A proton of energy 1.6 MeV approaches a gold nucleus ( Z =79 ). Find the distance of its closest approach.

$\displaystyle d_{0}=\frac{k Z e^{2}}{K_{p}}$ $\displaystyle =\frac{9 \times 10^{9} \times 79 \times\left(1.6 \times 10^{-19}\right)^{2}}{1.6 \times 1.6 \times 10^{-19} \times 10^{6}}$ $\displaystyle =711 \times 10^{-16} \mathrm{~m}$ $\displaystyle =7.11 \times 10^{-14} \mathrm{~m}$