1. The problem statement, all variables and given/known data The wavelength treshold for photoelectric emission from a sodium surface is 683 nm. Calculate the energy of the electrons which are ejected when a sodium surface is illuminated by light of wavelength 500 nm. If the intensity of the light is 2.0 W/m^2 and if 1 per cent of incident photons produce photoelectrons, estimate the number of electrons emitted per second from a sodium surface of area 2*10^-4 m^2 when it is illuminated by light of the wavelength 500 nm. 2. The attempt at a solution The treshold wavelength corresponds to a treshold frequency of 6E14 Hz, which in turn corresponds to a electron binding energy of 2.91E-19 J. This means that the energy of the electrons which are ejected is K = 1.1E-19J. Next, we observe that since the surface area is 2E-4 m^2, the total effect is 4E-4 W, and hence 4E-4 J of energy hit the surface each second. Since 1 per cent of this energy produces photoelectrons, the total photoelectron energy per second is 4E-6 J. Since each photoelectron has energy 1.1E-19, my estimated number of photoelectrons is 4E-6/1.1E-19 = 3.6E13 photoelectrons. My book says 1.0E13. What have I done wrong?