Molecular Beam Investigation of Rotational Transitions. I. The Rotational Levels of KCl and Their Hyperfine Structure

Abstract

The molecular beam electrical resonance method, which has heretofore been confined to the study of transitions between the $m=0\mathrm{}$ and $m=\pm 1$ states of diatomic polar molecules in the state $J=1\mathrm{}$, has been extended to the study of rotational transitions from $J=0\mathrm{}$ to $J=1\mathrm{}$. The inhomogeneous electric fields together with a stop wire are so arranged that molecules in the state $J=0\mathrm{}$ are selectively refocused on the detector. A transition to the state $J=1\mathrm{}$, under the influence of the applied rf field, is observed by a reduction in beam intensity at the detector. The spectra obtained in this way yield more information than with the older method. The hyperfine structure of the state $J=1\mathrm{}$ of the molecules ${\mathrm{K}}^{39}$ ${\mathrm{Cl}}^{35}$, ${\mathrm{K}}^{39}$ ${\mathrm{Cl}}^{37}$, and ${\mathrm{K}}^{41}$ ${\mathrm{Cl}}^{35}$ has been investigated at zero field for several vibrational states. The molecular constants obtained for these molecules are: The structure of the $J=1\mathrm{}$ state was attributable to a nuclear quadrupole interaction for each nucleus. The variations of ${\mathrm{}\left(\mathrm{eqQ}\right)\mathrm{}}_{\mathrm{K}}$ and ${\mathrm{}\left(\mathrm{eqQ}\right)\mathrm{}}_{\mathrm{Cl}}$ with vibrational state are The ratios of the quadrupole moments of the potassium isotopes is $\frac{Q_{{\mathrm{K}}^{41}}}{Q_{{\mathrm{K}}^{39}}}=1.220\mathrm{}\pm \mathrm{0.002.}$ From observation of line shifts in a weak electric field, the molecular dipole moment was determined in two vibrational states. For $v=0\mathrm{}$, $\mu =10.48\mathrm{}\pm 0.05$ Debye, and for $v=2\mathrm{}$, $\mu =10.69\mathrm{}\pm 0.05$ Debye. The ratio of the dipole moments is 1.020±0.004. The mass ratios of the chlorine and potassium isotopes calculated from the molecular constants are $\frac{{\mathrm{Cl}}^{35}}{{\mathrm{Cl}}^{37}}=0.9459803\mathrm{}\pm 0.0000015 \mathrm{and} \frac{{\mathrm{K}}^{39}}{{\mathrm{K}}^{41}}=0.9512189\mathrm{}\pm \mathrm{0.0000015.}$