Question

A fundamental particle of mass $m$ and charge $+e$ is projected, towards a heavy particle of charge $Ze$ (when $Z>0$) with velocity $v$, distance of nearest approach is:-

• A. $\frac{Z{e}^2}{2\pi {\epsilon }_{0}m{v}^2}$
• B. $\frac{Ze}{4\pi {\epsilon }_{0}m{v}^2}$
• C. $\frac{Z{e}^2}{8\pi {\epsilon }_{0}m{v}^2}$
• D. $\frac{Ze}{8\pi {\epsilon }_{0}m{v}^2}$

Answer 1

The correct option is A $\frac{Z{e}^2}{2\pi {\epsilon }_{0}m{v}^2}$

The potential energy of a particle of mass $m$, charge $e$ and particle of charge $Ze$ is,

$U=\frac{1}{4\pi \epsilon }\frac{Z{e}^2}{r}\{U=\frac{K{q}_1{q}_2}{r}\}$

During the process, when the particle of charge $e$, approaches charge $Ze$, kinetic energy is converted into potential energy.

$\Rightarrow \frac{1}{2}m{v}^2=\frac{1}{4\pi {\epsilon }_0}\frac{Z{e}^2}{r}$

Distance of closest approach is,

$r=\frac{Z{e}^2}{2\pi {\epsilon }_{0}m{v}^2}$