91Ó°ÊÓ

(a)

Considering the given information:

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

Apply the formula:

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

The\({{\rm{\eta }}^{\rm{0}}}\)particle decays by\({\eta ^0} \to \gamma + \gamma \)in this problem.

Rest energy of\({\eta ^0}\;{\rm{particle}} = 547.9\;{\rm{MeV}}\)

Rest energy of\(\gamma {\rm{ - particle}} = 0\;{\rm{MeV}}\)

Hence, energy release is,

\(\begin{array}{c}\Delta E = 547.9\;{\rm{MeV}} - 0\;{\rm{MeV}} - 0\;{\rm{MeV}}\\ = 547.9\;{\rm{MeV}}\end{array}\)

Therefore, the required energy released in the decay of \({{\rm{\eta }}^{\rm{0}}}\)particle is \(547.9\;{\rm{MeV}}\).

(b)

Considering the given information:

the given reaction is\({\eta ^0} \to \gamma + \gamma \).

the uncertainty principle is given as-

\({\rm{\Delta E}}{\rm{.\Delta t = }}\frac{{\rm{h}}}{{{\rm{4\pi }}}}\)

Where,\(\Delta E\)is uncertainty in energy measurement and\(\Delta t\)is uncertainty in time measurement.

The\({{\rm{\eta }}^{\rm{0}}}\)particle has a life time of\({\rm{2}}{\rm{.35 \times 1}}{{\rm{0}}^{{\rm{ - 19}}}}{\rm{\;s}}\).

As a result of the uncertainty in the energy released,\({\rm{\Delta E = }}\frac{{\rm{h}}}{{{\rm{4\pi \Delta t}}}}\).

\(\begin{array}{c}\Delta E{\rm{ = }}\frac{{{\rm{6}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 34}}}}\;{\rm{J}}{\rm{.s}}}}{{{\rm{4 \times 3}}{\rm{.14 \times }}\left( {{\rm{2}}{\rm{.35}}{\rm{.1}}{{\rm{0}}^{{\rm{ - 19}}}}\;{\rm{s}}} \right)}}\\{\rm{ = 2}}{\rm{.24}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 16}}}}{\rm{\;J}}\\{\rm{ = }}\frac{{{\rm{2}}{\rm{.24 \times 1}}{{\rm{0}}^{{\rm{ - 16}}}}\;{\rm{J}}}}{{{\rm{1}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 19}}}}}}\\{\rm{ = 1}}{\rm{.4}}\;{\rm{keV}}\end{array}\)

Therefore, the uncertainty in the energy released in the decay of a \({{\rm{\eta }}^{\rm{0}}}\) particle is \({\rm{1}}{\rm{.4}}\;{\rm{keV}}\).

(c)

Considering the given information:

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

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

Quark structure of\({\rm{\gamma = q\bar q}}\)

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

\(\begin{array}{l}{\eta ^0} \to \gamma + \gamma \\u\bar u + d\bar d \to q\bar q + q\bar q\end{array}\)

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

(d)

Considering the given information:

Charge on\(\gamma \)photon is\(0\)

Charge on\({\eta _o}\)meson is\(0\)

Baryon number on\(\gamma \)photon is\(0\)

Baryon number on\({\eta _o}\)meson is\(0\)

Lepton number on\(\gamma \)photon is\(0\)

Lepton number on\({\eta _o}\)meson is\(0\)

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

\(\begin{array}{l}{\rm{Charge: }}0 \to 0 + 0\\{\rm{Baryon}}\;{\rm{number:}}\;0 \to 0 + 0\\{\rm{Lepton}}\;{\rm{number:}}\;0 \to 0 + 0\end{array}\)

Thus, the reaction conserves charge, baryon number, and lepton number.

Hence, the conservation of charge, baryon number and lepton numbers for the reaction\({\eta ^0} \to \gamma + \gamma \)is verified.