Question

An $\alpha -$ particle of energy 5 $MeV$ is scattered through $18$$0$ by a fixed uranium nucleus. The distance of the closest approach is of the order of

• A. $1$
• B. ${10}^{-10}$ cm
• C. ${10}^{-12}$ cm
• D. ${10}^{-15}$ cm

Answer 1

The correct option is C ${10}^{-12}$ cm

On closest approach, the kinetic energy would be equal to the potential energy between the particle and uranium nucleus.

$⟹KE=\frac{1}{4\pi {ϵ}_0}\frac{2Z{e}^2}{r_0}$

Here, $Z=92,KE=5MeV=8\times {10}^{-13}$

Hence, $r_0\approx {10}^{-12}cm$