Question

If an $\alpha$ -particle of mass m, charge q and velocity v is incident on a nucleus of charge Q and mass m, then the distance of closest approach is

• A. $\frac{Qq}{4\pi {\epsilon }_0{m}^2{v}^2}$
• B. $\frac{Qq}{2\pi {\epsilon }_{0}m{v}^2}$
• C. $\frac{Qqm{v}^2}{2}$
• D. $\frac{Qq}{m{v}^2}$

Answer 1

Answer: (B)

The minimum distance from the nucleus up to which the $\alpha -$ particle approach, is called the distance of closest approach fro). At this distance the entire initial kinetic energy is converted into potential energy. So, $\frac{1}{2}m{v}^2=\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{r}$ $\Rightarrow$ $r=\frac{qQ}{2\pi {\epsilon }_{0}m{v}^2}$ None of the given option is correct.