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Chapter 14

Q. 14.6

Examine the lepton conservation laws in the decay of the \pi^{-} and \mu^{-}.

Step-by-Step

Verified Solution

The decays are

 

\begin{aligned}&\pi^{-} \rightarrow \mu^{-}+\bar{\nu}_{\mu} \\&\mu^{-} \rightarrow e^{-}+\bar{\nu}_{e}+\nu_{\mu}\end{aligned}

 

The \pi^{-} first decays to \mu^{-}, which in turn decays into e^{-}. However, we must also have neutrinos in order to conserve lepton number. The first reaction has L_{\mu}=0 on the left and L_{\mu}=+1-1=0 on the right. For the second reaction, we must conserve both L_{e} \text { and } L_{\mu}. On the left side we have L_{e}=0 \text { and } L_{\mu}=1. and on the right side we have L_{e}=1-1+0=0 and L_{\mu}=0+0+1=1 \text {, so both } L_{e} \text { and } L_{\mu} are conserved.