Question

An α particle having energy 10Mev collides with a nucleus of atomic mass 50 . Then distance of closest approach will be ?

Answer 1

$K=10MeV=10\times 10^6 \times 1.6\times 10^{-19} J=1.6\times 10^{-12} J$

$q_1=2e=2\times 1.6\times 10^{-19} C=3.2\times 10^{-19}C$ - charge on alpha particle

$q_2=50e=50\times 1.6\times 10^{-19} C=80\times 10^{-19}C$ - charge on nucleus

The kinetic energy of the alpha particle must be converted to electrical potential energy at the point of closest approach.

We have: $\frac{1}{2}mv^2=k\frac{q_1 q_2}{r}$

Distance of the closest approach is $r=k\frac{q_1 q_2 }{\frac{1}{2}mv^2}$

$r=9\times 10^9\times \frac{ 3.2\times 10^{-19}\times 80\times 10^{-19}}{1.6\times 10^{-12}}=1.44\times 10^{-14}\ m$

Answer: $1.44\times 10^{-14} \ m.$