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

TL;DR

This study investigates the neutron-shell structure of $^{25}$F using knockout reactions, revealing that its core significantly differs from a free $^{24}$O nucleus and contains a mixture of ground and excited states.

Contribution

It provides new experimental data on the spectroscopic factors and core composition of $^{25}$F, highlighting the difference from a free $^{24}$O core.

Findings

The spectroscopic factor for the $^{25}$F to $^{24}$O ground-state transition is 0.36.

The $^{24}$O core in $^{25}$F is approximately 35% in the ground state and 65% in excited states.

The sum of spectroscopic factors for the $ u$0d$_{5/2}$ orbital is 1.0 ± 0.3.

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.