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The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is
A
23.6 MeV
B
2.2 MeV
C
28.0 MeV
D
30.2 MeV
Detailed Solution
Reaction: $^2_1H + ^2_1H \rightarrow ^4_2He$ + energy
Binding energy of one deuterium nucleus (2 nucleons) = $2\times1.1 = 2.2$ MeV
Total binding energy of the two deuterium nuclei = $2\times2.2 = 4.4$ MeV
Binding energy of the helium nucleus (4 nucleons) = $4\times7.0 = 28.0$ MeV
Energy released = binding energy of product − binding energy of reactants = 28.0 − 4.4
= 23.6 MeV
Binding energy of one deuterium nucleus (2 nucleons) = $2\times1.1 = 2.2$ MeV
Total binding energy of the two deuterium nuclei = $2\times2.2 = 4.4$ MeV
Binding energy of the helium nucleus (4 nucleons) = $4\times7.0 = 28.0$ MeV
Energy released = binding energy of product − binding energy of reactants = 28.0 − 4.4
= 23.6 MeV