(i) $\mathit{K}\mathit{E}$ of $4.1\mathit{M}\mathit{e}\mathit{V}$ proton
$=4.1\times {10}^6\times 1.6\times {10}^{-19}\mathit{J}$
Mass of proton $=1.67\times {10}^{-27}\mathit{k}\mathit{g}$
${\mathit{v}}^2=\frac{2\mathit{K}\mathit{E}}{\mathit{m}}$
Or, ${\mathit{v}}^2=\frac{2\times 4.1\times {10}^6\times 1.6\times {10}^{-19}}{1.67\times {10}^{-27}}$
Or, ${\mathit{v}}^2=7.85\times {10}^{14}$
$\therefore \mathit{v}=2.8\times {10}^7\mathit{m}∕\mathit{s}$
(ii) Energy of proton $=4.1\mathit{M}\mathit{e}\mathit{V}$
Atomic number(Z) of lead $=82$
When the proton is at distance of closest approach $\left(\mathit{r}\right)$, then
$\mathit{K}\mathit{E}$ of the system $=0$
$\mathit{P}\mathit{E}$ of the system $=\frac{\mathit{k}\mathit{Z}{\mathit{e}}^2}{\mathit{r}}$
So, from the conservation of energy principle,
$4.1\mathit{M}\mathit{e}\mathit{V}=0+\frac{\mathit{k}\mathit{Z}{\mathit{e}}^2}{\mathit{r}}$
Or, ${\mathit{r}}_0=\frac{\mathit{k}\mathit{Z}{\mathit{e}}^2}{\left(4.1\times {10}^6\mathit{e}\right)}$
Or, ${\mathit{r}}_0=\frac{9\times {10}^9\times 82\times {\mathit{e}}^2}{4.1\times {10}^6}\mathit{m}$
$\therefore {\mathit{r}}_0=288\times {10}^{-16}\mathit{m}$
${\mathit{r}}_0=2.88\times {10}^{-14}$