Question

The ratio of the speed of the electron in the first Bohr's orbit of hydrogen atom and the speed of light is equal to (where $e,h$ and $c$ have their usual meaning)

• A. $\frac{e^2}{\pi h{\epsilon }_{0}c}$
• B. $\frac{e^2}{2\pi h{\epsilon }_{0}c}$
• C. $\frac{e^2}{h{\epsilon }_{0}c}$
• D. $\frac{e^2}{2h{\epsilon }_{0}c}$

Answer 1

The correct option is D $\frac{e^2}{2h{\epsilon }_{0}c}$

The orbital velocity of revolving electron in the $n^{th}$ orbit is,

$v_n=\frac{e^2}{2h{\epsilon }_0}\left(\frac{Z}{n}\right)$

For first orbit $n=1$ and for hydrogen, $Z=1$

$\therefore v_1=\frac{e^2}{2h{\epsilon }_0}$

$\Rightarrow \frac{v_1}{c}=\frac{e^2}{2h{\epsilon }_{0}c}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.

Why this question ? Key concept is: The orbital velocity of revolving electron in the $n^{th}$ orbit is, $v_n=\frac{e^2}{2h{\epsilon }_0}\left(\frac{Z}{n}\right)$