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JEE Main & Advanced Sample Paper JEE Main Sample Paper-25

  • 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)  1                    

    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 photoelectrons \[\frac{1}{2}m{{v}^{2}}_{\max }\,=e{{V}_{s}}\] Where \[{{V}_{s}}\]is the stopping potential \[{{V}_{s}}\,=\frac{mv_{\max }^{2}}{2e}=\frac{v_{\max }^{2}}{2(e/m)}\] \[=\frac{{{(1.2\,\times {{10}^{6}})}^{2}}}{2\times 1.8\times {{10}^{11}}}\,=0.4\,\times 10\,=4\,V\]


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