J. A. Brown

A two-proton knockout reaction from a ${}^{17}$C beam at 55 MeV/u was used to produce the unbound nucleus ${}^{15}$Be. No significant number of events were observed for the decay of ${}^{15}$Be into a neutron and a ${}^{14}$Be fragment. An upper limit for the ${}^{14}$Be production cross section of 0.079$\pm$0.038 mb was extracted, which is more than an order of magnitude smaller than the predicted two-proton knockout reaction cross section. Based on these results we conclude that any populated states in ${}^{15}$Be decay through three sequential neutron decays into ${}^{12}$Be, via the unbound first excited state of ${}^{14}$Be. Therefore, these states in ${}^{15}$Be must be neutron unbound by more than 1.54 MeV, which is the location of the first excited 2${}^{+}$ state in ${}^{14}$Be.

Neutron-unbound states in ${}^{25}$F and ${}^{26}$F were populated via the reactions ${}^9$Be(${}^{26}$Ne,${}^{24}$F + n) and ${}^9$Be(${}^{26}$Ne,${}^{25}$F + n), respectively. A resonance close to the neutron separation energy in ${}^{25}$F was identified with a decay energy of $28\pm 4$ keV. This resonance corresponds to an excited state in ${}^{25}$F at $4249\pm 116$ keV assuming it decays to the ground state of ${}^{24}$F. Guided by shell-model calculations, a spin and parity of 1/2${}^{-}$ can be assigned to this state. In the spectrum of ${}^{26}$F, which was produced in a nucleon-exchange reaction, there are indications for an excited state with a decay energy of $\sim$270 keV.

Neutron-unbound resonant states of ${}^{11}\mathrm{Be}$ were populated in neutron knockout reactions from ${}^{12}\mathrm{Be}$ and identified by ${}^{10}\mathrm{Be}$–$n$ coincidence measurements. A resonance in the decay-energy spectrum at 80(2) keV was attributed to a highly excited unbound state in ${}^{11}\mathrm{Be}$ at 3.949(2) MeV decaying to the $2^{+}$ excited state in ${}^{10}\mathrm{Be}$. A knockout cross section of 15(3) mb was inferred for this 3.949(2) MeV state, suggesting a spectroscopic factor near unity for this $0p3/2^{-}$ level, consistent with the detailed shell model calculations.

We have observed a resonance in neutron-fragment coincidence measurements that is presumably the first excited state of ${}^{23}\mathrm{O}$ at 2.8(1) MeV excitation energy which decays into the ground state of ${}^{22}\mathrm{O}$. This interpretation is consistent with theory. The reaction mechanism supports the assignment of the observed state as the $5/2^{+}$ hole state. This assignment and the recently observed $3/2^{+}$ particle state advance the understanding of ${}^{23}\mathrm{O}$.