If the binding energy per nucleon in _3Li^7 and _2He^4 nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction _1H^1 + _3Li^7 → 2 _2He^4 , - Physics MCQ AIEEE 2006 2026 | ExamTest.live

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If the binding energy per nucleon in_{3} Li^{7} and_{2} He^{4} nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction_{1} H^{1} +_{3} Li^{7} \to 2_{2} He^{4}, the energy released is:

Solution & Step-by-step Explanation

Energy released (Q -value) = (Total B.E. of products) - (Total B.E. of reactants)Reactants:_{1} H^{1} (B.E. is zero) and_{3} Li^{7} .B.E. of_{3} Li^{7} = 7 \times 5.60 = 39.2 MeV .Products: 2 nuclei of_{2} He^{4} .B.E. of products = 2 \times (4 \times 7.06) = 2 \times 28.24 = 56.48 MeV . Q = 56.48 - 39.2 = 17.28 MeV .

Try it yourself before checking the explanation above.

If the binding energy per nucleon in_{3} Li^{7} and_{2} He^{4} nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction_{1} H^{1} +_{3} Li^{7} \to 2_{2} He^{4}, the energy released is:

39.2 MeV

28.24 MeV

17.28 MeV

1.46 MeV

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