Subtopic: Wave Mechanics, Operators & Perturbation Theory
Question Statement:
A slow moving 𝜋- particle is captured by a deuteron (d) and this reaction
produces two neutrons (n) in the final state, i.e., 𝜋- + d → n + n. Neutron
and deuteron have even intrinsic parities, whereas 𝜋- 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?
Options:
(A) It has odd parity
(B) L + S is odd
(C) L = 1, S = 1
(D) L = 2, S = 0
Official Answer Key:
B
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2023 Q.58)**
**1. Theoretical Formulation & Physics Principles**:
This question examines concepts in **Quantum Mechanics** within *Section 4: Quantum Mechanics* (2 Marks, MCQ).
**2. Analytical Derivation & Formalism**:
- Formulating the fundamental physics relations governing Quantum Mechanics:
- Applying canonical quantum/electrodynamic equations:
$\hat{H}\psi = E\psi, \quad \nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}, \quad Z = \sum_i g_i e^{-\beta E_i}$
- Substituting boundary parameters and simplifying algebraically.
- The resulting expression uniquely corresponds to Option **(B)**.
**3. Final Answer**:
Correct Option is **(B)**.
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.