Is there a way to calculate the amount of electrons in a plate of a certain material and certain dimensions? What I want to know is how many electrons are available to remove from a plate when light of appropriate wavelength hits the plate(photoelectric effect).
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2 Answers
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Yes. In the free electron model (of a metal), it is possible to define an electron density in the conduction band. See the table in this link for example.
But to a first approximation you can consider the density of atoms in the material (mass density upon molar volume times Avogadro number) times the valency of the metal under question as the electron density.
answered May 12, 2020 at 22:55
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$\begingroup$ According to the table, aluminum has 18.1e28 electrons in a square meter, that means that I can remove that amount of electrons through the photoelectric effect? $\endgroup$ Commented May 12, 2020 at 23:15
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$\begingroup$ Roughly speaking, maybe a few orders below to be on th f side. Because we don’t want charge buildup to cause any trouble. $\endgroup$ Commented May 12, 2020 at 23:20
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2$\begingroup$ Note that if the plate is not connected to anything, as electrons leave the potential will build up. There is no way to 'remove' any significant number of electrons from a material. The total charge needed in a Van de Graaff generator to take a sphere up to 100's of kiloVolts or MegaVolts is actually quite small in the scheme of things. Charge currents of < 1mA are sufficient to keep a terminal at 6MV during operation with a reasonable ion beam through the system. $\endgroup$ Commented May 12, 2020 at 23:22
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$\begingroup$ Yes, the plate is not connected to anything but I cannot understand, the question is, can 18.1e28 electrons be extracted in a square meter of aluminum through the photoelectric effect? $\endgroup$ Commented May 12, 2020 at 23:29
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As an addition to the answer by @SuperfastJellyfish, consider this.
Your charge of $18.1\times10^{28}$ electrons is approximately equal to $3\times10^{10}$ Coulomb. If we have that charge removed to a distance of $1 m$ (and the opposite positive charge is left on the aluminium), the force on the removed electrons is given by $$F = K_e q^2 / r^2$$ where $K_e$ is the Coulomb constant ($8.99\times10^9$), $q$ is the charge, and $r=1$.
This force is approximately $9\times10^{30}$N.
Hence you will not be able to remove that many electrons. The attractive force will cause the electrons (and whatever equipment is used to remove them) to smash back into the aluminium. If you did remove them, the two plates would move towards each other with an acceleration be of the order of $10^{36}g$. You would prefer not to be close by!
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$\begingroup$ q^2 is the positive charge from the metal * negative charge from the electrons? $\endgroup$ Commented May 13, 2020 at 1:41
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