Question

A hydrogen atom initially in its ground state i.e., n= 1 level, absorbs a photon and ends up in n= 3 level. What must have been the frequency of the incident photon? (b) Now the electron makes spontaneous emission and comes back to the ground state. What are the possible frequencies of the photons emitted during this process?

Answer 1

For absorption of photon, the incident photon must have the energy equals to the energy difference between the two atomic states.

The energy difference between two atomic states is given by

  

We have , n₁ = 1 and n₂ = 3

So, .

NE -13.6 × | 32-12| = 12.09 eV

or,   

We know that,

  

Where, h = 6.63 x 10^-34 J-s is Planck's constant

is frequency

So, frequency of incident photon is

(b)  Now if the electron makes spontaneous emission it can go to ground state via two methods.

First : it can directly go to the ground state and

Second : the electron can jump to the n = 2 state and then from n= 2 to n = 1 state.

In first method :

In this method the electron will release energy equals to t he 12.09 eV (because this much energy electron had taken when it jumps from ground state to n = 3 state). And the frequency of the radiation (photon) will be 2.92 x 10¹⁵ Hz as calculated above.

In second method :

In this method photons of two different will get emitted from the transitions n =3 to n = 2 and n = 2 to n = 1 .

For n= 3 to n = 2

  

For n = 2 to n = 1

  

For any doubt please comment .