Question:hard

Which one of the following statements is true?

Show 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.
Updated On: Jul 28, 2026
  • In the decay \(\mu^+ \rightarrow e^+ + \nu_e + \bar\nu_\mu\), CPT is violated.
  • The decay \(\Lambda \rightarrow p^+ + \pi^-\) is allowed and strangeness is violated.
  • The decay \(p^+ \rightarrow e^+ + \gamma\) is allowed.
  • The decay \(\Omega^- \rightarrow \Xi^0 + K^-\) is allowed.
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Set up a quantum number table.
For each decay, list the baryon number $B$, lepton number $L$, strangeness $S$ and charge $Q$ on both sides. A decay through the strong or electromagnetic interaction must keep $B$, $L$, $S$ and $Q$ all exactly equal on both sides; a weak decay must still conserve $B$, $L$ and $Q$, but is allowed to change $S$ by exactly one unit. Any decay also needs the parent's rest mass to be at least as large as the sum of the daughters' rest masses.

Step 2: Statement (A), $\mu^+ \to e^+ + \nu_e + \bar\nu_\mu$.
Lepton numbers: the muon side carries muon lepton number $L_\mu = -1$ (since it's an antimuon) and electron lepton number $L_e = 0$; the products carry $L_e = -1 + 1 = 0$ (positron plus electron neutrino) and $L_\mu = -1$ (from the muon antineutrino). Both totals match, so lepton number is conserved separately in each flavor. Charge, energy and momentum all check out too. CPT is a general theorem that holds for any Lorentz-invariant local field theory, and nothing in this bookkeeping breaks it, so this decay does not violate CPT. Statement (A) fails.

Step 3: Statement (B), $\Lambda \to p^+ + \pi^-$.
$B$: $1 \to 1 + 0 = 1$ (conserved). $Q$: $0 \to 1 + (-1) = 0$ (conserved). $S$: $-1 \to 0 + 0 = 0$, a change of one unit. Since this is a weak process (matching the $\Lambda$'s known lifetime scale) and weak decays are allowed exactly this one-unit change in $S$, the decay proceeds and strangeness is indeed violated by it. Statement (B) holds up.

Step 4: Statement (C), $p^+ \to e^+ + \gamma$.
$B$: $1 \to 0 + 0 = 0$. This mismatch alone rules the decay out completely, regardless of any other quantum number, since baryon number must be conserved in every known interaction. Statement (C) fails.

Step 5: Statement (D), $\Omega^- \to \Xi^0 + K^-$.
Every quantum number matches: $Q: -1 \to 0+(-1)=-1$, $B: 1\to 1+0=1$, $S: -3 \to -2+(-1) = -3$. So this table alone would suggest the decay is fine. But quantum numbers matching is only a necessary condition, not a sufficient one; energy bookkeeping also has to work. Adding the masses of $\Xi^0$ (about $1315$ MeV) and $K^-$ (about $494$ MeV) gives about $1809$ MeV, which is more than the $\Omega^-$ mass of about $1672$ MeV. The parent is lighter than what the decay would need to produce, so it cannot happen. Statement (D) fails.

Final Answer:
Statement (B) is the only one that survives every check.\[ \boxed{\text{(B)}} \]
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