这是一份uwa西澳大学PHYS3100的成功案例

$$
\begin{aligned}
M_{o} c^{2} &=E^{\prime}+Q \
0 &=p^{\prime}-Q / c,
\end{aligned}
$$
where $E^{\prime}$ and $p^{\prime}$ label the relativistic energy and momentum of the final (unexcited) atom of mass $M_{a}^{\prime}$. Let us say that we detect the photon and measure its energy $Q$. So, we need to eliminate $E^{\prime}$ and $p^{\prime}$ . A slick way to do this is to use the energy-momentum relation $E^{\prime 2}=p^{2} c^{2}+M_{o}^{2} c^{4}$. Rearranging gives
$$
\begin{gathered}
E^{\prime}=M_{o} c^{2}-Q \
p^{\prime}=Q / c
\end{gathered}
$$
$\mathrm{SO}$,
$$
E^{\prime 2}-p^{\prime 2} c^{2}=\left(M_{o} c^{2}-Q\right)^{2}-Q^{2}=M_{o}^{\prime 2} c^{4}
$$
Finally,
$$
M_{o}^{2} c^{4}-2 M_{o} c^{2} Q=M_{0}^{22} c^{4}
$$

PHYS3100 COURSE NOTES ：

Let us write this in terms of the energy difference between the initial and final atoms, taken at rest,
$$
\Delta E \equiv M_{0} c^{2}-M_{0}^{\prime} c^{2} .
$$
We focus on $\Delta E$ because this would be the energy difference between discrete quantum energy levels of the atom. One could calculate such a difference from first principles in quantum mechanics. Solving for $M_{a}^{\prime} c^{2}$,
$$
M_{0} c^{2}-\Delta E=M_{0}^{\prime} c^{2}
$$
Squaring gives
$$
M_{0}^{2} c^{4}-2 M_{0} c^{2} \Delta E+(\Delta E)^{2}=M_{0}^{\prime 2} c^{4} .
$$
Combining this with Eq. (6.41) gives
$$
Q=\Delta E\left(1-\frac{\Delta E}{2 M_{0} c^{2}}\right)
$$