Question

One decay mode for the eta-zero meson is \({{\rm{\eta }}^{\rm{0}}} \to {{\rm{\pi }}^{\rm{0}}}{\rm{ + }}{{\rm{\pi }}^{\rm{0}}}\).

(a) Write the decay in terms of the quark constituents.

(b) How much energy is released?

(c) What is the ultimate release of energy, given the decay mode for the pi zero is\({{\rm{\pi }}^{\rm{0}}} \to {\rm{\gamma + \gamma }}\)?

Short answer

a. The equation \({{\rm{\eta }}^{\rm{0}}} \to {{\rm{\pi }}^{\rm{0}}}{\rm{ + }}{{\rm{\pi }}^{\rm{0}}}\) in terms of quarks is given by \(u\bar u + d\bar d \to (u\bar u + d\bar d) + (u\bar u + d\bar d)\).

b. The energy released in the decay of \({{\rm{\eta }}^{\rm{0}}}\) particle is \(277.9\;{\rm{MeV}}\).

c. The total release of energy in the decay process of \({\eta ^0}\) through the reactions\({\eta ^0} \to {\pi ^0} + {\pi ^0}\) and \({\pi ^0} \to \gamma + \gamma \) is \(547.9\;{\rm{MeV}}\).

Step-by-step solution

Definition of Concept

During the decay of a primary particle, some of its mass gets converted into energy. This energy adds to the energy of products either as\({\bf{\gamma }}\)-rays or as the kinetic energy of the product.

Find the decay in terms of the quark constituents

(a)

Considering the given information:

Given reaction is,\({\eta ^0} \to {\pi ^0} + {\pi ^0}\)

Quark structure of\({{\rm{\eta }}^{\rm{0}}}{\rm{ = u\bar u + d\bar d}}\)

Quark structure of\({{\rm{\pi }}^{\rm{0}}}{\rm{ = u\bar u + d\bar d}}\)

As a result, the equation in terms of quarks is,

\(\begin{aligned}{}{\eta ^0} \to {\pi ^0} + {\pi ^0}\\u\bar u + d\bar d \to (u\bar u + d\bar d) + (u\bar u + d\bar d)\end{aligned}\)

Therefore, the required equation\({\eta ^0} \to {\pi ^0} + {\pi ^0}\)in terms of quarks is given by\(u\bar u + d\bar d \to (u\bar u + d\bar d) + (u\bar u + d\bar d)\).

Find the energy released in the decay

(b)

Considering the given information:

The primary decay mode of negative pion is given by,\({\eta ^0} \to {\pi ^0} + {\pi ^0}\)

Apply the formula:

Energy released = Rest mass energy of initial particle - sum of energies of decay products

The\({{\rm{\eta }}^{\rm{0}}}\)particle is decayed as\({\eta ^0} \to {\pi ^0} + {\pi ^0}\),

The rest mass energy of the\({{\rm{\eta }}^{\rm{0}}}\)particle is\({E_\eta } = 547.9\;{\rm{MeV}}\)

The energy of the\({{\rm{\pi }}^{\rm{0}}}\)particle is\({E_\pi } = 135\;{\rm{MeV}}\).

As a result, energy released is-

\(\begin{aligned}{}\Delta E &= 547.9\;{\rm{MeV}} - 135\;{\rm{MeV}} - 135\;{\rm{MeV}}\\ &= 277.9\;{\rm{MeV}}\end{aligned}\)

Therefore, the required energy release in the decay of \({{\rm{\eta }}^{\rm{0}}}\) particle is \({\rm{277}}{\rm{.9}}\;{\rm{MeV}}\).

Find the ultimate release of energy

(c)

Considering the given information:

The decay product\({{\rm{\pi }}^{\rm{0}}}\)of the reaction\({\eta ^0} \to {\pi ^0} + {\pi ^0}\)is again decayed by\({\pi ^0} \to \gamma + \gamma \).

\({\rm{135MeV}}\)is the energy of the\({{\rm{\pi }}^{\rm{0}}}\)particle.

Because\({{\rm{\pi }}^{\rm{0}}}\)decays to\({\pi ^0} \to \gamma + \gamma \)and to conserve energy and momentum, each photon carries\({\rm{67}}{\rm{.5}}\;{\rm{MeV}}\)of energy.

As a result, the total energy carried by all four photons is\(4 \times 67.5\;{\rm{MeV}} = 270\;{\rm{MeV}}\).

The decay process has already released\({\rm{277}}{\rm{.9}}\;{\rm{MeV}}\)of energy.

So, the total release of energy is \(277.9\;{\rm{MeV}} + 270\;{\rm{MeV}} = 547.9\;{\rm{MeV}}\).

Therefore, the required total release of energy in the decay process shown through the reactions \({\eta ^0} \to {\pi ^0} + {\pi ^0}\) and \({\pi ^0} \to \gamma + \gamma \) is \(547.9\;{\rm{MeV}}\).