Two deuterium nuclei 12H fuse to form a helium-3 nucleus 23He and a free neutron according to the reaction
12H+12H→23He+n.
Given masses:
12H : 2.01355u23He : 3.01550u
Neutron : 1.00866u
Take 1u=931.5MeV/c2.
(a) Compute the mass defect for one fusion event.
(b) Find the energy released in MeV.
(c) Convert the energy obtained in part (b) to joules.
(d) Estimate the total energy, in kilowatt-hours, that would be released if 1.0g of deuterium were completely consumed in this reaction. (Avogadro number NA=6.02×1023mol−1.)