Inelastic (1(+)-->0(+)) electromagnetic form factor of 6Li from three-body models.

Abstract

Within the context of three-body (alpha particle plus two nucleons) models, and by assuming that the ground state of $^{6}\mathrm{He}$ (${0}^{+}$) and the lowest ${0}^{+}$ excited state of $^{6}\mathrm{Li}$ are members of the same isospin multiplet, the M1 transition form factor of $^{6}\mathrm{Li}$ for the inelastic electron scattering from the ground state (${J}^{\ensuremath{\pi}}$T${=1}^{+}$0) to the \ensuremath{\omega}=3.56-MeV excited state (${J}^{\ensuremath{\pi}}$T${=0}^{+}$1) is calculated. From the inelastic form factor evaluated at \ensuremath{\Vert}q\ensuremath{\Vert}=\ensuremath{\omega}, the radiative width for the deexcitation of the 3.56-MeV ${0}^{+}$ state of $^{6}\mathrm{Li}$ is given. It is suggested that the calculations of this work serve as a starting point for future investigations where refinements are made to the three-body models and more details of the electromagnetic interaction, e.g., meson exchange currents, are taken into account.