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NEET Sample Paper NEET Sample Test Paper-90

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