Disintegration of Lithium by Deuterons

Abstract

The energy distribution of the neutrons from the disintegration of lithium by 0.85 MEV deuterons has been determined by the method of recoil protons in a high pressure cloud chamber. The results of the experiment indicate that the neutrons come from the two reactions ${}_3{}^{}{}_{}{}^{}\mathrm{Li}_{}^7{}_{}{}^{}+{}_1{}^{}{}_{}{}^{}\mathrm{H}_{}^2{}_{}{}^{}\to 2{}_2{}^{}{}_{}{}^{}\mathrm{He}_{}^4{}_{}{}^{}+{}_0{}^{}{}_{}{}^{}n_{}^1{}_{}{}^{}; {}_3{}^{}{}_{}{}^{}\mathrm{Li}_{}^7{}_{}{}^{}+{}_1{}^{}{}_{}{}^{}\mathrm{H}_{}^2{}_{}{}^{}\to {}_4{}^{}{}_{}{}^{}\mathrm{Be}_{}^8{}_{}{}^{}+{}_0{}^{}{}_{}{}^{}n_{}^1{}_{}{}^{}.$ The neutrons from the first reaction (95 percent of the disintegrations) have a continuous energy distribution, while the neutrons from the second reaction (5 percent of the disintegrations) are nearly homogeneous, with a maximum energy of 13.3±0.5 MEV. From these data, the mass of ${}_4{}^{}{}_{}{}^{}\mathrm{Be}_{}^8{}_{}{}^{}$ was calculated to be 0.3±0.75 MEV greater than that of two alpha-particles.