Neutral dissociation of the I, I', and I" vibronic progressions of O2

J Chem Phys. 2010 May 28;132(20):204303. doi: 10.1063/1.3432199.

Abstract

It is suggested that the main mechanism for neutral dissociation of the I, I('), and I(") vibronic progressions in O(2) is due to their interaction with the vibrational continuum of the 1pi(u) (-1)(A (2)Pi(u))3ssigma(g) (3)Pi(u)(v(epsilon)) Rydberg state (J state) leading to the formation of the O(2p(4) (3)P)+O( *)(2p(3)((4)S)3s (3)S) fragments. In order to justify this, the O I 2p(3)((4)S)3s (3)S-->2p(4) (3)P fluorescence emission cross section following the neutral dissociation of the O(2) 1pi(u) (-1)(a (4)Pi(u))4ssigma(g)/3ddelta(g)/3dsigma(g) (3)Pi(u)(v) Rydberg states is simulated in the exciting-photon energy range of 14.636-16.105 eV. The results of high-resolution measurements (H. Liebel et al., J. Phys. B 34, 2581 (2001)) can be reproduced if a small adjustment of the computed potential curve of the J state is applied. Non-Franck-Condon resonant intensity distributions of the I, I('), and I(") progressions observed in the experiment are qualitatively explained by the presence of the O(2) 1pi(g) (-1)(X (2)Pi(g))npsigma(u)/nfsigma(u)/nfdelta(u) (3)Pi(u) perturber states. Present calculations allow to decide between two different assignments of the I, I('), and I(") states available in literature.