Question

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.

Answer 1

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 n^th orbit is given as:

v =

And, radius of the n^th 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.