利用镜像对电荷半径差异确定缺中子\(^{54}Ni\)的电荷半径及对称能约束

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

作者信息

Pineda Skyy V, König Kristian, Rossi Dominic M, Brown B Alex, Incorvati Anthony, Lantis Jeremy, Minamisono Kei, Nörtershäuser Wilfried, Piekarewicz Jorge, Powel Robert, Sommer Felix

机构信息

National Superconducting Cyclotron Laboratory, Michigan State University, East Lansing, Michigan 48824, USA.

Department of Chemistry, Michigan State University, East Lansing, Michigan 48824, USA.

出版信息

Phys Rev Lett. 2021 Oct 29;127(18):182503. doi: 10.1103/PhysRevLett.127.182503.

DOI:10.1103/PhysRevLett.127.182503

PMID:34767412

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 ΔR_{ch}=0.049(4) fm. Based on the correlation between ΔR_{ch} and the slope of the symmetry energy at nuclear saturation density (L), we deduced 21≤L≤88 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 1σ error bands. Our result indicates the neutron-skin thickness of ^{48}Ca as 0.15-0.21 fm.

摘要

利用共线激光光谱法测定了(^{54}Ni)的核均方根电荷半径为(R(^{54}Ni)=3.737(3))飞米。结合镜像核(^{54}Fe)的已知半径，提取出电荷半径差为(\Delta R_{ch}=0.049(4))飞米。基于(\Delta R_{ch})与核饱和密度下对称能斜率（(L)）之间的相关性，我们推断(21\leq L\leq88)兆电子伏特。目前的结果与双中子星合并GW170817得到的(L)值一致，支持软中子物质状态方程，并且在(1\sigma)误差带内与PREX - 2结果勉强一致。我们的结果表明(^{48}Ca)的中子皮厚度为(0.15 - 0.21)飞米。