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

Frequency of the emitted radiation,  ${\displaystyle v=v_0\left(\frac{1+\frac{u}{c}}{1-\frac{u}{c}}\right)}$

since u <<< c ,

${\displaystyle v=v_0\left(\frac{1+\frac{u}{c}}{1}\right)}$

${\displaystyle v=v_0(1+\frac{u}{c})}$

Ratio of frequencies of the emitted radiation,

${\displaystyle \frac{v}{v_0}=(1+\frac{u}{e})}$