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  4. 12. If the binding energy per nucleon in Li7 and He4 nuclei are respectively 5.60 MeV and 7.06 MeV, then energy of reaction Li7+p xrightarrow2 2He4 is:

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Q. 12. If the binding energy per nucleon in $ L{{i}^{7}} $ and $ H{{e}^{4}} $ nuclei are respectively 5.60 MeV and 7.06 MeV, then energy of reaction $ L{{i}^{7}}+p\xrightarrow{{}}2\,{{\,}_{2}}H{{e}^{4}} $ is:

JIPMERJIPMER 2000

Solution:

Binding energy of $ L{{i}^{7}}=39.20\,MeV $ Binding energy of $ H{{e}^{4}}=28.24\,MeV $ Therefore binding energy of $ 2H{{e}^{4}}=56.48\,MeV $ Now energy of reaction is $ =56.48-39.20=17.28\text{ }MeV $ $ =17.3\text{ }MeV $