An excited hydrogen atom in state n travels with a velocity v ( v c...

Let the mass of the hydrogen atom be $M$.

From the conservation of momentum:

$Mv = Mv' + \dfrac{h\nu}{c}$

where $v' is the velocity of the atom after emitting the photon.<br><br>$v' = v - \dfrac{h\nu}{Mc}$<br><br>From the conservation of energy:<br><br>$E_n + \dfrac{1}{2}Mv^2 = E_m + \dfrac{1}{2}Mv'^2 + h\nu$<br><br>$E_n - E_m = \dfrac{1}{2}Mv'^2 - \dfrac{1}{2}Mv^2 + h\nu$<br><br>The energy difference between the states is given by$E_n - E_m = h\nu_0$. Substituting this and the expression for$v'$:<br><br>$h\nu_0 = \dfrac{1}{2}M \left(v - \dfrac{h\nu}{Mc}\right)^2 - \dfrac{1}{2}Mv^2 + h\nu$<br><br>$h\nu_0 = \dfrac{1}{2}M \left(v^2 - \dfrac{2vh\nu}{Mc} + \dfrac{h^2\nu^2}{M^2c^2}\right) - \dfrac{1}{2}Mv^2 + h\nu$<br><br>$h\nu_0 = -\dfrac{vh\nu}{c} + \dfrac{h^2\nu^2}{2Mc^2} + h\nu$<br><br>Since the mass$M$of the atom is large, the recoil energy term$\dfrac{h^2\nu^2}{2Mc^2}$is negligibly small and can be ignored.<br><br>$h\nu_0 \approx h\nu \left($1 -$\dfrac{v}{c}\right)

\nu = \nu_0 \left(1 - \dfrac{v}{c}\right)^{-1}<br><br>Using the binomial expansion forv \ll c$:<br><br>$\nu $\approx \nu_0 \left(1 + \dfrac{v}{c}\right)$