The distance of the closest approach of an alpha particle fired at a nucleus with kinetic energy K is r₀. The distance of the closest approach when the $\mathrm{\alpha }$-particle is fired at the same nucleus with kinetic energy 2K will be 

At distance of closest approach, total energy of particle is converted into potential energy.

Let charge on $\mathrm{\alpha }$-particle is q₁ and charge on nucleus is q₂, then in first case

                                 $\mathrm{K}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}_0}$                                      …(i)

In second case, (let distance of closest approach is r₀' )

                                  $2\mathrm{K}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{r}{\text{'}}_0}$                                   …(ii)

On dividing Eq. (ii) by Eq. (i), we get

               $\frac{2\mathrm{K}}{\mathrm{K}}=\frac{{\displaystyle \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}_0\text{'}}}}{{\displaystyle \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}_0}}}=\frac{1}{{\mathrm{r}}_0\text{'}}\times \frac{{\mathrm{r}}_0}{1}=\frac{{\mathrm{r}}_0}{{\mathrm{r}}_0\text{'}}\Rightarrow {\mathrm{r}}_0\text{'}=\frac{{\mathrm{r}}_0}{2}$