Question

Which one of the following is an allowed process?

• A. \( \pi^- + p \to \pi^0 + n \)
• B. \( \pi^0 \to \gamma + \gamma + \gamma \)
• C. \( p + \bar{p} \to \Lambda^0 + \Lambda^0 \)
• D. \( p + \bar{p} \to \gamma \)

Hint:

Check charge, baryon number and strangeness on both sides of each reaction; also remember pi-zero to 3 photons breaks C-parity and p-pbar to a single photon breaks momentum conservation.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A particle reaction is allowed only if it obeys all the relevant conservation laws at once: electric charge, baryon number, strangeness (for the strong and electromagnetic interactions), and also more subtle rules like C-parity for photon decays, plus basic kinematics such as energy-momentum conservation. Checking each option against these rules tells us which one can actually happen.

Step 2: Key Formula or Approach:
Go through the four options one at a time, checking charge, baryon number, strangeness, and any special rule that applies.

Step 3: Check option (A).
\(\pi^- + p \to \pi^0 + n\): charge on the left is \(-1 + 1 = 0\), and on the right \(0 + 0 = 0\), so charge is conserved. Baryon number is \(0 + 1 = 1\) on the left and \(0 + 1 = 1\) on the right, conserved. Strangeness is zero on both sides. This is the well-known pion-nucleon charge-exchange reaction, and it goes through the strong interaction with nothing forbidding it. So (A) is ALLOWED.

Step 4: Check option (B).
\(\pi^0 \to \gamma + \gamma + \gamma\): the neutral pion has C-parity \(C = +1\). A system of \(n\) photons has \(C = (-1)^n\). For 3 photons, \(C = -1\), which does not match the pion's \(C = +1\). Since the electromagnetic interaction driving this decay conserves C-parity, three photons cannot come from a \(\pi^0\); only the observed 2-photon decay works. So (B) is FORBIDDEN.

Step 5: Check option (C).
\(p + \bar{p} \to \Lambda^0 + \Lambda^0\): baryon number on the left is \(1 + (-1) = 0\). On the right, each \(\Lambda^0\) carries baryon number \(+1\), so two of them give \(+2\). Since \(0 \ne 2\), baryon number is not conserved. The correct final state for proton-antiproton annihilation into strange baryons would need one \(\Lambda^0\) and one \(\bar{\Lambda}^0\), not two \(\Lambda^0\). So (C) is FORBIDDEN.

Step 6: Check option (D).
\(p + \bar{p} \to \gamma\): in the centre of mass frame the total momentum is zero, so a single photon in the final state would need zero momentum, but a photon with zero momentum also has zero energy, which cannot equal the rest energy of the annihilating pair. A single photon can never balance both energy and momentum for a massive system annihilating together, so annihilation into just one photon is kinematically forbidden. So (D) is FORBIDDEN.

Final Answer:
Only the pion-nucleon charge-exchange reaction respects every conservation law, so option (A) is the allowed process.\[ \boxed{\pi^- + p \to \pi^0 + n} \]