If the frequency of light incident on a photosensitive metal plate is doubled, then the KE of photoelectrons will be

Doubled

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Halved

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Quadrupled

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More than twice the previous value

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Solution

The correct option is B More than twice the previous value
$K E = h (V - V_{0})$
$V^{1} = 2 V$
$K E^{1} = h (2 V - V_{0})$
$K E^{1} = 2 h (V - V_{0}) + h V_{0}$
$K E^{1} = 2 K E + h V_{0}$
So, the answer is option (D).

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