Explanation

(c) According to law of conservation of energy, kinetic energy of $\alpha$-particle
= potential energy of $\ \alpha$-particle at distance of closest approach
$ie,$ $\frac{1}{2}mv^{2}\ = \frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}q_{2}}{r}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ 5$MeV=$\frac{9 \times \ 10^{9} \times (2e) \times \ (92e)}{r}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \because\frac{1}{2}{mv}^{2} = 5\ MeV\ \ \right)$
$\Rightarrow \ \ \ \ \ r = \frac{9 \times 10^{9} \times 2 \times 92 \times \left( 1.6 \times 10^{- 19} \right)^{2}}{5 \times 10^{6} \times 1.6 \times 10^{- 19}}$
$\therefore\ \ \ \ \ \ \ r\ = 5.3\ \times \ 10^{- 14}\ m \approx \ 10^{- 12}cm$