RAJASTHAN ­ PET Rajasthan PET Solved Paper-2007

  • question_answer
    If 200 MeV energy is released in the fission of 92 U235. What will be the required number of fission per second to produce 1 kW of power?

    A)  \[3.12\times {{10}^{13}}\]       

    B)  \[3.12\times {{10}^{3}}\]

    C)  \[3.1\times {{10}^{17}}\]        

    D)  \[3.12\times {{10}^{19}}\]

    Correct Answer: A

    Solution :

     Total energy/s = 1000 J Energy released/fission = 200 MeV \[=200\times 1.6\times {{10}^{-13}}J\] \[=3.2\times {{10}^{-11}}J\] \[\therefore \] Number of fission/s\[=\frac{1000}{3.2\times {{10}^{-11}}}\] \[=3.12\times {{10}^{13}}\]


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