Q. An alpha nucleus of energy $\frac{1}{2}mv^{2}$ bombards a heavy nuclear target of charge $Ze$. Then the distance of closest approach for the alpha nucleus will be proportional to
A) $1/m$
B) $1/v^{4}$
C) $1/Ze$
D) $v^{2}$
Explanation
(a) Distance of closest approach $r_{0} = \frac{Ze^{2}}{2mv^{2}\pi\varepsilon_{0}}$
$\Rightarrow r_{0} \propto \frac{1}{m}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.