When an $\alpha -$particle of mass 'm' moving with velocity 'v' bombards on a heavy nucleus of charge 'Ze', its distance of closest approach from the nucleus depends on ‘m’ as: 

At closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy. 

Kinetic energy $\mathrm{K}.\mathrm{E}.=\frac{1}{2}{\mathrm{mv}}^2$

Potential energy $\text{ P.E. }=\frac{\mathrm{KQq}}{\mathrm{r}}$

$\frac{1}{2}m{v}^2=\frac{\mathrm{KQq}}{r}\Rightarrow r\propto \frac{1}{\mathrm{m}}$