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JEE Main & Advanced Sample Paper JEE Main Sample Paper-15

  • question_answer
    Consider the nuclear reaction\[{{X}^{200}}\xrightarrow[{}]{{}}{{A}^{110}}+{{B}^{90}}\]. If the binding energy per nucleon for X, A and B is 7.4 MeV, 8.2. MeV and 8.2 MeV respectively, what is the energy released?

    A)  200 MeV            

    B)  160 MeV

    C)  110 MeV            

    D)  90 MeV

    Correct Answer: B

    Solution :

    \[\Delta E=(110\times 8.2+90\times 8.2)-200\times 7.4\] \[=160\,\text{MeV}\]


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