Atoms and Nuclei — JEE Main practice

1. Consider an $\alpha$-particle moving directly toward the center of a gold nucleus.

At a very large distance, its electrostatic potential energy is nearly zero, so its energy is entirely kinetic:

$$K_0=7.7\ \mathrm{MeV}.$$

2. As the positively charged $\alpha$-particle approaches the positively charged gold nucleus, electrostatic repulsion slows it down.

At the distance of closest approach $r_0$, the particle momentarily comes to rest. Therefore, by the law of conservation of energy,

$$\text{Initial Kinetic Energy} = \text{EPE}.$$

Hence,

$$\begin{aligned} K_0 &=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_0}. \end{aligned}$$

3. An $\alpha$-particle has charge

$$q_1=+2e,$$

and a gold nucleus ($Z=79$) has charge

$$q_2=+79e.$$

Substituting these values,

$$\begin{aligned} K_0 &=\frac{1}{4\pi\epsilon_0}\frac{(2e)(79e)}{r_0}, \end{aligned}$$

which gives

$$\begin{aligned} r_0 &=\frac{1}{4\pi\epsilon_0}\frac{158e^2}{K_0}. \end{aligned}$$

4. Convert the kinetic energy into SI units:

$$\begin{aligned} K_0 &=7.7\times10^6\ \mathrm{eV} \\ &=(7.7\times10^6)(1.6\times10^{-19})\ \mathrm{J} \\ &=12.32\times10^{-13}\ \mathrm{J}. \end{aligned}$$

5. Substitute the given values:

$$\begin{aligned} r_0 &=\frac{(9\times10^9)\times158\times(1.6\times10^{-19})^2} {12.32\times10^{-13}} \\ &=\frac{(9\times158\times2.56)\times10^{-29}} {12.32\times10^{-13}} \\ &=295\times10^{-16}\ \mathrm{m} \\ &=2.95\times10^{-14}\ \mathrm{m}. \end{aligned}$$

6. Therefore, the distance of closest approach is

$$r_0=2.95\times10^{-14}\ \mathrm{m}.$$

Hence, the correct option is (B).

1. In the Bohr model, the energy of an electron in the $n^{\text{th}}$ orbit is given by

$$E_n=-\frac{E_0}{n^2},$$

where $E_0$ is the ground-state energy.

2. The given energy is

$$-0.04E_0=-\frac{E_0}{n^2}.$$

Cancelling $-E_0$ from both sides,

$$0.04=\frac{1}{n^2}.$$

Therefore,

$$n^2=\frac{100}{4}=25,$$

so

$$n=\sqrt{25}=5.$$

Thus, the electron is in the fifth Bohr orbit.

3. According to Bohr's second postulate, the angular momentum is quantized as

$$L=\frac{nh}{2\pi},$$

where $n$ is the principal quantum number.

4. Hence,

$$\frac{2\pi L}{h}=n=5.$$

Therefore, the value of

$$\frac{2\pi L}{h}$$

is 5, so the correct option is (C).

1. At the distance of closest approach, the entire kinetic energy of the $\alpha$-particle is converted into electrostatic potential energy.

2. Using conservation of energy,

$$\begin{aligned} K &=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r}, \end{aligned}$$

where

$$q_1=2e,\qquad q_2=79e.$$

Thus,

$$\begin{aligned} K &=\frac{1}{4\pi\epsilon_0}\frac{158e^2}{r}. \end{aligned}$$

3. The kinetic energy of the $\alpha$-particle is

$$\begin{aligned} K &=7.9\ \mathrm{MeV} \\ &=7.9\times10^6\times1.6\times10^{-19} \\ &=12.64\times10^{-13}\ \mathrm{J}. \end{aligned}$$

4. Hence,

$$\begin{aligned} r &=\frac{(9\times10^9)\times158\times(1.6\times10^{-19})^2} {12.64\times10^{-13}} \\ &=2.88\times10^{-14}\ \mathrm{m}. \end{aligned}$$

5. The nucleus must lie within the distance of closest approach, so its estimated diameter is

$$\begin{aligned} d &=2r \\ &=2\times2.88\times10^{-14} \\ &=5.76\times10^{-14}\ \mathrm{m}. \end{aligned}$$

6. Therefore, the correct option is A.