The frequency of incident light falling on a photosensitive metal plate is doubled, the KE of the emitted photoelectrons is

The frequency of incident light falling on a photosensitive metal plate is doubled, the KE of the emitted photoelectrons is

  1. A

    double the earlier value

  2. B

    unchanged

  3. C

    more than doubled

  4. D

    less than doubled

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    Solution:

     Let hv0W0=K

    If frequency is doubled, let kinetic energy of photoelectrons be K1.

        2hv0W0=K1    2hvW0+W0=K1    2K+W0=K1

    i.e., kinetic energy is more than doubled.

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