NCERT Solutions for Class-12 Physics Chapter-12 Atoms

Question 1. Choose the correct alternative from clues given at end of each statement:
1, The size of the atom in Thomson’s model is the atomic size in Rutherford’s model.(much greater than/no different from/much less than.)
2. In the ground state of…………… electrons are in stable equilibrium, while 
in ……………..  electrons always experience a net force. (Thomson’s model/Rutherford’s model.)
3. A classical atom based on…………. is doomed to collapse. (Thomson’s model/Rutherford’s model.)
4. An atom has a nearly continuous mass distribution in a………….. but has 
a highly non­uniform mass distribution in (Thomson’s model/Rutherford’s model.)
5. The positively charged part of the atom possesses most of the mass in…………. (Rutherford’s model/both the models.)
Answer:
1. no different from
2. Thomson’s model; Rutherford’s model.
3. Rutherford’s model.
4. Thomson’s model, Rutherford’s model.
5. both the models.

Question 2. Suppose you are given a chance to repeat the alpha-particle 
scattering experiment using a thin sheet of solid hydrogen in place of the 
gold foil. (Hydrogen is a solid at a temperature 14 K). What results do you 
expect?
Answer: The nucleus of a hydrogen atom is a proton. The mass of it is 1.67 x 10-27 
kg, whereas the mass of an incident α-particle is 6.64 x 10-27 kg. Because the 
scattering particle is more massive than the target nuclei (proton). the α-particle 
won’t bounce back in even in a head-on collision. It is similar to a football colliding 
with a tennis ball at rest. Thus, there would be no large-angle scattering.

Question 3. What is the shortest wavelength present in the Paschen series 
of spectral lines?
Answer: The wavelength of the spectral lines forming the Paschen series is given by,

Question 4. A difference of 2.3 eV separates two energy levels in an atom. What 
is the frequency of radiation emitted when the atom transits from the upper level 
to the lower level?
Answer:

Question 5. The ground state energy of the hydrogen atom is -13.6 eV. What are the
kinetic and potential energies of the electron in this state?
Answer:
K.E. = -E (Total energy)
= -(-13.6) = 13.6 eV
P.E. = 2 X E = 2 X (-13.6)
= -27.2 eV

Question 6. A hydrogen atom initially in the ground level absorbs a photon which
excites it to the n = 4 level. Determine the wavelength and frequency of the photon.
Answer: We know, the energy of an electron in the nth orbit of a hydrogen atom is given by

Question 7. (a) Using Bohr’s model, calculate the speed of the electron in a hydrogen
atom in the n = 1, 2, and 3 levels.
(b) Calculate the orbital period in each of these levels.
Answer:
(a) Speed of an electron in nth orbit of a hydrogen atom is given by

Question 8. The radius of the innermost electron orbit of a hydrogen atom is 
5.3 x 10-11 m. What are the radii of the n = 2 and n = 3 orbits?
Answer: We know, the radius of the nth orbit of a hydrogen atom is given by 
rn = r0n2, where r0 = 5.3 x 10-u m is the radius of the innermost orbit of the hydrogen 
atom.
When n = 2,     r2 = 5.3 x 10-u x 4
= 2.12 x 10-10m
When n = 3,
= 5.3 x 10-11 x 9
= 4.77 x 10-10m.

Question 9. A 12.75 eV electron beam is used to bombard gaseous hydrogen 
at room temperature. What series of wavelengths will be emitted?
Answer:

Question 10. In accordance with Bohr’s model, And the quantum number that
characterizes the earth’s revolution around the sun in an orbit of radius 
1.5 x 1011 m with an orbital speed
3 x 1014 m s-1. (Mass of earth = 6.0 x 1024 kg.)
Answer: According to Bohr’s postulate of quantization of angular momentum

Question 11. Answer the following questions, which help you understand the
difference between Thomson’s model and Rutherford’s model better.
(a) Is the average angle of deflection of a-particles by a thin gold foil predicted
by Thomson’s model much less, about the same, or much greater than that
predicted by Rutherford’s model?
(b) Is the probability of backward scattering (i.e., scattering of a-particles at 
angles greater than 90°) predicted by Thomson’s model much less, about the 
same, or much greater than that predicted by Rutherford’s model?
(c) Keeping other factors fixed, it is found experimentally that for small thickness
t, the number of a-particles scattered at moderate angles is proportional to What 
clue does this linear independence on t provide?
(d) In which model is it completely wrong to ignore multiple scattering for the
calculation of the average angle of scattering of a-particles by a thin foil?
Answer:
(a) About the same
(b) Much less
(c) It suggests that the scattering is predominantly due to a single collision, because the
chance of a single collision increases linearly with the number of target atoms, and hence
linearly with thickness.
(d) In Thomson’s model, a single collision causes very little deflection. The observed
average scattering angle can be explained only by considering multiple scattering. So 
it is wrong to ignore multiple scattering in the Thomson model. In Rutherford’s model, most
of the scattering comes through a single collision and multiple scattering effects can be 
ignored as a first approximation.

Question 12. The gravitational attraction between electron and proton in a hydrogen
atom is weaker than the Coulomb attraction by a factor of about 10-40. An alternative
way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen
atom if the electron and proton were bound by gravitational attraction. You will find 
the answer interesting.
Answer: If electron and proton were bound by gravitational attraction, then

It is astonishing this value of r is much greater than the size of the universe.

Question 13. Obtain an expression for the frequency of radiation emitted when a
hydrogen atom de-excites from the level it to level (n – 1). For large it, shows that 
this frequency equals the classical frequency of revolution of the electron in the 
orbit.
Answer: The energy of an electron in the nth orbit of a hydrogen atom is given by,



i.e. frequencies are equal. This is called Bohr’s correspondence principle.

Question 14. Classically, an electron can be in any orbit around the nucleus of 
an atom. Then what determines the typical atomic size? Why is an atom not, say, 
a thousand times bigger than its typical size? The question had greatly puzzled 
Bohr before he arrived at his famous model of the atom that you have learned in 
the text. To stimulate what he might well have done before his discovery, let us 
play as follows with the basic constants of nature and see if we can get a quantity 
with the dimensions of length that is roughly equal to the known size of an atom 
(~ 10-10m).
(a) Construct a quantity with the dimensions of length from the fundamental 
constants e, me, and c. Determine its numerical value.
(b) You will find that the length obtained in («) many orders of magnitude smaller 
than the atomic dimensions. Further, it involves c. But energies of atoms are mostly 
in a non-relativistic domain where c is not expected to play any role. This is what 
may have suggested Bohr discard c and look for something else’ to get the 

right atomic size. Now, the Planck’s constant h had already made its appearance
elsewhere. Bohr’s great insight lay in recognizing that h, me, and e will yield the 
right atomic size. Construct a quantity with the dimension of length from h, 
mg, and e and confirm that its numerical value has indeed the correct order of magnitude.
Answer:
(a) Here, the dimensional formula of e is A1T1, the dimensional formula of me is M1,
dimensional formula

Question 15. The total energy of an electron in the first excited state of the hydrogen
atom is about -3.4 eV.
(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy of the electron in this state?
(c) Which of the answers above would change if the choice of the zero of potential
energy is changed?
Answer:
(a) K.E. = -E = – (-3.4 eV) = 3.4 eV
E. = 2E = 2 x (-3.4 eV)
= -6.8 eV
(b) Kinetic energy does not depend upon the choice of zero potential energy. Therefore,
its value remains unchanged. However, the potential energy gets changed with the change 
in the zero levels of potential energy.

Question 16. If Bohr’s quantisation postulate (angular momentum = nh/2π) is a 
basic law of nature, it should be equally valid for the case of planetary motion 
also. Why then do we never speak of quantization of orbits of planets around 
the sun?
Answer: Applying Bohr’s quantization postulate,

i.e., n is very large. Since n is very large, the difference between the two successive 
energy or angular momentum levels is very small and the levels may be considered continuous.

Question 17. Obtain the first Bohr’s radius and the ground state energy of a ‘muonic
hydrogen atom’ (i.e. an atom in which a negatively charged muon (μ-1) of mass 
about 207 me orbits around a proton).
Answer: Here the mass of the particle revolving around the proton is,