Question

The angular speed of the electron in the $n^{\text{th }}$ orbit of Bohr's hydrogen atom is

• A. directly proportional to $n$
• B. inversely proportional to $\sqrt{n}$
• C. inversely proportional to $n^2$
• D. inversely proportional to $n^3$

Answer 1

Answer: (D)

$\omega =\frac{v}{r}.$ Further $v\propto \frac{1}{n}$ and $r\propto n^2$,
hence $\omega \propto \left(1/n^3\right)$.