Question

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

Answer 1

At the distance $r_{0}$ of closest approach,
$K E$ of a proton $=P E$ of proton and the gold nucleus
$K=\frac{1}{2} m v^{2}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Z e \cdot e}{r_{0}}\left[q_{1}=Z e, q_{2}=e\right]$
Or $r_{0}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Z e^{2}}{K}$
But $K=5.0 MeV $
$=5.0 \times 1.6 \times 10^{-3} J$
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}} $
$=2.28 \times 10^{-14} m \simeq 2.3 \times 10^{-14} m$