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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2007

  • question_answer
    Consider the nuclear reaction \[{{X}^{200}}\to {{A}^{110}}+{{B}^{80}}\] binding energy per nucleon for\[X,\text{ }A\]and B are 7.4 MeV, 8.2 MeV and 8.1 MeV respectively, then the energy released in the reaction is

    A)  70 MeV

    B)  200 MeV            

    C)         190 MeV            

    D)         10 MeV

    E)  1480 MeV

    Correct Answer: A

    Solution :

    For\[X,\]energy \[=200\times 7.4=1480\text{ }MeV\] For A, energy \[=110\times 8.2=902\text{ }MeV\] For B, energy \[=80\times 8.1=648\text{ }MeV\] Therefore, energy released \[=(902+648)-1480\] \[=1550-1480=70\text{ }MeV\]                                 


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