Abstract
The angular dependence of the alignment of the \({}^{16}\) O(3 \({}^{-}\) , 6.13 MeV) nucleus formed in the \({}^{16}\) O( \(\alpha\) , \(\alpha\) ) \({}^{16}\) O(3 \({}^{-}\) ) and \({}^{15}\) N( \(\alpha\) , \(t\) ) \({}^{16}\) O(3 \({}^{-}\) ) reactions at \(\alpha\) -particle energies of 30.3 MeV is calculated taking into account the reactions’ mechanisms that determine the cross section for the formation of the \({}^{16}\) O(3 \({}^{-}\) ) nucleus and is compared with the experimental values of the alignments reconstructed from the measured functions of angular \(y\gamma\) correlations. It is established that the formation mechanisms of the \({}^{16}\) O(3 \({}^{-}\) ) nucleus significantly affect its orientation characteristics. The maximum amplitude alignment of the \({}^{16}\) O(3 \({}^{-}\) ) nucleus in both reactions is observed in the region of final particle emission angles in which direct mechanisms make the main contribution. At these angles, the calculated alignments are in satisfactory agreement with the experimental ones.
