Question

When an $\alpha$-particle of mass ${}^{\prime }{m}^{\prime }$ moving with velocity ${}^{\prime }{v}^{\prime }$ bombards on a heavy nucleus of charge ${}^{\prime }Z{e}^{\prime }$, its distance of closest approach from the nucleus depends on $m$ as:

• A. $\frac{1}{m}$
• B. $\frac{1}{\sqrt{m}}$
• C. $\frac{1}{m^2}$
• D. $m$

Answer 1

The correct option is A $\frac{1}{m}$

At closest distance of approach, the kinetic energy of the particle will completely convert into potential energy.

$\Rightarrow \frac{1}{2}m{v}^2=\frac{KQq}{d}$

$\Rightarrow d\propto \frac{1}{m}$