > λV
i.e., hydrogen atoms are excited to n = 4 level, i.e.,
transitions 4→1, 3→1, 2→1 are possible which 189 (d)
correspond to Lyman series, then transitions 4→2
and 3→2 are possible which correspond to Method 1: Memorisation
Balmer series, and then transition 4→3 is also In Lyman series, we get energy in UV region
possible which correspond to Paschen series
In Balmer series, we get energy in visible region
186 (a,b)
2
In Paschen/Brackett/Pfund series, we get energy
rn = n r1 in IR region

Method 2:

We know that
-
1
λ ( )
1 1
= R 2 - 2 for hydrogen atom
n1 n2
for e transition from n2→n1

For 4 to 3, E ∝
1
λ [ ]
1 1
=R -
9 16
=
7
9×16
R

() ( )
2
An πn 4
In = In n
= In n = 41n(n)
A1 πr1
2
For 2 to 1, 3 to 2, and 4 to 2, we get more value of
1/λ, i.e., more energy than 4→3
187 (b,c,d)
IR radiation has less energy than UV radiation
Statement (a) is false. The shortest wavelength of
Therefore, the correct option is (d)
P a g e | 62
190 (d) ∴ Wavelength of first line of Balmer series is
equal to 12375/1.89 Å = 6563 Å. It lies in visible
For hydrogen and hydrogen-like atoms, region
(z)2
En = -13.6 eV Energy of last line of Balmer series is equal to 3.4
(n)2
eV. Therefore, its wavelength is equal to
Therefore, ground state energy of doubly ionized 12375/3.4 Å = 3640 Å
lithium atom (z = 3, n = 1) will be
Since it is less than 4000 Å, therefore it lies in
(3)2 ultraviolet region. Hence, option (b) is wrong.
E1 = (-13.6) 2 = -122.4 eV
(2) Energy of last line of Paschen series is equal to
1.51 eV. It lies in infrared region. Since energy of
∴ Ionization energy of an electron in ground state
all the other lines of Paschen series is less than its
of doubly ionized lithium atom will be 122.4 eV
energy, therefore all the lines of Paschen series
191 (b,c) will lie in infrared region. Hence, option (c) is also
correct
2
E0z 1-( ) ( )
1
9
2 1 1 27 2
- E0z - = 3E0⇒ E0Z = 3E0
4 9 36 193 (a,b)

λ For the third line of Balmer series, n1 = 2, n2 = 5

⇒z = 2 1 ; = 3
λ2

( ) ( )
2
1 2 1 1 2 1 1 21 RZ
∴ = RZ 2 - 2 = RZ 2 - 2 =
( )
KE1 = E0 1-
1
9
- ϕ;
λ n1 n2 2 5 100

E = -13.6 eV

( )
21
KE2 = E0z 1- - ϕ
4 2
Z×
21
=
hc
=
1242 eVnm
100 λ 108.5 nm
1
KE ∝ 2 = 8.5 eV
λ 2 1242×100
Z = = 4⇒Z = 2
108.5×21×13.6
192 (a,c)
Binding energy of an electron in the ground state
First line of Lyman series is obtained during 2 2
of hydrogen-like ion = 13.6Z /n = 54.4 eV
transition of hydrogen atom from n = 2 to n = 1. (n = 1)
Hence, its energy is equal to
E2 - E1 = (13.6)eV - (-3.4 eV) = -1.2 eV 195 (a)

∴ Wavelength of the first line of Lyman series is The maximum number of electrons in an orbit is
2
equal to 2n . Since n > 4 is not possible, therefore the
maximum number of electrons that can be in the
12375 first four orbits are
Å = 1215 Å
10.2
2(1)2 + 2(2)2 + 2(3)2 + 2(3)2 + 2(4)
Therefore, it lies in ultraviolet region = 2 + 8 + 18 + 32 = 60

Since energy of all the other lines of Lyman series Therefore, possible elements are 60
is greater than that of first line, therefore all the
lines of Lyman series lie in ultraviolet region. 196 (a,c,d)
Hence, option (a) is correct
a. As v ∝ 1/n, so momentum = mv ∝ 1/n
First line of Balmer series is obtained during 2
b. is not true as radius r ∝ n
transition of hydrogen atom from n = 3 to n = 2.
Hence, its energy is equal to E3 - E2 = 1.89 eV 1 2
c. KE = mv ,
2

P a g e | 63
 -10
1
As v ∝ 1/n, so KE ∝ 2 (KE)1 = 14.4×10 -10 eV = 13.58 eV
n 2×0.53×10
-10
d. is true 14.4×10
(KE)2 = -10 eV = 3.39 eV
2×0.53×10 ×4
197 (a,b)
KE decreases by 10.2 eV
The correct choices are (a) and (b). The last two
statements are incorrect because they violate the -1 e
2
-14.4×10
-10
Now, PE = = eV
principle of conservation of charge. We always 4πε0 r r
have either an electron-positron ‘pair production’
-10
or an electron-positron ‘pair annihilation’. It is (PE)1 = -14.4×10-10 eV = -27.1 eV
only then that the total charge remains zero both 0.53×10
before and after reaction -10
(PE)2 = -14.4×10
-10 = -6.79 eV
198 (a,c) 0.53×10 ×4

Moseley’s law: λ ∝
1 λ (z -1)
2; z = 1 2
2
PE increases by 20.4 eV
(z-1) λ1 (z-1)
nh
Now, angular momentum, L = mvr =
2π
z1 - 1 = (z-1)2;
-34

( )
2 h 6.6×10 -34
λ 1 z -1 L2 - L1 = = = 1.05×10 Js
z1 = 2z - 1; z = = 2 2π 6.28
λ2 4 z-1
202 (c)
199 (a,c,d)

[ ]
-34 8
hc 6.63×10 ×3×10
1 Ke
2 E= = -9 J
pn = mvn, pn = vn ∝
2
, r ∝ n and Kn = i.e., λ 0.021×10
n n 2rn
-34 8
1 6.63×10 ×3×10
Kn ∝ 2 and L ∝ n = -9 -13 MeV
n 0.021×10 ×1.6×10
-4
200 (d) = 591.96×10 MeV = 59.196 keV

We know that ∴ (c) is the correct option

1
λ
2 1 1
[ ]1
= Rz 2 - 2 ⇒ ∝ z
n2 n1 λ
2 203 (d)

hc 12400
λmin = ⇒ Å = 0.155
λ is shortest when 1/λ is largest, i.e., when z is big. E 80×103
z is highest for lithium
204 (a,b,c)
201 (b,c,d)
When a stationary hydrogen atom emits a photon,
Ground state n = 1 then energy of the emitted photon will be equal to
the difference of the energy of the two levels
First excited state n = 2 involved in the transition. Hence, energy of
2 emitted photon will be equal to (Em - En)
1 e
KE = (z = 1)
4πε0 2r If a hydrogen atom is moving and a photon is
-10 emitted by it along the same direction in which it
14.4×10
KE = eV is moving, due to momentum of the emitted
2r
photon, the momentum of hydrogen atom will get
2
Now r = 0.53 n Å (z=1) decreased. Therefore, energy of the emitted
photon will be equal to (Em - En+ loss of KE of the

P a g e | 64
 hydrogen atom) mass number of the element concerned

But if the photon is emitted in a direction normal 207 (a)

to the motion of the hydrogen atom, then the
frequency of the emitted photon will be equal to In case of coolidge tube,
f0. Hence, option (a) is correct, obviously, option hc
(d) is wrong λmin = = λc (as given here)
eV

If the photon is emitted by the hydrogen atom in Thus, the cut-off wavelength is inversely
the direction opposite to its motion, then proportional to acceleration voltage. As V
frequency of the emitted photon will be less than increases, λc decreases
f0. Hence, option (c) is correct
λk is the wavelength of K∞line which is a
205 (a,c,d) characteristic of an atom and does not depend on
acceleration voltage of bombarding electron since
In Bohr model of hydrogen atom,
λk always refers to a photon wavelength of
2 -
R∝n transition of e from the target element from 2→1

1 The above two facts lead to the conclusion that

V∝
n λk - λc increases as accelerating voltage is
increased
3 1
T ∝ n and E ∝ 2
n 208 (a,b,c)

VR ∝ n 2
Since in hydrogen atom rn ∝ n , therefore graph
between rn and n will be a parabola through origin
TE ∝ n
and having increasing slope. Therefore, option (a)
2 2
T is correct. Since, rn ∝ n , therefore rn/r1 = n
∝n
R Hence, log (rn/r1) = 2log n
V
∴ ×n
E

206 (a,c)

Statement (a) is correct. The angular momentum

2
of the earth ( = mr ω) has to be equal to nh/2π.
2
This gives, n = 2πmr ω/h. Putting the numerical
values of the Earth’s mass, radius, and the angular
velocity, we get the given value of n. It means, graph between log (rn/r1) and log n will
be a straight line passing through origin and
Statement (b) is incorrect. The maximum number having positive slope (tan θ=2). Therefore, option
2
of electrons allowed in an orbit being 2n , the (b) is also correct. It radius of an orbit is equal to
required number is 2(1 +2 +3 +4 ) = 60
2 2 2 2
2
r, then area enclosed by it will be equal to A = πr
Statement (c) is correct. The ‘reduced mass’ of the 2
Since rn ∝ n , therefore An ∝ n
4

electron [μ = mM/(m + M)] being dependent on

the mass of the nucleus (M), the Rydberg constant
also varies with the mass number of the given
Hence,
An
A1
4
()
A
= n or log n = 4log n
A1
element
It means, graph between log (An/A1) and log n will
Statement (d) is incorrect. The ratio is not exactly be a straight line passing through origin and
equal to 4 but slightly different from 4 because of having positive slope (tan θ=4). Therefore, option
the dependence of the Rydberg constant on the
P a g e | 65
 (c) is also correct. 12375
λmin = Å
V
If frequency of revolution of electron is f, then its
angular velocity will be equal to ω = 2πf. Hence, Here, V is the applied voltage
its angular momentum will be equal to
2
Iω = mr ω. But according to Bohr’s theory, it is
equal to nh/2π, therefore,

nh nh
mr (2πf) =
2
or f = 2 2
2π 4π mr

2 1
Since r ∝ n , therefore f ∝ 3
n
212 (a,b,c)
Hence,
fn
f1
1 f
()
= 3 or log n = 3log n
n f1 Energy of K absorption edge

1242 eVnm 3
It means, graph between log (fn/f1) and log n will EK = = 72.21×10 eV = 72.21 KeV
0.0172 nm
be a straight line passing through origin and
having negative slope, tan θ=-3. Hence, it will be Energy of Kα line is
as shown in figure. Hence, the option (d) is wrong
he 1242 eVnm
Ek = = = 59.14 KeV
209 (b,d) α
eλα 0.021 nm

Line emission spectra can be obtained for atoms 1242

Similarly, Ek = = 64.69 KeV
and molecules both. An atom or molecule in an β
0.0192
excited state emits photons by making a
transition from excited state to ground state thus 1242
Ek = = 69 KeV
constituting line emission spectra
γ
0.0180

The wavelengths emitted by the molecular energy Energy of K shell = EK -EK ( α

)
levels which are generally grouped into several
= (59.14-72.21)KeV = -13.04 keV
bunches are also grouped and each group is well
separated from the other. The spectrum in this Energy of L shell = Ek - 72.21 keV
case looks like a band spectrum
β

= 64.69 keV - 72.21 keV = -7.52 keV

210 (a,b)
1242eVnm
Continuous spectrum is obtained from the bulk Energy of M shell = Ek - Ek = - 72.21
γ
0.018 nm
state of matter, it has no relation with the atomic
keV
or molecular state. It is produced by thermal
vibrations of atoms in the macroscopic matter = 69 keV - 72.21 keV = -3.21 keV
(not by the transition between the energy states
of atoms). Every vibrating atom emits light of 214 (a,c)
frequency of its vibrations. In a white hot matter,
hc
atoms vibrating with all frequencies within a EK - EL = (i)
λα
definite range are present, so this matter emits
frequencies of continuous range

211 (b)

The continuous X-ray spectrum is shown in figure

All wavelengths > λmin are found where

P a g e | 66
 hc Therefore, (c) and (d) are the correct options
EK - EM = (ii)
λβ
218 (b)
hc
EL - EM = ' (iii) 1. If Assertion is True, Reason is True, Reason is
λα
correct explanation of 1
hc hc hc
(ii)-(i) ⇒EL - EM = ' = - 2. If Assertion is True, Reason is True, Reason is
λα λβ λα
not correct explanation of 1
1 1 1
= + ' 3. If Assertion is True, Reason is False
λβ λα λα
4. If Assertion is False, Reason is True
Also, (EK-EM) > (EK-EL) > (EL - EM)
219 (c)
hc hc hc
> > '
λβ λα λα Lyman series: Its energy is in the ultraviolet
region
215 (a,b,d)

n(n-1)
= 6 ⇒n = 4
2

Balmer series: Its energy is in visible region

Now, frequency of energy of ultraviolet photon is

If the initial state were n = 3, in the emission much greater than frequency of visible region. So,
spectrum, no wavelengths shorter than λ0 would statement I is true. Statement II is false
have occurred 220 (a)
This is possible if initial state were n = 2 A H atom that drops from n = 2 level to n = 1
216 (c,d) level emits a photon of energy 10.2 eV and
+
wavelength 122 nm. A He ion emits a photon of
V = 6.6 kV = 6600 V the same energy and wavelength when it drops
-19
from n = 4 level to n = 2 level
eV 1.6×10 ×6600
Now, vmax = = -34
h 6.6×10 222 (b)
18
= 1.6×10 Hz An alpha particle carries 2 units of positive charge
and 4 units of mass. It is made up of protons and
Thus, the frequency of the X-rays cannot exceed 2 neutrons which make a nucleus of helium ie,
18
1.6×10 Hz. Hence, the correct choices are (c) helium atom is a deoid of 2 electrons ie, doubly
and (d) ionized helium atom.
217 (c,d) 223 (c)
In the case of hydrogen, It is difficult to excite nucleus by usual methods
employed for excitation for atoms because
Atomic number = mass number
difference in energy of allowed energy states for
In other atom, atomic number < mass number nucleus is of the order of tens to hundreds of MeV.

P a g e | 67
225 (d) c. eV0 = kE, where V0 is stopping potential. If V0
decreases, then KE decreases
As the energy of striking electron is increased, the
wavelength of the characteristic X-ray does not V0 is also decreased by increasing W0, hence λ0
change as characteristic X-rays are emitted when decreases as explained above
electrons are making transition from a higher
energy level to a lower energy and energy of 3
d. fk = Rc(z-1)2. If fk increases, then fk
characteristic X-ray is given by α
4 α α

decreases and hence Z decreases

hc
E= = | E f| - | E i | = E i - E f
λ 228 (d)

Which does not depend at all on the energy of the Burning candle gives continuous spectrum,
striking electron. The only dependence is that the sodium vapour gives line spectrum, Bunsen flame
striking electron should possess enough energy give band spectrum and dark lines in solar
to knock out an electron from the inner shell spectrum are due to absorption spectrum.

226 (c) 229 (b)

a. Moseley’s law is about characteristic X-rays

b. In photoelectric effect, from electromagnetic

From Moseley’s law: f =
1
λ ( )
1 1
= R(z-σ)2 2 - 2
n1 n2

radiation, electrons are ejected. In X-rays, the For Al, Z is highest and for Kβ, n1 = 1 and n2 = 3;
process is reverse; here, electromagnetic and for Kα, n1 = 1, and n2 = 2
radiation (X-ray) is produced from fast moving
electrons Hence, order of frequency:

Also X-rays are produced using high potential fk (Al) > fk (Al) > fk (Na) > fk (Be) > fk (Li)
β β
α α α

difference
Speed will be same (c) for all photons of any
c. Cut-off wavelength is related to potential frequency
difference applied in continuous X-rays
230 (a)
d. In continuous X-rays, electromagnetic
radiations are emitted a. Emission spectra is discrete as only those
wavelengths are emitted which correspond to
227 (d) energy difference of two energy levels

a. KE of electrons striking the target: KE = eV It is due to electronic transition, i.e., due to

transition of electron from one energy level
So, if V increases, KE increases
(higher) to the another energy level (lower). It is
12400 explained by quantum theory of light
Cut-off wavelength: λmin = Å
V
b. For energies of incident light less than
So, if V increases, λmin decreases ionization energy, the absorption spectra is
discrete. For energies of incident photon greater
b. If work function of target is increased in than ionization energy, the absorption spectra is
photoelectric effect, the KE of photoelectrons continuous. It occurs due to electronic transition
emitted decreases from: KE = hf - W0 which is explained by quantum theory of light

hc hc c. Continuous X-rays constitute continuous

Now, hf0 = W0⇒ = W0⇒λ0 =
λ0 W0 spectra and characteristic X-rays arise due to
electronic transition
It work function (W0) increase, then cut-off
wavelength (λ0) decreases d. Thermal radiation spectra is a continuous

P a g e | 68
 spectra and is explained by atomic transition and KEmax = hc/λ - ϕ
quantum theory
So, average KE is unpredictable
231 (a)
b. Minimum KE can be zero
2 4 2
mZ e Z
a. f = 2 3 3 ⇒f ∝ 3 c,d. In continuous X-rays, wavelength can vary
4ε0h n n
from λmin = hc/eV to ∞
nh
b. L = ⇒L ∝ n 235 (a)
2π
2
e .529 n
c. Magnetic moment: M = IA = vπr
2 rn = Å
2πr Z

e e e So, (a) → (q)

= vr = (mvr) = L
2 2m 2m 6
2.2×10 Z -1
Vn = ms
⇒M = ( )
e nh
2m 2π
⇒∝n
n

So, (d) → (p)

( )
2 2 2
ev e πmZe Ze Z
d. I = = ⇒I ∝ 3 1.06 Z
2
2πr 2π n2h2ε0 2ε0nh n I= 3 mA
n
232 (c)
So, (b) → (r)
2
.529n 3
rn = Å 12.5 Z
Z Magnetic field, B = 5 T
n
So, (a) → (s)
So, c. → s.
3
12.5Z
Magnetic field, B = 5 T 236 (b)
n

So, (b) → (t)

1
λ ( )
1 1
= R 2- 2
n1 n2
1
B∝ 5 Wavelength range from 950 Å to 1350 Å
n
corresponds to ultraviolet region of
6
2.2×10 Z electromagnetic spectrum, i.e., Lyman series
Vn =
n
For transition 4→1:
1
Vn ∝
n 1
λ1
= 1.1×10
7 15

16 ( )
n↑Vn↑
16
∴ λ1 = 7 = 970 Å
So, (c) → (q) 15×1.1×10
2
-13.6 Z For transition 3→1:
Total energy, En = 2 eV
n
9
λ2 = 7 = 1023 Å
n↑En↑ 8×1.1×10

So, d. → p. For transition 2→1:

234 (b) 4
λ3 = 7 = 1212 Å
3×1.1×10
a. Photoelectrons can have KE anywhere from 0 to
P a g e | 69
 2
For Li atom (Z = 3), for transition 2→1: 4 4
λ= = 7
3R 3(1.097×10 )
1
λ
''
2 1
= RZ 1- 2
2 ( ) -7
= 1.215×10 m = 121.5

'' 1 The frequency of the photon is

λ = 7 = 134 Å
1.1×10 ×9×0.75 8
c 3.00×10 15
+ f= = -7 = 2.47×10 Hz
For He atom (Z = 2) for transition 2→1: λ 1.215×10

1
λ
''
2 1
= RZ 1- 2
2 ( ) The energy of the photon is E = hf

= (4.136×10 )(2.47×1015) = 10.2 eV

-15

'' 1
λ = 7 = 303 Å 242 (b)
1.1×10 ×4×0.75

237 (b) As we know, energy of a photon is given by

hc
E=
h nh λ
On other planet: mvr = 2n ⇒v =
2π πmr 2
1500p
2 2 2 2 2 From the given condition, λ = 2
mv 1 e mn h 1 e p -1
= ⇒ =
r 4πε0 r2 n2m2r3 4πε0 r2

Putting n = 1, we get r =
2
4h ε0
Hence, E =
hc
λ
=
hc 1 10
1- ×10 J
1500 p2 ( )
2

)( )
mπe
hc 1 10
= 1- 2 ×10 eV
238 (c) (1500)(1.6×10-19 p

a0 =
h ε0
πme
2

2 and r =
n h ε0
πmee
2 =
2 2
2 h ε0
π(m/2)e
2 2

2 = 8a0
= 8.28 1- ( )1
p
2 eV

th
Hence, energy of n state is given by
239 (a)
8.28
Internal forces act between electron and proton, En = - 2 eV
n
then how can the atom get an acceleration

240 (c)

Given that E1 = -15.6 eV, E∞ = 0 eV

Ionization energy of the atom:

E∞ - E1 = 0 - (-15.6 eV) = 15.6 eV Maximum energy is released for transition from

p = ∞ to p = 1; hence, wavelength of most
So, ionization potential = 15.6 eV
energetic photon is 1500 Å
241 (c)
243 (b)
We can use the equation:
Consider energy level diagram of hydrogen atom.
1
λ
1 1
( ) ( )
1 1
= R 2- 2 = R 2- 2 =
n1 nu 1 2
3R
4
After absorbing photons of energy 10.2 eV, it
would reach the first excited state of -3.4 eV, since
energy difference corresponding to n = 1 and
n = 2 is 10.2 eV. When this excited hydrogen
atom de-excites, it would release 10.2 eV energy,
+ 2+
which is absorbed by He and Li

P a g e | 70
 th
Energy of the n state of a hydrogen-like atom Along x-axis: mu = 4 mv2cos θ (i)
with atomic number Z is given by
Along y-axis: mv1 = 4 mv2sin θ (ii)
2
13.6Z
En = eV
n
2 On squaring and adding equations (i) and (ii), we
get
+
After absorbing 10.2 eV, He electron moves from
2 2 2
2+
n = 2 to n = 4 and Li electron moves from u + v1 = 16v2
n = 3 to n = 6
1 2 1 2 1 2
mu + mv1 = 16 mv2 (iii)
2 2 2

The initial kinetic energy of neutron = 65 eV

Let kinetic energies of neutron and helium atom

be

1 2 1 2
+ Kn = mv1, KHe = (4m)v2
In the spectrum of He , there would be 4C2 = 6 2 2
lines
So, Eq. (iii) reduces to 65 + Kn = 4KHe

4KHe - Kn = 65 (iv)

a. Energy required to excite an electron from

+
ground state of singly ionized helium atom (He )
th 2 2 2
to n energy level is u + v1 = 16v2

If the neutron has sufficient energy to excite the

helium atom, then from conservation of energy,
λmin hc 1242 the energy of neutron must be equal to the sum of
= = = 24.4 nm
n=4→n=1 ∆E
[ ]
13.6 4-
4
16
kinetic energy of neutron, helium atom, and
excitation energy. So, we have

65 = Kn + KHe + 54.4 1= ( )
1
n
2

54.4
KHe + Kn = 10.6 + 2 (v)
n

On solving Eqs. (iv) and (v), we get

6C = 15 lines
2+
Similarly, in spectrum of Li there will be [
KHe = 15.52+
10.88
n
2 eV ] (vi)

[ ]
2
43.52
Kn = 2 -4.52 eV (vii)
λmin 1242 n
= = 10.4 nm
n=6→n=1 13.6 9-
9
36 [ ] The kinetic energy is always positive, so from
equation (vii) we have
244 (b)
43.52
Let the final speeds of neutron and singly ionized 2 > 4.52, n < 3.1
n
helium atom be v1 and v2, respectively. From
conservation of momentum, we have So, the only possible values of n are 2 and 3

Possible values of n
P a g e | 71
 2, 3 1 2 1
KE of electron = mv = k
2 2
Allowed values of neutron energy
Total energy of electron = KE + PE
KN
PE of electron = klog r
6.36 eV, 0.32 eV
1
= k + klog r
Allowed values of He atom energy 2

KHe
[ ]
2 2
K nh
E= 1+log 2
2 4π mk
17.84 eV, 16.33 eV
246 (d)
b. When the atom de-excites,
For Balmer series, n1 = 2;n2 = 3, 4,…
v=
6.63×10
-34 [ ]
(13.6)22×1.6×10-19 1 1
2- 2
nl nu (lower) (higher)

= 13.3
[ ]
1 1 15
2 - 2 ×10 Hz
nl nu
Therefore, in transition (VI), proton of Balmer
series is absorbed

The electron excited to n = 3 can make three 247 (c)

transitions:
Let n be the lower energy level of the given series
n=3 to n = 1, v1 = 11.67×10 Hz
15 of lines, then maximum wavelength is given by

n=3 to n = 2, v2 = 9.84×10 Hz
15 hc
λmax
2 1
= 13.6×Z 2 -
n ( [
1
n+1 ]
)2
15

[ ]
n=2 to n = 1, v3 = 1.83×10 Hz
hc 2 1 1
= 136×Z 2 - 2
245 (c) λmin n ∞

For a conservative force field, Where λmax = 41.02 nm and λmin = 22.79 nm

dU Solving above equations, we get n = 2, which

- =F
dr corresponds to Balmer series

Since U = klog r, Z=4

This force F = -k/r provides the centripetal force For next to longest wavelength, transition takes
for circular motion of electron place from n1 = 4 to n2 = 2

[ ]
2
mv k hc 2 1 1
= (i) So, = 13.6×4 -
r r λ 4 16

Applying Bohr’s quantization rule, ∴ λ = 30.47 nm

nh 248 (a)
mvr = (ii)
2π
The minimum energy required to remove the
From Eqs. (i) and (ii), we get -15
electron from K-level is 3×10 J

nh Let V be the potential difference required, then

r=
2π mk

From Eq.(i),

P a g e | 72
 -15
eV = 3×10 J 253 (a)
-15
3×10 Least intensity 2→1⇒Kα, X-rays
V= -19 = 18750 V
1.6×10
254 (b)
249 (d)

As photon energy of incident light is not equal to

∆E = 13.6 ZB { 2
( )
1 1
( )}
2 1 1
2 - 2 -13.6 ZA
1 2
2 - 2 eV
1 2
energy difference between two energy states, that
1.36×3 2 2
is why sample A does not absorb any photon and 81.6 eV =
4
(ZB-ZA)eV
therefore remains in ground state, and hence, no
emission spectra. Thus, sample B remains as it is, 2 2
ZB - ZA = 8 (i)
but it will de-excite itself to ground state by
emitting radiations of energy 10.2 eV Using conservation of momentum,

u
For A, mAu = MV1 - mA
2

When photon energy is replaced by electron 3

m u = MV1
beam, then atom of sample A can absorb either 2 A
st
10.2 eV (to reach 1 excited state) or 12.1 eV (to 3
nd
reach 2 excited state). In emission spectra of A, For B, mBu = MV2
2
we will have 3 lines corresponding to n = 2 to 1,
n = 3 to 2 and n = 3 to 1, having energies equal But, MV2 = 3MV1
to 10.2 eV, 1.9 eV and 12.1 eV, respectively
mB - 3mA
As least energy of this emission spectra is
corresponding to transition from n = 2 to 3 and Since both A and B carry same number of protons
st
sample B is in 1 excited state, so B can be excited and neutrons, we have
to some higher energy level and if B absorbs
ZB = 3ZA (ii)
energy corresponding to n = 1 to 2 and n = 1 to
3, then it may ionize 2
But ZB - ZA = 8
2

250 (a) 2 2
9ZA, ZBN = 8

∆E = -13.6z
2
[ ] 1 1
2- 2
2 3
ZA = 1, ZB = 3

[ ]
Hence, A is 21H and B is 64Li
1 1 2
47.2 = 13.6 z - ⇒z = 5
4 9 Now, the difference in energy between the first
251 (c) Balmer lines emitted by A and B

Characteristic X-ray arises due to transition of

electrons
∆E = 13.6 ( )
1 1 2
2 - 2 Z - 13.6
2 3 B ( )
1 1 2
2- 2 Z
2 3 A

5
×(ZB-ZA)
2 2
252 (a) = 13.6×
36
In photoelectric effect
13.6×5×8
= = 15.1 eV
E = W + Kmax 36

= 0.73 + 1.82 255 (c)

= 2.55 eV Number of atoms in the second excited state

P a g e | 73
 Nav 15 = n(n - 1)/2. But here transition takes place from
N2 = ×m× th
M 100 n state to state n1 = 2
23
6.023×10 15 23 Here, number of possible emission lines
= ×1.8× = 1.61×10
1.008 100 = n(n-1)(n - 2)/2 = 10 (given)

In the first excited state, On solving, n = 6

Nav 27 23 260 (9)

N1 = m× = 2.9×10
M 100
Number of revolutions before transition =
256 (d) frequency × time

E2 - E1 = ( )
E1
2 -E
2 1
Also, frequency ∝
1
n
3

hc
=
1242 eV.nm
= 13.14 eV (1/2)312.8×10-8
So required ratio = =9
λ2→1 94.54 nm (1/3)34.8×10-8
2
E1 = -17.52 eV = Z E01 261 (1)

hc 1242 eV nm Heat produced/sec =200 W

E3 - E1 = = = 15.57 eV
λ3→1 79.76 nm
200 200
⇒i = = 3 = 10 mA
E3 = -1.95 eV V 20×10

hc 1242 eV nm 262 (5)

E4 - E1 = = = 16.42 eV

[ ]
λ4→1 75.63 nm 2 4 2 4
mZ e mZ e 1 1
En = - 2 2 2 , so hf = + 2 2 -
257 (4) 8ε0n h 8ε0h 16 25

( ) [ ]
2 4
1 mZ e 9
Ephoton = 13.6 1- eV = 13.0 eV ∴f = 2 3 (i)
25 8ε0h 16×25
2 4 2 4
E/c = mv (momentum conserved) Zem Zem
and frequency f4 = 2 3 3 = 2 3 (ii)
-19
4ε0n h 4ε0(4)3h
E (13)(1.6×10 )
v= = = 4 m/s
mc (1.67)(10-27)(3)(108) ∴ f/f4 = 18/25, so m = 5

258 (2) 263 (2)

The shortest wavelength of Brackett series is T1 n 8
3
n n 2
3

corresponding to transition of electron between = 13 ⇒ 1 = 13 ⇒ 1 =

T2 n1 82 n1 n2 1
n1 = 4 and n2 = ∞
264 (6)
Similarly, the shortest wavelength of Balmer
(Z -1)
( )
2
series is corresponding to transition of electron λ1 1
= 2 2 since, ∝(Z-1)2
between n1 = 2 and n2 = ∞ λ2 (Z1-1) λ

( ) ( )
(Z)2 13.6 = 13.6 or Z = 2
16 4
1
4
=
(Z2-1) ; on solving, Z = 6
(11-1)2
2

259 (6) 265 (8)

th
When electron jumps from n state to ground Change in angular momentum = Iω
state, number of possible emission lines

P a g e | 74
 ⇒
mh nh
-
2π 2π
= Iω ⇒ω =
(m-n)h
6.28 I
1
λ
' = 4R ( )
1 1
2- 2
n2 n1

266 (4)

For the first line of Balmer series of hydrogen

'
Given λ = λ⇒4R
( )
1 1
2- 2 =
n2 n1
5R
36

( )
⇒n2 = 4
1 1 1 5R 36
= R 2- 2 = ⇒λ =
λ 2 3 36 5R

For singly ionized helium Z = 2

P a g e | 75