Question

Find the maximum angular speed of the electron of a hydrogen atom in a stationary orbit.

Answer 1

Let the mass of the electron be m.
Let the radius of the hydrogen's first stationary orbit be r.
Let the linear speed and the angular speed of the electron be v and ω, respectively.

According to the Bohr's theory, angular momentum (L) of the electron is an integral multiple of h/2$\mathrm{\pi }$, where h is the Planck's constant.
$\Rightarrow mvr=\frac{nh}{2\mathrm{\pi }}$ (Here, n is an integer.)
$$\begin{gathered} v=r\omega \\ \Rightarrow m{r}^2\mathrm{\omega }=\frac{nh}{2\mathrm{\pi }} \\ \Rightarrow \omega =\frac{nh}{2\mathrm{\pi }\times m\times r^2} \end{gathered}$$

$$\begin{gathered} \therefore \omega =\frac{1\times \left(6.63\times {10}^{-34}\right)}{2\times \left(3.14\right)\times \left(9.1093\times {10}^{-31}\right)\times {\left(0.53\times {10}^{-10}\right)}^2} \\ =0.413\times {10}^{17}\mathrm{rad}/\mathrm{s}=4.13\times {10}^{16}\mathrm{rad}/\mathrm{s} \end{gathered}$$