Lowest negative-parity states inBe12

Abstract

A very simple model is applied to the first four negative-parity states of 12 Be. Energies of the corresponding four states in 14 C are used to validate the model and to determine the doublet splitting parameters. Predictions for 12 Be are in remarkable agreement with excitation energies of the known 1 − at 2.70 MeV and the suspected (3 − ) at 4.56 MeV. Predicted excitation energies of 0 − and 2 − are 3.59 and 5.12 MeV, respectively.Introduction. In 12 Be, adding 2s 1/2 and 1d 5/2 neutrons to the first-excited state of 11 Be at 0.320 MeV [1] produces four negative-parity states, with J = 0 to 3. Of these, only the 1 − state at E x = 2.702 MeV [2] has been definitely identified. A candidate for the 3 − state is at 4.56 MeV. This state was initially suggested as 2 + [3], but it now appears likely [4,5] to be (3 − ) or a (3 − /2 + ) doublet. Millener [4] noted the excellent agreement between the 1.86 MeV energy difference between the 1 − and the probable (3 − ) in 12 Be and the 1/2 + -5/2 + splitting of 1.78 MeV in 11 Be. Comparison of heavy-ion-induced reactions leading to 12 Be [6,7] and 14 C [8] (where the 3 − is known) significantly strengthens the 3 − suggestion for the 4.56-MeV state. The suggestion [9] that the (3 − ) state might instead be 0 + has been addressed recently [10,11]. Nothing is known about possible 0 − and 2 − states, whose unnatural parity would have caused them to be very weak in the 10 Be(t,p) reaction. They were also not observed in two different investigations [12,13] of the 11 Be(d,p) reaction (in reverse kinematics), even though the 0 − spectroscopic factor is predicted [12] to be reasonably large. The possibility that the 0 − might be an isomer [14] depends on its energy relative to the neutron breakup threshold at 3.171 MeV [15].Various calculations have predicted the energy of the 0 − state. 16] produced a 0 − excitation energy of about 2.5 to 2.8 MeV. They had adjusted their potential to reproduce the ground state (g.s.) energy. They then found that they missed the energies of the second 0 + and first 2 + states by 0.84 and 0.92 MeV, respectively, which they fixed by adjusting a three-body force. They found they needed no three-body force for the 1 − state and hence used none for 0 − . Kanada-En'yo and Horiuchi [17] had predicted the 0 − to be very unbound, with E x ß 8-9 MeV, but they had the first 1 − above 5 MeV, and it is known at 2.7 MeV. Blanchon et al.[18] obtained an excitation energy of 2.91 MeV. Garrido et al. [9] have it at about 3.19 MeV. Kanungo et al.[12] calculated a 0 − energy at 5.59-5.91, but their energy of the 1 − state was 3.38-3.71 MeV. These predictions are summarized in Table I. A no-core shell-model calculation [19] considered only the first few states of 12 Be and thus provided no prediction for 0 − . Descouvement and Baye [20] calculated only natural-parity states and therefore had no 0 − prediction.In 14 C, four states with J π = 0 − to 3 − are known [21] and can be thought of as s and d neutrons coupled to the 1/2 − ground state (g.s.) of 13 C. Our aim h...

Observation of the near-threshold intruder $0^-$ resonance in $^{12}$Be

“…An ℓ = 1 assignment limits the possible spinparities to 0 − , 1 − or 2 − . The 1 − state has already been experimentally identified at 2.71 MeV [7], whereas the 2 − state should lie at much higher energy than the 0 − state, similar to the case of 14 C [34] and also considering the theoretical calculations shown in Fig. 3.…”

“…With one neutron in the intruder 2s 1/2 -orbital and the other in the normal 1p 1/2 -orbital, the 0 − and 1 − states in 12 Be are expected to have extended spatial distributions and should strongly couple to the continuum [1]. The 1 − state was observed at 2.715 MeV [7], whereas the 0 − state has been predicted by various theoretical approaches with an extremely large uncertainty, namely from 2.5 -9.0 MeV [24][25][26][27][28][29]34]. However, in spite of a large number of experimental studies, no bound 0 − state below the one-neutron separation energy (S n = 3.17 MeV) has been found [7-11, 35, 36].…”

“…However, in spite of a large number of experimental studies, no bound 0 − state below the one-neutron separation energy (S n = 3.17 MeV) has been found [7-11, 35, 36]. As outlined in [34], one neutron focused on the bound states in 12 Be and were insensitive to the coincident detection of Be residues related to a 12 Be resonance decay.…”