Charge Radius of Neutron-Deficient Ni 54 and Symmetry Energy Constraints Using the Difference in Mirror Pair Charge Radii

The nuclear root-mean-square charge radius of $^{54}\mathrm{Ni}$ was determined with collinear laser spectroscopy to be $R(^{54}\mathrm{Ni})=3.737(3)\text{ }\text{ }\mathrm{fm}$. In conjunction with the known radius of the mirror nucleus $^{54}\mathrm{Fe}$, the difference of the charge radii was extracted as $\mathrm{\ensuremath{\Delta}}{R}_{\mathrm{ch}}=0.049(4)\text{ }\text{ }\mathrm{fm}$. Based on the correlation between $\mathrm{\ensuremath{\Delta}}{R}_{\mathrm{ch}}$ and the slope of the symmetry energy at nuclear saturation density ($L$), we deduced $21\ensuremath{\le}L\ensuremath{\le}88\text{ }\text{ }\mathrm{MeV}$. 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 $1\ensuremath{\sigma}$ error bands. Our result indicates the neutron-skin thickness of $^{48}\mathrm{Ca}$ as 0.15--0.21 fm.