Question 705074

SCQMEDIUM

π−\pi^{-} has spin 0 and negative intrinsic parity. In a reaction a deuteron in its ground state (J=1J=1, parity is +1) captures a π−\pi^{-} in pp-wave to produce a pair of neutrons (intrinsic parity is +1). The neutrons will be produced in a state with

(A)

l=1,S=0l=1, S=0

(B)

l=0,S=1l=0, S=1

(C)

l=1,S=1l=1, S=1

(D)

l=0,S=0l=0, S=0

Detailed Solution

The reaction is π−+d→n+n\pi^{-} + d \rightarrow n + n. Initial parity: (−1)(+1)×(−1)1=+1(-1)(+1) \times (-1)^1 = +1. Final parity: (+1)(+1)×(−1)l(+1)(+1) \times (-1)^l. Total angular momentum conservation for the initial state: J=∣sπ+sd∣=∣0+1∣=1J = |s_{\pi} + s_d| = |0 + 1| = 1. Including orbital angular momentum l=1l=1, initial J=1J=1. For the final state, to conserve parity and angular momentum with l=0l=0, the two neutrons must be in a singlet spin state S=0S=0, so J=0J=0 and parity is +1+1.

Free Exam

Boost Your Exam Preparation!

Move beyond just reading solutions. Access our comprehensive Test Series, original Mock Tests, and interactive learning modules. Many premium tests are completely free!

  • Original Mocks & Regular Test Series
  • Real NTA-like Interface with Analytics
  • Many Free Tests Available