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AMU Medical AMU Solved Paper-2011

  • question_answer
    Work function of potassium metal is 2.30 eV. When light of frequency \[8\times 1014\text{ }Hz\]is incident on the metal surface, photoemission of electrons occurs. The stopping potential of the electrons will be equal to

    A)  0.1 V                                    

    B)  1.0 V

    C)  5.0 V                                    

    D)  3.3 V

    Correct Answer: D

    Solution :

                                    \[E=e{{V}_{0}}\]                 \[E=hv=e{{V}_{0}}\] Stopping potential,                 \[{{V}_{0}}=\frac{hv}{e}\] \[=\frac{6.6\times {{10}^{-34}}\times 8\times {{10}^{14}}}{1.6\times {{10}^{-19}}}=3.3\,V\]


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