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) $v^{2}$
B) 1/m
C) $1/v^{4}$
D) 1/$Ze$
Explanation
(b) At distance of closest approach relative velocity of two particles is $v$. Here target is considered as stationary, so $\alpha$-particle comes to rest instantaneously at distance of closest approach. Let required distance is $r,$ then from work energy-theorem.
$0 - \frac{{mv}^{2}}{2} = - \frac{1}{{4\pi\varepsilon}_{0}}\frac{Z_{e} \times Z_{e}}{r}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r \propto \frac{1}{m}$
$\propto \frac{1}{v^{2}}$
$\propto {Ze}^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.