When an α - 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 $\mathrm{P}.\mathrm{E}.=\frac{\mathrm{KQq}}{\mathrm{r}}$
$\frac{1}{2}{\mathrm{mv}}^2=\frac{\mathrm{KQq}}{\mathrm{r}}\Rightarrow \mathrm{r}\propto \frac{1}{\mathrm{m}}$