The 15 N(d, τ) 14 C reaction and high-lying T = 1 states in mass-14 nuclei: A determination of 1 p 1 2 2 and 1 p 1 2 1p 3 2 diagonal matrix elements
197114 pages
Published in:
- Nucl.Phys.A 178 (1971) 275-288
- Published: 1971
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Abstract: (Elsevier)
Energy spectra up to E x = 20 MeV and angular distributions of τ-particles from the 15 N(d, τ) 14 C reaction have been measured at an incident energy of 52 MeV and have been analysed by the DWBA. Five strong l = 1 transitions to 1p proton hole states occur up to E x = 11.29 MeV, which is 3 MeV above the particle emission threshold. The absence of direct l = 0 and l = 2 transitions demonstrates that 15 N is a good shell-model nucleus with respect to protons. The isobaric analogue 1p hole states in 14 C and 14 N are identified by a comparison with T = 1 states excited in the 15 N(p, d) reaction. This implies the following assignments for 14 C levels: J π = 2 + for the 8.32 MeV level; J π = 1 + for a new level at 11.29±0.04 MeV with a width of 150±50 keV; J π = (0, 1, 2) + for one of the 10.433 and 10.453 MeV levels. This state corresponds to the formerly unknown T = 1, T z = 0 isobaric analogue at 12.52 MeV in 14 N. The J π = 0 + and 1 + , T = 1 states of mass 14 are rather pure 1p hole states, while the (2 + , 1) states are strongly mixed with (s, d) 2 configurations. The complete 1p −2 spectrum could be observed. Combined with neutron pick-up data the six diagonal matrix elements of the 1 p 1 2 −2 and the 1 p 1 2 −1 1 p 3 2 −1 interaction have been determined and compared to theoretical results. The eight multipole coefficients α κτ ( j 1 j 2 ) have been calculated by the use of the energy weighted sum rules. The 1 p 1 2 −1 1 p 3 2 −1 spectrum is nearly identical to one produced by a δ-force without exchange terms.- Nuclear Reactions
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