CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2008

  • question_answer
    On bombarding \[{{U}^{235}}\] by slow neutron, 200 MeV energy is released, if the power output of atomic reactor is 1.6 MW, then the rate of fission will be

    A)  \[5\times {{10}^{22}}/s\]            

    B)  \[5\times {{10}^{16}}/s\]

    C)  \[8\times {{10}^{16}}/s\]            

    D)  \[20\times {{10}^{16}}/s\]

    Correct Answer: B

    Solution :

    Energy released on bombarding \[{{U}^{235}}\] by neutron = 200 MeV Power output of atomic reactor =1.6 MW \[\therefore \] Rate of fission \[=\frac{1.6\times {{10}^{6}}}{200\times {{10}^{6}}\times 1.6\times {{10}^{-19}}}\]                 \[=5\times {{10}^{16}}/s\]


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