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To first order, my guess is that room temperature vs cryogenic temperature doesn't matter. The energy needed to knock an atom out of place is significantly higher than the thermal energy of the atoms at either temperature, so the temperature of the crystal shouldn't play a direct role. (If you want to heal Si in a reasonable timeframe, you've got to heat it ...

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Since a hole is just an abstraction for an absent electron, when the hole enters a metal, this can be understood as an electron below the semiconductor's Fermi level vanishing from of the metal (so the current is understood to be carried by an electron on the metal side of things). Reversely, an electron appearing in the metal below to semiconductors Fermi ...

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There is no hole current in conductors because they have overlapping valence and conduction bands. In conductors, electrons are loosely bound to the nucleus hence, can detach easily at room temperature. Also, a large number of free electrons thus, available are conduction electrons. When the covalent bond breaks, electrons are freed from the atom. The ...

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The colder the crystal gets, the more difficult become diffusion processes in the crystal. Since radiation damage to crystalline materials often involves atoms getting knocked out of their positions in the lattice, and since healing that damage requires diffusion to be enabled, exposing the crystal to radiation at cryogenic temperatures should lock in the ...

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In a solid (semiconductor), an electron hole is the absence of an electron where one is expected. In this sense you can think of an electron hole as if it were an air bubble (the absence of water) in a body of water- a local spot where there should be water but there isn't. In the case of water in the presence of gravity, water falls down- but bubbles in the ...

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I'm assuming this is talking about semiconductors that above 0K. This means that there is heat energy in this system. The heat energy causes some electrons to get excited and break free from the regular silicon lattice. Thats why electron are always bouncing around and creating and filling holes. There is nothing that is compelling them to from the bond to ...

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Nothing pushes a bonding electron to fill a vacancy, but nothing holds it back either - the orbital that it occupies and the empty one are identical.

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OP is asking for the math behind the limit. So $$f(E)=\frac{1}{e^{\beta(E-\mu)}+1}$$ $$\lim_{\beta \rightarrow \infty}f(E)=\lim_{\beta\rightarrow \infty}\frac{1}{e^{\beta(E-\mu)}+1}$$ $$\lim_{\beta\rightarrow \infty}\frac{1}{e^{-\beta|E-\mu|}+1} =1\ \ \text{for} \ \ E<\mu$$ $$\lim_{\beta\rightarrow \infty}\frac{1}{e^{\beta|E-\mu|}+1} =0\ \ \text{for} \ \ ... 0 One approach is to consider the translation from a single covalent bond to a bulk semiconductor with multiple covalent bonds. The picture below shows the energy diagram. The left image shows the molecular orbitals as the linear combinations of atomic orbitals. A single Si-Si bond (blue) would fill the highest occupied molecular orbital (HOMO) with two ... 0 Since small energy gaps are generally associated with large dielectric constants, it is almost always the case that the binding energy of an electron to a donor impurity is small compared with the energy gap of the semiconductor. Since this binding energy is measured relative to the energy of the conduction band levels from which the bound impurity level is ... 0 The idea is similar to what happens in some molecular bonds. In the water molecule for example, the electrons from the 2 hydrogen atoms are closer to the oxigen atom, forming a dipole. It happens due to the energy of atomic orbitals of each atom, what favors a migration from H to O. In the p-n junction, band structures of the silicon crystals plays the role ... 0 The reason why electrons move from n to p and holes move from p to n is not that there is an electric field that draws them there. The actual reason is that in thermodynamics, particles (for example electrons or holes, but also more generally, hydrogen molecules or diesel vapor) constantly move, and especially they move back and forth between two mutually ... 1 Basics Initially, at a metallurgical junction, there is a very large density gradient in both electron and hole concentrations. The majority of carrier electrons in the n region will begin diffusing into the p region and the majority of carrier holes in the p region will begin diffusing into the n region. The diffusion process can not continue indefinitely. ... 2 For aan absorption process, the electron absorbed a photon of energy E_\omega and transition to conduction band E_c(k_c) from valance band E_v(k_v). This process satisfies both energy and momentum conservation: energy conservation$$ E_\omega = E_c - E_v, $$and momentue conservation$$ \hbar k_\omega = \hbar k_c - \hbar k_v. $$The dispersion of a ... 0 ITO is a "degenerately doped" semiconductor. Essentially it's a poor metal, has the Fermi level in the conduction band. Is black colored in sintering targets. It functions as the elctrode in a solar cell by allowing 90%+ of the light to get through, while giving good conducivity for collecting current. (The units of interest are called Seimens ... 1 As @Quantumwhisp has correctly pointed out, the semiconductor junctions described in the classical textbooks are usually treated within the semiclassical approximation. However, nowadays we are able to fabricate devices where the wave function is coherent across the junction is everyday reality - these range from tunnel diodes (one tunnel junction) to ... 2 The magic words are "semiclassical model" (you can read about it for example in Ashcroft / Mermin, chapter 12), https://www.fzu.cz/~knizek/literatura/Ashcroft_Mermin.pdf). You are right in that the solution of an electron with crystal momentum \vec{k} for a lattice is itself periodic. You can however superimpose wave functions with differing \... 1 A formal form of writing:$$ \rho(E) = \sum_n \delta(E - E_n). $$where$$ E_n = \frac{n^2\hbar^2\pi^2}{2 m a^2}. $$8 Kittel (at least my 5th edition) goes through this derivation. Refer to the diagram below and remember that in semiconductor physics the chemical potential \mu is also the Fermi level E_F below. The derivation essentially involves calculating the concentration of electrons and holes at temperature T in the conduction and valence bands respectively, ... 0 In this resource there is at least a little bit more explanation. Especially the power of T factor is explained as being the consequence of scattering as an influence on charge mobility. Maybe also the reference cited therein guides you further. 0 Intrinsic carrier concentration is the concentration of electrons or holes in a pure, undoped, semiconductor. Doping a semiconductor changes the concentration of electrons and holes but it doesnt change the intrinsic concentration. It just stops being an example of an intrinsic semiconductor. Its the same as heating water above 0 C doesnt change the freezing ... 1 "How exactly do you measure the roughness in terms of interatomic crystal spacing?" Perhaps this image will help. It shows the surface of a silicon wafer: (Taken from NIST.) You can see a series of plateaus and steps. Each of these plateaus is a crystal plane, and each step between them has a height equal to the interatomic crystal spacing. So, if ... 0 One can also imagine this quantum mechanically as a transition between individual Landau levels. Then from the energy difference between two Landau levels cyclotron frequency will be:$$ \omega_c = \frac{ e B}{m^*} $$More details can be found, for example, in Solid State Physics from John J. Quinn and Kyung-Soo Yi (p. 395 - ...) 1 The electrons of the conduction band of the n-side of the junction can lower its energy by occupying the available states of the valence band in the p-side. But if we are at the equilibrium situation, other electrons have already migrated, creating an E-field in the region. The work done (in the equilibrium state) against the E-field to migrate equals the ... 1 This is very similar in principle to a Brownian ratchet. The naïve expectation that this gives net energy output is based on an "ideal diode" that only allows current through in one direction, just as a Brownian ratchet is based on an "ideal pawl" that only allows motion through in one direction. The resolution here is the same- if the ... 2 Yes, resistance is R=V/I, even if V/I is not constant. You would just have something like R=R(I)=V(I)/I, where R(I) and V(I) are functions of I (or you could do a similar expression with functions of V). To be nitpicky, V=IR is not Ohm's law. Ohm's law is that R=V/I is constant. So by using R=V/I you are just using the definition of ... 1 Well semiconductors behave quite different than conductors, the I-V characteristics aren't linear like conductors and in fact, they even differ when they are forward biased and reverse biased. So to avoid all these, instead of resistance, we define dynamic Resistance which is given as$$R_{dynamic}=\frac{\Delta V}{\Delta I}$$3 I think the key assumption here is that the semiconductor as a whole is electrically neutral. Electrons carry charge -e while the wholes carry charge +e (which is really the charge of the underlying lattice ions.) The net charge is then:$$Q = ep - en = 0 \Rightarrow n=p. Having said that, I do agree that the statement is somewhat trivial, and asking ...

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To summarize the answers given so far: In any finite system, the energy levels are discrete, and the number of energy levels is proportional to the number of atoms in your system. So if you have a particle only a few atoms across, your energy levels are noticeably distinct, even within the same band. But for a macroscopic crystal, the number of levels is in ...

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