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JIPMER Jipmer Medical Solved Paper-2007

  • question_answer
    In the reaction \[_{1}^{2}H+_{1}^{3}H\to _{2}^{4}He+_{0}^{1}n,\] if the binding energies of \[_{\text{1}}^{\text{2}}\text{H,}_{\text{1}}^{\text{3}}\text{H}\]and \[_{2}^{4}\text{He}\] 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 \[{}_{2}^{4}\text{He}\,\text{=}\,\text{c}\] In a nuclear reaction the resultant nucleus is more stable than the reactants. Hence, binding energy of \[{}_{2}^{4}\text{He}\] will be more than that of\[({}_{1}^{2}\text{H}\,\text{+}\,{}_{1}^{3}\text{H)}\text{.}\] Thus, energy released per nucleon = resultant binding energy \[=c-(a+b)=c-a-b\]


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