Question

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

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

Answer 1

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

$\frac{m{v}^2}{r}=\frac{KZ{e}^2}{r^2}$
For $H:Z=1$
$\frac{m{v}^2}{r}=\frac{K{e}^2}{r^2}(K=\frac{1}{4\pi {\epsilon }_0})$
$\frac{m{v}^2}{r}=\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{r^2}$
$v^2=\frac{e^2}{4\pi {\epsilon }_{0}rm}$
$v=\frac{e}{\sqrt{4\pi {\epsilon }_{0}rm}}$