Two deuterons undergo nuclear fusion to form a Helium nucleus. The energy released in this process is (given binding energy per nucleon for deuteron = 1.1 MeV and for helium = 7.0 MeV )

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∵ Q = ∑ B . E . of products - ∑ B . E . of reactants H 2 + 1 H 2 → 2 H 4 Energy released = Q = 4 B . E . 2 H 4 - 4 B . E . 1 H 2 = 4 × 7 − 4 × 1.1 = 28 − 4.4 = 23.6 MeV

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