Question

A slow moving \(\pi^{-}\) particle is captured by a deuteron (\(d\)) and this reaction produces two neutrons (\(n\)) in the final state, i.e., \(\pi^{-} + d \to n + n\).
Neutron and deuteron have even intrinsic parities, whereas \(\pi^{-}\) has odd intrinsic parity. \(L\) and \(S\) are the orbital and spin angular momenta, respectively of the system of two neutrons. Which of the following statements regarding the final two-neutron state is/are CORRECT?

• A. It has odd parity
• B. \(L + S\) is odd
• C. \(L = 1\), \(S = 1\)
• D. \(L = 2\), \(S = 0\)

Hint:

The parity of a system is determined by the intrinsic parities of the particles involved and the orbital angular momentum. Always check the parity conservation rules when determining the total parity.

Answer 1

The Correct Option is A, C, D

- The parity of the final state is determined by the parity of the \(\pi^{-}\) meson, which is odd, and the intrinsic parities of the neutron and deuteron, which are even. The final state parity will therefore be odd (option (A)).
- Given that the total spin \(S\) and orbital angular momentum \(L\) combine to produce an odd total parity, \(L + S\) must be odd.
- The correct quantum numbers for the system are \(L = 1\), \(S = 1\), leading to the appropriate two-neutron state (option (C)).