Say you have a vacuum tube, such as the kind used in old amplifiers, wherein electrons are accelerated from the cathode to the anode through an electric field. Presuming, for the sake of argument, that electrons coming off of the cathode have a zero energy, and the anode, made of a material with a work function of 4eV, is at a potential of +24V relative to the cathode, then wouldn't a photon with an energy of 28eV be emitted upon impact of the electron on the anode? Using E=h*f and c=lambda*f, then wouldn't the wavelength of this emitted photon be ~44.28nm? That's beyond the highest reaches of ultraviolet light and into X-ray territory, so why are vacuum tubes safe? Is there something wrong with my understanding of how all this electron/photon business works?
Second side question if you care to answer it:
I turn off the filament in my vacuum tube and let the cathode get cold, but I keep it plugged into the circuit, with the anode still at +24V relative to cathode, then I shine a light at 635nm on the cathode. Supposing my cathode has a work function of 4eV as well, do electrons hop off of the cathode due to photoelectric effect or not? Without the anode there, I would say no, because h*f-4eV is negative (-2.05eV), meaning the electron doesn't have enough energy to escape the metal plate. With the anode however, there's an electric field pulling the electron as well, so can it make the jump? I calculate it would reach the anode with 25.95eV of energy if it did.