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AMU Medical AMU Solved Paper-2015

  • question_answer
    Light of wavelength \[\lambda =5893\,\overset{o}{\mathop{A}}\,\] is incident on potassium surface. The stopping potential for the emitted electrons is 0.36 V. What is the maximum kinetic energy of the photoelectrons?

    A)  0.18eV                

    B)  0.36eV

    C)  0.74eV                

    D)  2.1eV

    Correct Answer: B

    Solution :

                    In case of photoelectric effect, \[e{{V}_{0}}=\frac{1}{2}m{{v}^{2}}\]    [where, \[{{V}_{0}}\] is stopping potential] \[\Rightarrow \] Maximum kinetic energy of photoelectric is 0.36 eV.


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