Question

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

Answer 1

We know $mvr=\frac{nh}{2\pi }$

$\Rightarrow m{r}^2\omega =\frac{nh}{2\pi }$

$\Rightarrow \omega =\frac{nh}{2\pi \times m\times r^2}$

= $\frac{1\times 6.63\times {10}^{-34}}{2\times 3.14\times 9.1\times {10}^{31}\times (053{)}^2\times {10}^{-20}}$

= $0.413\times {10}^{17}rad/s=4.13\times {10}^{17}rad/s$