The work functions for potassium and caesium are `2.25` eV and `2.14`eV respectively. Will the photoelectric effect occur for either of these elements with incident light of wavelength `5650 Å`?

Explanation

Data: `Phi` (potassium) `=2.25 eV, Phi` (caesium) =2.14 eV, `lambda= 5.650 xx 10^(-7) m, h=6.63 xx 10^(-34) J.s,c=3 xx 10^(8)` m/s
`Phi` (potassium) `=2.25 eV = 2.25 xx 1.6 xx 10^(-19) J = 3.6 xx 10^(-19)` J
`Phi`(caesium) `=2.14 eV = 2.14 xx 1.6 xx 10^(-19) J = 3.424 xx 10^(-19)` J
The energy of the incident photons,
`(hc)/lambda = (6.63 xx 10^(-34) xx 3 xx 10^(8))/(5.650 xx 10^(-7)) = 3.52 xx 10^(-19)` J
The energy is greater than `Phi` (caesium), but less than `Phi` (potassium).
Hence, photoelectric effect will occur in case of caesium, but in case of potassium.