A Band of Unusual Structure Probably Due to a Highly Unstable Calcium Hydride Molecule

Abstract

In the spectrum of the calcium are in hydrogen at low pressure (best 3-20 mm) an isolated band occurs in association with the lines of neutral calcium. The wide spacing of the lines and the absence of appreciable isotope effect practically assure its hydride origin. The simple structure indicates a diatomic molecule. Reproductions and measurements of the lines are given. The band appears to consist of a P and an R branch with a single missing line in the middle. The two branches do not form a single series. From the abnormally weak central lines, the intensity of each branch rises to a maximum, then begins to fall slowly, until the lines are cut off sharply at the tenth line in the R and the eleventh in the P branch. These two lines both show a perturbation of -0.60 wave-number units. Assuming that these last perturbed lines have a common initial state, the band is analyzed in accordance with Kratzer's theory, applicable to molecules which have net electronic angular momentum in the plane of molecular rotation. For the initial state, the energy may be expressed by: $\frac{E^{\prime }}{\mathrm{hc}}=C+28,353.38+0.039\mathrm{} (j^{\prime }-\frac{3}{4})+4.798\mathrm{} {(j^{\prime }-\frac{3}{4})}^2-0.000900\mathrm{} {(j^{\prime }-\frac{3}{4})}^4$; for the final state, by: $\frac{E^{\prime \prime }}{\mathrm{hc}}=C-0.043\mathrm{} (j^{\prime \prime }-\frac{1}{2})+4.232\mathrm{} {({(j}^{\prime \prime }-\frac{1}{2})}^2-0.000192\mathrm{} {(j^{\prime \prime }-\frac{1}{2})}^4$, where $j$ is the rotational quantum number. The last line in each branch corresponds to $j^{\prime }=11\mathrm{}$. This limitation of the $j^{\prime }$ values, and the limitation to the value 0 of the vibrational quantum numbers $n^{\prime }$ and $n^{\prime \prime }$, indicated by the isolated character of the band, are explained as due to extreme molecular instability, such that the force holding the Ca and H atoms together begins to fall off very rapidly with distance if the atoms depart somewhat from their (rotationless) equilibrium position. Such a rapid falling off is possible only for a non-polar molecule. The emitting molecule is probably, then, a compound of hydrogen with an excited (not ionized) calcium atom, which may be of an unstable ${\mathrm{H}}_2$-like type analogous to the emitters of the Zn, Cd, Hg, Cu, Mg and other hydride bands.