Second answer

[Guillaume Smet]

Without Hyperfine Interaction
For a system with J=3/2, g=1, and B_0=2 T, the energy levels are:

E(m_J)=−gμ_B B_0 m_J
For transitions ΔmJ= 1;
ΔE=g_μ_B B_0 = 1×5.788×10−5 eV/T×2 T = 1.16×10−4 eV

This applies to both mJ=−3/2→−1/2 and mJ​=−1/2→+1/2 transitions.

With Hyperfine Interaction
Adding hyperfine field B_{hf}=10 T introduces the term A*m_I*m_J where:

A = \frac{g\mu_N B_{hf}}{J} = \frac{3.152 \times 10^{-8} \text{ eV/T} \times 10 \text{ T}}{3/2} = 2.1 \times 10^{-7} eV
Each transition splits into two:

For m_I = +1/2:\Delta E = g\mu_B B_0 – A(m_J – m_J’) = 1.16 \times 10^{-4} – 2.1 \times 10^{-7} eV

For m_I = -1/2:\Delta E = g\mu_B B_0 + A(m_J – m_J’) = 1.16 \times 10^{-4} + 2.1 \times 10^{-7}eV

The hyperfine interaction splits each transition by approximately 4.2×10^{−7} eV.

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