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Q.

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

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a

Double the earlier value

b

More than doubled

c

Unchanged

d

Less than doubled

answer is C.

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Detailed Solution

$E = W_{o} + K_{m a x}$

$\Rightarrow K_{m a x} = E - W_{o} = h v - W_{o}$

$\Rightarrow K_{1} = h v - W_{o} a n d K_{2} = 2 h v - W_{o}$

$\Rightarrow K_{2} > 2 K_{1}$

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