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NEET Physics Nuclear Physics And Radioactivity NEET PYQ-Nuclear Physics

  • 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:                                                                           [AIPMT (S) 2005]

    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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