Solution
= Solution
Let the photon energies be $E_1,E_2$ and their momentum angle be $\theta$. Conservation of <four-momentum> and the null photon momenta give
$$
M^2c^2=(p_1+p_2)^2
=2p_1\mathbin\cdot p_2
=\frac{2E_1E_2}{c^2}(1-\cos\theta).
$$
Using $1-\cos\theta=2\sin^2(\theta/2)$,
$$
\boxed{\sin\frac\theta2
=\frac{Mc^2}{2\sqrt{E_1E_2}}}.
$$
\b[Thus the opening angle depends only on the parent rest energy and the two photon energies.]