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NEET Sample Paper NEET Sample Test Paper-89

  • question_answer
    Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is: (given binding energy per nucleon for deuteron =1.1 MeV and for helium=7.0MeV)

    A) 30.2 MeV         

    B)        32.4 MeV

    C) 23.6 MeV         

    D)        25.8 MeV

    Correct Answer: C

    Solution :

    [c] \[{}_{1}{{H}^{2}}+{}_{1}{{H}^{2}}\to {}_{2}H{{e}^{4}}\] Total binding energy of two deuterium nuclei \[=1.1\times 4=4.4\,\,MeV\] Binding energy of a \[({}_{2}H{{e}^{4}})\] nuclei \[=4\times 7=28\,\,MeV\] Energy released in this process \[=28-4.4=23.6\,\,MeV\]


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