Thank you for reporting, we will resolve it shortly
Q.
A beam of beryllium nucleus $(z=4)$ of kinetic energy $5.3\, MeV$ is headed towards the nucleus of gold atom $(Z=79) .$ What is the distance of closest approach?
Atoms
Solution:
Distance of closest approach $d=\frac{1}{4 \pi \varepsilon_{0}} \frac{z e Z e}{K}$
where, $Z=$ atomic number of target nucleus
$z=$ atomic number of incident beam
$K=$ kinetic energy of incident beam
$e=$ electronic charge
Given : $z=4 ; Z=79 ; K=5.3 \,MeV$ and
$e=1.6 \times 10^{-19} \,C$
$d=\frac{9 \times 10^{9} \times 4 \times 1.6 \times 10^{-19} \times 79 \times 1.6 \times 10^{-19}}{5.3 \times 1.6 \times 10^{-19} \times 10^{6}}=8.58 \times 10^{-14} \,m$