Why does the \({\eta ^0}\) meson have such a short lifetime compared to most other mesons?

Short Answer

Expert verified

As, the \({\eta ^0}\) meson consists of quark-antiquark pairs of the same flavour, its lifetime is short.

Step by step solution

01

Meson

Mesons are hadronic subatomic particles made up of an equal number of quarks and antiquarks, generally one of each, and linked together by strong interactions in particle physics.

02

Lifetime of meson

Unlike the other mesons, that consists of a quark and its appropriate antiquark of the same flavour, such as \({\rm{u\bar u}}\). These quark-antiquark pairs tend to annihilate, which causes its shorter lifetime.

Therefore, due to quark and antiquark, lifetime of meson is short.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Accelerators such as the Triangle Universities Meson Facility (TRIUMF) in British Columbia produce secondary beams of pions by having an intense primary proton beam strike a target. Such "meson factories" have been used for many years to study the interaction of pions with nuclei and, hence, the strong nuclear force. One reaction that occurs is\({{\rm{\pi }}^{\rm{ + }}}{\rm{ + p}} \to {{\rm{\Delta }}^{{\rm{ + + }}}} \to {{\rm{\pi }}^{\rm{ + }}}{\rm{ + p}}\), where the \({{\rm{\Delta }}^{{\rm{ + + }}}}\)is a very short-lived particle. The graph in Figure \({\rm{33}}{\rm{.26}}\)shows the probability of this reaction as a function of energy. The width of the bump is the uncertainty in energy due to the short lifetime of the\({{\rm{\Delta }}^{{\rm{ + + }}}}\).

(a) Find this lifetime.

(b) Verify from the quark composition of the particles that this reaction annihilates and then re-creates a d quark and a \({\rm{\bar d}}\)antiquark by writing the reaction and decay in terms of quarks.

(c) Draw a Feynman diagram of the production and decay of the \({{\rm{\Delta }}^{{\rm{ + + }}}}\)showing the individual quarks involved.

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.

Explain how conservation of baryon number is responsible for conservation of total atomic mass (total number of nucleons) in nuclear decay and reactions.

An antibaryon has three antiquarks with colors\[{\rm{\bar R\bar G\bar B}}\]. What is its color?

Verify the quantum numbers given for the \({{\rm{\Omega }}^{\rm{ + }}}\) in Table \(33.2\) by adding the quantum numbers for its quark constituents as inferred from Table \(33.4\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free