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

  • question_answer
    If the binding energy per nucleon in \[_{3}^{7}Li\] and \[_{2}^{4}He\] nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction \[p+_{3}^{7}Li\xrightarrow{{}}2_{2}^{4}He\] energy of proton must be

    A)  28.24 MeV

    B)  17.28 MeV

    C)  1.46MeV

    D)  39.2MeV

    Correct Answer: B

    Solution :

    [b] Let E be the energy of proton, then \[E+7\times 5.6=2\times [4\times 7.06]\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,E=56.48-39.2=17.28\,MeV\]


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