Question

What is the distance of closest approach (in $\times 10^{-14}m$ ) when a $5.0MeV$ proton approaches a gold nucleus?

Answer 1

Answer: 2.3

At the distance $r_0$ of closest approach,
$KE$ of a proton $=PE$ of proton and the gold nucleus
$K=\frac{1}{2}m{v}^2=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Ze\cdot e}{r_0}\left[q_1=Ze,q_2=e\right]$
Or $r_0=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Z{e}^2}{K}$
But $K=5.0MeV$
$=5.0\times 1.6\times 10^{-3}J$
For gold, $Z=79$
$\therefore r_0=\frac{9\times 10^9\times 79\times {\left(1.6\times 10^{-19}\right)}^2}{5.0\times 1.6\times 10^{-13}}$
$=2.28\times 10^{-14}m\simeq 2.3\times 10^{-14}m$