The dominant interaction of ${\mathrm{O}}_2$ with a magnetic field is through the electronic spin magnetic moment. However, a precise comparison with experiment of the results of calculating the microwave paramagnetic spectrum, assuming only this interaction, shows a systematic discrepancy. This discrepancy is removed by introducing two corrections. The larger (approximately 0.1 percent, or 7 gauss) is a correction for the second-order electronic orbital moment coupled in by the spin-orbit energy. Its magnitude is proportional to the second-order term ${\mu }^{\prime \prime }$ in the spin-rotation coupling constant. The smaller (approximately 1 gauss) is a correction for the rotation-induced magnetic moment of the molecule. Since the dependence of this contribution on quantum numbers is quite unique, this coefficient can also be determined by fitting the magnetic spectrum. A total of 120 $X$-band and 78 $S$-band lines were observed. The complete corrections have been made on 26 lines with a mean residual error of roughly 0.5 Mc/sec. This excellent agreement confirms the anomalous electronic moment to 60 parts per million (ppm) and also confirms the validity of the Zeeman-effect theory.

A new result is the rotational magnetic moment of -0.25±0.05 nuclear magnetons per quantum of rotation. Knowledge of this moment allows the electronic contribution to the effective moment of inertia to be determined. Making this correction of 65 ppm, and using the latest fitting of the universal atomic constants, the equilibrium internuclear distance is recomputed to be $R_e=1.20741\mathrm{}\pm 0.00002$ A. We can also deduce that the magnitude of ${\lambda }^{\prime \prime }$, the second-order spin-orbit contribution to the coupling of the spin to the figure axis, is 465±50 Mc/sec, or less than one percent of the total coupling constant $\lambda$.

Theoretical intensities of a number of the microwave transitions are calculated and successfully compared with experiment over a range of 100 to 1 in magnitude. It turns out that $\Delta M=0\mathrm{}$ transitions are over a hundred times weaker than the $\Delta M=\pm 1$ transitions and thus are too weak to observe. Also, $J$ breaks down as a quantum number in the presence of a magnetic field. This allows $\Delta J=\pm 2$ transitions to comprise roughly half of all lines observed.

• Received 9 June 1954

DOI:https://doi.org/10.1103/PhysRev.97.951