Q 12-12-011JEE MainAIEEE 2006Medium
An alpha nucleus of energy $\dfrac{1}{2}mv^2$ bombards a heavy nuclear target of charge $Ze$. Then the distance of closest approach for the alpha nucleus will be proportional to
Answer: (D) $\dfrac{1}{m}$
Concept. At closest approach in a head-on collision all kinetic energy has become electric potential energy: $\tfrac12mv^2=\dfrac{k(2e)(Ze)}{d}$.
Step 1. Distance. $d=\dfrac{4kZe^2}{mv^2}\propto\dfrac{1}{m}$, $\dfrac1{v^2}$ and $Z$.
Answer. $\dfrac1m$ (Option D).
Solution by Sreeraj P, M.Sc Physics
