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

Double the earlier value

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Unchanged

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More than doubled

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Less than doubled

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Solution

The correct option is C
More than doubled

$E = W_{0} + K_{m a x} \Rightarrow K_{m a x} = E - W_{0} = h v - W_{0} \Rightarrow K_{1} = h v - W_{0} a n d K_{2} = 2 h v - W_{0} \Rightarrow K_{2} > 2 K_{1}$

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

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