Physics - Modern Physics Question with Solution | TestHub

Moment of Inertia $\mathrm{I}={\mathrm{MR}}^2$ Radius of the ${\mathrm{n}}^{\text{th }}$ Bohr orbit is, $r_n=\frac{{\epsilon }_0{h}^2{n}^2}{\pi {\mathrm{me}}^2}$ For $\mathrm{n}=2$, $r_2=\frac{4{\epsilon }_0{h}^2}{\pi m{e}^2}$ $\therefore$ Moment of inertia of the electron in the $2^{\text{nd }}$ orbit is $\begin{array}{rl}\text{ M.I }& =\mathrm{m}\times {\left[\frac{4{\epsilon }_0{\mathrm{h}}^2}{\pi {\mathrm{me}}^2}\right]}^2\\ & =\mathrm{m}\times \frac{16{\epsilon }_0^2{\mathrm{h}}^4}{{\pi }^2{\mathrm{m}}^2{\mathrm{e}}^4}\\ & =\frac{16{\epsilon }_0^2{\mathrm{h}}^4}{{\pi }^2{\mathrm{me}}^4}\end{array}$