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?