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NEET AIPMT SOLVED PAPER SCREENING 2005

  • question_answer
                    In the reaction \[_{1}^{2}H+\,_{1}^{3}H\,\to \,_{2}^{4}He+\,_{0}^{1}n\], if the binding energies of \[_{1}^{2}H\,,\,_{2}^{3}H\] and \[_{2}^{4}He\] are respectively a, b and c (in MeV), then the energy (in MeV) released in-this reaction is:                                                         

    A)                 \[c+a-b\]            

    B)                 \[c-a-b\]

    C)                 \[a+b+c\]           

    D)                 \[a+b-c\]

    Correct Answer: B

    Solution :

                              Key Idea: The energy released per nuclear reaction is the resultant binding energy.                 Binding energy of (\[_{1}^{2}H\] + \[_{1}^{3}H\]) = a + b                 Binding energy of \[_{2}^{4}He\] = c                 In a nuclear reaction the resultant nucleus is more stable than the reactants. Hence, binding energy of \[_{2}^{4}He\] will be more than that of \[\left( _{1}^{2}H+\,_{1}^{3}H \right)\]                 Thus, energy released per nucleon                 = resultant binding energy                 \[=c-(a+b)=c-a-b\]                 Note:    Out of the elements present in periodic table, Fe(56) has maximum binding energy. Hence it is most stable.


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