RAJASTHAN ­ PET Rajasthan PET Solved Paper-2002

  • question_answer
    The kinetic energy of neutron at temperature 300 K will be\[(m=1.67\times {{10}^{-27}}kg)\]

    A)  \[6.21\times ~{{10}^{-21}}J\]    

    B) \[8.2\times {{10}^{-21}}J\]

    C)  \[9\times {{10}^{-21}}J\]       

    D)  None of these

    Correct Answer: A

    Solution :

     \[E=\frac{3}{2}KT\] [where, K = Botzmann constant\[=1.38\times {{10}^{-23}}J/K]\] \[\therefore \] \[E=\frac{3}{2}\times 1.38\times {{10}^{-23}}\times 300\] \[=6.21\times {{10}^{-21}}J\]


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