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Q11Q

Expert-verifiedFound in: Page 1363

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Consider the neutrino whose symbol is $\overline{{\mathbf{\nu}}_{\mathbf{\tau}}}$ . (a) Is it a quark, a**

**lepton, a meson, or a baryon? (b) Is it a particle or an antiparticle? (c) Is it a boson or a fermion? (d) Is it stable against spontaneous decay?**

(a) It is a neutrino which is unchanged leptons therefore this is a lepton.

(b) neutrinos have a positive charge therefore it is an antiparticle.

(c) The antiparticle is a lepton that has a 1/2 spin therefore the particle is fermion.

(d) Yes, it is stable against spontaneous decay.

**Properties of Leptons **

** **

**Three of the leptons (electron, muon, and tauon) have an electric charge equal to -1e .****There are also three uncharged neutrinos (also leptons), one corresponding to each of the charged leptons.****The antiparticles for the charged leptons have a positive charge.****lepton has half-integer spin quantum numbers.**

Given neutrino is $\overline{{\nu}_{\tau}}$

Since it is a neutrino which is unchanged leptons therefore this is a lepton.

We know that leptons have an electric charge equal to -1e and uncharged leptons (neutrinos), one corresponding to each of the charged leptons. The antiparticles for the charged leptons have a positive charge.

We know If the spin is a one-half integer, then the particle is a fermion. If the spin is an integer, such as zero or one or two, then the particle is a boson.

Since the antiparticle is a lepton that has a 1/2 spin therefore the particle is a fermion.

Since it is a lepton which is a fundamental particle, and all the fundamental particles are light and stable they cannot interact by any strong force. So, it is stable against spontaneous decay.

Hence, yes, it is stable against spontaneous decay.

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