# Tag Info

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I am not going to solve this completely for you, but note just a few things: The charge of an electron is negative In a simple harmonic oscillator, the force must be in a direction opposite to the displacement You had earlier set $x=A\sin\omega t$ so it seems to me that $\frac{x}{A\sin\omega t}=1$ I think that if you ponder the above you will see where ...

1

The band gaps are always present even in lonely atoms. But the allowed bands are something which forms when atoms meet and form a periodic lattice. What happens is that the energy levels of each atom split because of overlapping of electrons' wavefunctions. Thus adding a new atom to your lattice effectively adds yet another level to each band. When there're ...

2

Then, there is the case that such an operator is defined on the full interval I assume that by "full interval" you mean the whole real line. First question: Do we then need any boundary conditions? Yes, as noted by Sam Bader, boundary conditions are part of the Hamiltonian. In my physics lecture we used so-called Born von Karmann boundary ...

0

Diffraction of electron waves inside a periodic crystal structure, i guess? Superposition of wave functions. That is the intuitive way. One more thing, you can get this with perturbation theory too.Or with tight binding approximation. Anyways for electrons that satisfy special condition there will be constructive interference and they will interact strongly ...

2

If you consider a single energy level in an atom, e.g. the $1s$ state of hydrogen, then if you bring two atoms together the level will split into two states (often called the bonding and anti-bonding orbitals). If you bring in a third atom the two states will split again giving four states, and so on. Grouping $n$ atoms will split the original distinct level ...

3

Consider your equation $$\frac{P}{Ka}\sin(Ka) + \cos(Ka) = \cos(ka) \, .$$ The right hand side can only ever attain values in the range $[-1,1]$. Therefore, if there is an energy $E$ which causes the left hand side to take a value outside the range $[-1,1]$, then that $E$ can never be realized for any value of $k$. In other words, that $E$ is forbidden and ...

1

It's good that you're considering questions like this; I find that this type of questions really forces a student to a deeper understanding of the math involved. Do we then need any boundary conditions? Yes, boundary conditions should be considered as part of the definition of the Hamiltonian and its domain. Different boundary conditions can result in ...

4

any eigenfunction to this Schrödinger operator is automatically periodic with the potential's period, is this true? No!! The eigenfunctions are Bloch waves $\psi(x) = u(x)e^{ikx}$, where $u$ is periodic (with the period of the lattice). But the product $\psi$ is not periodic (with the period of the lattice) unless $k=0$. I put up an example on Wikipedia ...

1

The hole is a quasiparticle, i.e. something that behaves as if, to some extent, there were a particle instead of the system that we are dealing with. The positron is a real particle, that exists on its own. The hole only exists in the presence of other electrons, it cannot exists on its own. This is because a hole is the absence of an electron, but the ...

2

To have the right picture in mind, you need to also take into account the Pauli exclusion between the electrons, being fermions, but also more importantly, do not exclude the nucleus from the picture here! Now, Why rule one holds you may ask? Well it clearly cannot be due to dipole dipole interaction between electrons as it's so insanely small (let's say ...

1

Step 0: Outline. We are going to define a candidate Hamiltonian. We know that we get the right answer if it has all the right behavior. While checking everything the Hamiltonian might possibly do seems daunting, things are simplified by two important facts: The Hilbert space is small -- only 4-dimensional. In fact, we're going to group three basis ...

14

Graphene is only transparent because it is very thin (one atom thick). If it absorbs 2% per layer then just a few hundred layers would absorb almost all light and that would still be a very thin sheet of graphite. The question should be why does graphene absorb so much light compared to diamond which really is transparent? A simplified answer is that ...

3

I assume the biggest factor is the thickness. Graphene is a layer of carbon one atom thick. Light is absorbed/reflected by the top layers of a material and if you make any material into a layer one atom thick you'll find it increases transparency a lot. The thing that is special about graphene is that it forms bonds in a 2D layer where most materials ...

0

What makes Graphene so strong is its electrostatic forces resulting from delocalized electrons flowing through positively charged carbon atoms. This diffrence in charge creates a strong electrostatic attraction that holds Graphene together. This phenomenon also explains why it is such a strong conductor.

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Even in an intrinsic material, $\tau$ is different for electrons and holes. This is because in the equation for $\mu$, the value of $\tau$ is not lifetime but rather the average time between scattering events. Since holes aren’t an actual particle, the movement of holes is really the movement of electrons in the valence band. Because there are many more ...

1

The concentration of donors comes into first formula as Fermi level depends on donor concentration(see Neamen, Semiconductor Physics and Devices page 123). For usual doped semiconductors at room temperature all donor electrons are excited, that is the region where n is not dependent on T and you get third formula. I am not sure whether the model predicts ...

3

If the valence band maximum and the conduction band minimum are on the same position in k-space, this means that you have a direct gap semiconductor (GaAs, InAs, ...). If the CB minimum is at a finite k-value, it would be an indirect gap semiconductor (like Si, Ge, AlAs, ...) The bands appear as parabolas due to the dispersion of a quasi-free electron/hole ...

0

Consider two models: A wave packet of a free electron with $m_e$ with negligible mean energy relative to rest state A wave packet with same parameters of an electron in crystal with $m^*$ with negligible mean energy relative to band edge Assuming that wave packet is large enough for the effective mass approximation to hold (i.e. its uncertainty of ...

0

Second derivative of kinetic energy with respect to momentum equals inverse mass of a particle. In a metal, you have a band structure defined through the dispersion relation of the form E(k) where k is wave vector of electron. Second derivative of this expression can be also taken to be some sort of inertia of a particle, as you can see by analogy with a ...

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It implies that the band in question would have a narrower bandwidth than would be expected from an electron with free electron mass. In turn, this also means that the electron finds it harder to hop from site to site meaning that the electron is more localized that would be an electron with free electron mass.

2

Note that Sommerfeld's model simply generalizes Drude's theory of metals by taking into account the fact that electrons are fermions, so Pauli exclusion becomes a very important factor. In Sommerfeld's model, there's no effective mass to talk about, as one basically ignores the atoms(nuclei) in the system and considers free moving fermions. So there, your ...

2

The bigger the crystal is, the closer-together those discrete values of k are. A real crystal might be 1cm long, or 100 million atoms. Then there would be 100 million little dots between the vertical bars on your figure. Those dots would be equally spaced with respect to the horizontal axis. It's too many dots to see, they blend together to look just like a ...

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The short answer is that BCS theory is derived bottom-up from quantum mechanics (you assume that there is some local attractive interaction between electrons, and perform a mean field approximation), while the older Ginzburg-Landau theory is derived top-down from thermodynamics (you assume that superconductivity can be described by some order parameter, and ...

0

The ambiguity is resolved by choosing the coordinates first. So: 1)Set up a cartesian system with x,y,z coordinates. 2)Pick the group you would like to study. Let's say $P4_32_12$ from page 1151 of this document: http://mcl1.ncifcrf.gov/dauter_pubs/284.pdf that DavePhD recommended. 3)You can then see that the asymmetric unit is given by 0\leq ...

2

The built-in potential, $V_{ic}$, is obtained by analyzing the equilibrium situation, where drift and diffusion currents cancel each other out in the depletion region. This built-in potential is necessary for current not to flow in the system. When you apply an external voltage you expect some current to flow, so the built-in potential must be altered, ...

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Why do minority carriers form a diffusion current not a drift current after they cross the potential barrier? Because the electric field is zero outside the depletion region, so a drift current cannot be driven.

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1) If the n and p doped regions are externally connected using a perfectly conducting wire, why will not any current flow? In thermal equilibrium no current can flow if one connects the two sides of the junction using a perfectly conducting wire. The built-in potential existing at the junction will remain the same, drift and diffusion currents will ...

0

The voltage of a Galvanic cell can both increases with temperature or decreases with temperature. You can prove using thermodynamics or using your intuition that: for Galvanic cells that get hot when working, the voltage decreases as the temperature increases; for Galvanic cells that get cold when working, the voltage increases as the temperature increases. ...

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$k$ is the wave vector of the valence band that the hole resides. $m_V$ is usually denoted as an effective mass of the hole. The effective mass is described in the band structure of the semiconductor. $E_{0,v}$ is most likely a first-approximation of the electrostatic energy of the lattice. In the first approximation, it just shifts the allowed quantized ...

0

The exciton is a quasi-particle. It can be thought of just the interaction energy between the hole and electron. This pair of particles obeys the Coulomb field and is a bound state. Much like the hydrogenic bound states, the exciton's energy is quantized through discrete levels. I suppose in the context of a many-body system, a hole will have local ...

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