How much energy is released when a ${\mu }^{-}$muon at rest decays into an electron and two neutrinos? Neglect the small masses of the neutrinos.

A${\mu }^{-}$particle decays into an electron plus a muon neutrino and an electron antineutrino. They have relatively long lifetimes because their decays are mediated by weak interaction. Despite their zero charges, a neutrino is distinct from an antineutrino; the spin angular momentum of a neutrino has a component that is opposite to its linear momentum, while for an antineutrino, that component is parallel to its linear momentum.

${\mu }^{-}$is initially at rest, so all of its energy is in the form of its rest energy.Once the decay begins, energy has to be conserved.So, any energy over the amount required for the rest energies of electrons will contribute to the kinetic energy of particles.

• The rest energy ofthe particle is$105.7MeV$.
• The rest energy ofthe particle is$0.511MeV$.

Thus, the total energy released is,

$$\begin{gathered} E_{r}eleased=E_{(}rest,{\mu }^{-})-E_{(}rest,e^{-}) \\ {E}_{r}eleased=105.7-0.511 \\ {E}_{r}eleased=105.19\hspace{0.33em}MeV \end{gathered}$$

Therefore, Energy released is$105.19MeV$when${\mu }^{-}$at rest, decays into an electron and two neutrinos.