The Band Spectrum of HS

Abstract

The absorption spectrum of HS is obtained by passing repeated flashes from a source of continuous background through a discharge tube in which HS radicals are formed from ${\mathrm{H}}_{2}$S by pulses of radiofrequency current synchronized to precede immediately the flashes. The spectrograms obtained show only one band at 3237A. The rotational structure indicates that the ground state is an inverted $^{2}\ensuremath{\Pi}$, and the excited state a $^{2}\ensuremath{\Sigma}$. Because of the strong $\ensuremath{\Lambda}\ensuremath{\Sigma}$ coupling in the $^{2}\ensuremath{\Pi}$ state the doublet is wide, and since the Boltzmann factor greatly favors $^{2}\ensuremath{\Pi}_{\frac{3}{2}}$, only the strongest branch, ${Q}_{2}$, from the $^{2}\ensuremath{\Pi}_{\frac{1}{2}}$ state is found. From $^{2}\ensuremath{\Pi}_{\frac{3}{2}}$ the $^{Q}P_{21}$ and $^{R}Q_{21}$ satellite branches as well as ${P}_{1}$, ${Q}_{1}$ and ${R}_{1}$ are found. For the $^{2}\ensuremath{\Pi}$ ground state the best values of the constants obtained are $B_{0}^{\ensuremath{'}\ensuremath{'}}=9.47$; $D_{0}^{\ensuremath{'}\ensuremath{'}}=\ensuremath{-}0.001$ and $A=\ensuremath{-}378.6$ ${\mathrm{cm}}^{\ensuremath{-}1}$. The constants for the excited $^{2}\ensuremath{\Sigma}$ state are $B_{0}^{\ensuremath{'}}=8.30$; $D_{0}^{\ensuremath{'}}=\ensuremath{-}0.00078$, and $\ensuremath{\gamma}=0.32$ ${\mathrm{cm}}^{\ensuremath{-}1}$. The origin of the 0,0 band is at 30,659.1 ${\mathrm{cm}}^{\ensuremath{-}1}$.