The binding energy per nucleon of deuteron ( ^2_1H ) and helium nucleus ( ^4_2He ) is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to - Physics MCQ AIEEE 2004 2026 | ExamTest.live

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The binding energy per nucleon of deuteron (_{1}^{2} H) and helium nucleus (_{2}^{4} He) is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is:

Solution & Step-by-step Explanation

Total binding energy = (B.E. per nucleon) \times (Number of nucleons) .B.E. of one deuteron = 1.1 \times 2 = 2.2 MeV .B.E. of two deuterons = 2 \times 2.2 = 4.4 MeV .B.E. of one helium nucleus = 7 \times 4 = 28 MeV .Energy released = B.E. of product - B.E. of reactants Energy released = 28 - 4.4 = 23.6 MeV .

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The binding energy per nucleon of deuteron (_{1}^{2} H) and helium nucleus (_{2}^{4} He) is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is:

13.9 MeV

26.9 MeV

23.6 MeV

19.2 MeV

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