Physics - Modern Physics Question with Solution | TestHub

According to the Bohr's atomic model.
$\text{Angular momentum, }L=mv=\frac{nh}{2\pi }....(i)$
since, angular velocity $\omega =\frac{v}{r}$
$\Rightarrow v=r\omega$
From Eq. (i), we get
$\begin{array}{rl}m(r\omega )r& =\frac{nh}{2\pi }\\ m\omega r^2& =\frac{nh}{2\pi }\\ \omega & =\frac{nh}{2\pi m{r}^2}....(ii)\end{array}$
since, the radius of the electron in $n^{th}$ orbit of
Bohr's atomic model is given as, $r=\frac{n^2{h}^2{\epsilon }_0}{\pi mz{e}^2}\dots$ (iii)
Squaring the Eq. (iii) and substituting its value in Eq. (ii), we get
$\omega =\frac{nh(\pi nze{)}^2}{2\pi m\left(n^4{h}^4{\epsilon }_0^2\right)}\Rightarrow \omega \propto \frac{1}{n^3}$