Question

The angular speed of the electron in the $n^{th}$ 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
$\omega \propto \left(1/n^3\right)$