$4$ Formula $3$ points

Alpha particles $($a$)$ were shot toward a gold foil. Gold has an atomic number of $79.$ An a had initial kinetic energy of about $6\mathrm{.}9$ MeV $.$ When an a was closest to a gold nucleus, it stopped completely by the electric repulsive force between the two nuclei. Assume all the a kinetic energy was converted to electric potential energy The closest distance $($center to center$)$ between the two nuclei is ${10}^{-15}$ meter.