An -particle accelerated through volt is fired towards a nucleus. It distance of closest approach is — Atomic Structure Chemistry Question

💡 Solution & Explanation

When accelerated by the same potential $V$, K.E. equals $qV$. Distance of closest approach $r_0 = \frac{k \cdot q_{nucleus} \cdot q}{K.E.} = \frac{k \cdot Z \cdot e \cdot q}{qV} = \frac{kZe}{V}$. This value is independent of the particle's initial charge or mass. Thus, it remains $r$ for the proton. Therefore, correct answer is A.