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AMU Medical AMU Solved Paper-2009

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

    A)  33.0 MeV                           

    B)  28.4 MeV

    C)  23.8 MeV                           

    D)  4.6 MeV

    Correct Answer: C

    Solution :

                     In the reaction                 \[_{1}^{2}H+_{1}^{2}H\xrightarrow{{}}_{2}^{4}He+Q\] Energy of reactants is \[=4\times 1.15\] (for 2 deuterons)                 = 4.60 MeV Energy of product \[=4\times 7.1=28.4\,\,MeV\] \[\therefore \] Q of reaction = 28.4 - 4.6 = 23.8 MeV


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