JEE Main & Advanced Physics Nuclear Physics And Radioactivity Question Bank Critical Thinking

  • question_answer
    An atomic power nuclear reactor can deliver 300 MW. The energy released due to fission of each nucleus of uranium atom \[{{U}^{238}}\] is 170 MeV. The number of uranium atoms fissioned per hour will be       [Pb. PET 2004; Kerala PET 2005]

    A)            \[30\times {{10}^{25}}\]  

    B)            \[4\times {{10}^{22}}\]

    C)            \[10\times {{10}^{20}}\]  

    D)            \[5\times {{10}^{15}}\]

    Correct Answer: B

    Solution :

               \[P=\frac{nE}{t}\]\[\Rightarrow 300\times {{10}^{6}}\]\[=\frac{n\times 170\times {{10}^{6}}\times 1.6\times {{10}^{-19}}}{t}\]                                                               \ Number of atoms per sec \[\frac{n}{t}=1.102\times {{10}^{19}}\]                    Number of atoms per hour = 1.02 ´ 1019 ´ 3600                    = 3.97 ´ 1022.


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