Abstract
Using the Doppler-broadened 4.43-MeV gamma line from the reaction N15(p, α)C12* as the exciting radiation, resonance fluorescence from the 4.431-MeV level in Na23 has been observed. The 4.431-MeV ground-state transition, accounting for (95±3)% of the de-excitations, exhibits an isotropic angular distribution. This fact and the absence of branching to the 0.44-MeV 52+ level make ½ the most probable spin assignment. With this spin value, a self-absorption study leads to a partial width Γ0=(2.14±0.10)eV for the ground-state transition. When this width is combined with the observed branching, a total width Γ=(2.25±0.20)eV is obtained for the 4.431-MeV level in Na23, which is found to populate also the 2.39 MeV excited state in (5±3)% of its decays. A comparison of scattering and self-absorption data for different values of the ratio of natural width to Doppler width confirms the spin assignment of ½ to the 4.431-MeV level, which might be the first member of the K=12 band based on the Nilsson orbit 14. For a value η=+4 of the deformation parameter, the E1 transition probability calculated with Nilsson wave functions agrees with the experimental value.