# NCERT Solutions for Class 12 Physics Chapter 12 - Atoms

**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 positiv****ely 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, (n_{1} = 3 and n_{2} = ∞ 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^{–7}m

= 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^{–19}J

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 × 10^{14} 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 n_{1} = 1 to n_{2} = 4

Energy absorbed E = E_{2} – E_{1}

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

= 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 × 10^{15} 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 × 10^{18} 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 × 10^{5} 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^{–10}m

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

= 1.52 × 10^{–16}s

For n = 2,

∴ r_{2} = 2^{2}.r_{1} = 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^{–15}s

For n = 3, radius r_{3} = 3^{2}, r_{1} = 9r_{1} = 9 × 0.53 × 10^{-10}m

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

= 4.1 × 10^{–15} s

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

**Sol.** Given, the radius of the innermost electron orbit of a hydrogen r_{1} = 5.3 × 10^{–11}m

As we know that r_{n} = n^{2}r_{1}

For n = 2, r_{2} = 2^{2}r_{1} = 4 × 5.3 × 10^{–11} = 2.12 × 10^{–10} m

For n = 3, r_{3} = 3^{2}r_{1} = 9 × 5.3 × 10^{–11} = 4.77 × 10^{–10}m

**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^{–19}J

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

= 0.993 × 10^{–7}m = 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 × 10 ^{11} m with orbital speed 3 × 10^{4} m/s. (Mass of earth = 6.0 × 10^{24} kg.)**

**Sol.** Given, radius of orbit r = 1.5 × 10^{11}m

Orbital speed v = 3 × 10^{4 }m/s; Mass of earth

M = 6 × 10^{24 }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 × 10^{74}

or n = 2.6 × 10^{74}

Thus, the quantum number is 2.6 × 10^{74} 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.