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JEE Main & Advanced Sample Paper JEE Main Sample Paper-18

  • question_answer
    The minimum energy required to overcome the attractive forces electron and surface of Ag metal is \[7.52\times {{10}^{-19}}J\]. What will be the maximum K.E. of electron ejected out from Ag which is being exposed to U.V. light of\[\lambda =360\overset{\text{o}}{\mathop{\text{A}}}\,\]?

    A)  \[36.38\times {{10}^{-19}}J\]    

    B)  \[6.92\times {{10}^{-19}}J\]

    C)  \[57.68\times {{10}^{-19}}J\]    

    D)  \[67.68\times {{10}^{-19}}J\]

    Correct Answer: B

    Solution :

     Energy absorbed\[=\frac{hc}{\lambda }\] \[=\frac{6.625\times {{10}^{-27}}\times 3.0\times {{10}^{10}}}{360\times {{10}^{-8}}}=5.52\times {{10}^{-11}}erg\] \[=5.52\times {{10}^{-18}}Joule\] \[=(7.52\times {{10}^{-19}})-(0.552\times {{10}^{-19}})\] \[=6.92\times {{10}^{-19}}\,Joule\].


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