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WB JEE Medical WB JEE Medical Solved Paper-2010

  • question_answer
    The energy released by the fission of one uranium atom is 200 MeV. The number of fissions per second required to produce 3.2 W of power is (Take, \[1\,eV=1.6\times {{10}^{-19}}J\])

    A)  \[{{10}^{7}}\]                                   

    B)  \[{{10}^{10}}\]

    C)  \[{{10}^{15}}\]                                 

    D)  \[{{10}^{11}}\]

    Correct Answer: D

    Solution :

                     We have the energy released by fission of one uranium atom is 200 MeV. So,     \[E=200\times {{10}^{6}}\times 1.6\times {{10}^{-19}}\]                 \[=3.2\times {{10}^{-11}}\,J\] The power required = 3.2 W Thus the number of fissions required is equal to\[\frac{3.2}{3.2\times {{10}^{-11}}}={{10}^{11}}\]fissions.


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