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(a) For the first and second case (labeled 1 and 2) we have

$eV_{01}=hc/λ_{1}−Φ,eV_{02}=hc/λ_{2}−Φ$

from which $h$ and $Φ$ can be determined.

Thus,

$h=c(λ_{1}−λ_{2})e(V_{1}−V_{2}) =(3.00×10_{17}nm/s)[(300nm)_{−1}−(400nm)_{−1}]1.85eV−0.820eV =4.12×10_{−15}eV⋅s$

(b) The work function is

$Φ=λ_{1}−λ_{2}3(V_{2}λ_{2}−V_{1}λ_{1}) =300nm−400nm(0.820eV)(400nm)−(1.85eV)(300nm) =2.27eV$

(c) Let $Φ=hc/λ_{max}$ to obtain

$λ_{max}=Φhc =2.27eV1240eV⋅nm =545nm$

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