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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2006

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


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