Answer
(a) The electrical potential is 4.79 MeV and in Joules is $7.66 \times 10^{-13} \mathrm{J}$.
(b) $R = 51.7 \times 10^{-15} \mathrm{m}$
Work Step by Step
(a) Using the conservation law we could get \begin{gather} K_{1}+U_{1}=K_{2}+U_{2}\\ 0+U_{1}=K_{2}+0\\ U_{1}= K_2= \boxed{4.79 \mathrm{Mev}} \end{gather} And in Joules will be \begin{aligned} U_{1} &=(4.79 \times 10^{6} \mathrm{eV} )( 1.60 \times 10^{-19} \mathrm{J/eV})\\ &=\boxed{7.66 \times 10^{-13} \mathrm{J}} \end{aligned} (b) From the value of $U_1$ we could get the radius of the nucleus by \begin{aligned} R &=k \frac{q q_{r}}{U_{1}} \\ &=\left(9.0 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{\left( 3.2\times 10^{-19} \mathrm{C}\right)\left( 137.6 \times 10^{-19} \mathrm{C}\right)}{7.66 \times 10^{-13} \mathrm{J}} \\ &=51.7 \times 10^{-15} \mathrm{m} \end{aligned}
