Consider an excited hydrogen atom in state `n` moving with a velocity `v(v lt lt 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 principle to calculate the frequency v of the emitted radiation, compare this with the frequency `v_(0)` emitted if the atom were at rest.

Explanation

KEY IDEA
Normally, ignored the recoil of the atom that is limiting the photon. but here the atom that is limiting the photon will also change its momentum due to recoil.
Calculation : Let us assume that the atom emits a photon of frequency and has a velocity of `v_(0)` after the emission
By the law of conservation of energy. we have
`(1)/(2)mv^(2)+E_(n)=(1)/(2)mv_(0)^(2)+E_(m)+hv.`
By the law of conservation of momentum, we have
`mv=mv_(0)+h(v)/(c)`.
In absence of recoil, the atom would have emitted a photon of requency `v_(0)`.
`E_(n)-E_(m)=hv_(0)`.
So, we have the equations as
`(1)/(2)m(v^(2)-v_(0)^(2))=h(v_(0)-v)`
and `m(v-v_(0))=h(v)/(c)`
By dividing the two equations, we get
`(v+v_(0))/(2c)=(v_(0)-v)/(v)`.
Since v and `v_(0) lt ltc`, we get
`v~~v_(0)(1+(v)/(c))`
Learn : If you notice, there is a striking between this expression and the expression for the frequency of the wave emitted by a moving source. There we get the formula for the frequency as
`v=v_(0)(1-(v)/(c))^(-1)`
By applying binomial approximation, we get
`v~~v_(0)(1+(v)c)`