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MGIMS WARDHA MGIMS WARDHA Solved Paper-2009

  • question_answer
    Maximum velocity of photoelectron emitted is \[4.8m{{s}^{-1}}.\] If e/m ratio of electron is \[1.76\times {{10}^{11}}Ck{{g}^{-1}},\]then stopping potential is given by

    A) \[5\times {{10}^{-10}}J/C\]                         

    B) \[3\times {{10}^{-7}}J/C\]

    C) \[7\times {{10}^{-11}}J/C\]

    D) \[2.5\times {{10}^{2}}J/C\]

    Correct Answer: C

    Solution :

                     \[e{{V}_{s}}=\frac{1}{2}mv_{m}^{2}\] \[{{V}_{s}}=\frac{mv_{m}^{2}}{2e}=\frac{v_{m}^{2}}{2(e/m)}\] \[=\frac{{{(4.8)}^{2}}}{2\times 1.76\times {{10}^{11}}}\] \[=7\times {{10}^{-11}}J/C\]                                


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