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Yes,the photoelectric effect can be explained without photons! One can read it in L. Mandel and E. Wolf, Optical Coherence and Quantum Optics, Cambridge University Press, 1995, a standard reference for quantum optics. Sections 9.1-9.5 show that the electron field responds to a classical external electromagnetic radiation field by emitting ...


3

In a small metal bar (let's say 1cm of length, a few grams of weight) there are about $10^{24}$ protons, amounting to a total charge of $10^{14}$ e.s.u. The interaction energy between these protons is $Q^2/r$, which amounts to $10^{28}$ erg, or $10^{21}$ J. This is repulsive energy, of course, so that the value we just estimated is the work we would need to ...


3

The configuration that you describe (all electrons leaving the metal) would be impossible to generate by using a light source. The electrostatic repulsion of two point charges is given by Coulomb's law: $F=\frac{k_e q_1 q_2}{r^2}$ If the positive (protons) and negative (electrons) charges occur in roughly the same quantity, then the material is ...


1

This can't happen. As you remove electrons the metal will get more and more positively charged. Either it becomes so imbalanced it takes too much energy to remove the electrons anymore, or, the metal tears itself apart from the repulsive force of too much charge in one place.


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Sure, why not? Any process which imparts sufficient energy to the electron to kick it free of the nucleus is an ionization process. Heck, hitting the electron with another particle (while statistically unlikely) counts as ionization.


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Referring to the Watson 1928 paper you cited (see footnote 31 on page 737) I infer the following. The most probable direction of initial photo-electron ejection is a little forward of perpendicular to the incoming un-polarized X-ray beam (based on ejection being parallel to the electric vector of the x-ray photons which vector is perpendicular to the beam ...


0

They talk about contact voltages because it affects “stop voltage” as measured by their instruments, but it doesn’t affect $\Delta U_{\rm stop} / \Delta\nu$ since aforementioned contact voltages are assumed independent of illumination conditions. Because a photovoltaic cell made of a homogeneous piece of material won’t work. A piece of conductor will not ...


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I believe the the reason the photoelectric experiment is performed in a vacuum is simply that the electrons produced close to the threshold have a relatively low energy. Low energy electrons will not travel very far in the air before they are defected by collisions and this makes then very difficult to detect without a vacuum.


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Well it may help to first simply consider the famous the photoelectric effect, and see how photon frequency $\nu$ is related to the photoelectron's kinetic energy, once you are sure you understand up to this point, then we can move on to your question on photoelectric current. The picture is a typical diagram used to elaborate on the photoelectric effect, ...



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