CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    If e/m of electron is\[1.76\times {{10}^{11}}C\text{ }k{{g}^{-1}}\] and the stopping potential is 0.71 V, then the maximum velocity of the photo-electron is

    A)  \[150\text{ }km\,\,{{s}^{-1}}\]       

    B)         \[200\text{ }km\,\,{{s}^{-1}}\] 

    C)         \[500\text{ }km\,\,{{s}^{-1}}\]       

    D)         \[250\text{ }km\,\,{{s}^{-1}}\]

    E)  \[100\text{ }km\,\,{{s}^{-1}}\]

    Correct Answer: C

    Solution :

    From condition of stopping potential \[\frac{1}{2}m{{v}^{2}}=eV\] \[v=\sqrt{\frac{2eV}{m}}=\sqrt{2\times 1.76\times {{10}^{11}}\times 0.71}\] \[=5\times {{10}^{5}}m{{s}^{-1}}=500\text{ }km{{s}^{-1}}\]


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