Question

Which one of the following statements is true?

• A. In the decay \(\mu^+ \rightarrow e^+ + \nu_e + \bar\nu_\mu\), CPT is violated.
• B. The decay \(\Lambda \rightarrow p^+ + \pi^-\) is allowed and strangeness is violated.
• C. The decay \(p^+ \rightarrow e^+ + \gamma\) is allowed.
• D. The decay \(\Omega^- \rightarrow \Xi^0 + K^-\) is allowed.

Hint:

Check charge, baryon number and strangeness for each decay, then also check whether the parent particle actually has enough mass to produce the listed final particles.

Answer 1

The Correct Option is B

Step 1: Check statement (A).
The decay \(\mu^+ \rightarrow e^+ + \nu_e + \bar\nu_\mu\) is the standard weak decay of the muon. It conserves charge, lepton number for each generation (an electron neutrino and a muon antineutrino appear together with the muon disappearing), energy and momentum. CPT symmetry is a property that every local, Lorentz-invariant quantum field theory obeys automatically, and the Standard Model weak interaction that drives this decay is exactly such a theory. Nothing about this decay breaks CPT, so statement (A) is FALSE.

Step 2: Check statement (B).
The decay \(\Lambda \rightarrow p^+ + \pi^-\) proceeds through the weak interaction (it is the dominant decay mode of the \(\Lambda\) baryon, with a relatively long lifetime typical of weak processes). The \(\Lambda\) carries strangeness \(S=-1\), while both the proton and the pion carry \(S=0\). So strangeness changes from \(-1\) to \(0\), a change of one unit. Weak interactions are allowed to change strangeness by exactly \(\pm 1\), unlike strong or electromagnetic interactions which must conserve it exactly. So this decay is allowed, and strangeness is indeed violated (changed) in the process. Statement (B) is TRUE.

Step 3: Check statement (C).
The decay \(p^+ \rightarrow e^+ + \gamma\) would take a baryon (baryon number \(B=1\)) to a positron and a photon, both of which have baryon number \(0\). Baryon number conservation is an extremely well-tested law (it is why the proton, the lightest baryon, is stable), and this decay would break it outright. This decay has never been observed and is forbidden in the Standard Model, so statement (C) is FALSE.

Step 4: Check statement (D).
For \(\Omega^- \rightarrow \Xi^0 + K^-\), first check the simple conservation laws: charge is \(-1 \rightarrow 0 + (-1) = -1\) (conserved), baryon number is \(1 \rightarrow 1 + 0 = 1\) (conserved), and strangeness is \(-3 \rightarrow -2 + (-1) = -3\) (also conserved). Even though every quantum number balances, a decay also needs enough rest-mass energy to work with. The \(\Omega^-\) has a mass of about \(1672\) MeV, while the \(\Xi^0\) (about \(1315\) MeV) and the \(K^-\) (about \(494\) MeV) together add up to about \(1809\) MeV, more than the \(\Omega^-\) itself weighs. Since the products would need more energy than the parent particle provides, this decay cannot happen. Statement (D) is FALSE.

Final Answer:
Only statement (B) correctly describes an allowed decay, since weak decays are permitted to violate strangeness by one unit.\[ \boxed{\text{(B)}} \]