When an $α$-particle of mass $M$ moving with velocity $v$ bombards a heavy nucleus of charge $Ze$, its distance of closest approach from the nucleus depends on $M$ as

$∝\frac{1}{M}$

The correct answer is Option (3) → $∝\frac{1}{M}$

At distance of closest approach, initial kinetic energy of $\alpha$-particle is completely converted into electrostatic potential energy.

$ \frac{1}{2} M v^{2} = \frac{1}{4 \pi \epsilon_{0}} \frac{(2e)(Ze)}{r} $

$ r = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{2Ze^{2}}{\frac{1}{2} M v^{2}} $

$ r \propto \frac{1}{M} $

Thus, the distance of closest approach is inversely proportional to the mass $M$ of the $\alpha$-particle (for same $v$).