Question

The decay mode of the negative muon is \({{\rm{\mu }}^{\rm{ - }}} \to {{\rm{e}}^{\rm{ - }}}{\rm{ + }}{{\rm{\bar \nu }}_{\rm{e}}}{\rm{ + }}{{\rm{\nu }}_{\rm{\mu }}}\). (a) Find the energy released in \({\rm{MeV}}\). (b) Verify that charge and lepton family numbers are conserved.

Short answer

a) The energy released in the decay is\({\rm{105}}{\rm{.19 MeV}}\).

b)The charge and lepton number are conserved in the decay process.

Step-by-step solution

Concept Introduction

A decay process must satisfy all the conservation laws of particle physics, which include conservation of mass, charge, energy, momentum, strangeness, parity baryon number, lepton number, etc.

Calculating the energy

a)

The released energy\({\rm{\Delta E}}\)is the energy difference between the\({{\rm{\mu }}^{\rm{ - }}}\)particle's rest mass energy and the rest mass energies of the three particles\({e^ - },{\bar \nu _e}\), and\({{\rm{\nu }}_{\rm{\mu }}}\).

\(\begin{array}{c}\Delta E = {E_\mu } - {E_e} - {E_\nu } - {E_{\bar \nu }}\\ = 105.7\;{\rm{MeV}} - 0.511\;{\rm{MeV}} - 2(0)\\ = 105.19\;{\rm{MeV}}\end{array}\)

Therefore, the required solution is\({\rm{105}}\;{\rm{MeV}}\).

Verify that charge and lepton family numbers are conserved.

b)

Both the\({e^ - }\)and\({{\rm{\mu }}^ - }\)have charge equal to\( - e\), whereas for neutrino\(\left( {{\nu _\mu }} \right)\)and for antineutrino\(\left( {{{\bar \nu }_e}} \right)\), its value is zero. The conservation of the charge is shown below:

\(\begin{array}{c}{Q_1} = - e\\{Q_2} = - e + 0 + 0 = - e\\{Q_1} = {Q_2}\end{array}\)

Here\({{\rm{Q}}_{\rm{1}}}\)is the total charge on the left side and\({{\rm{Q}}_2}\)is the total charge on the right side of the reaction (1) and e is the elementary charge. As they are equal, the charge is conserved in the reaction.

The\(e\)- lepton number\(L\left( e \right)\) for\({e^ - }\)and\({\bar \nu _e}\)is\( + 1\;{\rm{and}} - 1\)respectively and for others it is zero. The\(\mu \)- lepton number\(L\left( \mu \right)\)for\({\mu ^ - }\)and\({\nu _\mu }\)is\( + 1\)and for others it is zero. Thus, if subscript 1 represents the left side of the reaction and subscript 2, the right side of the reaction. Then, we get

\(\begin{array}{c}L{(\mu )_1} = L{(\mu )_2}\\ = + 1\\L{(e)_2} = L{(e)_1}\\ = 0\end{array}\)

As,\(e\)- lepton number and\(\mu \)- lepton number are same on both the sides of the reaction, hence, they are conserved.

Therefore, we can say that lepton numbers are also conserved in this reaction.