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A source emits monochromatic light of frequency `5.5xx10^(14)` Hzat a rate of 0.1 W. Of the photons given out, 0.15% fall on the cathode of a photocell which gives a current of `6muA` in an extrnal circuit. (a) Find the enrgy of a photon. (b) Find the number of photons leaving the source per second. (C) Find the percentage of the photons falling on the cathode which produce photoelectrons. |
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Answer» Correct Answer - A::B::C (a)`E = hf= (6.6xx10^(-034))(5.5xx10^(14))` `=36.3xx10^(-20) J` =2.75 eV (b) Number of photons leaving the source per second, n = (p)/(E ) = (0.1)/(36.3xx10^(-20) `=2.75xx10^(-17)` (c ) Number of photons falling on cathode per second `n_1 = (0.15)/(100)xx2.75xx10^(17)` `=4.125x10^(-14)` Number of photonelectrons emitting per second, `n_2 = (6xx10^(-6))/(1.6xx10^(-19)) = 3.75xx10^(13)` `:. % = (n_2)/(n_1)xx100 = (3.75)/(4.125x10^(14)xx100` =9%. |
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