Punjab Medical Punjab - MET Solved Paper-2001

  • question_answer
    The binding energy of deuterium and helium atom is \[1.1\,\,MeV\] and\[7.0\,\,MeV\]. If two deuterium nuclei fuse to form helium atom, the energy released is

    A) \[19.2\,\,MeV\]                               

    B) \[23.6\,\,MeV\]

    C) \[26.9\,\,MeV\]                               

    D) \[13.9\,\,MeV\]

    Correct Answer: B

    Solution :

    Total binding energy of                 \[_{1}{{H}^{2}}=2\times 1.1=2.2\,\,MeV\] Total binding energy of two deuterium nuclei\[=2\times 2.2=4.4\,\,MeV\] Total binding energy of                 \[_{2}H{{e}^{4}}=4\times 7=28\,\,MeV\] Energy released in fusion\[=28-4.4\]                                                  \[=23.6\,\,MeV\]


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