How different is the core of ${}^{25}$F from ${}^{24}$O${}_{g.s.}$?

Abstract

The neutron-shell structure of ${}^{25}$F was studied using quasi-free (p,2p) knockout reaction at 270A MeV in inverse kinematics. The sum of spectroscopic factors of $\pi$0d${}_{5/2}$ orbital is found to be $1.0\pm 0.3$. However, the spectroscopic factor for the ground-state to ground-state transition (${}^{25}$F, ${}^{24}$O${}_{g.s.}$) is only $0.36\pm 0.13$, and ${}^{24}$O excited states are produced from the 0d${}_{5/2}$ proton knockout. The result shows that the ${}^{24}$O core of ${}^{25}$F nucleus significantly differs from a free ${}^{24}$O nucleus, and the core consists of 35% ${}^{24}$O${}_{g.s}$. and 65% excited ${}^{24}$O.

Comments: 5 pages, 5 figures