Relativistic energy of emitted photon from decay

Question

I've been stuck on a problem for a while now and need a bit of help. The question states that we have a nucleus in an excited state of mass $M_*$ which decays to the ground state by photon emission, where the ground state has mass $M$.

We are asked to show through conservation that the emitted photon has an energy:

$$ E^{em}_\gamma = \frac{(M_*^2-M^2)c^2}{2M_*}. $$

I have used the fact that the energy of a photon is $E=pc$ and also momentum conservation, $p_* = p + p_\gamma$ (not sure if this is valid) and get close to the right answer, but struggle to express it only in terms of the mass; I get a momentum factor in the denominator and slightly off with some other stuff.

I have also used the fact that $E_{before} = E_{after}$ which I have then using the relativistic energies that, $E_*^2 = E_\gamma^2 + E^2$. I feel like this is not right, and that it should be $E_*^2 = (E_\gamma + E)^2$, but using this I am much further away from the answer.

Sorry for the ramblings, I think my grasp on what to do for this problem is not that strong so I may have made some wrong assumptions.

Thanks

Answer 1

The initial excited nucleus is at rest. So its energy ($E_*$) is just the rest energy $M_*c^2$, and its momentum ($p_*$) is zero.
The final nucleus in ground state is not at rest because it recoiled. So its energy ($E$) is not $Mc^2$ but somewhat larger, and its momentum ($p$) is not zero.
The emitted photon has energy $E_\gamma$ and momentum $\frac{E_\gamma}{c}$.

Using all these pieces, conservation of energy gives $$M_*c^2 = E + E_\gamma \tag{1}$$ and conservation of momentum gives $$0 = p + \frac{E_\gamma}{c} \tag{2}$$

Furthermore, energy and momentum of the final nucleus are connected by the relativistic energy-momentum relation $$E^2 = (pc)^2 + (Mc^2)^2 \tag{3}$$

Now you have three equations (1, 2, 3) for three unknowns ($E_\gamma$, $E$, $p$). You should be able to eliminate $E$ and $p$ from these and come up with a formula for $E_\gamma$.