$\mathrm{\alpha }$-particles of energy 400 KeV are bombarded on nucleus of ${}_{82}Pb$. In scattering of $\mathrm{\alpha }$-particles, its minimum distance from nucleus will be

Suppose closest distance is r, according to conservation of energy

$\begin{array}{l}400\times {10}^3\times 1.6\times {10}^{-19}=9\times {10}^9\frac{\left(Ze\right)\left(2\mathrm{e}\right)}{\mathrm{r}}\\ \Rightarrow 6.4\times {10}^{-14}=\frac{9\times {10}^9\times \left(82\times 1.6\times {10}^{-19}\right)\times \left(2\times 1.6\times {10}^{-19}\right)}{\mathrm{r}}\\ \Rightarrow \mathrm{r}=5.9\times {10}^{-13}\mathrm{m}=0.59\mathrm{pm}\end{array}$