26.(a) For the ﬁrst and second case (labeled 1 and 2) we haveeV01=hc/λ1−ΦandeV02=hc/λ2−Φ,from whichhand Φ can be determined. Thus,h=e(V1−V2)c(λ−11−λ−12)=1.85 eV−0.820 eV(3.00×1017nm/s)[(300 nm)−1−(400 nm)−1]=4.12×10−15eV·s.(b) The work function isΦ=
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.