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Calculate the energy liberated when a single helium nucleus is formed by the fusion of two deuterium nuclei.

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Calculate the energy liberated when a single helium nucleus is formed by the fusion of two deuterium nuclei. Given 

Mass of H2 = 2.01478 u 

Mass of 2He4 = 4.00388 u

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The reaction can be written as

1H2 + 1H2 → 2He4 + Energy

Mass defect AM = [2 x 2.01478 - 4.00388] amu = 0.02568 u

Hence energy released is = 0.2568×931 MeV ≈ 24 MeV

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