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Chapter 1: Problem 7

State which of the following reactions are allowed by the conservation laws and which are forbidden, giving the reasons: $$ \begin{aligned} \pi^{0} & \rightarrow e^{+}+e^{-} \\ e^{-}+p & \rightarrow n+v_{e} \\ \mu^{+} & \rightarrow e^{+}+e^{-}+e^{+} \\ K^{0}+n & \rightarrow \Lambda+\pi^{0} \\ \Xi^{0} & \rightarrow \Lambda+\pi^{0} . \end{aligned} $$

Short Answer

Expert verified
Reactions 1, 2, 4, and 5 are allowed; Reaction 3 is forbidden.

Step by step solution

01

Analyze Reaction 1

For the reaction \( \pi^{0} \rightarrow e^{+}+e^{-} \), check for conservation of charge, lepton number, and energy. The initial charge is 0 and the final charge is also 0 (\( +1 + -1 = 0 \)), so charge is conserved. The lepton numbers are also 0 initially and finally, so lepton number is conserved. Energy and momentum are also conserved in this decay, thus this reaction is allowed.
02

Analyze Reaction 2

Consider the reaction \( e^{-}+p \rightarrow n+v_{e} \). Check conservation laws: charge conservation (initial charge is 0 and final charge is 0), baryon number conservation (both proton and neutron are baryons with baryon number 1), and lepton number conservation (initial lepton number is 1 and final is 1 due to neutrino). All conservation laws are respected, so the reaction is allowed.
03

Analyze Reaction 3

For \( \mu^{+} \rightarrow e^{+}+e^{-}+e^{+} \), verify conservation of charge, lepton number, and energy. The initial charge is +1 and the final charge is +1, so charge is conserved. However, the initial lepton number (\( L_\mu = 1 \), \( L_e = 0 \)) is not conserved in the final state (\( L_\mu = 0 \), \( L_e = 1 \)), hence this reaction is forbidden due to lepton number conservation violation.
04

Analyze Reaction 4

For \( K^{0}+n \rightarrow \Lambda+\pi^{0} \), check baryon, charge, and strangeness conservation. The baryon numbers are conserved (1 and 1 initially and finally). The initial and final strangeness is also conserved (0 at both sides). Charge is 0 on both sides. Therefore, the reaction is allowed.
05

Analyze Reaction 5

Examine \( \Xi^{0} \rightarrow \Lambda+\pi^{0} \). Verify the conservation laws: baryon number (1 initially and finally), hypercharge and isospin (since strangeness decreases by 1 but the baryon changes), and charge (0 both initially and finally), so all are conserved. Thus, this reaction is allowed.

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Key Concepts

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

Charge Conservation
In particle physics, charge conservation is a fundamental principle. It states that the total electric charge remains constant in any isolated physical process. This means the sum of electric charges before and after a reaction must be the same. For instance:
  • In the reaction \(e^{-} + p \rightarrow n + v_{e}\), the initial total charge is 0. This is because the electron has a charge of \(-1\) and the proton has \(+1\), canceling each other out.
  • In the final state, the neutron (which is neutral) and the neutrino (also neutral) together add up to a total of 0 as well.
Charge conservation is respected, showing how it acts as a rulebook to determine if certain reactions can happen in nature.
Lepton Number Conservation
Lepton number conservation refers to the principle that the total number of leptons remains constant in a reaction. Leptons include particles like electrons, muons, and neutrinos, and each has their own 'family' lepton number:
  • For electrons and electron neutrinos, the electron lepton number \(L_e\) is 1.
  • For muons and muon neutrinos, the muon lepton number \(L_\mu\) is 1.
Let's consider the reaction \(\mu^{+} \rightarrow e^{+}+e^{-}+e^{+}\). Before the reaction, the muon lepton number \((L_\mu)\) is 1, but after the reaction, the electron lepton number \((L_e)\) becomes 1 and the muon lepton number \((L_\mu)\) is 0. This discrepancy shows lepton number conservation is violated, making this reaction forbidden.
Baryon Number Conservation
Baryon number conservation is an essential law in particle physics that states the total number of baryons remains unchanged in any interaction. Baryons include protons and neutrons, each carrying a baryon number of 1. Let's break it down:
  • In the reaction \(e^{-} + p \rightarrow n + v_{e}\), the initial baryon number is 1, due to the proton.
  • After the reaction, the final baryon number is also 1, because of the neutron.
Baryon number conservation ensures that processes like particle decays and collisions cannot create or destroy baryons without corresponding anti-baryons, which keeps the balance in our universe.
Strangeness Conservation
Strangeness conservation involves a quantum number related to particles known as strange particles, which were first observed in cosmic rays. Strangeness is conserved in strong and electromagnetic interactions but not always in weak interactions. For instance:
  • In the reaction \(K^{0}+n \rightarrow \Lambda+\pi^{0}\), strangeness is initially contributed by \(K^{0}\), which has a strangeness of +1.
  • On the right side, the \(\Lambda\) has a strangeness of -1, maintaining the balance since the system involves weak interactions where total strangeness decreases by 1.
While the weak force can change strangeness, it's still a crucial factor in understanding particle interactions across different forces.
Isospin Conservation
Isospin conservation is a concept analogous to angular momentum conservation, and it applies in particle processes. It helps to predict possible outcomes of particle interactions. Isospin \(I\) and its third component \(I_3\) play significant roles:
  • In the reaction \(\Xi^{0} \rightarrow \Lambda + \pi^{0}\), the \(\Xi^{0}\) particle has its isospin component that needs to be consistent with the final products \(\Lambda\) and \(\pi^{0}\).
  • The conservation of isospin dictates that the sum of the isospin components of the outgoing particles matches those of the initial particles.
Isospin conservation helps predict and validate reactions, ensuring we understand how particles switch between states post-collision or decay.

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Most popular questions from this chapter

(a) A negative muon, when brought to rest in liquid hydrogen, can form a molecular ion \(\mathrm{H}_{2}^{+}\)by displacing an electron. Why? (b) Hydrogen contains a small amount of the heavier isotope deuterium, and it is found that negative muons stopping in hydrogen eventually form molecular ions HD \(^{+}\). Why? (c) What is the typical internuclear distance in such an ion? (d) If the two nuclei fuse to form \({ }^{3} \mathrm{He}\), what may happen to the muon?

An electron of energy \(20 \mathrm{GeV}\) is deflected through an angle of \(5^{\circ}\) in an elastic collision with a stationary proton. What is the value of the square of the 4 -momentum transfer, \(q^{2}\), and down to what approximate distance does such a collision probe the internal structure of the proton? (The mass of the electron can be neglected compared with the energies involved. The proton mass \(M c^{2}\) is \(0.938 \mathrm{GeV}\).)

In the following reaction in hydrogen, $$ \pi^{-}+p \rightarrow X^{-}+p $$ a boson \(X\) is observed with mass \(2.4 \mathrm{GeV}\). (a) If the incident pion beam momentum is \(12 \mathrm{GeV} / c\), calculate the maximum angle of emission of the recoil proton with respect to the beam direction, and its momentum. (b) Calculate the angle and momentum of the proton when the 4 -momentum transfer is a maximum, and compute the value of \(q_{\max }^{2}\)

The flux of primary cosmic rays averaged over the Earth's surface is approximately \(1 \mathrm{~cm}^{-2} \mathrm{~s}^{-1}\), and their average kinetic energy is \(3 \mathrm{GeV}\). Show that the power transferred to the Earth from cosmic rays is about \(2.5\) gigawatt. (Earth radius \(=6400 \mathrm{~km}\).)

(a) Show that a negative muon captured in an \(S\)-state by a nucleus of charge \(Z e\) and mass \(A\) will spend a fraction \(f \simeq 0.25 A(Z / 137)^{3}\) of its time inside the nuclear matter, and that in time \(t\) it will travel a total distance \(f c t(Z / 137)\) in the nuclear matter. (b) The law of radioactive decay of free muons is \(d N / d t=-\lambda_{d} N\), where \(\lambda_{d}=1 / \tau\) is the decay constant and the lifetime is \(\tau=2.16 \mu \mathrm{s}\). For a negative muon captured in an atom \(Z\) the decay constant is \(\lambda=\lambda_{d}+\lambda_{c}\), where \(\lambda_{c}\) is the probability per unit time of nuclear capture. For aluminium \((Z=13, A=27)\) the mean lifetime of negative muons is \(0.88\) \mus. Calculate \(\lambda_{c}\) and, using the expression for \(f\) in (a), compute the interaction mean free path \(\Lambda\) for a muon in nuclear matter. (c) From the magnitude of \(\Lambda\) estimate the magnitude of the weak coupling constant in the reaction \(\mu^{-}+p \rightarrow n+v\), assuming that a coupling constant of unity corresponds to a mean free path equal to the range ( \(1 \mathrm{fm}\) ) of the nuclear forces.
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