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EAMCET Medical EAMCET Medical Solved Paper-2004

  • question_answer
    K1 and K2 are the maximum kinetic energies of the photoelectrons emitted, when light of wavelengths\[{{\lambda }_{1}}\]  and  \[{{\lambda }_{2}}\] respectively are incident on a metallic surface. If \[{{\lambda }_{1}}=3{{\lambda }_{2}},\]then:

    A)                  \[{{K}_{1}}>\frac{{{K}_{2}}}{3}\]                              

    B)                  \[{{K}_{1}}<\frac{{{K}_{2}}}{3}\]

    C)                  \[{{K}_{1}}=3{{K}_{2}}\]                              

    D)                  \[{{K}_{2}}=3{{K}_{1}}\]

    Correct Answer: B

    Solution :

                      Energy of each photon is                 \[E=\frac{hc}{\lambda }-\frac{hc}{{{\lambda }_{0}}}\] Case \[{{K}_{1}}=\frac{hc}{{{\lambda }_{1}}}-\frac{hc}{{{\lambda }_{0}}}\]                 \[{{K}_{1}}=\frac{hc}{3{{\lambda }_{2}}}-\frac{hc}{{{\lambda }_{0}}}\]                    Case II                                                                                                                   ?(1)                 \[{{K}_{2}}=\frac{hc}{{{\lambda }_{2}}}-\frac{hc}{{{\lambda }_{0}}}\]                                                                                                       ?(2)                 From Eq. (1)                 \[3{{K}_{1}}=\frac{hc}{{{\lambda }_{2}}}-\frac{3hc}{{{\lambda }_{0}}}\]                 \[3{{K}_{1}}={{K}_{2}}+\frac{hc}{{{\lambda }_{0}}}-\frac{3hc}{{{\lambda }_{0}}}\]                 \[3{{K}_{1}}={{K}_{2}}-\frac{2hc}{{{\lambda }_{0}}}\] \[3{{K}_{1}}<{{K}_{2}}\]                 \[{{K}_{1}}<\frac{{{K}_{2}}}{3}\]


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