Knowledge Bank

The very large isotope shift $\sigma (N16O2)-\sigma (N18O2)=-19cm-1$ of the $(000)\leftarrow (000)$ band at $\Sigma 491\backslash AA$ can be explained only in terms of a very small frequency in the antisymmetrical mode of the excited electronic state. Photographs of the absorption spectra yield the isotope shifts $\Delta \nu 1\prime ,\Delta \nu 2\prime ,\Delta \nu 1\prime \prime$ and $\Delta \nu 2\prime \prime$. The frequency of the antisymmetrical mode of the ground state is $\nu 3\prime \prime =1618cm-1$ and the isotope shift of the zero-point energy in this mode is $\Delta G3\ast (0)=15.5cm-1$. Hence sufficient information is available for calculating the zero-point energy in the antisymmetrical mode of the excited state. The preliminary result is $\Delta G31(0)=3.5cm-1$. A strong semidiffuse band $714cm-1$ toward the violet side of the $(000)\leftarrow (000)$ band cannot be explained in terms of symmetrical $frequencies.1$ We have assigned this interval to $2\nu 31\equiv 1+-0+$ of a double-minimum potential. The double-minimum potential allows a quantitative explanation of the band interval, its isotope shift and the relative intensity of the band. The tentative value for the barrier is $600cm-1$. This work was supported by the U.S. Air Force Office of Scientific Research. $\mathrm{<mrow\; data-mjx-texclass="ORD">1</mrow>}$ R. K. Ritchie, A. D. Walsh and P. A. Warsop, Conference on Spectroscopy, Edited by M. J. Wells, Pergamon Press Ltd., London, 1962, p. 289.