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Asked by Piyush Last Modified
Answer 
Shivam Sharma
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 (E1) of radiation at level n as:
Now, the relation for energy (E2) of radiation at level (n − 1) is givenas: 
Energy (E) released as a result of de-excitation:
E = E2−E1
hν = E2 − E1 … (iii)
Where,
ν = Frequency of radiation emitted
Putting values from equations (i) and (ii) in equation (iii), we get:
For large n, we can write
Classical relation of frequency of revolution of an electron is given as: 
Where,
Velocity of the electron in the nth orbit is given as:
v =
And, radius of the nth orbit is given as:
r = 
Putting the values of equations (vi) and (vii) in equation (v), we get:
Hence, the frequency of radiation emitted by the hydrogen atom is equal to its classical orbital frequency.
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