Abstract
The two-proton pick-up reaction 12C(12C,14O)10Be has been measured at 211.4 MeV incident energy to study the structure of 10Be states. This reaction populates most strongly 0+ and 2+ two-proton hole-configurations in the Ip-shell, but also odd-parity states. A two-step mechanism is needed in the latter case with particle-hole excitations of the type (lp)−1(2sld)1 in inelastic excitation (including neutron excitations) in the one step, and the pick-up of a pair of protons from 12C in the other step. For all observed states the oscillatory structure of the angular distributions is characteristic of their spins, as can be verified for states with known spins. In this way we can make the firm spin assignments of 4+ and 3− for the states at 11.8 MeV and 10.55 MeV, respectively. From the reaction mechanism, which offers optimum conditions to populate the 4+ configuration in the Ip-shell, and from the J(J+l)-rule for excitation energies within a rotational band we conclude, that the assigned 4+ state is most probably a member of the ground state band of 10Be.
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