1. The problem statement, all variables and given/known data A beam of light has two wavelengths λ1 (A°, i.e. angstrom) and λ2 with total intensity of I (W/m2) equally distributed amongst the two wavelengths. The beam falls normally on an area A m2 of a clean metallic surface of work function φ (eV). Assume that there is no loss of light by reflection and that each photon has enough energy to eject one electron. Calculate the number of photoelectrons liberated in 2 seconds. 2. Relevant equations E = nhc/λ 3. The attempt at a solution Number of electrons liberated due to light of wavelength λ1 : I/2×A×2 = n1 hc/λ1 n1 = ( IAλ1 )/hc Similarly, n2 = ( IAλ2 )/hc And answer will be n1 + n2 Which is indeed correct. However, I don't understand something. Shouldn't the number of photoelectrons ejected be independent of the wavelength of incident light (cos here n ∝ λ)? Isn't that what we've always heard, that number of photoelectrons ejected is only dependent on the intensity of incident light, as long as the light has enough energy to supersede the work function?