Answer
$8.6875MeV$ work is done by electric force
Work Step by Step
Copper nucleus carries $Z=29$ protons
charge on $1$ proton $=+e$
so charge on gold nucleus $q_{Cu}=+29e=+29\times1.6\times10^{-19}C$
$q_{Cu}=+4.64\times10^{-18}C$
radius of Copper nucleus is $r=4.8\times10^{-15}m$
We want to put a proton on surface of this nucleus . For the calculation we can assume nucleus as point charge { $q_{Cu}=+4.64\times10^{-18}C$ }
and we want to put another charge { $q_{p}=+1.6\times10^{-19}C$ } at a distance $r=4.8\times10^{-15}m$
so potential energy of the system will be
$V=-k\frac{|q_{Cu}|\times |q_{p}|}{r}$
putting
$r=4.8\times10^{-15}m$
$q_{p}=+1.6\times10^{-19}C$
$q_{Cu}=+4.64\times10^{-18}C$
$k=8.99\times10^{9}N.m^2/C^2$
$V=-(8.99\times10^{9}N.m^2/C^2)\frac{4.64\times10^{-18}C\times1.6\times10^{-19}C}{4.8\times10^{-15}m}$
$V=-13.90\times10^{-13}J$
$1.6\times10^{-19} J$ is equal to $1eV$
$1J$ is equal to $\frac{1}{1.6\times10^{-19}}eV$
so $13.90\times10^{-13} J$ is equal to $\frac{13.90\times10^{-13}}{1.6\times10^{-19}}eV$= $8.6875\times10^{6}eV$
so $V=-13.90\times10^{-13}J=-8.6875\times10^{6}eV=-8.6875MeV$
so 8.6875MeV work is done by electric force
