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Electromagnetic radiation of wavelength lambda = 0.30 mu m falls on a photocell operating in the saturation mode. The corresponding spectra sensitivity of the photocell is J = 4.8mA//W. Find the yield of photoelectrons, i.e. the number of photoelectrons produced by each incident photon. |
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Answer» Solution :Suppose `N` photons fall on the photocell PER sec. Then the power incident is `N(2picancelh c)/(LAMBDA)` This will GIVE rise to a photocurrent of `N(2picancelh c)/(lambda).J` which means that `N(2picancelh c)/(elambda).J` electrons have been emitted. THUS the number of photoelectrons produced by each photon is `w = (2picancelh c)/(elambda) = 0.0198~~0.02` |
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