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

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    If 200 MeV energy is released in the fission of a single \[{{U}^{235}}\]nucleus, the number of fissions required per second to produce 1 kilowatt power shall be (Given \[1\ eV=1.6\times {{10}^{-19}}J\]) [AMU 1995; MP PMT 1999]

    A)            \[3.125\times {{10}^{13}}\]   

    B)            \[3.125\times {{10}^{14}}\]

    C)            \[3.125\times {{10}^{15}}\]   

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

    Correct Answer: A

    Solution :

                       \[P=n\,\left( \frac{E}{t} \right)\Rightarrow 1000=\frac{n\times 200\times {{10}^{6}}\times 1.6\times {{10}^{-19}}}{t}\]                    \[\]\[\Rightarrow \frac{n}{t}=3.125\times {{10}^{13}}.\]


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