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Question

One milliwatt of light of wavelength λ=4560 ˚A is incident on a cesium metal surface. Calculate the electrons liberaated. Assume a quantum efficiency of η=0.5%. [work function for cesium=1.89 eV] Take hc=12400 eV˚A

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Solution

We know that

E=hv=hcλ=6.6×1034×3×1084560×1010J=4.34×1019J

1mw of light energy is equivalent to 1034.34×1019=2.3×1015Photons/s

The quantum efficiency is 0.5

This means that only 0.5 of these photons release photoelectrons.
So, the number of electrons released from the surface per second

=2.3×1015×0.5100=1.15×1013electrons/s
The electrons curcent =1.15×1013×1.6×1018A

=1.84μA

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