Abstract
We have recorded the resonance-enhanced multiphoton ionization spectrum for NO (A) products from photodissociation of the N-2-NO complex. We made measurements at excitation energies ranging from 28 to 758 cm(-1) above the threshold to produce NO (A) + N-2 (X) products, and the resulting spectra reveal the NO (A) rotational states formed during dissociation, allowing us to determine the rotational state distribution. At the lowest available energies, 28 and 50 cm-1 above threshold, we observed contributions from NO (A) rotational states that exceed the available energy and must originate from excitation due to hotbands of the complex. At all higher energies, we did not observe any energetically disallowed NO (A) rotational states, and for all available energies above 259 cm(-1) the observed rotational transitions do not extend to the maximum allowed by energy conservation. Furthermore, the observed distributions were typically biased toward low rotational states, in contrast with expectations from vibrational predissociation. From the rotational state distributions, we determined the average fraction of energy partitioned into NO (A) rotation, f (NO rot, ave), to be 0.088 at the highest available energy, and this fraction increased as the available energy decreased. By combining the average NO (A) rotational energy along with the average center-of-mass translational energy from our previous work, we determined the average rotational energy for the undetected N-2 (X) photoproduct. The results showed that the N-2 fragment has a higher average rotational energy relative to the NO fragment. Finally, we found that the NO (A) rotational state distribution was colder than expected for a statistical dissociation.