An electron $($charge $$e$=1\mathrm{.}601019$C$$e$=1\mathrm{.}601019$C$)$ is at rest at a distance $1\mathrm{.}001010$m$1\mathrm{.}001010$m from the center of a nucleus of lead $($charge $+82$e$=+1\mathrm{.}311017$C$+82$e$=+1\mathrm{.}311017$C$).$ In a nuclear physics experiment, a proton $($mass $1\mathrm{.}671027$kg$1\mathrm{.}671027$kg$,$ charge $+$e$=+1\mathrm{.}601019$C$+$e$=+1\mathrm{.}601019$C$)$ is fired directly at a target nucleus of unknown charge. $($You can treat both objects as point charges, and assume that the nucleus remains at rest.$)$ When it is far from its target, the proton has speed $2\mathrm{.}50106$m$/$s$2\mathrm{.}50106$m$/$s$.$ The proton comes momentarily to rest at a distance $5\mathrm{.}291013$m$5\mathrm{.}291013$m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are $5\mathrm{.}291013$m$5\mathrm{.}291013$m apart?