Question Statement
Which of the following statements concerning the deuteron ($^2\text{H}$) ground state is/are CORRECT?
(A)
The total nuclear spin and parity of the ground state is $J^P = 1^+$.
(B)
The ground state is an admixture of predominantly $^3S_1$ ($L = 0$) and a small percentage of $^3D_1$ ($L = 2$).
(C)
The non-zero positive electric quadrupole moment of the deuteron is direct experimental evidence for a non-central (tensor) nuclear force.
(D)
The isospin of the ground state is $T = 1$.
Step-by-Step KaTeX Derivation & Detailed Explanation
### Step-by-Step Multi-Select Evaluation: 1. **Statement A (Nuclear Spin & Parity)**: The deuteron is a bound state of one proton and one neutron with total spin $J = 1$ and positive parity $\pi = +1$. (CORRECT) 2. **Statement B (Orbital Angular Momentum Admixture)**: Due to the tensor component of the nuclear potential, the ground state wave function is not a pure spherically symmetric $S$-state ($L=0$), but a superposition of $^3S_1$ ($L = 0, \approx 96\%$) and $^3D_1$ ($L = 2, \approx 4\%$). (CORRECT) 3. **Statement C (Quadrupole Moment & Tensor Force)**: A pure $S$-state has zero quadrupole moment. The measured positive quadrupole moment ($Q = +0.286 e\text{ fm}^2$) conclusively demonstrates the presence of a non-central tensor force mixing states of different $L$ while preserving $J$. (CORRECT) 4. **Statement D (Isospin)**: Generalized Pauli principle requires the total nucleon wave function to be antisymmetric: $(-1)^{L+S+T+1} = -1 \implies L + S + T = \text{odd}$. For $L=0, S=1$: $0 + 1 + T = \text{odd} \implies T = 0$ (isospin singlet). The $T = 1$ state is unbound. (INCORRECT) 5. **Conclusion**: Options A, B, and C are correct.