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JEE Main & Advanced Physics EM Waves Question Bank Mock Test - Electromagnetic Waves and Dual Nature

  • question_answer
    Given that a photon of light of wavelength 10,000 \[\overset{\text{o}}{\mathop{\text{A}}}\,\] has an energy equal to 1.23 eV. When light of wavelength 5000 \[\overset{\text{o}}{\mathop{\text{A}}}\,\] and intensity \[{{I}_{0}}\] falls on a photoelectric cell, the saturation current is \[0.40\times {{10}^{-6}}\]\[\overset{\text{o}}{\mathop{\text{A}}}\,\] and the stopping potential is 1.36 V; then the work function is

    A) \[0.43\,eV\]       

    B) \[1.10\,eV\]

    C) \[1.36\,eV\]       

    D) \[2.47\,eV\]

    Correct Answer: B

    Solution :

    [b] Since \[E\propto \frac{1}{\lambda }\], so energy corresponding to 5000 \[\overset{\text{o}}{\mathop{\text{A}}}\,\] is \[E=2.46eV\] Now, \[hv-W=e{{V}_{s}}\] Or \[2.46eV-W=1.36eV\] \[W=(2.46-1.36)eV=1.1eV\]


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