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MGIMS WARDHA MGIMS WARDHA Solved Paper-2008

  • question_answer
    Maximum velocity of the photoelectrons emitted by a metal surface is \[1.2\times {{10}^{6}}m{{s}^{-1}}.\] Assuming the specific charge of the electron to be \[1.8\times {{10}^{11}}C\,k{{g}^{-1}},\]the value of the stopping potential in volt will be

    A) 2                                             

    B) 3

    C) 4                                             

    D) 6

    Correct Answer: C

    Solution :

    Specific charge of electron, \[\frac{e}{m}=1.8\times {{10}^{11}}C\,k{{g}^{-1}}\] Maximum kinetic energy of photoelectron \[\frac{1}{2}mv_{\max }^{2}=e{{V}_{s}}\] where\[{{V}_{s}}\]is the stopping potential. \[\Rightarrow \]               \[{{V}_{s}}=\frac{mv_{\max }^{2}}{2e}=\frac{v_{\max }^{2}}{2(e/m)}\]                 \[=\frac{{{(1.2\times {{10}^{6}})}^{2}}}{2\times 1.8\times {{10}^{11}}}\] \[=0.4\times 10=4V\]


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