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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    A photon of energy 8 eV is incident on metal surface of threshold frequency\[1.6\times {{10}^{15}}Hz.\] The kinetic energy of the photoelectrons emitted (in eV): (Take\[h=6\times {{10}^{-34}}J-s\])

    A)  1.6                        

    B)         6

    C)  2                            

    D)         1.2

    E)  2.6

    Correct Answer: C

    Solution :

    \[KE=hv-h{{v}_{0}}\] \[=8eV-\left( \frac{6\times {{10}^{-34}}\times 1.6\times {{10}^{15}}}{1.6\times {{10}^{-19}}}eV \right)\] \[=8-6=2eV\]


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