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

  • question_answer
    The work function of a metallic surface is 5.01 eV. The photoelectrons are emitted when light of wavelength 2000 \[\overset{\text{o}}{\mathop{\text{A}}}\,\] falls on it. The potential difference applied to stop! The fastest photoelectrons is \[\left[ h=4.14\times {{10}^{-15}}eVs \right]\]

    A) 1.2 V   

    B) 2.24 V

    C) 3.6 V   

    D) 4.8 V

    Correct Answer: A

    Solution :

    [a] \[eV=hv-{{\phi }_{0}}=\left( \frac{12375}{2000}-5.01 \right)eV\] \[V=(6.1875-5.01)V=1.18V\approx 1.2V\]


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