Two-neutron transfer in the He6+Sn120 reaction

A large yield of $\ensuremath{\alpha}$ particles produced in the $^{120}\mathrm{Sn}(^{6}\mathrm{He},\ensuremath{\alpha})$ reaction was measured at 20.3-, 22.2-, 22.4-, and 24.5-MeV bombarding energies. The $\ensuremath{\alpha}$ particles are distributed over a broad energy range in the vicinity and below the elastic scattering $^{6}\mathrm{He}$ peak. Energy integrated $\ensuremath{\alpha}$-particle cross sections have been obtained at ${\ensuremath{\theta}}_{\text{lab}}={36}^{\ensuremath{\circ}},{40}^{\ensuremath{\circ}}$, and ${60}^{\ensuremath{\circ}}$. The $\ensuremath{\alpha}$ energy distributions have been analyzed at a fixed laboratory angle $(\ensuremath{\approx}{60}^{\ensuremath{\circ}})$ in terms of the reaction $Q$ value, considering the $2n$-transfer reaction kinematics $^{120}\mathrm{Sn}(^{6}\mathrm{He},\ensuremath{\alpha})^{122}\mathrm{Sn}$. A kinematical analysis of the $Q$-value distribution shows that the recoil system $^{120}\mathrm{Sn}+2n$ is formed in highly excited states in the continuum, at increasing excitation energies as the bombarding energy increases. It is shown that by using Brink's formula, the excitation energy depends on the transferred angular momentum following a linear relation with the square of the angular momentum, indicating that some kind of dinuclear rotating system is formed after the reaction.