(i) $KE$ of $4.1\ \text{MeV}$ proton $= 4.1 \times 10^6 \times 1.6\times10^{-19}\ \text{J}$ Mass of proton $= 1.67 \times 10^{-27}\ \text{kg}$ $v^2 = \frac{2KE}{m}$ Or, $v^2 = \frac{2\times4.1\times10^6\times1.6\times10^{-19}}{1.67\times10^{-27}}$ Or, $v^2 = 7.85\times10^{14}$ $\therefore v= 2.8\times10^7\ \text{m/s}$ (ii) Energy of proton $= 4.1\ \text{MeV}$ Atomic number(Z) of lead $= 82$ When the proton is at distance of closest approach $(r)$, then $KE$ of the system $= 0$ $PE$ of the system $= \frac{kZe^2}{r}$ So, from the conservation of energy principle, $4.1\ \text{MeV} = 0 + \frac{kZe^2}{r}$ Or, $r_0 = \frac{kZe^2}{4.1\times10^6 e}$ Or, $r_0 = \frac{9\times10^9 \times 82 \times e^2}{4.1\times10^6}\ \text{m}$ $\therefore r_0 = 288\times10^{-16}\ \text{m}$ $r_0 = 2.88\times10^{-14}$