# NCERT Solutions for Class 12 Physics Chapter 12 Atoms

• Q. Answer the following questions, which help you to understand the difference between Thomson’s model and Rutherford’s model better.
• (i) Is the average angle of deflection of α-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 ?
• (ii) Is the probability of backward scattering (i.e., scattering of α-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 ?
• (iii) 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 t. What clue does this linear dependence on t-provide ?
• (iv) In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering to α-particles by a thin foil ?
• Ans. (i) The average angle of deflection of α-particles by a thin gold foil predicted by Thomson’s model is of same size as predicted by Rutherford’s model. This is because the average angle was taken in both models.
• (ii) The probability of scattering for a-particles at angles greater than 90° predicted by Thomson’s model is much less than that predicted by Rutherford’s model.
• (iii) Scattering is mainly due to single collisions. The chances of a single collision increases linearly with the number of target atoms. As the number of target atoms increase with an increase in thickness, the collision probability depends linearly on the thickness of the target.
• (iv) It is wrong to ignore multiple scattering in Thomson’s model for the calculation of average angle of scattering of a-particles by a thin foil. This is because a single collision causes very little deflection in this model. Hence, the observed average scattering angle can be explained only by considering multiple scattering.

Q. In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a 7.7 MeV a-particle before it comes momentarily to rest and reverses its direction ?

Ans The key idea here is that throughout the scattering process, the total mechanical energy of the system consisting of an a-particle and a gold nucleus is conserved. Let system’s initial mechanical energy is Ei, before the particle and nucleus interact, and it is equal to its mechanical energy Ef when the a-particle momentarily
stops. The initial energy Ei is just the kinetic energy K of the incoming a-particle. The final energy Ef is just the electric potential energy U of the system.

Let d be the centre-to-centre distance between the a-particle and the gold nucleus when the α-particle is at its stopping point. Then we can write the conservation of energy Ei = Ef as

$$\text{K}=\frac{1}{4\pi\epsilon_0}\frac{(2e)(Ze)}{d}=\frac{2Ze^2}{4\pi\epsilon_0d}$$

Thus, the distance of closest approach d is given by

$$\text{d} =\frac{2Ze^{2}}{4\pi\epsilon_0K}\\\text{The maximum kinetic energy found in}\space\alpha-\text{particles of natural origin is 7.7 MeV or}\\1.2×10^{-12}\text{J}.\text{Since}\frac{1}{4\pi\epsilon_0}=9.0×10^{9}\text{Nm}^{2}/\text{C}^{2}.\\\text{Therefore ,with e}=1.6×10^{-19}\text{C},\text{we have},\\\text{d}=\frac{(2)(9.0×10^{9}\text{Nm}^{2}/\text{C}^{2})(1.6 ×10^{\normalsize -19}\text{C}^{2})Z}{1.2×10^{\normalsize -12}\text{J}}\\=3.84×10^{\normalsize -16}\text{Z}m\\\text{The atomic number of foil material gold is Z = 79, so that}\\\text{d}=(Au)=3.0×10^{\normalsize -14}m\\=30\text{fm}\\(1\text{fm}(i.e, \text{fermi)})=10^{\normalsize -15}m$$

The radius of gold nucleus is, therefore, than 3.0 × 10–14 m. This is not in very good agreement with the observed result as the actual radius of gold nucleus is 6 fm. The cause of discrepancy is that the distance of closest approach is considerably larger than the sum of the radii of the gold nucleus and the α-particle. Thus, the α-particle reverses its motion without ever actually touching the gold nucleus.

Q. A 10 kg satellite circles earth once every 2 h in an orbit having a radius of 8000 km. Assuming that Bohr’s angular momentum postulate applies to satellites just as it does to an electron in the hydrogen atom. Find the quantum number of the orbit of the satellite.

• Ans. As we know

$$mv_{n}r_{n}=\frac{nh}{2\pi}\\\text{Here}\space m=10\space\text{kg}\space\text{and}\space r_n=8×10^{6}\text{m}.\\\text{The time period T of the circling satellite,}\\\text{T}=2h.=7200 s.\\\text{Thus the velocity}\\\text{v}_n=\frac{2\pi r_n}{\text{T}}\\\text{The quantum number of the orbit of satellite}\\\text{n}=\frac{(2\pi r_n)^{2}×m}{(T×h)}\\\text{Substituting the values}\\\text{n}=\frac{(2\pi×8×10^{6}\text{m})^{2}×10}{(7200s×6.64×10^{-34})}\\=5.3×10^{45}$$

Note that the quantum number for the satellite motion is extremely large. In fact, for such large quantum numbers the results of quantisation conditions tend to those of classical physics.

Q. The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10–11m. What are the radii of the n = 2 and n = 3 orbits ?

Ans Given,

The radius of the innermost orbit of a hydrogen atom,

r1 = 5.3 × 10–11 m.

Consider r2 be the radius of the orbit at n = 2. It is related to the radius of the innermost orbit.

r2 = (n)2r1

= 4 × 5.3 × 10–11

= 2.12 × 10–10 m

For n = 3, r3 = (n)2 r1

= 9 × 5.3 × 10–11

= 4.77 × 10–10 m

Thus, the radii of an electron for n = 2 and n = 3 orbits are 2.12 × 10–10 m and 4.77 × 10–10 m respectively.

Q. In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 1011 m with orbital speed 3 × 104 m/s.

Ans Given, Radius of the orbit of the Earth around the Sun (r) = 1.5 × 1011 m.

Orbital speed of the Earth (v) = 3 × 104 m/s.

Mass of the Earth (m) = 6.0 × 1024 kg.

From Bohr’s model, angular momentum is quantised and given by

$$\text{mvr}-\frac{nh}{2\pi}\\\text{Here},h=\text{Planck’s constant}=6.62×10^{-34}\text{Js}n=\text{Quantum number}\\\therefore n =\frac{mvr×2\pi}{h}\\=\frac{2\pi×6×10^{24}×3×10^{4}×1.5×10^{11}}{6.62×10^{-34}}\\=25.61×10^{73}\\=2.6×10^{74}\\\text{Therefore, the quanta number that characterises that Earth’s revolution is} 2.6 × 10^{74}.$$

Q. A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level ?

Ans Separation of two energy levels in an atom,

E = 2.3 eV

= 2.3 × 1.6 × 10–19

= 3.68 × 10–19 J

If v be the frequency of radiation emitted when the atom transits from the upper level to the lower level.

Energy associated with it is

E = hv

Here, h = Planck’s constant = 6.62 × 10–34 Js

$$\therefore\space v=\frac{E}{h}=\frac{3.68×10^{-19}}{6.62×10^{-34}}\\=5.55×10^{14}\text{Hz.}$$

Q. It is found experimentally that 13.6 eV energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius and the velocity of the electron in a hydrogen atom.

Ans Total energy of the electron in hydrogen atom is

E = – 13.6 eV

= – 13.6 × 1.6 × 10–19 J

= – 2.2 × 10–18 J.

Thus, from energy electron formula we have

$$-\frac{e^{2}}{8\pi\epsilon_0r}=-2.2×10^{-18}\text{J}.\\\text{This gives the orbital radius,}\\r=-\frac{e^{2}}{8\pi\epsilon_0E}\\=\frac{-(9×10^{9}\text{Nm}^{2}/\text{C}^{2})(1.6×10^{-19}\text{C})^{2}}{(2)(-2.2×10^{-18}\text{J})}\\=5.3×10^{-11}\text{m}.\\\text{The velocity of the revolving electron can be computed}\\\text{v}=\frac{e}{\sqrt{4\pi\epsilon_0 mr}}\\=2.2×10^{6}\text{m/s}$$

Q. What is the shortest wavelength present in the Paschen series of spectral lines ?

Ans. For calculating wavelength, formula is,

$$\frac{1}{\lambda}=1.097×10^{+7}\bigg[\frac{1}{n_1^{2}}-\frac{1}{n_2^{2}}\bigg]\\\text{Here, h = Planck’s constant} = 6.6 × 10^{–34}\text{Js}\\c=\text{Speed of light}=3×10^{8}\text{m/s}(n_1\space\text{and}\space n_2\text{are integers})\\\text{The shortest wavelength present in the Paschen series of the spectral lines is given by values}\space n_1=3\space\text{and}\space n_2=\infty.\\\frac{1}{\lambda}=1.097×10^{+7}\bigg[\frac{1}{(3)^{2}}-\frac{1}{(\infty)^{2}}\bigg]\\\lambda=\frac{9}{1.097×10^{7}}\\= 8.189 × 10^{\normalsize –7} m\\=818.9\space\text{nm.}$$

Q. Using the Rydberg formula, calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum.

Ans. The Rydberg formula is

$$\frac{hc}{\lambda_{if}}=\frac{me^{4}}{8\epsilon_0^{2}h^{2}}\bigg(\frac{1}{n_f^{2}}-\frac{1}{n_i^{2}}\bigg)\\\text{The wavelengths of the first four lines in the Lyman series correspond to transitions from ni} = 2, 3, 4, 5\space\text{to n}_f=1.\text{We know that}\\\frac{me^{4}}{8\epsilon_0^{2}h^{2}}=13.6eV=21.76×10^{-19}\text{J}\\\text{Therefore},\space\lambda_{i1}=\frac{hc}{21.76×10^{-19}\bigg(\frac{1}{1}-\frac{1}{n_{1}^{2}}\bigg)}m\\\frac{6.625×10^{-34}×3×10^{8}×n_i^{2}}{21.76×10^{-19}×(n_i)^{2}-1}m\\=\frac{0.9134 n_i^{2}}{n_i^{2}-1}×10^{-7}m\\=\frac{913.4n_i^{2}}{(n_i^{2}-1)}\text{Å}\\ \text{Substituting}\space n_i=2, 3, 4,5,\text{we get}\space\lambda_{21}=1218\text{Å},\lambda_{31}=1028\text{Å},\space\lambda_{41}=974.3\space\text{Å},\text{and}\space\lambda_{51}=951.4\space\text{Å}.$$

Q. 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 temperatures below 14 K). What results do you expect ?

Ans. In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough, because the mass of hydrogen (1.67 × 10–27 kg) is less than the mass of incident α-particles. Thus, the mass of the scattering particle is more than the target nucleus
(hydrogen).

As a result, the a-particles would not bounce back if solid hydrogen is used in the α-particle scattering experiment.

Q. In the Rutherford’s nuclear model of the atom, the nucleus (radius about 10–15 m) is analogous to the sun about which the electron move in orbit (radius ≈ 10–10 m) like the earth orbits around the sun. If the dimensions of the solar system had the same proportions as those of the atom, would the earth be closer to or farther away from the sun than actually it is ? The radius of earth’s orbit is about 1.5 × 1011 m. The radius of sun is taken as 7 × 108 m.

Ans. The ratio of the radius of electron’s orbit to the radius of nucleus is 10-10 m 10-15 m = 105,i.e.,is, the radius of the electron’s orbit is 105 times larger than the radius of nucleus.If the radius of the Earth’s orbit around the sun were  105 times larger than the radius of the Sun, the radius of the Earth’s orbit would be 105 × 7 × 108 m = 7 × 1013 m. This is more than 100 times greater than the actual orbital radius of Earth. Thus, the Earth would be much farther away from the Sun.

It implies that an atom contains a much greater fraction of empty space than our solar system does.

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

Ans. Quantisation of orbits of planets around the Sun never comes in discussion because the angular momentum associated with planetary motion is largely relative to the value of Planck’s constant (h). The angular momentum of the Earth in its orbit is of the order of 1070 h. This leads to a very high value of quantum levels n of the order of 1070. For large values of n, successive energies and angular momenta are relatively small. Hence, the quantum levels for planetary motion are considered continuous.