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Find the frequency of light which ejects electrons from a metal surface, fully stopped by a retarding potential of 3V. The photoelectric effect begins in this metal at frequency 6xx10^(14)s^(-1) |
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Answer» `10^(5)` If `V_(s)` is retarding POTENTIAL then `E_(k)=eV_(s) and w=hv_(0)` `:.eV_(s)=hv-hv_(0)` or `v=(eV_(s))/(H)+v_(0)` Given `V_(s)=3V, v_(0)=6xx10^(-14)s^(-1)` `:.` REQUIRED frequencyv is `v=(1*6xx10^(-19)xx3)/(6*62xx10^(-34))+6xx10^(14)` `v=1*324xx10^(15)` per sec. |
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