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JEE Main & Advanced Physics Nuclear Physics And Radioactivity Question Bank Nucleus, Nuclear Reaction

  • question_answer
    200 MeV of energy may be obtained per fission of \[{{U}^{235}}\]. A reactor is generating 1000 kW of power. The rate of nuclear fission in the reactor is         [MP PET 1995]

    A)            1000                                         

    B)            \[2\times {{10}^{8}}\]

    C)            \[3.125\times {{10}^{16}}\]   

    D)             931

    Correct Answer: C

    Solution :

                       Power \[\Rightarrow {{\left( \frac{1}{2} \right)}^{4}}={{\left( \frac{1}{2} \right)}^{t/48}}\Rightarrow t=192\ hour.\]                    Rate of nuclear fission \[=\frac{{{10}^{6}}}{200\times 1.6\times {{10}^{-13}}}\]= 3.125 ´ 1016.


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