Question

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.

Answer 1

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`