1.

Photo electrons are liberated by ultraviolet light of wavelength `3000 Å` from a metalic surface for which the photoelectric threshold wavelength is `4000 Å`. Calculate the de Broglie wavelength of electrons emitted with maximum kinetic energy.

Answer» Correct Answer - `lambda_(d)=sqrt((hlambda.lambda_(th))/(2m.c(lambda_(th)-lambda)))=sqrt((6xx10^(-7)h)/(m_(e )c))m=12.08Å`


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