If the frequency of light in a photoelectric experiment is doubled then maximum kinetic energy of photo electron.

be doubled

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be halved

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becomes more than double

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become less than double

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Solution

The correct option is A be doubled
$h V = \frac{1}{2} m v^{2} + w o r k f u n c t i o n$
Wen V frequency in when kinetic energy is maximum work function = 0
So if we double the frequency then kinetic energy is doubled.

Hence (A) option is correct

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