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Question

One milliwatt of light of wavelength λ=4560 oA is incident on a cesium metal surface. Calculate the electron current liberated. Assume a quantum efficiency of η=0.5%. [Work function for cesium =1.89 eV] Take hc=12400 eVA.

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Solution

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

1 mW 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 photo electrons.

So,the number of electrons released from the surface per second =2.3×1015×0.5100=1.15×1013electrons/s

The electron current =1.15×1013×1.6×1019A=1.84μA


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