The distance of closest approach of an $\alpha$-particle fired towards a nucleus with momentum p, is r. If the momentum of the $\alpha$-particle is 2p, then the corresponding distance of closest approach is

At the distance of closed approach r,

$K=\frac{1}{4\pi {\epsilon }_0}\frac{\left(2e\right)\left(Ze\right)}{r}$

$\Rightarrow r=\frac{2Z{e}^2}{4\pi {\epsilon }_{0}K}$

where, Ze = charge of the nucleus,
2e = charge of the alpha particle
and K = kinetic energy of the alpha particle.

$\because K=\frac{p^2}{2m}$

where, p is the momentum of the $\alpha$-particle and m is the mass of the electron

$\therefore r=\frac{2Z{e}^{2}2m}{4\pi {\epsilon }_0{p}^2}\text{ or }r\propto \frac{1}{p^2}$

$\frac{r\text{'}}{r}={\left(\frac{p}{p\text{'}}\right)}^2={\left(\frac{p}{2p}\right)}^2=\frac{1}{4}\Rightarrow r\text{'}=\frac{r}{4}$