Question

The angular speed of the electron in the $n^{th}$ Bohr orbit of the hydrogen at om is proportional to

• A. $n$
• B. $n^3$
• C. $\frac{1}{n}$
• D. $\frac{1}{n^3}$

Answer 1

The correct option is D $\frac{1}{n^3}$

We know that $mvr=\frac{nh}{2\pi }$...... (1)

$v=\omega r$ ........................................... (2)

$r=0.526n^2$ ............................................... (3)

From above three equation we can see angular speed is proportional to $\frac{1}{n^3}$.