A metal surface is illuminated by a light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one-fourth of its original value, then the maximum KE of emitted photoelectons will becomes

(1 / 16)^{t h}

of original value

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Unchanged

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Twice the original value

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Four times the original value

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Solution

The correct option is A Unchanged
We know that the Einstein's equation for photoelectric effect is $K E_{m a x} = h ν - W$
So maximum kinetic energy is depends on frequency $(ν)$ of incident light and work function $(W)$ of metal.
Thus, maximum kinetic energy does not depend on the intensity. Hence the maximum kinetic energy remains unchanged.

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