Question

The distance of closest approach for an alpha nucleus of velocity $v$ bombarding a stationary heavy nucleus target of charge $Ze$ is directly proportional to

• A. $v$
• B. $m$
• C. $\frac{1}{v^{2}}$
• D. $\frac{1}{Ze}$

Answer 1

Answer: (C)

$\frac{1}{v^{2}}$

Answer 2

At the distance of closest approach $(d)$,
Kinetic energy of $\alpha$-particle
= Potential energy of $\alpha$- particle and target nucleus
$\therefore \frac{1}{2}mv^{2}=\frac{1}{4\pi\varepsilon_{0}} \frac{\left(2e\right)\left(ze\right)}{d}$
or $d=\frac{1}{4\pi\varepsilon_{0}} \frac{4Ze^{2}}{mv^{2}}$
$\therefore d \propto\frac{1}{v^{2}}$