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Is the decay \({{\rm{\mu }}^{\rm{ - }}} \to {{\rm{e}}^{\rm{ - }}}{\rm{ + }}{{\rm{\nu }}_{\rm{e}}}{\rm{ + }}{{\rm{\nu }}_{\rm{\mu }}}\)possible considering the appropriate conservation laws? State why or why not.

Short Answer

Expert verified

The decay\({\mu ^ - } \to {e^ - } + {\nu _e} + {\nu _\mu }\)is not possible as per the conservation laws.

Step by step solution

01

Definition of Concept

The conservation laws that must be satisfied by a decay process are- mass, charge, momentum, angular momentum, baryon number and lepton number.

02

Explain is the decay \({{\rm{\mu }}^{\rm{ - }}} \to {{\rm{e}}^{\rm{ - }}}{\rm{ + }}{{\rm{\nu }}_{\rm{e}}}{\rm{ + }}{{\rm{\nu }}_{\rm{\mu }}}\) possible considering the appropriate conservation laws

Considering the given information:

Given reaction is\({\mu ^ - } \to {e^ - } + {\nu _e} + {\nu _\mu }\)

We can check the following conservation laws to see if the decay\({\mu ^ - } \to {e^ - } + {\nu _e} + {\nu _\mu }\)is possible.

Decay:\({\mu ^ - } \to {e^ - } + {\nu _e} + {\nu _\mu }\)

Charge:\( - 1 \to - 1 + 0 + 0\)

\(\therefore \)the charge is conserved.

\({\rm{Baryon}}\;{\rm{number}}:0 \to 0 + 0 + 0\)

\(\therefore \)Baryon number is conserved

\({\rm{Electron}}\;{\rm{Lepton}}\;{\rm{number}}\;\left( {{L_e}} \right):0 \to + 1 + 1 + 0\)

\(\therefore \)electron lepton number is not conserved

\({\rm{Muon}}\;{\rm{Lepton}}\;{\rm{number}}\;\left( {{L_\mu }} \right): + 1 \to 0 + 0 + 1\)

\(\therefore \)muon lepton number is conserved

As a result, even though charge, muon lepton number\(\left( {{{\rm{L}}_{\rm{\mu }}}} \right)\), and baryon number are all conserved, the electron lepton number\(\left( {{{\rm{L}}_e}} \right)\)is not.

For the given decay,\(\left( {{{\rm{L}}_e}} \right)\)is not conserved. As a result, according to conservation laws, decay is not possible.

Therefore, the required decay\({\mu ^ - } \to {e^ - } + {\nu _e} + {\nu _\mu }\)is not possible as per the conservation laws.

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