Subtopic: Optoelectronic Devices (LED, Photodiode, Solar Cell)
Question Statement:
In a nucleus, the interaction V0l⃗ ⋅ s⃗ is responsible for creating spin-orbit doublets.
The energy difference between p1/2 and p^3/2 states in units of V0
\hbar^2
2 is
____________ (Round off to the nearest integer)
Official Answer Key:
3 to 3
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.60)**
**1. Nuclear Physics Principles (Spin-Orbit Interaction)**:
- Spin-orbit potential: $V_{so} \vec{l} \cdot \vec{s}$.
- Using total angular momentum $\vec{j} = \vec{l} + \vec{s}$:
$\vec{l} \cdot \vec{s} = \frac{1}{2}\left[ j(j+1) - l(l+1) - s(s+1) \right]\hbar^2$
- For $p$-orbital nucleons: $l = 1, s = 1/2$.
$l(l+1) = 2, \quad s(s+1) = \frac{3}{4}$
**2. Detailed Step-by-Step Evaluation**:
- For $p_{3/2}$ state ($j = 3/2$):
$\langle \vec{l}\cdot\vec{s} \rangle_{3/2} = \frac{1}{2}\left[\frac{3}{2}\left(\frac{5}{2}\right) - 2 - \frac{3}{4}\right]\hbar^2 = \frac{1}{2}\left[\frac{15}{4} - \frac{11}{4}\right]\hbar^2 = \frac{1}{2}(1)\hbar^2 = +\frac{1}{2}\hbar^2$
- For $p_{1/2}$ state ($j = 1/2$):
$\langle \vec{l}\cdot\vec{s} \rangle_{1/2} = \frac{1}{2}\left[\frac{1}{2}\left(\frac{3}{2}\right) - 2 - \frac{3}{4}\right]\hbar^2 = \frac{1}{2}\left[\frac{3}{4} - \frac{11}{4}\right]\hbar^2 = \frac{1}{2}(-2)\hbar^2 = -1\hbar^2$
- Energy difference between the doublet states:
$\Delta E = E(p_{1/2}) - E(p_{3/2}) = V_{so}\left[ \frac{1}{2}\hbar^2 - (-1\hbar^2) \right] = \frac{3}{2} V_{so}\hbar^2 = 3 \left(\frac{V_{so}\hbar^2}{2}\right)$
- Expressed in units of $\frac{V_{so}\hbar^2}{2}$, the energy difference is exactly **3**.
**3. Official Verification**:
The verified official IIT Master Answer range is **3 to 3**.
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