An α-particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of the closest approach is of the order of ___ .

The distance of closest approach is given by

$\begin{array}{r}{\mathrm{r}}_0=\frac{\mathrm{ze}2\mathrm{e}}{4\pi {ϵ}_0\mathrm{E}}\end{array}$

Equating kinetic energy and potential energy, we get:

$\begin{array}{r}\frac{1}{2}{\mathrm{mv}}^2=\frac{1}{4\pi {ϵ}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{r}}\end{array}$

$\begin{array}{r}\mathrm{r}=\frac{9\times {10}^9\times 2\times 92\times {\left(1.6\times {10}^{-19}\right)}^2}{5\times {10}^6\times 1.6\times {10}^{-19}}\\ \end{array}$

$\mathrm{r}=5.3\times {10}^{-14} \mathrm{m}\approx {10}^{-12} \mathrm{cm}$