Physics - Modern Physics Question with Solution | TestHub

According to Bohr's quantization condition, the angular momentum of an electron in the $n$-th orbit is given by:

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

For the first Bohr orbit, $n=1$:

$mvr=\frac{h}{2\pi }$

The velocity of the electron is:

$v=\frac{h}{2\pi mr}$

The number of revolutions per second (frequency $f$) is given by the ratio of velocity to the circumference of the orbit:

$f=\frac{v}{2\pi r}$

Substituting the value of $v$ into the frequency equation:

$f=\frac{\frac{h}{2\pi mr}}{2\pi r}=\frac{h}{4{\pi }^{2}m{r}^2}$

Answer: $h/4{\pi }^{2}m{r}^2$