Past exam of the atomic and laser physics course of the University of Oxford / 2022 / 2 / d / Solution

The remaining six positions give the other isotope's sequenceThe adjacent differences divided by $4,5,6,7,8$ are $0.3125,0.3140,0.3133,0.3143,0.3138\mathrm{GHz}$, respectively. Thus $a_{\nu ,153}\simeq -0.314\mathrm{GHz}$ and the same endpoint argument gives $\boxed{I_{153}=5/2}$.

For the same electronic energy level, the magnetic-dipole hyperfine structure constant is proportional to ${\mu }_I/I$, to leading order in the small isotope-dependent change of the electronic wavefunction. Since the two nuclear spins are equal,Using the full spans averages over the approximate Landé interval rule: ${\sum }_{F=4}^{8}F=30$, soThe constants have the same sign, so the nuclear magnetic dipole moments have the same relative sign. Their absolute sign cannot be obtained from their ratio without the electronic field's sign. Finite nuclear-size corrections to the hyperfine interaction, often called the hyperfine anomaly, and electric-quadrupole corrections are neglected at this accuracy.