Q. An elementary particle of mass $m$ and charge $+ e$ is projected with velocity $v$ at a much more massive particle of charge $Ze,$ where $Z > 0.$ What is the closest possible approach of the incident particle
A) $\frac{Ze^{2}}{2\pi\varepsilon_{0}mv^{2}}$
B) $\frac{Ze}{4\pi\varepsilon_{0}mv^{2}}$
C) $\frac{Ze^{2}}{8\pi\varepsilon_{0}mv^{2}}$
D) $\frac{Ze}{8\pi\varepsilon_{0}mv^{2}}$
Explanation
(a) Suppose distance of closest approach is $r,$ and according to energy conservation applied for elementary charge.
Energy at the time of projection = Energy at the distance of closest approach
$\Rightarrow \frac{1}{2}mv^{2} = \frac{1}{4\pi\varepsilon_{0}}.\frac{(Ze).e}{r} \Rightarrow r = \frac{Ze^{2}}{2\pi\varepsilon_{0}mv^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.