1. The problem statement, all variables and given/known data The work function for potassium is 2 eV. Consider light with wave length 3.6*10^-7 m hitting the potassium surface. a) what is the stopping potential? b) what is the kinetic energy and velocity of the fastest electrons emitted? 2. Relevant equations KE_max = hf - Φ q_e V_s = E_photon - q_e Φ where q_e is the charge of electron, V_s is stopping potential, and Φ is work function 3. The attempt at a solution for finding the stopping potential, V_s = (hf/q_e) - Φ f = c/λ , from the given wavelength, I found f = 8.33*10^14 then hf = 6.6*10^-34 * f = 5.49*10^-19 then, V_s = (5.49*10^-19 / 1.6*10^-19) - 2 eV = 1.43 volts Correct? And as for finding the kinetic energy, I need to use the first equation, but apparently the units of Φ should not be in eV because that wouldn't make sense, how should I calculate KE_max then? Edit: ok I converted the hf into eV, then I got the answer 1.43 eV, which is the same as the answer for a), did I do something wrong? or is it actually suppose to be the same?