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Found in: Page 1363

### Fundamentals Of Physics

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, alepton, 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.

See the step by step solution

## Step 1: Concept used to solve the question

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.

## Step 2:(a) Finding the neutrino Is it a quark, a lepton, a meson, or a baryon

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

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

## Step 3:(b) Finding the neutrino Is a particle or antiparticle

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.

## Step 4:(c) Finding the neutrino Is a particle or antiparticle

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.

## Step 5:(d) Finding stability

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.