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JEE Main & Advanced Physics Nuclear Physics And Radioactivity Question Bank Self Evaluation Test - Nuclei

  • question_answer
    The binding energy of deuteron \[(_{1}^{2}H)\] is 1.15 A MeV per nucleon and an alpha particle \[(_{2}^{4}He)\]has a binding energy of 7.1 MeV per nucleon. Then in the reaction \[_{1}^{2}He+_{1}^{2}He\to _{2}^{2}He+Q\] the energy released Q is:

    A)  5.95 MeV

    B)  26.1 MeV

    C)  23.8 MeV

    D)  289.4 MeV

    Correct Answer: C

    Solution :

    [c] Given,  \[_{1}{{H}^{2}}{{+}_{1}}{{H}^{2}}{{\to }_{2}}{{H}^{4}}+Q\] The total binding energy of the deutrons \[=4\times 1.15=4.60\,MeV\] The total binding energy of alpha particle \[=4\times 7.1=28.4\,MeV\] The energy released in the process \[=28.4-4.60=23.8\,MeV\].


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