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As pointed in the comment by @CuriousOne, there is indeed a paper providing such explanation ("The photoelectric effect without photons" by Lamb and Scully). It was however later found to be faulty. A very good review and references on this can be found in these answers to a different question posted on this website, but apparently there is no consensus on ...


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This is because usually a electron can ONLY stay in certain energy states in a given atom, this is because of the quantum mechanical forces such as strong force, and Heisenberg uncertainty principles. Let me explain further, as electron gets closer to an atom the electrostatic & other quantum mechanical forces between the electron starts pulling it, and ...


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You are right - the electron might make it all the way to the other plate. But then it will have zero velocity, and it still feels the electric field. So it will "fall back" to the plate it came from - and the net current is zero. In reality the velocity distribution of the electrons is continuous (they don't all travel directly to the other plate with the ...


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The number of electrons ejected per photon is around $10^{-6}$ to $10^{-5}$ depending on the metal. The quantum efficiency for the initial creation of the photoelectron is around 0.1 to 1. This is the process used in photomultiplier tubes, and the high quantum efficiency is responsible for the very high sensitivity of PMTs. The trouble is that the ...


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The energy of the photon used to reap the electron from the atom and give it some kinetic energy, thus - $h\nu =\frac{hc}{\lambda}=\frac{1}{2}mv^2+W_0$, where $W_0$, the energy needed to reap the electron from the atom, is constant per type of atom. Thus $ v=\sqrt{\frac{2(h\nu-W_0)}{m}}=\sqrt{\frac{2(\frac{hc}{\lambda}-W_0)}{m}}$ As long as ...


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In the photoelectric effect, photons incident on the cathode cause the emission of electrons. Assuming there is a sufficient electric field, these electrons will make their way across to the anode, contributing current. For simplicity, let's assume every photon generates a photo-electron. Then if $N$ photons per second hit the cathode, the current will be ...


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One way to look at current is "the total number of electrons passing a particular plane per unit time, multiplied by their charge". How fast they are going doesn't matter - if they are going faster, they will appear to be further apart. A given amount of light (above the critical frequency) will knock a given number of photo-electrons into space. That is ...



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