91Ó°ÊÓ

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.

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\({\rm{105}}\;{\rm{MeV}}\).

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.