The following charged particles accelerated from rest, through the same potential difference, are pr — Atomic Structure Chemistry Question

💡 Solution & Explanation

Kinetic energy gained by accelerating through potential $V$ is $K.E. = qV$. The distance of closest approach $r_0 = \frac{kQq_{nucleus}}{K.E.} = \frac{k \cdot q \cdot Z_{Au}e}{qV} = \frac{k \cdot Z_{Au}e}{V}$. The charge $q$ cancels out, meaning $r_0$ depends only on the accelerating potential $V$ and the target nucleus, making it the same for all particles. Therefore, correct answer is D.