Question

47. A pion at rest (m_p = 273m_e) decays to a muon (m_m = 207m_e) and an antineutrino (m~0). Find (a) the kinetic energy of the muon
and (b) the energy of the antineutrino in electron volts.

Answer 1

I'm going to work in a system of units such that c = h-bar = 1
This means all masses, momenta and energies have to be in eV
The Q-value of the decay is:
pion mass - muon mass = 66m_e
This is all the available kinetic energy in the system of decay products, so to conserve energy
KE_muon + KE_neutrino = 66m_e
Conservation of momentum
|p_muon| = |p_neutrino| = p
Since the neutrino is being treated as massless
E_neutrino = KE_neutrino = p_neutrino = p
The kinetic energy of the muon is the total energy minus the rest energy
\(\sqrt{p^2+(207m_e)^2}-207m_e + p = 66m_e\)

I'll leave rearranging this as an exercise for you to do.
You can check your answer against this:
http://www.wolframalpha.com/input/?i=Rearrange+sqrt {p^2%2B%28m%29^2}+-+m+%2B+p+%3D+Q+for+p

Answer 2

Since the URL didn't parse, here's a shortened one: http://tinyurl.com/ntpptbu