Charge Radius of Neutron-deficient ${}^{54}$Ni and Symmetry Energy Constraints Using the Difference in Mirror Pair Charge Radii

Abstract

The nuclear root-mean-square charge radius of ${}^{54}$Ni was determined with collinear laser spectroscopy to be $R{(}^{54}$Ni) = 3.737\,(3)~fm. In conjunction with the known radius of the mirror nucleus ${}^{54}$Fe, the difference of the charge radii was extracted as $\Delta R_{\mathrm{ch}}$ = 0.049\,(4)~fm. Based on the correlation between $\Delta R_{\mathrm{ch}}$ and the slope of the symmetry energy at nuclear saturation density ($L$), we deduced $20\le L\le 70$\,MeV. The present result is consistent with the $L$ from the binary neutron star merger GW170817, favoring a soft neutron matter EOS, and barely consistent with the PREX-2 result within 1$\sigma$ error bands. Our result indicates the neutron-skin thickness of ${}^{48}$Ca as 0.15\,-\,0.19\,fm.

Comments: 6 total pages, 3 figures