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NEET Sample Paper NEET Sample Test Paper-79

  • question_answer
    Consider the nuclear reaction\[{{\operatorname{D}}^{200}} \to  {{A}^{110}} + {{B}^{80}}\]. If the binding energy per nucleon for D, A and B are 7.4 MeV. 8.2 MeV and 8.1 MeV respectively, the amount of energy released in the reaction is

    A)  200 MeV                     

    B)  140 MeV

    C)  70 MeV                       

    D)  None of these

    Correct Answer: C

    Solution :

    For D, \[\operatorname{energy} = 200 \times 7.4\] \[=\,\,\,1480\,\,MeV\] For A, \[\operatorname{energy} = 110 \times  8.2 MeV\] \[=\,\,\,9.2\,\,MeV\] For B, \[\operatorname{energy} = 80 \times  8.1\] \[=\,\,\, 648 MeV\] \[\therefore ~Energy released = \left( 902 + 648 \right) -\left( 1480 \right)\] \[=\,\,\,70\,\,MeV\]


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