Question

Light of wavelength 4000 Å is incident on two metals A and B. Which metal will emit photoelectrons, if their work functions are 3.8 e V and 1.6 e V respectively?

Answer 1

Given: λ = 4000 Å = 4000 × 10^-10 m

Energy of photon, E = hv = `(hc)/lambda`

`= ((6.6 xx 10^-34) xx (3.0 xx 10^8))/(4000 xx 10^-10)`

`= (4.95 xx 10^-19)/(1.6 xx 10^-19)`eV

= 3.1 eV

The incident light will yield photoelectrons if its energy is greater than the work function of the metal.
Thus, metal B will emit photoelectrons.