Question

The primary decay mode for the negative pion is \[{\pi ^ - } \to {\mu ^ - } + {\bar \nu _\mu }\]. What is the energy release in MeV in this decay?

Short answer

The released energy in the decay process is \[33.9\;{\rm{MeV}}\].

Step-by-step solution

Definition of Energy

If we subtract the rest mass energy of a particle from the total energy of the particle, we will get the kinetic energy of the particle. During the decay process of a particle, some amount of mass is converted into energy and is released as the product of the reaction.

Finding required energy

The primary decay mode for the negative pio is \[{\pi ^ - } \to {\mu ^ - } + {\bar \nu _\mu }\]. The energy release\(\left( {\Delta E} \right)\) in MeV in this decay is

\[\begin{array}{c}\Delta E = {E_\pi } - {E_\mu } - {E_{\bar \nu }}\\ = 139.6\;{\rm{MeV}} - 105.7\;{\rm{MeV}} - 0\\ = 33.9\;{\rm{MeV}}\end{array}\]

The release energy is 33.9 MeV.