Abstract

The H atom production channels in the ultraviolet (UV) photochemistry of m-pyridyl radical (m-C5H4N) were investigated at excitation wavelengths from 224 nm to 248 nm by high-n Rydberg atom time-of-flight (HRTOF) technique. The photofragment yield (PFY) spectrum of the H atoms indicates a broad UV absorption feature near 234 nm. The product kinetic energy release is small; the average product kinetic energy at the wavelengths from 224 nm to 248 nm is 0.12 to 0.19 of the maximum excess energy (assuming the lowest energy product channel, H+cyanovinylacetylene). The product kinetic energy distributions are consistent with the H+cyanovinylacetylene, H+3,4-pyridyne, and H+2,3-pyridyne product channels, with H+cyanovinylacetylene as the main H-loss channel. The angular distributions of the H-atom products are isotropic. After the UV electronic excitation, the m-pyridyl radical undergoes internal conversion to the ground electronic state and then unimolecular dissociation to the H+cyanovinylacetylene, H+3,4-pyridyne, and H+2,3-pyridyne products. The dissociation mechanism of the m-pyridyl radical is similar to that of the o-pyridyl radical reported in the early study.

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