Explanation

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