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One milliwatt of light of wavelength 456...

One milliwatt of light of wavelength 4560 A is incident on a cesium surface. Calculate the photoelectric current liberated assuming a quantum efficiency of 0.5 %. Given Planck's constant `h=6.62xx10^(-34)J-s` and velocity of light `c=3xx10^(8)ms^(-1)`.

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The correct Answer is:
`84 xx 10^(-6) Amp`
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