Abstract
The ultraviolet photolysis of Ar–HBr(v=1) is studied through wave packet dynamics simulations, focusing on the fragmentation pathway Ar–HBr +ℏω→H+Ar–Br. Photolysis starts from two initial states of Ar–HBr(v=1) with a different angular shape, namely the ground and the first excited van der Waals (vdW) states, corresponding to the Ar–H–Br and Ar–Br–H isomers, respectively. It is found that the yield of Ar–Br radical products is substantially higher for the initial excited vdW state of the cluster, where H dissociation is less hindered. In addition, the yield of radical formation is much higher in the Ar–HBr(v=1) photolysis than that previously found in the Ar–HCl(v=0) case, even for the ground vdW state, where the initial angular distribution of both clusters is similar. Another unexpected difference is that Ar–HCl(v=0) photolysis exhibits strong manifestations of quantum interference, while these effects are much weaker in Ar–HBr(v=1). A lower probability of the first collision between the recoiling hydrogen and the Ar atom in the case of Ar–HBr(v=1), due to geometrical differences between its initial state and that of Ar–HCl(v=0), is suggested to explain the different photolysis behavior of both clusters. The implications of the present findings in the photolysis of other related precursor clusters are discussed.
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