Punjab Medical Punjab - MET Solved Paper-2011

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

    A) \[1000\]                              

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

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

    D) \[931\]

    Correct Answer: C

    Solution :

    Energy    \[=200MeV\]                 \[=200\times 1.6\times {{10}^{-19}}\times {{10}^{6}}\]                 \[=200\times 1.6\times {{10}^{-13}}\] Power\[P=1000kW\]                  \[=1000\times {{10}^{3}}W\] Rate of fission\[=\frac{Power}{Energy}\]                           \[=\frac{{{10}^{6}}}{200\times 1.6\times {{10}^{-13}}}\]                          \[=3.125\times {{10}^{16}}/s\]


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