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

(a) Find the energy released.

(b) What is the uncertainty in the energy due to the short lifetime?

(c) Write the decay in terms of the constituent quarks.

(d) Verify that baryon number, lepton numbers, and charge are conserved.

Short Answer

Expert verified

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

(b) The uncertainty in the energy released in the decay of a \({{\rm{\eta }}^{\rm{0}}}\) particle is \(1.4\;{\rm{keV}}\).

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

Step by step solution

01

Definition of Concept

(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 \(547.9\;{\rm{MeV}}\).

02

Find the uncertainty in the energy

(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 \({\rm{1}}{\rm{.4}}\;{\rm{keV}}\).

03

Find the decay in terms of the constituent quarks

(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\(u\bar u + d\bar d \to q\bar q + q\bar q\).

04

Verify the baryon number, lepton numbers, and charge are conserved

(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.

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