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Chapter 44: Problem 21

In which of the following decays are the three lepton numbers conserved? In each case, explain your reasoning. (a) \(\mu^{-} \rightarrow\) \(\mathrm{e}^{-}+\nu_{\mathrm{c}}+\overline{\nu}_{\mu} (\mathrm{b}) \tau^{-} \rightarrow \mathrm{e}^{-}+\overline{\nu}_{\mathrm{c}}+\nu_{\tau} ;(\mathrm{c}) \pi^{+} \rightarrow \mathrm{e}^{+}+\gamma ;\) (d) \(\mathrm{n} \rightarrow\) \(\mathrm{p}+\mathrm{e}^{-}+\overline{\nu}_{\mathrm{c}} .\)

Short Answer

Expert verified
The decays (a) and (b) conserve lepton numbers.

Step by step solution

01

Understanding Lepton Number Conservation

Lepton number conservation implies that the total number of leptons and antileptons must remain constant in a decay process. Each lepton family (electron, muon, tau) has its associated lepton number, which must be conserved separately.
02

Analyzing Muon Decay (a)

For the decay \( \mu^{-} \rightarrow \mathrm{e}^{-} + u_{\mathrm{c}} + \overline{u}_{\mu} \), we check each lepton number. The initial muon has a muon number of +1, and the rest have to also sum up to zero. \( \mathrm{e}^{-} \) has an electron number of +1, \( u_{\mathrm{c}} \) has a charge lepton number of +1, and \( \overline{u}_{\mu} \) has a muon number of -1, thus conserving the total lepton number for each type.
03

Analyzing Tau Decay (b)

For the decay \( \tau^{-} \rightarrow \mathrm{e}^{-} + \overline{u}_{\mathrm{c}} + u_{\tau} \), we check each lepton number. The tau has a tau number of +1 at the start. The sum of electron lepton number is +1 for \( \mathrm{e}^{-} \), and -1 for \( \overline{u}_{\mathrm{c}} \). The neutrino \( u_{\tau} \) has a tau lepton number of +1; lepton number is conserved for each lepton family.
04

Analyzing Pion Decay (c)

The decay \( \pi^{+} \rightarrow \mathrm{e}^{+} + \gamma \) doesn't preserve any lepton numbers since there are no leptons in the initial state. A pion has no lepton number, while the positron \( \mathrm{e}^{+} \) has an electron number of -1, which disrupts lepton conservation.
05

Analyzing Neutron Decay (d)

For the decay \( \mathrm{n} \rightarrow \mathrm{p} + \mathrm{e}^{-} + \overline{u}_{\mathrm{c}} \), we check the collective numbers. The initial neutron has no lepton number. The electron has a lepton number of +1 and the antineutrino \( \overline{u}_{\mathrm{c}} \) has a charge lepton number of -1, so the total lepton number is conserved as zero.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Muon Decay
In particle physics, muon decay is an important process to understand the concept of lepton number conservation. Understanding this decay helps illustrate how different lepton numbers must be conserved. Consider the decay process: \( \mu^{-} \rightarrow \mathrm{e}^{-} + u_{\mathrm{c}} + \overline{u}_{\mu} \).In this reaction, we start with a muon \(\mu^{-}\), which has a muon lepton number of +1.
  • The electron \(\mathrm{e}^{-}\) in the decay process has an electron lepton number of +1.
  • The electron neutrino \(u_{\mathrm{c}}\) also has an electron number of +1.
  • The muon antineutrino \(\overline{u}_{\mu}\) has a muon lepton number of -1.
These numbers add up in such a way that preserves the total lepton numbers: +1 for electrons and 0 for muons, exhibiting perfect lepton number conservation. This process ensures no arbitrary creation or destruction of lepton numbers, maintaining the sums across all particle subprocesses.
Tau Decay
Tau decay is another fascinating example showcasing lepton number conservation. The decay of the tau occurs as follows: \( \tau^{-} \rightarrow \mathrm{e}^{-} + \overline{u}_{\mathrm{c}} + u_{\tau} \).Here, we begin with a tau particle, which has a tau lepton number of +1.
  • When the tau particle decays to an electron \(\mathrm{e}^{-}\), it introduces an electron lepton number of +1.
  • The presence of an electron antineutrino \(\overline{u}_{\mathrm{c}}\) adds an electron lepton number of -1.
  • The tau neutrino \(u_{\tau}\) carries a tau lepton number of +1.
In evaluating these numbers, we see that lepton number is preserved exactly: +1 for both electron and tau lepton numbers, while muon numbers remain unchanged at zero. Tau decay thus upholds the fundamental principles of lepton number conservation, where specific lepton types balance out perfectly.
Pion Decay
Pion decay is a unique example of a process where lepton number conservation is violated. Consider the given reaction: \( \pi^{+} \rightarrow \mathrm{e}^{+} + \gamma \).It is essential to note that:
  • The pion \(\pi^{+}\) itself does not possess any lepton number initially, being a meson.
  • After the decay, the positron \(\mathrm{e}^{+}\) has an electron lepton number of -1, which now exists without an initial lepton counterpart.
  • The photon \(\gamma\) does not contribute to lepton numbers, as photons are neutral with respect to lepton count.
This presents a situation where the electron lepton number is not balanced: starting with none and ending with -1, indicating a violation of lepton number conservation. Hence, pion decay showcases a reaction type where initial states lack lepton-type particles, leading to an outcome where conservation laws are breached.
Neutron Decay
Neutron decay offers an insight into lepton number conservation in weak interactions of baryons. The decay is represented as: \( \mathrm{n} \rightarrow \mathrm{p} + \mathrm{e}^{-} + \overline{u}_{\mathrm{c}} \).In this case:
  • The initial neutron, similar to a pion, carries no lepton number.
  • During decay, the produced electron \(\mathrm{e}^{-}\) contributes an electron number of +1.
  • The antineutrino \(\overline{u}_{\mathrm{c}}\) contributes an electron number of -1.
  • The proton \(\mathrm{p}\), being a baryon, does not affect lepton numbers.
Upon evaluating the total electron lepton numbers (+1 from the electron, -1 from the antineutrino), they sum to zero, perfectly conserving lepton numbers.Neutron decay, therefore, exemplifies how lepton number conservation holds in interactions involving baryons and leptons, maintaining an overall balance throughout the decay process.

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Most popular questions from this chapter

The starship Enterprise, of television and movie fame, is powered by combining matter and antimatter. If the entire \(400-\mathrm{kg}\)\ antimatter fuel supply of the Enterprise combines with matter how much energy is released? How does this compare to the U.S. yearly energy use, which is roughly 1.0 \(\times 10^{20} \mathrm{J} ?\)

The \(\mathrm{K}^{0}\) meson has rest energy 497.7 MeV. A \(\mathrm{K}^{0}\) meson moving in the \(+x\) -direction with kinetic energy 225 MeV decays into a \(\pi^{+}\) and a \(\pi^{-}\) , which move off at equal angles above and below the \(+x\) -axis. Calculate the kinetic energy of the \(\pi^{+}\) and the angle it makes with the \(+x\) -axis. Use relativistic expressions for energy and momentum.

A proton and an antiproton collide head-on with equal kinetic energies. Two \(\gamma\) rays with wavelengths of 0.780 fm are produced. Calculate the kinetic energy of the incident proton.

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The weak force may change quark flavor in an interaction. Explain how \(\beta^{+}\) decay changes quark flavor. If a proton undergoes \(\beta^{+}\) decay, determine the decay reaction.
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