Abstract
Calculations for the 40Ar(p,n) 40K reaction to the 0 + anti-analog state (AAS) are given along with experimental angular distributions for an incident proton energy of 22.8 MeV. The results of the calculations show little sensitivity to the parameters. The large angle data and the relative strengths of the analog state and AAS cross sections indicate that the two-step (p,d) (d,n) process plays a dominant role in the AAS transition.
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