J & K CET Medical J & K - CET Medical Solved Paper-2008

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

    A) \[5\times {{10}^{16}}{{/}_{s}}\]                

    B) \[10\times {{10}^{16}}{{/}_{s}}\]

    C) \[15\times {{10}^{16}}{{/}_{s}}\]                              

    D) \[20\times {{10}^{-16}}{{/}_{s}}\]

    Correct Answer: A

    Solution :

                     \[Rate\text{ }of\text{ }fission=\frac{P}{energy\text{ }for\text{ }fission}\] \[=\frac{1.6\times {{10}^{6}}}{200\times 1.6\times {{10}^{-13}}}\] \[=5\times {{10}^{16}}{{s}^{-1}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner