second task

[DMauricio34]

For the first case, without hyperfine interaction we use the EPR resonance condition E = -gJ*μB*B0*mJ, with gJ = 1, B0 = 2 T, and the Bohr magneton μB = 5.7883×10^-5 eV/T, we calculat the energy for levels with mJ = -3/2, -1/2, and +1/2. This gives us a transition energy of 0.116 meV between adjacent levels. When we turn on the hyperfine interaction, an extra energy term A*mI*mJ appears, where A = (2/3) μN Bhf. With Bhf = 10 T and μN = 3.1525×10^-8 eV/T, we get A ≈ 2.1×10^-7 eV. This introduces small splittings to the energy levels depending on the nuclear spin. So, each EPR line splits into two levels — one slightly higher and one slightly lower.

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