Answer
$\alpha$ particle can come close to a distance of $d=7.2725\times10^{-14}m$
Work Step by Step
$\alpha$ particle carries a charge $q_{\alpha}=+2e=+2\times1.6\times10^{-19}C$
$q_{\alpha}=3.2\times10^{-19}C$
gold nucleus carries $Z=79$ protons
charge on $1$ proton $=+e$
so charge on gold nucleus $q_{Au}=+79e=+79\times1.6\times10^{-19}C$
$q_{Au}=+1.264\times10^{-17}C$
Kinetic energy of $\alpha$ particle $(KE)_{\alpha}=5.0\times10^{-13}J$
this kinetic energy is completely converted to potential energy of $\alpha$ particle and gold nucleus system once $\alpha$ comes to halt in close vicinity of gold nucleus.
suppose closest distance between $\alpha$ particle and gold nucleus is $d$
so potential energy of the system will be
$V=-k\frac{|q_{\alpha}|\times |q_{Au}|}{d}$
when $\alpha$ particle comes to halt numerical value of kinetic energy and potential energy are equal
so $(KE)_{\alpha}=k\frac{|q_{\alpha}|\times |q_{Au}|}{d}$
$d=k\frac{|q_{\alpha}|\times |q_{Au}|}{(KE)_{\alpha}}$
putting
$q_{Au}=+1.264\times10^{-17}C$
$q_{\alpha}=3.2\times10^{-19}C$
$(KE)_{\alpha}=5.0\times10^{-13}J$
$k=8.99\times10^{9}N.m^2/C^2$
$d=(8.99\times10^{9}N.m^2/C^2)\frac{3.2\times10^{-19}C\times 1.264\times10^{-17}C}{5.0\times10^{-13}J}$
$d=7.2725\times10^{-14}m$
