Question

If ${\in }_0$ is the permittivity of vacuum and $r$ is the radius of the orbit of H – atom in which the electron is revolving, then the velocity of the electron is given by:

• A. $v=\frac{e}{\sqrt{4\pi {\in }_{0}rm}}$
• B. $v=e\times \sqrt{4\pi {\in }_{r}m}$
• C. $v=\frac{\sqrt{4\pi {\in }_{0}rm}}{e}$
• D. $v=\frac{\sqrt{4\pi {\in }_{0}rm}}{e^2}$

Answer 1

The correct option is A $v=\frac{e}{\sqrt{4\pi {\in }_{0}rm}}$

For an electron that is present in an H-atom, the centripetal force and the centrifugal force are equal. The centripetal force exists due to the coulombic attraction between the nucleus and electron. Hence, equating the centrifugal and coulombic forces we get
$\frac{m{v}^2}{r}=\frac{e^2}{4\pi {\in }_0{r}^2}$
$\Rightarrow v=\frac{e}{\sqrt{4\pi {\in }_{0}rm}}$