What is the moment of inertia of the electron moving in second Bohr orbit of hydrogen atom? [ $\mathrm{h}=$…
What is the moment of inertia of the electron moving in second Bohr orbit of hydrogen atom?
[ $\mathrm{h}=$ Planck's constant, $\mathrm{m}=$ mass of electron, $\varepsilon_0=$ permittivity of free space, $\mathrm{e}=$ charge on electron]
$\frac{4 \varepsilon_0^2 h^4}{\pi^2 m e^4}$
$\frac{8 m \varepsilon_0^2 h^4}{\pi^2 e^4}$
$\frac{16 \varepsilon_0^2 h^4}{\pi^2 m e^4}$
$\frac{\varepsilon_0^2 h^4}{16 \pi^2 m e^4}$
Solution
Moment of Inertia $\mathrm{I}=\mathrm{MR}^2$
Radius of the $\mathrm{n}^{\text {th }}$ Bohr orbit is,
$r_n=\frac{\varepsilon_0 h^2 n^2}{\pi \mathrm{me}^2}$
For $\mathrm{n}=2$,
$r_2=\frac{4 \varepsilon_0 h^2}{\pi m e^2}$
$\therefore \quad$ Moment of inertia of the electron in the $2^{\text {nd }}$ orbit is
$\begin{aligned}
\text { M.I } & =\mathrm{m} \times\left[\frac{4 \varepsilon_0 \mathrm{~h}^2}{\pi \mathrm{me}^2}\right]^2 \\
& =\mathrm{m} \times \frac{16 \varepsilon_0^2 \mathrm{~h}^4}{\pi^2 \mathrm{~m}^2 \mathrm{e}^4} \\
& =\frac{16 \varepsilon_0^2 \mathrm{~h}^4}{\pi^2 \mathrm{me}^4}
\end{aligned}$
.