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EAMCET Medical EAMCET Medical Solved Paper-1999

  • question_answer
    The threshold frequency for photo electric effect for a metal surface is found to be \[4.8\times {{10}^{16}}Hz,\] the stopping potential required when the metal is irradiated by radiation of frequency \[5.6\times {{10}^{16}}Hz\]is: (taking \[h=6.6\times {{10}^{-34}}J-s\]and\[e=6.6\times {{10}^{-19}}C\])

    A)  22.4 V                                  

    B)  33 V

    C)  66 V                                     

    D)  198 V

    Correct Answer: B

    Solution :

                     The equation for photoelectric effect given by Einstein is expressed as \[{{E}_{K}}=hv-W\]                 or            \[hv=h{{v}_{o}}+e{{V}_{o}}\]                 Given: \[{{v}_{o}}=4.8\times {{10}^{16}}Hz,v=5.6\times {{10}^{16}}Hz,\]                                 \[h=6.6\times {{10}^{-34}}Js,e=1.6\times {{10}^{-19}}C.\]                                 \[6.6\times {{10}^{-34}}\times 5.6\times {{10}^{16}}\]                 \[=6.6\times {{10}^{-34}}\times 4.8\times {{10}^{16}}+{{V}_{0}}\times 1.6\times {{10}^{-19}}\]                                 \[36.96\times {{10}^{-18}}\]                 \[=31.68\times {{10}^{-18}}+{{V}_{0}}\times 1.6\times {{10}^{-19}}\]                 \[{{V}_{0}}=\frac{36.96\times {{10}^{-18}}-31.68\times {{10}^{-18}}}{1.6\times {{10}^{-19}}}\]                                 \[=\frac{523\times {{10}^{-18}}}{1.6\times {{10}^{-19}}}\] \[=3.3\times {{10}^{19}}\times {{10}^{-18}}=3.3\times 10=33\,\text{volt}\]


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