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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2001

  • question_answer
    If a\[{{H}_{2}}\]nucleus is completely converted into energy, the energy produced will be around:

    A)  1 MeV                                 

    B)  939 MeV            

    C)         9.39 MeV           

    D)         238 MeV

    E)  200 MeV

    Correct Answer: B

    Solution :

    Mass of\[{{H}_{2}}\]nucleus = mass of a proton \[=1.67\times {{10}^{-27}}kg\] Equivalent energy \[=\frac{1.67\times {{10}^{-27}}\times {{(3\times {{10}^{8}})}^{2}}}{1.6\times {{10}^{-13}}}MeV\]                 \[=939\,MeV\]


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