Light of wavelength 5.4 * 10^-7 m falls on a photosurface and causes the emission of electrons of maximum kinetic energy 2.1 eV at a rate of 10^15 per second. The light is emitted by a 60 W light bulb.
Find the work function of the surface.
The Attempt at a Solution
Ok I actually have an answer. I'm not sure if it's right though. I just want you guys to check if my reasoning and answer makes sense.
So first I thught since P= W/t, 60W = 60 J /s
That means more than one photon is hitting the surface. Then, E = W + Ek
60 J - ((2.1eV)(1.60*10^-19)(10^15 electrons)) = W
Then I thought that the work function means the amount of energy required to remove just one electron so, I divided 59.999664 J by (1 * 10^15 electrons) and got 5.9999664 * 10^-14 J.
Does this make sense?