Physics - Modern Physics Question with Solution | TestHub

The radius of the $n$-th Bohr orbit of a hydrogen atom is given by $r=\frac{{\epsilon }_0{n}^2{h}^2}{\pi m{e}^2}$.

For the second Bohr orbit, $n=2$.

$r=\frac{4{\epsilon }_0{h}^2}{\pi m{e}^2}$

The moment of inertia of the electron is $I=m{r}^2$.

Substituting the value of $r$:

$I=m{\left(\frac{4{\epsilon }_0{h}^2}{\pi m{e}^2}\right)}^2$

$I=\frac{16{\epsilon }_0^2{h}^4}{{\pi }^{2}m{e}^4}$

Answer: $\frac{16{\epsilon }_0^2{h}^4}{{\pi }^{2}m{e}^4}$