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

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 $Delta R_{rm ch}$ = 0.049,(4)~fm. Based on the correlation between $Delta R_{rm ch}$ 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.