What is the distance of closest approach(in x10^-14 m) when a 5.0 MeV proton approaches a gold nucleus?

At the distance r₀ of closest approach,
KE of a proton = PE of proton and the gold nucleus

$K=\frac{1}{2}m{v}^2=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Ze\cdot e}{r_0}\left[q_1=Ze,q_2=e\right]$

$\begin{array}{rcl}or r_0& =& \frac{1}{4\pi {\epsilon }_0}\cdot \frac{Z{e}^2}{K}\\ But K& =& 5.0\mathrm{MeV}=5.0\times 1.6\times {10}^{-13}\mathrm{J}\end{array}$

$\text{ For gold, }Z=79$

$\begin{array}{rcl}\therefore r_0& =& \frac{9\times {10}^9\times 79\times {\left(1.6\times {10}^{-19}\right)}^2}{5.0\times 1.6\times {10}^{-13}}\\ & =& 2.28\times {10}^{-14}\mathrm{m}\simeq 2.3\times {10}^{-14}\mathrm{m}\end{array}$