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 (OH)-O-16 v " = 0 and v " = 1 and of 1600 K for (OH)-O-18 v " = 0. A quantity of 29% (OH)-O-16 exists in the v = 1 state, while (OH)-O-18 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 O-3/H2O system.