Question
Consider an excited hydrogen atom in state n moving with a velocity u(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

Velocity of hydrogen atom in state ‘n’ = u
Also the velocity of photon = u
But u << C
Here the photon is emitted as a wave.
So its velocity is same as that of hydrogen atom i.e. u.
$\therefore$ Accounding to Doppler’s effect,
Frequency $\text{v}=\text{v}_0\bigg(\frac{1+\frac{\text{u}}{\text{c}}}{1-\frac{\text{u}}{\text{c}}}\bigg)$
as u <<< C
$1-\frac{\text{u}}{\text{c}}=\text{q}$
$\therefore\ \text{v}=\text{v}_0\bigg(\frac{1+\frac{\text{u}}{\text{c}}}{1}\bigg)=\text{v}_0\Big(1+\frac{\text{u}}{\text{c}}\Big)$
$\text{v}=\text{v}_0\Big(1+\frac{\text{u}}{\text{c}}\Big)$

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