Answer
${\bf 34.18 } \, \text{MeV}$
Work Step by Step
In this experiment, a proton (charge $ q = e $) is fired at a stationary $^{207}\text{Pb}$ nucleus (charge $ q = 82e $). To study nuclear reactions, the proton needs to have 20 MeV of kinetic energy upon reaching the surface of the lead nucleus. We need to find the initial kinetic energy $ K_i $ (in MeV) required for the proton to achieve this upon impact.
To find the initial kinetic energy $ K_i $ of the proton, we’ll apply the conservation of energy principle, which states that the initial total energy (kinetic + potential) must equal the final total energy (kinetic + potential).
$$
K_i + U_i = K_f + U_f\tag 1
$$
The proton is far away initially, so the initial potential energy $ U_i $ is approximately zero. The initial kinetic energy is $ K_i $, which is what we are trying to find.
Upon reaching the surface of the lead nucleus, the proton has a final kinetic energy $ K_f = 20 \, \text{MeV} $.
The final potential energy $ U_f $ due to the Coulomb interaction between the proton and the lead nucleus is $ U_f = \frac{1}{4 \pi \epsilon_0} \frac{(e)(82e)}{r_f}$
The radius of the proton, $ r_p = (1.2 \, \text{fm})(1)^{1/3} = 1.20 \, \text{fm} $
The radius of the lead nucleus, $ r_{\text{Pb}} (1.2 \, \text{fm})(207)^{1/3} = 7.10 \, \text{fm} $
Thus, the distance between the centers $ r_f $ when the proton just touches the lead nucleus is $
r_f = r_p + r_{\text{Pb}} = 1.20 \, \text{fm} + 7.10 \, \text{fm} = 8.30 \, \text{fm} $
Substitute all of that into (1)
$$
K_i = 20 \, \text{MeV} + \frac{1}{4 \pi \epsilon_0} \frac{(e)(82e)}{(r_p + r_{\text{Pb}} )}
$$
Rearranging to solve for $ K_i $:
\begin{align}
K_i + 0 = (20\times 10^6\times 1.6\times 10^{-19})+ \frac{1}{4 \pi \epsilon_0} \frac{82e^2}{(r_p + r_{\text{Pb}} )}
\end{align}
Plug the known;
\begin{align}
K_i &= (20\times 10^6\times 1.6\times 10^{-19})\\\\
&+ \frac{(9.0 \times 10^9 ) (82)(1.60 \times 10^{-19} )^2}{(8.30 \times 10^{-15} )} =\bf5.48 \times 10^{-12} \;\rm J
\end{align}
$$
K_i = \color{red}{\bf 34.18 } \, \text{MeV}
$$
