CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    The energy released in the fission of 1 kg of \[_{92}{{U}^{235}}\] is (energy per fission = 200 MeV)

    A)  \[5.1\times {{10}^{26}}eV\]

    B)  \[5.1\times {{10}^{26}}J\]

    C)  \[8.2\times {{10}^{13}}J\]

    D)  \[8.2\times {{10}^{13}}MeV\]

    E)  \[5.1\times {{10}^{23}}MeV\]

    Correct Answer: C

    Solution :

    Energy released from 1 kg of uranium \[=\frac{200\times {{10}^{6}}\times 1.6\times {{10}^{-19}}\times 6.023\times {{10}^{26}}}{235}\] \[=8.2\times {{10}^{13}}J\]


You need to login to perform this action.
You will be redirected in 3 sec spinner