Hello! First of all, this is my first post here. I hope it's on the right thread. I managed to answer most of the questions, but I think at least some of them are wrong (for example, d)). Any help would be really appreciated. 1. The problem statement, all variables and given/known data A monochromatic light beam of wavelength λ=500nm hits a clean sodium (Na) surface. The work function of sodium is W=3.7e-19 J = 2.3 eV. a) What is the magnitude of the momentum of one photon of this beam? b) What are the frequency and the energy of such photon? c) What is the cutoff frequency of sodium? Graph it. d) Will there be emission of electrons? If so, what is the maximum kinetic energy of an emitted electron? e) What is the cutoff potential for a cesium plate illuminated by photons of wavelength λ=400nm, considering that the maximum kinetic energy of the photoelectrons is 2.23 eV? 2. Relevant equations p = h/λ f = c/λ E = hf = K+W 3. The attempt at a solution a) p = h/λ p = h/500e-9 p = 1.32e-27 (kg*m)/s b) f = c/λ f = c/500e-9 f = 6e14 Hz E = hf E ≈ 3.97e-19 J c) f = E/h f = 3.7e-19/h f ≈ 5.6e14 Hz I think the graph is something like this, V0 being equal to W (I guess), and v0 being equal to f. d) Since E = 3.97e-19 J is greater than W = 3.7e-19 J, there will be electron emission. The kinetic energy will be: E = K + W 3.97e-19 J = K + 3.7e19 J K ≈ - 2.3 eV I think this is wrong, because the kinetic energy shouldn't be negative, right? Is the formula wrong? Or am I using the wrong one? (Edit: I think I just forgot the minus in the second exponent.... The result, then, should be: 3.97e-19 J = K + 3.7e-19 J ≈ 2.7e-20 J ≈ 0.1685 eV) e) I'm not sure how to solve this one.... Is the cutoff potential the same thing as the work function? If so, I'll run into the same problem as the previous one. If not, I have no idea how to solve this. Thank you in advance!