Question

The primary decay mode for the negative pion is $\pi^- \rightarrow \mu^- + \overline{\nu_\mu}$. What is the energy release in MeV in this decay?

$33.9\textrm{ MeV}$

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This is College Physics Answers with Shaun Dychko. The negative pion decays into a muon and the muon anti-neutrino and the question is how much energy is released in this decay? So the energy released will be the difference in mass between the reactant and the products times

*c squared*. So that is the mass of the negative pion minus the mass of the muon times*c squared*; the neutrino is massless. So we look up in table [33.2] for the masses here and the mass of the pi negative meson is 139.6 megaelectron volts per*c squared*and the mass of the muon is 105.7 megaelectron volts per*c squared*and so we write those numbers here multiplied by*c squared*and we get 33.9 megaelectron volts of energy released.