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) $\dfrac{1}{m}$
At closest approach all the kinetic energy has become potential energy:
$$\frac{1}{2}mv^2 = \frac{1}{4\pi\epsilon_0}\frac{(2e)(Ze)}{r_0} \;\Rightarrow\; r_0 = \frac{4kZe^2}{mv^2}$$
So $r_0 \propto \dfrac{1}{m}$ (and also $\propto Z$, $\propto \dfrac{1}{v^2}$).
Solution by Sreeraj P, M.Sc Physics