Second answer

[SanderDW]

With no hyperfine interaction: There is only a shift because of the nuclear spin so not important for transitions. So that means we use the Zeemann splitting of the electron cloud which are equidistant levels with a distance of g_J \mu_B B_0. Using g_J = 0.8, B_0 = 2 T and \mu_B = 57.88 \mu eV/T we find an energy splitting of 92.608 \mu eV which are microwaves.

With hyperfine interaction: Generally the levels now get an extra meaningful shift of A m_I m_J. Because of the selection rule m_I doesn’t change so the difference in shift between relevant levels is caused by their difference in m_J. A m_I m_J – A m_I m_J’ = A m_I (m_J – m_J’). The selection rule for m_J changes gives us that the peaks shift with A m_I (where we assumed m_J to be the level with larger m_J value). For a nucleus with I = 1/2 this means we get two peaks that are shifted by +-A/2. A = \mu B_hf/IJ so using B_hf = 10 T, I = 1/2, J = 3/2, g_N = 1 and \mu_N = 3.152 * 10^-8 eV/T so that A = 0.42 \mu eV. Shifting the previous photon energy to two energies, namely 92.398 \mu eV and 92.818 \mu eV

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