Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level $n$ to level $(n-1)$. For large $n$,show that this frequency equals the classical frequency of revolution of the electron in the orbit.

(N/A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{m e^4}{8 \epsilon_0^2 h^2 n^2}$.
When the atom de-excites from level $n$ to $(n-1)$,the energy of the emitted photon is $h \nu = E_n - E_{n-1}$.
$h \nu = \frac{m e^4}{8 \epsilon_0^2 h^2} \left( \frac{1}{(n-1)^2} - \frac{1}{n^2} \right)$.
$\nu = \frac{m e^4}{8 \epsilon_0^2 h^3} \left( \frac{n^2 - (n-1)^2}{n^2(n-1)^2} \right) = \frac{m e^4}{8 \epsilon_0^2 h^3} \left( \frac{2n-1}{n^2(n-1)^2} \right)$.
For large $n$,$2n-1 \approx 2n$ and $(n-1) \approx n$,so $\nu \approx \frac{m e^4}{8 \epsilon_0^2 h^3} \left( \frac{2n}{n^4} \right) = \frac{m e^4}{4 \epsilon_0^2 h^3 n^3}$.
The classical frequency of revolution $\nu_c$ is $\frac{v}{2 \pi r}$.
Using $v = \frac{e^2}{2 \epsilon_0 h n}$ and $r = \frac{\epsilon_0 h^2 n^2}{\pi m e^2}$,we get $\nu_c = \frac{v}{2 \pi r} = \frac{e^2 / (2 \epsilon_0 h n)}{2 \pi (\epsilon_0 h^2 n^2 / \pi m e^2)} = \frac{m e^4}{4 \epsilon_0^2 h^3 n^3}$.
Thus,for large $n$,the quantum frequency equals the classical frequency.