12.13 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.

Hint: The energy of the radiation is equal to the energy difference between  two levels of hydrogen atom.

Step 1: Find the energy in nth level.
It is given that a hydrogen atom de- excites from an upper level (n)
to a lower level (n-1).
We have the relation for energy (E) of radiation at level n as:
$$E_{1}=-\frac{m e^{4}}{8n^2 \epsilon_{0}^{2} {h}^2}$$            . . . . . . . . . . . . .(i)
where,
h = Planck's constant
m = Mass of the hydrogen atom
e = Charge on an electro
${\in }_{0}$ = Permittivity of free space

Step 2: Find the energy in (n-1)th level.
Now, the relation for energy (${\mathrm{E}}_{2}$) of radiation at level (n -1) is given as:
$$E_{2}=-\frac{m e^{4}}{8(n-1)^2 \epsilon_{0}^{2} {h}^2}$$          . . . . . . . . . . . . . .(ii)
Energy (E) released as a result of de-excitation:
$$E=E_{1}-E_{2}\\h\nu=E_{1}-E_{2}$$
where
$$\nu$$= Frequency of radiation emitted

Step 3: Find the frequency of the radiation.
Putting values from equations (i) and (ii) in equation (iii), we get;
$$\nu=\frac{m e^{4}}{8 \epsilon_{0}^{2} {h}^3} [\frac{1}{(n-1)^{2}}-\frac{1}{n^{2}}]=\frac{m e^{4} (2n-1)}{8 \epsilon_{0}^{2} {h}^3 n^{2} (n-1)^{2}}$$
$$\therefore \nu=\frac{m e^{4}}{4 \in_{0}^{2} h^3 n^{3}}$$                         . . . . .  . . . . (iii)

Step 4: Find the frequency of revolution of an electron in the orbit.
Classical relation of the frequency of revolution of an electron is given as:
$$\nu_{c}=\frac{v}{2 \pi r}$$                         . . . . . . . . . . . (iv)

Velocity of the electron in ${\mathrm{n}}^{\mathrm{th}}$ orbit is given as:
$$v=\frac{e^{2}}{4 \pi \epsilon_{0} \frac{h}{2 \pi} n}$$  . . .  . . . .  . . . . (v)
And, radius of the ${\mathrm{n}}^{\mathrm{th}}$ orbit is given as
Putting the values of equations (vi) and (vii) in equation (v), we get:

$$\nu_{c}=\frac{m e^{4}}{4 \epsilon_{0}^{2}{h^{3}} n^{3}}$$
Hence, the frequency of radiation emitted by the hydrogen atom is equal to its classical orbital frequency.