Question

The angular speed of the electron in n^th orbit of Bohr's hydrogen atom is proportional to –

• A. n
• B. $\frac{1}{\sqrt{n}}$
• C. $\frac{1}{n^{2}}$
• D. $\frac{1}{n^{3}}$

Answer 1

Answer: (D)

$\frac{1}{n^{3}}$

Answer 2

w = $\frac{v}{r}$

v µ $\frac{1}{n}$ and r µ n²

hence w µ $\frac{1}{n^{3}}$