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)

Inversely proportional to $n^{3}$

Answer 2

$\omega = \frac{v}{r}$ Further $v \propto \frac{1}{n}$ and $r \propto n^{2}$, hence

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