Question

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 ν₀ emitted if the atom were at rest.

Answer 1

Let the frequency emitted by the atom at rest be ν₀.

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
$$\begin{gathered} \mathrm{Frequency}\mathrm{of}\mathrm{the}\mathrm{emitted}\mathrm{radiation},\nu ={\nu }_0\left(\frac{1+{\displaystyle \frac{u}{c}}}{1-{\displaystyle \frac{u}{c}}}\right) \\ \mathrm{Since}u<<<c, \\ \nu ={\nu }_0\left(\frac{1+{\displaystyle \frac{u}{c}}}{1}\right) \\ \nu ={\nu }_0\left(1+\frac{u}{c}\right) \end{gathered}$$
Ratio of frequencies of the emitted radiation, $\frac{\nu }{{\nu }_0}=\left(1+\frac{u}{c}\right)$