The Rotational Analysis of the First Negative Group of Oxygen (O2+) Bands

The ultraviolet $\mathrm{O}_2^{}{}_{}{}^{+}$ bands have been produced by a hollow cathode discharge, or with greater intensity from a mixture of helium with a small amount of oxygen in a large discharge tube. Photographs were taken in the second order of a 21 foot Rowland grating (dispersion = 1.32A/mm). The following bands were used for a rotational analysis: 1-8; 0-8; 1-9; 0-9. As was expected, since NO and $\mathrm{O}_2^{}{}_{}{}^{+}$ have the same number of electrons, these bands correspond to a ${}_{}{}^2{}_{}{}^{}\Pi \to {}_{}{}^2{}_{}{}^{}\Pi$ transition like the double headed $\beta$ bands of NO. The lower ${}_{}{}^2{}_{}{}^{}\Pi$ is case $a$ as in NO. Unlike the case of NO, however, the upper ${}_{}{}^2{}_{}{}^{}\Pi$ is case $b$. The lower ${}_{}{}^2{}_{}{}^{}\Pi$ is regular, with a doublet separation $A^{\prime \prime }$ (corresponding to zero rotation) of 195 ${\mathrm{cm}}^{-1\mathrm{}}$. In the upper ${}_{}{}^2{}_{}{}^{}\Pi$ the value of the spin-orbital coupling coefficient $A^{\prime }$ is +8.2. Each band consists of eight branches (four $P$ and four $R$), the $Q$ branches apparently being too weak to appear on the plates. This is in agreement with theory. $\Lambda$-type doubling is negligible in the upper ${}_{}{}^2{}_{}{}^{}\Pi$ but is present in the lower ${}_{}{}^2{}_{}{}^{}\Pi$ state. Alternate levels in each successive $\Lambda$-type doublet are missing. This is as prediced by the quantum theory of homopolar molecules, since it is known that the nuclear spin of the oxygen atom is zero. The probable electronic configuration of the upper ${}_{}{}^2{}_{}{}^{}\Pi$ state is given as... $2p{\pi }^{3}3d{\pi }^2$, and that of the lower ${}_{}{}^2{}_{}{}^{}\Pi$ state as... $2p{\pi }^{4}3d\pi$. The constants of the molecule in the two electronic states are given by: $B^{\prime }=1.048-0.014\mathrm{} v^{\prime }$; $B^{\prime \prime }=1.610-0.009\mathrm{} v^{\prime \prime }$; ${r_e}^{\prime }=1.41\mathrm{}\times {10}^{-8\mathrm{}}$ cm; ${r_e}^{\prime \prime }=1.14\mathrm{}\times {10}^{-8\mathrm{}}$ cm; $A^{\prime }=+8.2\mathrm{}$ ${\mathrm{cm}}^{-1\mathrm{}}$; $A^{\prime \prime }=+195\mathrm{}$ ${\mathrm{cm}}^{-1\mathrm{}}$.