second task

[DanteBouckhout]

In the absence of hyperfine interactions, the energy levels of a lanthanum atom under an external magnetic field can be calculated straightforwardly. Each energy level E(mJ) depends on the magnetic quantum number mJ​ according to the formula E=−gμNB0mJE=−gμN​B0​mJ​, where g is the Landé g-factor, μN​ is the nuclear magneton, and B0​ is the external magnetic field strength.

E(−32)=−1×3.15×10−8 eV/T×2 T×−32=9.45×10−8 eV E(−23​)=−1×3.15×10−8eV/T×2T×−23​=9.45×10−8eV
E(−12)=−1×3.15×10−8 eV/T×2 T×−12=3.15×10−8 eV E(−21​)=−1×3.15×10−8eV/T×2T×−21​=3.15×10−8eV
E(12)=−1×3.15×10−8 eV/T×2 T×12=−3.15×10−8 eV E(21​)=−1×3.15×10−8eV/T×2T×21​=−3.15×10−8eV

With hyperfine interactions considered, additional terms involving the hyperfine coupling constant AA and the nuclear magnetic quantum number mImI​ are added to the energy expressions.

E(−32)=−1×3.15×10−8 eV/T×2 T×−32+42×10−8 eV×12×(−32)=−22.05×10−8 eVE(−23​)=−1×3.15×10−8eV/T×2T×−23​+42×10−8eV×21​×(−23​)=−22.05×10−8eV
E(−12)=−1×3.15×10−8 eV/T×2 T×−12+42×10−8 eV×12×(−12)=−7.35×10−8 eVE(−21​)=−1×3.15×10−8eV/T×2T×−21​+42×10−8eV×21​×(−21​)=−7.35×10−8eV
E(12)=−1×3.15×10−8 eV/T×2 T×12+42×10−8 eV×12×12=7.35×10−8 eVE(21​)=−1×3.15×10−8eV/T×2T×21​+42×10−8eV×21​×21​=7.35×10−8eV

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