Light of intensity 2.0 W/ m^2 is incident on an aluminim surface. The wavelength of the light is 160nm. For aluminium the work function is 4.2eV. Find the kinetic energy of the slowest moving electrons So then Kmin would be the wavelength at which the cutoff frequency occurs, yes? e = hf -> e = hc / lambda and E = 4.2eV = hc / lambda then find the cutoff wavelength? At this point im stumped still... please do help thank you in advance! Heres another one part b of this question Fidn the average rate per unit area at which photons strike the aluminium surface P = E / t = hf / t = hc / lambda t and lambda = 160nm. I = P / A = hc / lambda t A. I know intensity, h ,c , lambda and all i need is something over t A giving units of per second per sq m Am i right? input would be greatly appreciated!