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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2006

  • question_answer
    Maxium 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 \[1.8\times {{10}^{11}}C\text{ }k{{g}^{-1}}\], 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. \[\Rightarrow \]               \[{{V}_{s}}=\frac{mv_{\max }^{2}}{2\,e}=\frac{v_{\max }^{2}}{2\,(e/m)}\]                 \[=\frac{{{(1.2\times {{10}^{6}})}^{2}}}{2\times 1.8\times {{10}^{11}}}\]                 \[=0.4\times 10=4V\]


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