The distance of closest approach of an -particle fired towards a nucleus with momentum 'P' is . What — Atomic Structure Chemistry Question

💡 Solution & Explanation

Kinetic energy $K = \frac{P^2}{2m}$. Distance of closest approach $r \propto \frac{1}{K} \propto \frac{1}{P^2}$. If momentum is doubled ($2P$), the new distance is proportional to $\frac{1}{(2P)^2} = \frac{1}{4P^2}$. The new distance becomes $r/4$. Therefore, correct answer is D.