Answer
See explanation.
Work Step by Step
a. From Table 44.3, the rest energy of the $\Sigma^0$ is 1193 MeV.
From Table 44.3, the rest mass of the $\Lambda^0$ is 1116 MeV.
Assuming the $\Lambda^0$ is produced at rest, the energy of the photon is the difference, 77 MeV.
b. $p=\frac{E}{c}=4.1\times10^{-20}kg\cdot m/s$.
If the $\Lambda^0$ had this much momentum, its KE would be $\frac{p^2}{2m}$, which is only about 2.8 MeV. This is much less than the 77 MeV of the photon’s energy, justifying the original assumption.
