Second task: Photon energy

[WannesB]

First we calculate the energy without hyperfine interaction:
for the transition energy, we calculate the separate energy levels E(mJ=-1/2) and E(mJ=-3/2)
E(m=-1/2) = -g * mu_N * B * mJ (with g = 1, mu_N = 3.15e-8 eV/T, B = 2T, mJ=1/2)
E(mJ=-1/2) = 3.15e-8 eV
Same for E(mJ=-3/2) gives 9.45e-8 eV, so we got a transition energy of 6.3e-8 eV.

Now with the hyperfine interaction, we need to adapt our formula for E to:
E(mJ) = g * mu_N * B * mJ + A * mJ * mI, with A = mu_N * Bhf/(I * J) = 4.2e-7 eV.
From the answers of the previous section, we then get:
E(mJ=-3/2) = 9.45e-8 eV + 4.2e-7 * (1/2) * (-3/2) eV = -22.1e-8 eV
E(mJ=-1/2) = 3.15e-8 eV + 4.2e-7 * (1/2) * (-1/2) eV = -7.35e-8 eV
E(mJ=1/2) = -3.15e-8 eV + 4.2e-7 * (1/2) * (1/2) eV = 7.35e-8 eV
E(mJ=-3/2) – E(mJ=-1/2) = -1.47e-7 eV
E(mJ=-1/2) – E(mJ=1/2) = -1.47e-7 eV
(the value of E(mJ=1/2) without hyperfine is the negative as E(mJ=-1/2)(only the negative sign of mJ changes))

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