# Tag Info

5

Can this frozen form freeze further? Or can it become more solid? (for example, by exposing to colder temperatures and/or a higher pressure). Can ice freeze further by transforming into a different crystalline form? The ice will remain solid while lowering temperature or pressure but might change in state, or phase, as you mention. But you should ...

5

First, for clarification: As Jon Custer already mentioned, THz radiation is not difficult to generate. It's simply part of the black body radiation. What is in fact difficult is to generate coherent or at least narrow-band THz radiation. Regarding emission from semiconductor material: There is not so many materials, which would offer a bandgap in the THz ...

4

The mean free path can be meaningful quantity in quantum mechanics, although usually only in a semi-classical regime. It is particularly useful in the kinetic theory of quantum liquids at low temperature, where the excitations of the system can be described as quasiparticles that propagate approximately ballistically and interact only rarely. You can define ...

4

Imagine the single energy-bands as a set of "places" ($\rightarrow$ states) which can be ingested by an electron for example. Between these sets of places there are gaps (at least for non-free electrongas models). So if the band is not filled up, electrons can "swap seats" and conductivity is given. If your band is full you need a quite "huge" amount of ...

3

You are in luck. Alexander Gavriliuk, Ivan Trojan, and Viktor Struzhkin observed the anticipated transition in 2012. Their paper is in Physical Review Letters 109 086402 (2012) (link to the paper). They mention Mott's prediction in the abstract, and discuss the various prediction of where it would occur in comparison to what they observed. Now, they had ...

2

In a semiconductor, there are contributions to the total electric current from both electrons in the conduction band and holes in the valence band. For N-type material, there are far more electrons in the conduction band than there are holes in the valence band. Thus, almost all of the electric current is due to drift of conduction band electrons. ...

2

The issue here is how much the refractive index $n$ tells you about dissipation. As you rightly said, the imaginary part of $n$, which depends on both real and imaginary parts of $\epsilon$, leads to an imaginary part in k which describes an exponentially decaying electric field. However, this doesn't necessarily correspond to dissipation (i.e. a drop in ...

2

The density of states $\rho(E) \propto E^{1/2}$ is valid only for free electrons. Electrons in a solid are certainly not free, and the density of states is complicated. Certainly: the density of states is zero inside the band gap.

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From this step, $$U_G=\frac1a\int_{-a/2}^{a/2}\mathrm{d}x U(x)e^{-iGx}=\frac1a\int_{-a/2}^{a/2}\mathrm{d}x \sum^\infty_{n=-\infty}A\cdot \delta(x-na)e^{-iGx}$$ note that the summation is an impulse train with spacing $a$. Since the integral is from $-\frac{a}{2}$ to $\frac{a}{2}$, just the impulse at $x = 0$ is integrated over so only the $n=0$ term ...

1

In some sense, you have answered your own question. All this means is that you cannot determine $a$ from a series of reflections along the same axis (e.g. ($001$), ($002$), ($003$), etc.). These are all along that same direction in reciprocal space, so therefore they are all on the same axis. If you measure an off-axis reflection, you will able to calculate ...

1

Measuring an off-axis peak simply means that you're looking for crystal planes, which are not parallel to the sample surface. Therefore, they have an in-plane component (in addition to the perpendicular lattice plane spacing, which is commonly measured), which is the part, you're looking for.

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How atomic orbitals merge into crystal band structure is, well, complicated to capture in simple models. As seen in, say, Ashcroft and Mermin (chapter 28), the energy surfaces for Si have symmetry along the <100> directions. In contrast, the surface for Ge have symmetry along the <111> directions, with the band minimum at the zone edge. A comparison ...

1

This is a very good question. It turns out that the phase transition occurs precisely when the chemical potential becomes equal to zero (assuming that the ground state energy is at zero). The order parameter in the BEC is the "macroscopic wave function" or rather the square root of the single-particle reduced density matrix. The broken symmetry is usually ...

1

Those other (lower) bands will be filled. The most interesting bands are the ones that are only just barely full, or only partially filled, as well as the empty bands just above the filled ones. Usually when people speak of the band gap, they mean between the highest filled band and the lowest empty band.

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Recombination A useful way to think about recombination rate in semiconductors follows from the equation, $$R = An + Bnp + Cpn^2 - G + I/q$$ where $n$ and $p$ are the electron and hole density. The first term deals with the non-radiative recombination. In your question you mentioned Shockley-Reed-Hall (SRH), think of this term as a very simple ...

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I think that you make a mistake. The operator (i) $\ \hat U = −\hat {\vec {\mu}} \cdot \vec B = −g\ \mu _B \ \hat {\vec J} \cdot \vec B$ doesn't give energy levels, but an additional energy term in the Hamiltonian. Let's take for simplicity the direction $z$ of our system of axes in the direction of $\vec B$. Then my formula (i) becomes (ii) \$ \ \hat U = ...

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There is a very nice demonstration that you can treat holes as positively charged carriers in the Hall effect. As you may know, to observe the Hall effect we place a semiconductor in a magnetic field and pass a current through it. We observe a polarization (voltage) at right angles to both the current and the magnetic field as a consequence of the Lorentz ...

1

The poster above is incorrect: FCC metals do not have a ductile to brittle transition temperature and instead remain ductile at low temperatures. This is because the stress required to move dislocations is not strongly temperature-dependent in FCC metals, and thus failure occurs by plastic flow instead of crack propagation. In BCC metals, the stress ...

1

because the holes somehow will attract electrons and get them from conduction band to valence band The reason that a p-type semiconductor is p-type is that it contains acceptor impurities. These are atoms that tend to capture electrons in localized states around their nucleus. For example, group III boron is a typical acceptor impurity in silicon. ...

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