A proton and an $\alpha$-particle are situated at r distance apart . At very large, distance apart when released, the kinetic energy of proton will be

The potential energy between proton and $\mathrm{\alpha }$-particle is given by
$\mathrm{U}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{\mathrm{e}\left(2\mathrm{e}\right)}{\mathrm{r}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{2{\mathrm{e}}^2}{\mathrm{r}}$
K.E. of proton and a-particle is given by
$$\begin{gathered} \text{ K.E. }=\frac{{\mathrm{p}}^2}{2\mathrm{m}}+\frac{{\mathrm{p}}^2}{2\left(4\mathrm{m}\right)}=\frac{5}{4}\times \frac{{\mathrm{p}}^2}{2\mathrm{m}} \\ \therefore \frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{2{\mathrm{e}}^2}{\mathrm{r}}=\frac{5}{4}\times \frac{{\mathrm{p}}^2}{2\mathrm{m}} \\ \mathrm{or} \frac{{\mathrm{p}}^2}{2\mathrm{m}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{8{\mathrm{e}}^2}{5\mathrm{r}} \\ \therefore \text{ K.E. of proton }=\frac{{\mathrm{p}}^2}{2\mathrm{m}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{8{\mathrm{e}}^2}{5\mathrm{r}} \end{gathered}$$