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One milliwatt of light of wavelength 4560Å is incident on a cesium surface. Calculate the photoelectric current liberated, assuming a quantum efficiency of 0.5%. Given work function for cesium = 1.93 eV : |
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Answer» 1.856 mA `E=HV=(hv)/(lambda)=(6.62xx10^(-34)xx3xx10^(8))/(4560xx10^(-10))` `=4.32xx10^(-19)J` Power of source `=10^(3)W(=P)` NUMBER of photons liberarted per sec. `N=(P)/(E)=(10^(-3))/(4.32xx10^(-19))=2.32xx10^(15)` per sec. As quantum efficiency is `0.5%` of EFFECTIVE number of incident photons is only `0.5%` which ejecte photoelectron. `:.` Number of electrons ejected as `N_(e)= 2.32xx10^(15)xx(0.5)/(100)=1.16xx10^(13)//sec` `:.` CORRESPONDING photo electric current proudced, `i=(q)/(t)=(n.N_(e))/(t)=1.6xx10^(-19)xx1.16xx10^(13)` `=1.856xx10^(-6)A` `u=1.856 muA`. |
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