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The binding energy per nucleon of ${}_{3}^{7}L i$ and ${}_{2}^{4}H e$ Nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reaction
${}_{3}^{7}{Li} + {}_{1}^{1}H \to {}_{2}^{4}{He} + {}_{2}^{4}{He} + Q$
the value of energy Q released is

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19.6 MeV

-2.4 MeV

8.4 MeV

17.3 MeV

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detailed solution

Correct option is B

Binding energy of Li 37 nucleus =7×5.60MeV=39.2MeV Binding energy of Henucleus 24=4×7.06MeV=28.24MeVthe reaction isLi 37+H 11→2He 24+Q∴Q=2BE of He 24-BE of Li 37=2×28.24MeV-39.2MeV=56.48MeV-39.2MeV=17.28MeV≈17.3MeV

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The binding energy per nucleon of L 37iand H 24e Nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reactionLi 37+H 11→He 24+He 24+Qthe value of energy Q released is

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The binding energy per nucleon of ${}_{3}^{7}L i$ and ${}_{2}^{4}H e$ Nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reaction
${}_{3}^{7}{Li} + {}_{1}^{1}H \to {}_{2}^{4}{He} + {}_{2}^{4}{He} + Q$
the value of energy Q released is