A free, stationary $Ir^{191}$ nucleus with an excitation energy $E$ transitions to the ground state

Question

A free, stationary $Ir^{191}$ nucleus with an excitation energy $E$ transitions to the ground state by emitting a gamma quantum. If $m$ is the rest mass of the nucleus and $c$ is the speed of light, what is the fractional change in the energy of the gamma quantum due to the recoil of the nucleus? (Assume the recoil velocity is non-relativistic).

Professor PC Thomas Classes · Answer

When the photon is emitted, the nucleus recoils. By conservation of momentum:
$p_{	ext{nucleus}} = p_{\gamma} \approx \frac{E}{c}$
The recoil kinetic energy of the nucleus is $K = \frac{p_{	ext{nucleus}}^2}{2m} = \frac{E^2}{2mc^2}$.
By conservation of energy, the actual energy of the emitted gamma photon $E'$ is slightly less than the transition energy $E$:
$E' = E - K = E - \frac{E^2}{2mc^2}$.
The absolute change in energy is $\Delta E = K = \frac{E^2}{2mc^2}$.
The fractional change is $\frac{\Delta E}{E} = \frac{E}{2mc^2}$.