An alpha nucleus of energy $\frac{1}{2} m v^2$ bombards a heavy nuclear target of charge $Z e$. Then the distance of closest approach for the alpha nucleus will be proportional to

Since, here nuclear target is heavy it can be assumed safely that it will remain stationary and will not move due to the Coulombic interaction force. At distance of closest approach relative velocity of two particles is $v$. Here target is considered as stationary, so $\alpha$-particle comes to rest instantaneously at distance of closest approach. Let required distance is $r$, then from work energy-theorem. $\begin{aligned} 0-\frac{m v^2}{2} & =-\frac{1}{4 \pi \varepsilon_0} \frac{Z e \times 2 e}{r} \\ \Rightarrow \quad r & \propto \frac{1}{m} \\ & \propto \frac{1}{v^2} \\ & \propto Z e^2\end{aligned}$