A sigma0 particle with mass 1193 MeV/c^2 decays into a lambda0 particle with mass 1116 MeV/c^2 a photon. Find the energy and momentum of the photon, assuming that the kinetic energy of the lambda0 particle is negligible.

This is a standard A-level particle physics question. To solve it, we need to use three equations given in the formula sheet of the paper. To find the energy of the photon, we use Einsteins famous energy-mass equation, E=mc² to find the energy difference of the two particles, which is in turn the energy of the photon, since we take the kinetic energy of the resultant particle to be zero. The units are already in a convenient MeV/c² form, so the energies of the particles are 1193 MeV and 1116 MeV. The energy of the photon is just the difference between them, so 77 MeV. The next part is trickier. We need to use both the equation of the energy of the photon (E=hf) and the momentum of the photon (p=h/lambda) to find the momentum of the photon. We can plug our values in to the first one and then use the second one to find the momentum, or we can rearrange these to find that E=p/c. So our momentum of the photon is 77 Mev/c, which is already an accepted unit so there is no need to convert to SI units. I really like this example because it tests understanding rather than calculation skills. If you understand what you are doing, this example barely takes any time to solve since it doesn't require calculation.