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in Physics by (70.7k points)

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

Mass of 1H2 = 2.01478 u 

Mass of 2He4 = 4.00388 u

1 Answer

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Best answer

The reaction can be written as

Mass defect AM = [2×2.01478 – 4.00388] amu = 0.02568 u 

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

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