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Photo electrons are liberated by ultravi...

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.

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The correct Answer is:
`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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