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 $\omega \propto\left(1 / n^{3}\right)$.