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NEET Physics Nuclear Physics And Radioactivity NEET PYQ-Nuclear Physics

  • question_answer
    The binding energy of deuteron is 2.2 MeV and that of \[_{2}^{4}He\] is 28 MeV. If two deuterons are fused to form one \[_{2}^{4}He\] then the energy released is:                                                          [AIPMT (S) 2006]

    A)        25.8 MeV       

    B)       23.6 MeV         

    C)  19.2 MeV

    D)                   30.2 MeV         

    Correct Answer: B

    Solution :

    The reaction can be written as:
       \[_{1}{{H}^{2}}+{{\,}_{1}}{{H}^{2}}\,\xrightarrow[{}]{{}}\,{{\,}_{2}}H{{e}^{4}}\,+\text{energy}\]
       The energy released in the reaction in difference of binding energies of daughter and parent nuclei.
       Hence, energy released
       = binding energy of \[_{2}H{{e}^{4}}\]
                \[=28-2\times 2.2=23.6\,MeV\]


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