At the distance of closest approach ($d$), the initial kinetic energy of the \alpha particle is completely converted into electrostatic potential energy. Kinetic Energy ($K$) = $\frac{1}{2}mv^2$. Potential Energy ($U$) = $\frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{d}$ . Equating $K = U$: $\frac{1}{2}mv^2 = \frac{2Ze^2}{4\pi\epsilon_0 d}$. Solving for $d$: $d = \frac{4Ze^2}{4\pi\epsilon_0 m v^2}$. From this expression, it is clear that the distance of closest approach is inversely proportional to the mass ($m$) of the \alpha particle ($d \propto \frac{1}{m}$) and inversely proportional to the square of the velocity ($d \propto \frac{1}{v^2}$). It is directly proportional to the charge of the target ($Z$). Therefore, $1/m$ is the correct proportionality.