Consider an excited hydrogen atom in state *n* moving with a velocity υ(ν<<*c*). It emits a photon in the direction of its motion and changes its state to a lower state *m*. Apply momentum and energy conservation principles to calculate the frequency ν of the emitted radiation. Compare this with the frequency ν_{0} emitted if the atom were at rest.

#### Solution

Let the frequency emitted by the atom at rest be ν_{0}.

Let the velocity of hydrogen atom in state 'n' be u.

But u << c

Here, the velocity of the emitted photon must be u.

According to the Doppler's effect,

The frequency of the emitted radiation, ν is given by

Frequency of the emitted radiation, `v = v_0 ((1 + u/c)/(1 - u/c))`

since u <<< c ,

`v = v_0 ((1+u/c)/1)`

`v = v_0 (1 + u/c)`

Ratio of frequencies of the emitted radiation,

`v/v_0 = (1 + u/e)`