QB_PH_SEC9_006: Nuclear Models & Decays

In nuclear beta decay, a transition with nuclear spin change $\Delta J = 1$ (with no change in nuclear parity $\Delta \pi = \text{no}$) where the emitted electron and antineutrino have their spins aligned parallel ($S = 1$) is classified as

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Allowed Beta Decay ($L = 0$)**: The emitted leptons carry zero orbital angular momentum relative to the nucleus, so parity is unchanged ($\Delta \pi = (-1)^L = +1$). 2. **Lepton Spin Coupling**: - **Fermi Type**: Leptons couple to spin singlet state $S_{\text{lepton}} = 0$. Hence $\Delta J = 0$. - **Gamow-Teller Type**: Leptons couple to spin triplet state $S_{\text{lepton}} = 1$. The total angular momentum change is governed by $\vec{J}_i = \vec{J}_f + \vec{S}_{\text{lepton}}$, yielding: $\Delta J = 0, \pm 1 \quad (0^+ \to 0^+ \text{ strictly forbidden})$ 3. **Given Parameters**: $\Delta J = 1$, no parity change, leptons in parallel spin ($S = 1$). 4. **Conclusion**: This is an allowed Gamow-Teller transition. Option (B) is correct.