# NCERT Solutions for Class 12 Physics Chapter 12 - Atoms

## NCERT Solutions for Class 12 Physics Chapter 12 Free PDF Download

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1. Choose the correct alternative from the clues given at the end of the each statement:

(a) 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.)

(b) In……the ground
state of the ground state of ………electrons are in stable equilibrium, while in ………electrons always experience it net force.

(Thomson’s model/Rutherford’s model)

(c) A classical atom based on ………is doomed to collapse.
(Thomson’s model/Rutherford’s model)

(d) An atom has a nearly continuous mass distribution in a ………but has a highly non-uniform mass distribution in ………..
(Thomson’s model/Rutherford’s model)

(e) The positively charged part of the atom possesses most of the mass in ………
(Rutherford’s model/both the models)

Sol. (a) No different from

(b) Thomson’s model, Rutherford’s model

(c) Rutherford’s model

(d) Thomson’s model, Rutherford’s model

(e) Both the models

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

Sol. The purpose of scattering experiment is not completed because solid hydrogen will be a much lighter target then the alpha particle acting as a projectile. By using the conditions of elastic collisions, the hydrogen will move much faster than alpha after the collision. We cannot determine the size of hydrogen nucleus.

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

Sol. For Paschen series, (n1 = 3 and n2 = ∞ for shortest wavelength)

$$\frac{hc}{\lambda}=\text{R}\bigg[\frac{1}{n_1^{2}}-\frac{1}{n_2^{2}}\bigg]\\\text{[where R is the Rydberg constant]}\\\frac{hc}{\lambda}=13.6×1.6×10^{\normalsize-19}\bigg[\frac{1}{3^{2}}-\frac{1}{\infty^{2}}\bigg]\\\text{or}\space\frac{hc}{\lambda}=\frac{21.76×10^{\normalsize-19}}{9}\\\text{or}\space\lambda=\frac{9×6.63×10^{\normalsize-34}×3×10^{8}}{21.76×10^{\normalsize-19}}$$

= 8.2265 × 10–7m

= 822.65 nm

Thus, the shortest wavelength present is 822.65 nm.

4. 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?

Sol. Given, difference in energy levels E = 2.3 eV, = 2.3 × 1.6 × 10–19J

Let v be the frequency, then using formula E = hv

$$\text{or}\space v=\frac{\text{E}}{h}=\frac{2.3×1.6×10^{\normalsize-19}}{6.63×10^{\normalsize-34}}$$

= 5.6 × 1014 Hz

5. The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?

Sol. Given, the ground state energy of hydrogen atom

E = – 13.6 eV

We know that

Kinetic Energy (KE) = – E = 3.6 eV

Potential Energy (PE) = – 2KE = – 2 × 13.6

= –27.2 eV

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 photon.

Sol. For ground state n1 = 1 to n2 = 4

Energy absorbed E = E2 – E1

$$=+13.6\bigg(\frac{1}{n_1^{2}}-\frac{1}{n_2^{2}}\bigg)×1.6×10^{\normalsize-19}\text{J}\\=13.6\bigg(\frac{1}{1}-\frac{1}{4^{2}}\bigg)×1.6×10^{\normalsize-19}\\=13.6×1.6×10^{\normalsize-19}\bigg(\frac{15}{16}\bigg)$$

= 20.4 × 10–19

or E = hv = 20.4 × 10–19

$$\text{Frequency, v =}\frac{20.4×10^{\normalsize-19}}{h}\\=\frac{20.4×10^{\normalsize-19}}{6.63×10^{\normalsize-34}}=3.076×10^{15}$$

= 3.1 × 1015 Hz

Wavelength of photon,

$$\lambda=\frac{c}{v}\\=\frac{3×10^{8}}{3.076×10^{15}}=9.74×10^{\normalsize-8}\text{m}$$

Thus, the wavelength is 9.7 ×10–8 m and frequency is 3.1 × 1018 Hz.

7. (a) Using the 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.

Sol. (a) Speed of the electron in Bohr’s nth orbit

$$v=\frac{c}{n}\alpha\\\text{where,}\space\alpha=\frac{2\pi \text{Ke}^{2}}{ch}\\\alpha=0.0073\\v=\frac{c}{n}×0.0073\\\text{For n = 1, v}_1 =\frac{c}{1}×0.0073=3×10^{8}×0.0073\\\text{= 2.19 × 10}^6 {\text{m/s}}\\\text{For n = 2, v}_{2}=\frac{c}{2}×0.0073=\frac{3×10^{8}×0.0073}{2}\\\text{= 1.095 × 10}^{6}{\text{ m/s}}\\\text{For n = 3, v}_3=\frac{c}{3}×0.0073=\frac{3×10^{8}×0.0073}{3}$$

= 7.3 × 105 m/s

(b) Orbital period of electron

$$\text{T}=\frac{2\pi r}{v}\\\text{Radius of n}^{th} {\text{orbit}}\\r_n=\frac{n^2h^2}{4\pi^{2}Kme^{2}}\\\therefore r_1=\frac{(1)^{2}×(6.63×10^{\normalsize-34})^{2}}{4×9.87×(9×10^{9})×9×10^{\normalsize-31}×(1.6×10^{\normalsize-19})}$$

= 0.53 ×10–10m

$$\text{For n = 1, T}_1=\frac{2\pi r_1}{v_1}\\=\frac{2×3.14×0.53×10^{\normalsize-10}}{2.19×10^{6}}$$

= 1.52 × 10–16s

For n = 2,

∴ r2 = 22.r1 = 4 × 0.53 × 10–10

$$\text{and Velocity, v}_n=\frac{v_1}{n}\\v_2=\frac{v_1}{2}=\frac{2.19×10^{6}}{2}\\\text{Time period, T}_{2}=\frac{2×3.14×4×0.53×10^{-10}×2}{2.19×10^{6}}$$

= 1.216 × 10–15s

For n = 3, radius r3 = 32, r1 = 9r1 = 9 × 0.53 × 10-10m

$$\text{and velocity, v}_3=\frac{v_1}{3}=\frac{2.19×10^6}{3}\text{m/s}\\\text{Time period T}_3=\frac{2×3.14×9×0.53×10^{\normalsize-10}×3}{2.19×10^{6}}$$

= 4.1 × 10–15 s

8. 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?

Sol. Given, the radius of the innermost electron orbit of a hydrogen r1 = 5.3 × 10–11m

As we know that rn = n2r1

For n = 2, r2 = 22r1 = 4 × 5.3 × 10–11 = 2.12 × 10–10 m

For n = 3, r3 = 32r1 = 9 × 5.3 × 10–11 = 4.77 × 10–10m

9. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Sol. Energy of electron beam E = 12.5 eV = 12.5 × 1.6 × 10–19J

$$\text{Using the relation E =}\frac{hc}{\lambda}\\\Rarr\space\lambda=\frac{hc}{\text{E}}\\=\frac{6.62×10^{\normalsize-34}×3×10^{8}}{12.5×1.6×10^{\normalsize-19}}$$

= 0.993 × 10–7m = 993 × 10–10 m

λ = 993Å

This wavelength falls in the range of Lyman series (912Å to 1216Å )

10. In accordance with the Bohr 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. (Mass of earth = 6.0 × 1024 kg.)

Sol. Given, radius of orbit r = 1.5 × 1011m

Orbital speed v = 3 × 104 m/s; Mass of earth

M = 6 × 1024 kg.

$$\text{Angular momentum, mvr =}\frac{nh}{2\pi}\text{N/L}\qquad\text{or}\\n=\frac{2\pi vrm}{h}\\=\frac{2×3.14×3×10^{4}×1.5×10^{11}×6×10^{24}}{6.63×10^{\normalsize-34}}$$

= 2.57 × 1074

or n = 2.6 × 1074

Thus, the quantum number is 2.6 × 1074 which is too large.

The electron would jump from n = 1 to n = 3.

$$\text{E}_{3}=\frac{-13.6}{3^{2}}=-1.5\text{eV}$$

Thus, they belong to Lyman series.