Abstract
A discussion on the nature of the 0+ states in 68Ni (Z=28, N=40) is presented and a comparison is made with its valence counterpart 90Zr (Z=40, N=50). Evidence is given for a 0+ proton intruder state at only ~2.2 MeV excitation energy in 68Ni, while the analogous neutron intruder states in 90Zr reside at 4126 keV and 5441 keV. The application of a shell-model description of 0+ intruder states reveals that many pair-scattered neutrons across N=40 have to be involved to explain the low excitation energy of the proton-intruder configuration in 68Ni.
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