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

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double the earlier value

unchanged

more than doubled

less than doubled

answer is C.

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Detailed Solution

$Let h v_{0} - W_{0} = K$

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

$\begin{array}{l} 2 h v_{0} - W_{0} = K_{1} \\ \Rightarrow 2 (h v - W_{0}) + W_{0} = K_{1} \\ \Rightarrow 2 K + W_{0} = K_{1} \end{array}$

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

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