CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    The temperature at which protons in proton gas would have enough energy to overcome Coulomb barrier of\[4.14\times {{10}^{-14}}J\]is (Boltzmann constant\[=1.38\,\times {{10}^{-23}}\,J{{K}^{-1}}\])

    A)  \[2\times {{10}^{9}}K\]

    B)         \[{{10}^{9}}K\]

    C)  \[6\times {{10}^{9}}K\]

    D)         \[3\times {{10}^{9}}K\]

    E)  \[4.5\times {{10}^{9}}K\]

    Correct Answer: A

    Solution :

    Given: \[k=1.38\times {{10}^{-23}}J{{K}^{-1}}\] The energy of proton gas \[=4.14\times {{10}^{-14}}J\] \[\therefore \]  \[E=\frac{3}{2}kT\] or \[4.14\times {{10}^{-14}}=\frac{3}{2}\times 1.38\times {{10}^{-23}}\times T\]                 \[T=2\times {{10}^{9}}K\]


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