Question

For an electron, revolving in $n^{\text{th}}$ orbit of a hydrogen atom, the angular momentum is proportional to,

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

Answer 1

The correct option is D $\frac{1}{n^3}$


According to the second postulate of Bohr's theory,
$m{v}_n{r}_n=\frac{nh}{2\pi }$
$\Rightarrow m{r}_n^2{\omega }_n=\frac{nh}{2\pi }$
$\Rightarrow {\omega }_n=\frac{nh}{2\pi m{r}_n^2}.....1$

$\because r_n=r_1{n}^2\text{for hydrogen atom}$

$\therefore {\omega }_n=\frac{nh}{2\pi m\left(r_1^2{n}^4\right)}=\frac{h}{2\pi m\left(r_1^2{n}^3\right)}$

From the above equation, we can write, ${\omega }_n\propto \frac{1}{n^3}$


Hence, $\left(D\right)$ is the correct answer.