The distance of the closest approach of an alpha particle fired at nucleus with momentum $P$ is $r_0$. The distance of the closest approach when the alpha particle is fired at the same nucleus with momentum $2P$ will be

The correct option is A $\frac{r_0}{4}$
By conservation of energy,

$K.E$ of the proton = Electrostatic $P.E$ of proton and target nucleus at closest distance.

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

$\Rightarrow r=\frac{1}{4\pi {\epsilon }_0}\frac{Z{e}^2}{K.E}---\left(1\right)$

We know that,

$K.E=\frac{P^2}{2m}$

As all the other factors are constant,

$r\propto \frac{1}{P^2}$

$\Rightarrow \frac{r_2}{r_1}={\left[\frac{P_1}{P_2}\right]}^2$

Given that, $\left[\begin{array}{cc}{P}_1=P;& r_1=r_0\\ P_2=2P;& r_2=?\end{array}\right]$

$\Rightarrow \frac{r_2}{r_0}={\left[\frac{P}{2P}\right]}^2$

$\Rightarrow \begin{array}{c}{r}_2=\frac{r_0}{4}\end{array}$

Hence, option $\left(A\right)$ is correct.