An $\mathrm{\alpha }$-particle accelerated through $\mathrm{V}$ volt is fired towards a nucleus. Its distance of closest approach is $\mathrm{r}$. If a proton accelerated through the same potential is fired towards the same nucleus, the distance of closest approach of the proton will be:

Given, a particle is shot in the direction of the nucleus while being accelerated by V volts.
We can conclude from the question that Kinetic Energy is equivalent to Potential Energy.

$\mathrm{K}\cdot \mathrm{E}=\mathrm{P}\cdot \mathrm{E}$

$K\cdot E=1/2m{v}^2$

$\mathrm{P}\cdot \mathrm{E}= \frac{1}{4\mathrm{\pi ϵ}} \frac{\left(q_1{q}_2\right)}{b}$

Equating both, we get

$2\mathrm{eV}=\frac{1}{4\pi {ϵ}_0}\frac{2\mathrm{e}\times {\mathrm{q}}_2}{\mathrm{r}}$

$\text{ Here }{\mathrm{q}}_2=4\pi {ϵ}_0\mathrm{rV}$ (1)

For Proton

$\mathrm{eV}=\frac{1}{4\pi {ϵ}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{b}}$

$\mathrm{eV}=\frac{1}{4\pi {ϵ}_0}\frac{{\mathrm{eq}}_2}{\mathrm{b}}$

${\mathrm{q}}_2=4\pi {ϵ}_0\mathrm{rb} \left(2\right)$

Comparing equation: (1) and (2) we get $b=r$

Consequently, r is the distance of the closest approach.