Question

A beam of beryllium nucleus $(z=4)$ of kinetic energy $5.3MeV$ is headed towards the nucleus of gold atom $(Z=79).$ What is the distance of closest approach?

• A. $10.32\times 10^{-14}m$
• B. $8.58\times 10^{-14}m$
• C. $3.56\times 10^{-14}m$
• D. $1.25\times 10^{-14}m$

Answer 1

Answer: (B)

Distance of closest approach $d=\frac{1}{4\pi {\epsilon }_0}\frac{zeZe}{K}$
where, $Z=$ atomic number of target nucleus
$z=$ atomic number of incident beam
$K=$ kinetic energy of incident beam
$e=$ electronic charge
Given : $z=4;Z=79;K=5.3MeV$ and
$e=1.6\times 10^{-19}C$
$d=\frac{9\times 10^9\times 4\times 1.6\times 10^{-19}\times 79\times 1.6\times 10^{-19}}{5.3\times 1.6\times 10^{-19}\times 10^6}=8.58\times 10^{-14}m$