Question

An elementary particle of mass $m$ and charge $+e$ is projected with velocity $v$ towards a much more massive particle of charge $Ze$, where $Z>0$. What is the closest possible distance of approach of the incident particle?

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

Answer 1

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

$\frac{1}{2}m{V}^2=\frac{1}{4\pi {ϵ}_0}\frac{\left(Ze\right).e}{r}$

$R=\frac{Z{e}^2}{2\pi {ϵ}_{0}m{v}^2}$