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JEE Main & Advanced AIEEE Solved Paper-2006

  • 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 \[P+_{3}^{7}Li\to 2_{2}^{4}He\] energy of proton must be     AIEEE  Solved  Paper-2006

    A) 28.24 MeV     

    B)        17.28 MeV         

    C)        1.46 MeV

    D) 39.2 MeV

    Correct Answer: B

    Solution :

    According to given reactions, Energy of proton\[=7\times 5.60-2\times [4\times 7.06]\] \[\therefore \]Energy of proton =17.28 MeV


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