I came across a rather simple question asking the recoil speed of hydrogen atom when it releases photon while electron goes from n=2 to n=1.
The photon emitted will have energy released during the change in orbit. So where does energy for recoil speed obtained from conservation of momentum(mv=E/c) came from?
Does the atom release more energy other than the photon?
The difference ($E_2-E_1$)in electronic energy levels gives the energy that the photon would have if the atom were held stationary. If the atom is allowed to recoil, the photon energy will be $(E_2-E_1)$ minus the recoil energy.
But the recoil energy is only about $10^{-9}$ of the photon energy, so the reduction in photon energy is pretty negligible!
Since the atom's recoil momentum, $p$, is equal and opposite to the photon momentum, $h/\lambda$,
$$E_{k\ atom}=\frac{p^2}{2m_{atom}} = \frac{1}{2m_{atom}} {\left(\frac {h} {\lambda}\right)}^2=\frac{1}{2m_{atom}\ c^2} {\left(\frac {hc} {\lambda}\right)}^2=\frac {E_{phot}^2}{2m_{atom}\ c^2} $$ So $$\frac{E_{k\ atom}}{E_{phot}}=\frac {E_{phot}}{2m_{atom}\ c^2} \approx \frac {1.9\ \text{eV}}{2 \times 931\ \text{ MeV}}$$
It comes from the emitted photon.
For example, in the Mossbauer effect, this recoil does not take as much from the photon, because the mass of the entire crystal lattice reduces the recoil speed by 1/M and hence the energy transfer by 1/M.
https://en.wikipedia.org/wiki/M%C3%B6ssbauer_effect
Related question here: Question about transfer of momentum in photons
