Work functions of three metals A, B and C are 1.92 eV, 2.0 eV and 5. eV respectively. Which metal will emit photoelectrons when a light of wavelength `4100Å` is incident on the metal surfaces?

Explanation

Energy of incident photon,
`E = hf = h.(c)/(lamda) = (6.625 xx 10^(-27) xx 3 xx 10^(10))/(4100 xx 10^(-8))` erg
`= (6.625 xx 10^(-27) xx 3 xx 10^(10))/(1.6 xx 10^(-12) xx 4100 xx 10^(-8)) eV ~~ 3.03 eV`
Hence, this photon will be able to emit photoelectrons from metals A and B but not from C.