Physics

Physics

Problem

For the fusion reaction 12H+13H  →  24He+01n^{2}_{1}\text{H}+^{3}_{1}\text{H}\;\rightarrow\;^{4}_{2}\text{He}+^{1}_{0}\text{n}, calculate the energy released per reaction event in MeV. Atomic masses: 12H=2.0141 u^{2}_{1}\text{H}=2.0141\,\text{u} 13H=3.0160 u^{3}_{1}\text{H}=3.0160\,\text{u} 24He=4.0026 u^{4}_{2}\text{He}=4.0026\,\text{u} 01n=1.0087 u^{1}_{0}\text{n}=1.0087\,\text{u} Take 1 u=931.5 MeV/c21\,\text{u}=931.5\,\text{MeV}/c^{2}. a) 14.0 MeV14.0\,\text{MeV} b) 1.76 MeV1.76\,\text{MeV} c) 17.6 MeV17.6\,\text{MeV} d) 10.4 MeV10.4\,\text{MeV}

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