Question

The decay mode of the positive tau is\({{\bf{\tau }}^ + } \to {\rm{ }}{{\bf{\mu }}^ + }{\rm{ }} + {\rm{ }}{{\bf{\nu }}_{\bf{\mu }}}{\rm{ }} + {\rm{ }}{{\bf{\bar \nu }}_{\bf{\tau }}}\).

(a) What energy is released?

(b) Verify that charge and lepton family numbers are conserved.

(c) The \({\tau ^ + }\)is the antiparticle of the \({\tau ^ - }\). Verify that all the decay products of the \({\tau ^ + }\)are the antiparticles of those in the decay of the \({\tau ^ - }\) given in the text.

Short answer

a) The energy released is\({\rm{1671}}\;\;{\rm{MeV}}\).

b) It is verified that\(L{(\mu )_1} = L{(\mu )_2}{\rm{ and }}L{(\tau )_1} = L{(\tau )_2}\). Thus, lepton numbers are conserved. The charge is also conserved for the reaction.

c) From the reaction\({\tau ^ - } \to {\mu ^ - } + {\bar \nu _\mu } + {\nu _\tau }\), we can see that all the particles are antiparticles of the products of decay reaction of \({\tau ^ + }\).

Step-by-step solution

Concept Introduction

In particle physics, only those reactions take place which satisfies the conservation laws such as conservation of energy, charge, baryon number, lepton number, parity, mass, etc.

Given Data

Principal decay mode of \({\tau ^ + }\) is: \({\tau ^ + } \to {\mu ^ + } + {\nu _\mu } + {\bar \nu _\tau }\;\; \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)\)

Calculating the energy

a)

The released energy is the difference between the\({\tau ^ + }\)particle's rest mass energy\(\left( {{E_\tau }} \right)\) and the three rest mass energies of the\({{\rm{\mu }}^{\rm{ + }}}\)particle\(\left( {{E_\mu }} \right)\),\({\nu _\mu }\)\(\left( {{E_\nu }} \right)\) and\({\bar \nu _\tau }\)particle\(\left( {{E_{\bar \nu }}} \right)\).

\(\begin{array}{c}\Delta E = {E_\tau } - {E_\mu } - {E_\nu } - {E_{\bar \nu }}\\ = 1777\;{\rm{MeV}} - 105.7\;{\rm{MeV}} - 2(0)\\ = {\rm{1671}}\;\;{\rm{MeV}}\end{array}\)

Therefore, the required solution is \({\rm{1671}}\;\;{\rm{MeV}}\).

Verify that charge and lepton family numbers are conserved

b)

Both the\({\tau ^ + }\)and\({{\rm{\mu }}^{\rm{ + }}}\)have charge equal to\( + e\), whereas for neutrino\(\left( {{\nu _\mu }} \right)\)and for antineutrino\(\left( {{{\bar \nu }_\tau }} \right)\), 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\(\tau \)- lepton number\(L\left( \tau \right)\) for\({\tau ^ + }\)and\({\bar \nu _\tau }\)is\( - 1\)and for others it is zero. The\(\mu \)- lepton number\(L\left( \mu \right)\)for\({\mu ^ + }\)and\({\nu _\mu }\)is\( - 1\)and\( + 1\)respectively. Thus, if subscript 1 represents the left side of the reaction and subscript 2, the right side of the reaction. Then, we get

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

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

Verifying the decay products

c)

If we analyze the decay process of\({\tau ^ - }\), given as-

\({\tau ^ - } \to {\mu ^ - } + {\bar \nu _\mu } + {\nu _\tau }\)

We can clearly that each particle in this reaction is the antiparticle of the corresponding particle in reaction 1.

Therefore, it is verified that all the products of \({\tau ^ - }\) decay are the antiparticles of the products in reaction (1).