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
= Permittivity of free space
Step 2: Find the energy in (n-1)th level.
Now, the relation for energy () 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 orbit is given as:
\(v=\frac{e^{2}}{4 \pi \epsilon_{0} \frac{h}{2 \pi} n}\) . . . . . . . . . . . (v)
And, radius of the 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.
