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NEET AIPMT SOLVED PAPER SCREENING 2006

  • 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:

    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:                 \[_{1}{{H}^{2}}+{{\,}_{1}}{{H}^{2}}\,\xrightarrow[{}]{{}}\,{{\,}_{2}}H{{e}^{4}}\,+energy\]                 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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