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9

If there was nothing in the way then an electron leaving the anode of a 4 volt battery would have a kinetic energy of 4 electronvolts by the time it reached the cathode. However the mean free path of electrons in metal wires is exceedingly short so electrons never build up anything like this velocity. The end result is that the electron velocities are ...


5

This can be computed for small changes in the pressure by considering the partial derivative of the temperature w.r.t. pressure at constant entropy. If we suddenly raise the pressure a bit, then this is to a good approximation an isentropic process as no heat is exchanged and it is not a violent process causing large irreversible effects. So, we want to ...


4

Introduction (BCS Theory): Cooper pairs in BCS theory are explained like this: the energy that can break cooper pairs in material is for example $10^{-4}\text{ }eV$ and thermal energy of that material is $E=kT$ where $k$ is Boltzman's Constant, so if thermal energy is lower than the energy that can break cooper pairs (in this example $kT<10^{-4}\text{ ...


3

The neodymium magnet (Nd$_2$Fe$_{14}$B) has a tetragonal crystal structure with the $c$ axis much longer than the $a$ and $b$ axes. I did google for an image of the unit cell but couldn't find a nice one - you may have more luck. Anyhow, the crystal can be magnetised in any direction, but it is much more easily magnetised in the direction of the $c$ axis and ...


3

Hints: Define difference $\delta:=\Delta-\Delta_0$. Deduce from $|\delta|\ll |\Delta_0|$ that the lhs. of eq. (1) is $$\tag{A}\text{lhs}~\approx~ -\frac{\delta}{\Delta_0}.$$ Substitute $\xi=x\Delta $ in the integral on the rhs. of eq. (1). Deduce using $\hbar \omega_D \gg \Delta$ that the rhs. is $$\tag{B} \text{rhs}~\approx~ \int_{\mathbb{R}} \! ...


2

Your question concerning the scaffolding that makes up various metals, in relation to their appearance is a unique one. As far as I know, what determines the appearance of various metals is determined by the demarcation of the atoms that they are composed of. Various material have a statistically higher chance of reflecting the photons of a wide range of ...


2

Well, we are not only looking at the electrons of an object when we look at it. What I understood your basic question to be is why we see different objects having different color. Well the reason for that is because different materials are able to reflect only certain frequencies of light. The reason for this is a little more complex. Color in itself is ...


2

The colors depend on the frequency of light. Let me explain. In atoms, there are various discrete energy levels that electrons can occupy. A photon that has energy (which remember depends of its frequency) which matches exactly the difference between the electron and the next excited state, will get absorbed by that electron and get excited to the next ...


2

I found the very last paragraph of the following answer quite explanatory: http://physics.stackexchange.com/a/8324/46100 Interpretation of the Partial Wave Expansion: In the literature, you will often come across terms such as $s$-wave scattering. The partial wave expansion decomposes the scattering process into the scattering of incoming waves with ...


2

For starters, when we talk about voltage as energy per unit charge, is this energy manifest simply as the kinetic energy of the electron? Voltage is potential energy per unit charge. An analogy: voltage is to charge as altitude (as on the surface of Earth) is to mass. So if you lift a 1kg rock off the ground by a height of 1m, you've added some ...


2

Neutrons with the angstrom-scale wavelengths appropriate for diffraction from ordinary crystals have kinetic energies of a few milli-eV. Since neutron detection always involves a nuclear reaction with an energy of a few mega-eV, it's more or less impossible to directly measure small changes in a neutron's energy due to scattering in a crystal. So neutron ...


1

In principle, it is very simple and straightforward. The problem is to map out the region where the integer filling state is the ground state. Suppose you have $L$ sites. Take $N=L$ particles, find its ground state energy, which is denoted as $E_g(L)$. Note that here the Hamiltonian does not contain the $\mu $ term. Do it again for $N=L+1$, the ground ...


1

Do not equate potential with kinetic energy. How fast electrons flow in a conductor has very little to do with their potential. You need to consider the current and the charge carrier density for that. Depending on the material you can have a few fast electrons or many more slower ones. In semiconductors the carrier velocity will be higher - which is why the ...


1

The BiSb type topological insulators are also protected by particle-number/charge conservation symmetry which a superconductor would break. http://arxiv.org/pdf/0901.2686v2.pdf


1

I think Adam's answer is excellent, and I'd rather make this a comment but can't as I just signed up in order to answer. While I agree with most of what Adam said, there are cases where large-N works well for $N=1$. The case I'm familiar with is the large-N expansion for the "Spin Ices" Dy$_2$Ti$_2$O$_7$ and Ho$_2$Ti$_2$O$_7$, which have Ising spins on a ...


1

For starters, when we talk about voltage as energy per unit charge, is this energy manifest simply as the kinetic energy of the electron? No, not at all. Recall that, in electric circuits especially, voltage is not measured at a point (in general, the potential at a point is not physically meaningful - only the difference in potential of two points ...


1

Let's assume, you take a one-dimensional chain of atoms and compress it. In order to investigate the bandstructure, you will need to determine the electronic wavefunctions of quantum-mechanically allowed states. If you know your wavefunctions for the initial condition, before you compress your chain of atoms, you need to also scale the solution in order to ...


1

Note that "$v_y$" is not velocity. Physically it is the local curvature of the Fermi surface. Also it is possible to treat one amongst the 4 constants (sic) $\eta, v_x, v_y, e$ as an overall scale and drop it from the action. Here I do this with "$v_x$", and write the propagator as $$ G(k)^{-1} = i\eta k_0 + k_x + C k_y^2,$$ where $\eta$ and $C$ are ...


1

Practically yes, but the true answer is a little bit complexer. Below the level of the crytals, there are the magnetic domains. In a domain have the atoms the same orientation (and thus, the same magnetic dipole moment). In the case of a magnetized ferromagnetic material, these magnetic pole of these domains are directing to the same (or to the nearly ...


1

I can answer that yes, momentum is definitely transferred to the surrounding lattice, though not through a direct scatting of electrons with protons. This is important in increasingly small electronics. When the cross-section of the wire or metal is extremely small, the collisions can be enough to displace the metal, and wreck a circuit. Tungsten is ...


1

The Drude model is fine for thinking about some things. (It is still taught.) The electrons collide with the atoms. (or a bit more precisely, the outer electrons of the atoms.) Since this is a classical picture perhaps it's OK to have a classical picture of the atoms. Imagine they are little sphere's all joined to the other atoms by little springs. ...


1

Wait a minute. Surrounding the protons are a matrix of electrons. These are the electrons not "cool enough" to exist in the conduction band. I imagine these are what collide with the electrons in the Drude model approach. The closer you get to those electrons, the more they're going to push back by the Coulomb potential. Furthermore, I imagine some Pauli ...


1

According to my limited understanding of density functional theory. Coulomb interaction is one of the correlation effects. Besides Coulomb interaction, there are interaction due to Pauli exclusion principle and change of kinetic energy compared with that of non-interacting electron gas.


1

There is some evidence that cellular structure exists in C-4 and TNT. In particular, cells in C-4 were reported by Dunne (Dunne B B 1970 Science 167 1124–1126) and in TNT by Howe et al. (Howe P, Frey R and Melani G 1976 Combust. Sci. Technol. 14 63–74). You can find a review article (open access) I wrote that covers cellular instability in condensed phase ...


1

Note that $J_{xy}$ and $J_z$ have the dimensions of energy while $J$ and $K$ have dimension of energy/(area x time). In the process of making the continuum Hamiltonian into a discrete one, you will need to choose short distance scales for both space as well as (Euclidean) time. You can choose a square lattice of length say, $a$ for space and split the time ...



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