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JEE Main & Advanced Physics Photo Electric Effect, X- Rays & Matter Waves Question Bank Critical Thinking

  • question_answer
    Photoelectric emission is observed from a metallic surface for frequencies \[{{\nu }_{1}}\] and \[{{\nu }_{2}}\] of the incident light rays \[({{\nu }_{1}}>{{\nu }_{2}})\]. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of \[1:k\], then the threshold frequency of the metallic surface is                    [EAMCET (Engg.) 2001]

    A)             \[\frac{{{\nu }_{1}}-{{\nu }_{2}}}{k-1}\]                      

    B)            \[\frac{k{{\nu }_{1}}-{{\nu }_{2}}}{k-1}\]

    C)            \[\frac{k{{\nu }_{2}}-{{\nu }_{1}}}{k-1}\]                     

    D)            \[\frac{{{\nu }_{2}}-{{\nu }_{1}}}{k}\]

    Correct Answer: B

    Solution :

               By using \[h\nu -h{{\nu }_{0}}={{K}_{\max }}\]                    \[\Rightarrow h({{\nu }_{1}}-{{\nu }_{0}})={{K}_{1}}\]                                  ?.(i)                    And \[h({{\nu }_{2}}-{{\nu }_{0}})={{K}_{2}}\]                                  ?.(ii)                    \[\Rightarrow \frac{{{\nu }_{1}}-{{\nu }_{0}}}{{{\nu }_{2}}-{{\nu }_{0}}}=\frac{{{K}_{1}}}{{{K}_{2}}}=\frac{1}{K}\], Hence \[{{\nu }_{0}}=\frac{K{{\nu }_{1}}-{{\nu }_{2}}}{K-1}\].


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