Alright, I am having some trouble with the following problem, because my solution I arrive at seems to be way too large. The problem is: What is the maximum photelectron speed if potassium is illuminated with light of 250 nm? From that I have λ = 250 nm f = c/λ = 3x10^8 / 250x10^-9 = 1.2x10^15 Eo of potassium = 2.3 eV From there I tried getting the velocity by obtaining the kinetic energy. Eelec = hf = (4.14x10^-15)(1.2x10^15) = 4.968 Kmax = Eelec - Eo = 4.968 - 2.3 = 2.668 V = √2K/m = √2(2.668)/(9.11x10^-31) = 2.42x10^15 m/s So my calculated answer is 2.42x10^15 m/s which is much faster than the speed of light, so I know it must be wrong. I can't seem to figure out another way to arrive at an answer though. Any help is much appreciated.