Low-lying proton intruder state in 13B

Abstract

The neturon rich nucleus 13 B was studied via the proton transfer reaction 4 He( 12 Be, 13 Bγ) at 50A MeV. The known 4.83-MeV excited state was strongly populated and its spin and parity were assigned to 1/2 + by comparing the angular differential cross section data with DWBA calculations. This low-lying 1/2 + state is interpreted as a proton intruder state and indicates a deformation of the nucleus.The existence of intruder states in light neutron-rich unstable nuclei is often considered to be evidence for one or morehω configurations in the low-lying states around the psd shell. The ground state of 11 Be is 1/2 + which is lower in energy by 0.3 MeV than the 1/2 − state [1]. In 12 Be, there is a 1 − intruder state at 2.7-MeV excitation energy [2]. The energies of these low-lying, nonnormal parity states indicate 1hω configurations. Furthermore, the presence of low-lying 2 + 1 [3] and 0 + 2 [4,5] states in 12 Be suggests a 2hω configuration.Three theoretical interpretations have been proposed for these one or morehω configurations in the low-lying states of neutron-rich unstable nuclei: (1) the monopole interaction of the tensor force [6], (2) the loosely bound nature of some orbitals [7], and (3) nuclear deformation [5,8,9]. In Ref.[6],the effective interaction was determined so that the model reproduces the energy levels in light, neutron-rich nuclei including intruder states, and the importance of monopole interaction due to the tensor force was pointed out. Reference[7] discusses the fact that in the psd shell, the 2s 1/2 orbital gains its energy relative to the other orbitals due to its loosely bound nature. The non-zero

Structure ofC14andB14from theBedoor1, Wuosmaa2, Albers3 et al. 2016Phys. Rev. C17130View full textAdd to dashboardCite

“…Assuming J π = 1/2 + , the deduced spectroscopic factor is C 2 S = 0.13±0.02, or about 5% of that of the ground state. A J π = 1/2 + assignment agrees with the results of Ota et al, where the 4.83 MeV state was populated with a proton-stripping spectroscopic strength of C 2 S = 0.2 ± 0.02 [15]. These observations imply that π(1s) 2 (0p) −2 admixtures are present in the 14 C ground-state wave function.…”

“…A recent lifetime measurement has suggested that the lowest negative-parity excited state at 3.53 MeV had J π =3/2 − and possessed neutron 2p − 2h character [14]. Finally, a J π = 1/2 + proton-"intruder" state was reported in 13 B from the 14 Be(α, t) 13 B reaction [15] at 4.83 MeV excitation energy. Population of this state in the 14 C(d, 3 He) 13 B reaction may suggest some admixtures of proton-(sd)…”

Spectroscopy of ^{<span class="cmr-7">13</span>}B via the (^{<span class="cmr-7">18</span>}O,^{<span class="cmr-7">16</span>}O) Two Neutron Transfer Reaction

“…1 is another evidence of the selectivity of the ( 18 O, 16 O) reaction. Indeed, states excited in the one nucleon transfer reactions are scarcely populated, as for example the 4.8 MeV state which has been recently identified as a proton intruder state [2]. Finally, a qualitative comparison with the 14 C excitation energy spectrum populated via the ( 18 O, 16 O) at 85 MeV is reported in Fig.…”

“…The N = 8 nucleus 13 B is, therefore, expected to be on the borderline between these two opposite regimes. Recent results have demonstrated such peculiarity, showing that the 13 B isotope exhibits both low lying proton and neutron intruder states indicating a reduced shell gap between p and sd shells [2,3].…”

12B(n,g) : the influence on r-process nucleosynthesis of light elements

“…Rauscher et al [10] estimated the uncertainties of the reaction rates for the 11 B(n,γ) and 12 B(n,γ) reactions to be a factor of 2, based on the limited spectroscopic information on 12 B and 13 B and considering both direct and resonant captures. Although there have been several investigations of the poorly known structure of 13 B [11], such as single-nucleon knockout [12], a proton intruder study [13], and a search for isobaric analog states [14], as well as theoretical efforts in cluster studies [15], the spin-parities of most excited states are not well determined [16].…”