Explanation

(d) At the distance of closest approach, $r$
$K = \frac{1}{4\pi\varepsilon_{0}}\frac{(2e)(Ze)}{r}$
$r = \frac{2Ze^{2}}{4\pi\varepsilon_{0}K}$
Where, $Ze =$ charge of the nucleus
$2e =$ charge of the alpha particle
$K =$ kinetic energy of the alpha particle
$\therefore K = \frac{p^{2}}{2m}$
Where $p$ is the momentum of the $\alpha$-particle and m is the mass of the electron
$\therefore r = \frac{2Ze^{2}2m}{4\pi\varepsilon_{0}p^{2}}$
Or $r \propto \frac{1}{p^{2}}$
$\frac{r^{'}}{r} = \left( \frac{p}{p^{'}} \right)^{2} = \left( \frac{p}{2p} \right)^{2} = \frac{1}{4} \Rightarrow r^{'} = \frac{r}{4}$