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 ($J^{\pi }$T${=1\mathrm{}}^{+}$0) to the ω=3.56-MeV excited state ($J^{\pi }$T${=0\mathrm{}}^{+}$1) is calculated. From the inelastic form factor evaluated at ‖q‖=ω, 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.

• Received 2 December 1988

DOI:https://doi.org/10.1103/PhysRevC.39.1685