Abstract
The rotational and vibrational energy distributions have been determined for the OH radicals resulting from the 193 nm photolysis of the van der Waals complex N2O·H2O. Laser-induced fluorescence was used to probe the OH radicals. The rotational distributions of OH are characterized by the Boltzmann temperatures of 1700 K for 16OH v‘ ‘ = 0 and v‘ ‘ = 1 and of 1600 K for 18OH v‘ ‘ = 0. A quantity of 29% 16OH exists in the v = 1 state, while 18OH is produced almost exclusively in the v = 0 state. A possible reaction scheme is discussed based on the comparison of the present results with the corresponding bimolecular reaction and with the O3/H2O system.
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