Two deuterium nuclei fuse to form a helium-3 nucleus and a free neutron according to…

PHYSICS · Multi-part · A Level · Physics · Nuclear physics · Nuclear physics

Problem

Two deuterium nuclei 12H{}^{2}_{1}\text{H} fuse to form a helium-3 nucleus 23He{}^{3}_{2}\text{He} and a free neutron according to the reaction 12H+12H→23He+n{}^{2}_{1}\text{H}+{}^{2}_{1}\text{H}\rightarrow{}^{3}_{2}\text{He}+n. Given masses: 12H{}^{2}_{1}\text{H} : 2.01355 u2.01355\,\text{u} 23He{}^{3}_{2}\text{He} : 3.01550 u3.01550\,\text{u} Neutron : 1.00866 u1.00866\,\text{u} Take 1 u=931.5 MeV/c21\,\text{u}=931.5\,\text{MeV}/c^{2}. (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 1.0 g1.0\,\text{g} of deuterium were completely consumed in this reaction. (Avogadro number NA=6.02×1023 mol−1N_{A}=6.02\times10^{23}\,\text{mol}^{-1}.)

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