Physics - Elecrostatics Question with Solution | TestHub

At closest approach, all the kinetic energy of the $\alpha$-particle will converted into the potential energy of the system, K.E. = P.E. $$\begin{array}{rl}& 5\mathrm{MeV}=\frac{1}{4\pi {\epsilon }_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{r}}\\ & 5\times 10^6\times \mathrm{e}=9\times 10^9\frac{{\mathrm{Z}}_1\times {\mathrm{Z}}_2{\mathrm{e}}^2}{\mathrm{r}}\\ & \mathrm{r}=\frac{9\times 10^9\times 92\times 2\times 1.6\times 10^{-19}}{5\times 10^6}\\ & \therefore \mathrm{r}=5.3\times 10^{-14}\mathrm{m}=5.3\times 10^{-12}\mathrm{cm}\end{array}$$