This is a question on the photoelectric effect: For potassium, the work function is 3.65E-19J. Find the maximum wavelength of light that will cause photoelectrons to be emitted when shone onto potassium.

We know thatphoton energy = minimum energy needed to free an electron + max kinetic energy of emitted photoelectronAs we have been asked for the maximum wavelength , we know these photoelectrons just have enough energy to leave the surface of the potassium, but they will not have any kinetic energy. Therefore:photon energy = min energy needed to free electron (work function) E = hf = work function as v = f * wavelengthwork function = (hv)/wavelength, therefore rearranging we get wavelength = (hv) / work function wavelength = (6.63x10^-34 x 3.00x10⁸)/(3.65x10^-19)= 5.4493x10^-7m = 5.45x10^-7m (to 3s.f)