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Photoelectrons are liberated by ultra light of wavelength `2000 Å` from a metallic surface for which the photoelectric threshold is 4000Å. Calculate the de Broglie wavelenth of electrons emitted with maximum kinetic energy. |
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Answer» K.E. = Quantum Energy - Threshold energy `=(6.626xx10^(-34)xx3xx10^(8))/(2000xx10^(-10))-(6.626xx10^(-34)xx3xx10^(8))/(4000xx10^(-10))` `(6.626xx10^(-34)xx3xx10^(8))/10^(-10)((1)/(2000)-(1)/(4000))=4.969xx10^(-19)` Joule `(1)/(2) mv^(2)=4.969xx10^(-19)implies m^(2)v^(2)=2xx4.969xx10^(-19)xx9.1xx10^(-31)` `mv = 9.51xx10^(-25)implies lambda=(h)/(mv)=(6.626xx10^(-34))/(9.51xx10^(-25))=0.696xx10^(-9)m` |
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