Question

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 distance of approach of the incident particle

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

Suppose distance of closest approach is r, and according to the principle of conservation of energy,
Energy at the time of projection = Energy at the distance of closest approach
$\Rightarrow \frac{1}{2}m{v}^2=\frac{1}{4\pi {ϵ}_0}.\frac{\left(Ze\right).e}{r}\Rightarrow r=\frac{Z{e}^2}{2\pi {ϵ}_{0}m{v}^2}$