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
0 = Permittivity of free space

Step 2: Find the energy in (n-1)th level.
Now, the relation for energy (E2) 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;
\(\nu=\frac{m e^{4}}{8 \epsilon_{0}^{2} {h}^3} [\frac{1}{(n-1)^{2}}-\frac{1}{n^{2}}]=\frac{m e^{4} (2n-1)}{8 \epsilon_{0}^{2} {h}^3 n^{2} (n-1)^{2}}\)
For large n, we can write (2n-1)2n and (n-1)n.
\(\therefore \nu=\frac{m e^{4}}{4 \in_{0}^{2} h^3 n^{3}}\)                         . . . . .  . . . . (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 nth orbit is given as:
\(v=\frac{e^{2}}{4 \pi \epsilon_{0} \frac{h}{2 \pi} n}\)  . . .  . . . .  . . . . (v)
And, radius of the nth orbit is given as
r=4π0h2π2me2n2                             ...(vi)
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.