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First, webcams do use CCD or CMOS sensors, usually whichever chip is cheapest at the time. You can catch photons, but not reliably. In other words, for every photon that you catch, you will miss several. There will also be a noise signal, typically equivalent to many photons, Consider a CCD sensor. When a photon arrives, it may successfully excite an ...


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$\renewcommand{ket}[1]{|#1\rangle}$ The basic logical connection here is $$\text{symmetry} \rightarrow \text{degeneracy} \rightarrow \text{avoided crossing} \rightarrow \text{band gap} \, .$$ $\textrm{symmetry}\rightarrow \textrm{degeneracy}$ Consider an operator $S$ and let $T(t) = \exp[-i H t / \hbar]$ be the time evolution operator. If $$ [ T(t), S] = 0 ...


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What you said is right. The incident field creates the electron-hole pair, then they oscillate normal to the surface. Since the motion is normal to the surface, it radiates in every direction except normal to the surface. (Dipole radiation pattern.) So if you want to avoid that effect, you want the outgoing light you measure to be normal to the surface. If ...


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If there is only one band maximum in the BZ, this point is one of the high-symmetry points of the BZ. However, there can be cases where there are many points which are a band maximum and they are not at one of the high-symmetry points of the BZ. These points however are all connected by a symmetry operation. An example of a system with band minima away ...


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I will restrict my answer to bipolar junction transistors. Let us take an npn transistor. Lots of electrons in the emitter, and if we forward bias the emitter-base junction they will head into the base. However, in the base they are minority carriers and are quickly gobbled up through recombination as the base attempts to keep $np = n_{i}^{2}$. None of the ...


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Electron and holes are Fermions (particles with spin 1/2). This means that no two particles can share the same microstate. The Fermi-Dirac distribution describes how Fermions fill the available states consistent with this property. Bosons on the other hand (particles with integer spin) can occupy the same state. The Bose-Einstein distribution describes how ...


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As boyfarrell pointed out, the IR peak that you observe is due to second order reflection from the diffraction grating of the spectrometer. Strong spectral lines are often seen in the second order (with the advantage of better resolution). You can easily verify it by inserting a long-pass filter that blocks 370 nm (some orange glass or even sunglasses) ...


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No. We had 'solid state' devices long before we understood them. The selenium rectifier and the 'cats whisker' diode we commercially available long before we understood them at the quantum level...19th century if I am correct.


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First off, let's answer the question: what part of that equation would change? $f\left(\epsilon\right)$ is always the same if you're in equilibrium. Scattering won't change that, because scattering alone won't move you out of equilibrium. If anything, scattering has the opposite effect. What can change is $Z\left(\epsilon\right)$. How it would change ...


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I'll comment on the second issue that $E_F$ should always be greater than zero, which leads to a finite scattering rate. However, that's beside the point because in an ideal Fermi liquid at $T\rightarrow 0, |E_F-E|\rightarrow0$, the scattering rate becomes zero, which leads to an infinite lifetime. If in graphene, as you say, the independent particle picture ...



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