# Tag Info

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In the dark you have a single Fermi level which is spatially dependent on the net free charges $E_f(z)$. When the semiconductor is perturbed by light the Fermi level splits into two quasi fermi levels for electrons $\mu_e(z)$ holes holes $\mu_h(z)$. The quasi fermi level separation (or electrochemical potential) is, $\Delta E_f(z) = \mu_e(z) - \mu_h(z)$.

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The rest mass of an electron is estimated to be about 9.1 * 10^-31 Kg. The Schwarzschild radius defines the maximum size sphere which any mass may occupy in order to be a black hole. The Schwarzschild radius for an object with the rest mass of an electron would be about 1.4 * 10^-58 meters. But the Planck length is about 1.6 * 10^-35 meters. Quantum ...

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The electron having mass will exhibit gravitational attraction and begin moving again, however; unless there is a larger object near the electron to it's nucleus it will orbit around it's original host and eventually regain it's original velocity as long as the Lagrange point is not breached.

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The electron doesn't absorb the photons, it scatters them. The energy absorbed by an interaction depends on the scattering angle - this can be determined using the Compton formula for the wavelength of the scattered photon. And the electron tends to scatter more at different angles (proportional to the Thomson differential cross section). From this I found ...

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The idea of electrons orbiting the nuclus is called the Bohr model and has now been replaced with a quantum model. Electrons exhibit wave-particle duality, which means they sometime act like a wave and sometimes like a particle. They don't actually orbit the nucleus but have areas around it where you are more likely to find them called orbitals. The ...

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So an electron can only orbit a nucleus where its wavelength makes a standing wave, leading to discrete energy levels in atoms. That is the Bohr model which has been superseded by quantum mechanics. In quantum mechanics the electron occupies definite energy levels which arise because of the potential well that is generated by the nucleus. As the other ...

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Well, you answered your question partly yourself. Only a finite amount of states (= energy and wavelength of the electron) are allowed, this is the basis of quantum mechanics. If an electron is in a certain state (ie. it has a certain energy and is in a certain orbital), you can then try to calculate its energy and wavefunction and from that its momentum (as ...

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Please completely abandon the idea of an electron orbiting a nucleus. That is only believed in the now long defunct Bohr model of the hydrogen atom. In modern Quantum Mechanics, in order to describe the electron in a hydrogenic (mono-electronic) atom, we need to solve the atom’s Schrödinger Equation which yields the wave functions ...

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If you were holding some charge there with some force and always had then an equal charge would distribute throughout the surface of the conductor so that an equal but opposite charge could be right where you are holding your charge. So it is just like the charge was always distributed on the surface. If however you inserted some charge somewhere really ...

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When there is a resonance in electron response, both refractive index and absorption change rapidly: the refractive index has a "jiggle" in the vicinity of the resonance, like this sketch (adapted from this earlier answer by John Rennie - but I disagree with the "n=1" label so I cut it off...: As you can see there is higher refractive index at the low ...

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In the semi-classical approach you treat the electromagnetic field of the incoming light as a (sinusoidal, with angular frequency $\omega$) perturbation $H^{\prime}$ in the Hamiltonian of the atom. The wavefunction of the atom can be expressed in terms of a linear combination of its eigenstates, each with a multiplying coefficient. If we start on the ground ...

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The source and sink phenomenological description of charge - how realistic is it? It isn't realistic at all. To be perfectly honest it's totally misleading. I've heard over and over an electron described as a source of the electric field, but that is a misleading term. Yes it is, because like Timaeus said, it's the electromagnetic field. See ...

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There is a grand tradition in electromagnetism to talk about the electric fields using the same terminology as we use for velocity fields. For instance we talk about the flux which rightly is a flow per area (and sometimes we multiply by the area and still call it a flux, which is even more confusing to call two things a flux) but it isn't a flow because it ...

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Due to gravitational time dilation, for an observer of the planet, the frequency of electromagnetic radiation would be slower. Visible light emitted from the planet would appear as infrared or micro-waves. The amplitude of the radiation would not change. Since frequency decreases while amplitude remains constant, the radiometer would receive less ...

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I'm not sure if I can help you on the part concerning the Weyl fermions. But your question seems to deal rather with what is a geometrical phase. Parallel transport and geometrical phase Maybe the more intuitive thing to do first is to draw a parallel between geometrical phase and parallel transport. As shown in the image of this wikipedia article, ...

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If you mean that the photon energy can be converted to excitation of the nuclear motion of the atoms that make up the molecule than the answer to the first part of your question is yes. Molecules rotate and vibrate at discrete energies and by absorption of a photon of the correct energy (typically in the infrared or millimeter to microwave range of the ...

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How quickly discharge will occur in the situation you sketch depends entirely on the surface properties of the negative electrode. For current to flow, electrons need to be released from the negative electrode; once they are free, they will accelerate unimpeded to the positive electrode. They will arrive there with 1.5 eV of energy, causing a small amount of ...

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Electrons (and other charge carriers, e.g., ions) in vacuum travel without resistance. However, as pointed out, correctly, in the other answers, there are no charge carriers in vacuum. Nevertheless, electrons can escape from the terminals if they have a kinetic energy which is bigger than the potential barrier of the terminal surface, i.e., the work ...

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Lines of electrostatic force exist between the positive and negative poles of the battery, even though they're separated by a vacuum. Vacuum permittivity is ε0 = 8.854 * 10^-12 farads per meter. By convention, this is called the dielectric constant of 1, a baseline against which the dielectric permittivities of other materials are compared. ...

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Typically electromagnetic radiation starts with movement of an electron, a charged particle. Either as a varying current, say in an antenna, or within an atom when an electron drops to a lower energy state and changes shell within the atom. In either case there is change in movement of the charge, which emits energy as a photon, aka a quantum of ...

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In vacuum there are no charge carriers like ions or electrons. With nothing to carry charge, i.e. current, such a battery would discharge much, much slower than when the battery poles are connected by something that can carry charge like a conductor or an imperfect insulator.

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That's a give and take. The emission of photons is always based on an energetic level of particles higher than the surrounded world. To reach this higher level it needs an receive of photons. A moving electron will be deflected in a non parallel to the movement magnetic field and emit photons. How the electron reach the kinetic energy for this propagation? ...

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There is no such thing as classical motion of an electron in an atom. The quantum states electrons in an atom are in are atomic orbitals, which possess a definite energy, but not a definite position. The Bohr model of the electron, in which electrons are thought of as classical particles orbiting the nucleus, is false. The question whether or not two ...

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Why don't electrons collide among themselves Because they aren't anything like billiard balls. Check out the wave nature of matter. And take a look at the Wikipedia atomic orbitals article: "The electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves". The Heisenberg principle states that we ...

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Yes. There are loads of physical processes in which photons are created. It won't be possible to list them all out but well known examples are matter-antimatter annihilation (e.g. electron-positron annihilation, at lower energies.), the acceleration of charged particles, radioactive decay (notably, Gamma Decay), etc.

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First of all, there is no need to worry over this. The reason we can get away with using conventional flow even though electron flow is what is really happening is that the two are indistinguishable except for in some experiments that are really hard to do. On to your questions. The GND is not an "electron reservoir". First of all, in most circuits there ...

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You are having some misconceptions Wire(more pricisely axle) is not rotated between magnetic field in a turbine generator.Wire(more pricisely axle) is used to rotate an coil (present between magnetic fields) present in generator. Magnetic field itself can not force electrons in a conductor to move untill it is at rest. We need an moving conductor so that a ...

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Yes, and a good example of this is (or rather was) the LEP collider that preceded the LHC. This collided electrons with positrons, so it was a matter-antimatter collider just as you say, and the collisions created all sorts of particles including electrons and protons. However, there are several symmetries that, as far as we know, have to be obeyed. Two of ...

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A few years ago the XUV physics group at the AMOLF Institute in Amsterdam were (to my knowledge the first to be) able to directly image the orbitals of excited hydrogen atoms using photoionization microscopy. For more details see the paper, Hydrogen Atoms under Magnification: Direct Observation of the Nodal Structure of Stark States. A.S. Stolodna et al. ...

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The first images of hydrogen s orbitals were obtained in 2013 by physicists in the Netherlands.

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The main issue with running a fluorescent light bulb backwards is that there are some irreversibility in the way that it convert electricity to light. The Mercury vapour in the tube emits UV rays. Theses rays are absorbed by the phosphor coating on the tube which emits white light. However, if we illuminate the phosphor with white light (reversing the ...

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No, it is not possible. That would violate the conservation of lepton and baryon number.

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If I understand your questions correctly: Yes, it can be somehow the other way around. But we do "know": There are two sort of particles in here, one of them has a certain charge and is light, the other has the opposite charge and is heavy. You can then claim that the heavy ones rather stay in place and the light ones sprint around and make up the current. ...

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Generally it is said that current is due to the flow of electrons; how can we make this claim? If the wire is in a magnetic field the moving charges will move in a circle based on the magnetic force. This happens until enough charge imbalance develops on the edges of the wire to produce an equal and opposite electric force. But measuring the voltage ...

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From classical models, the electron and a proton revolve around their mutual center of mass, which approximately lies on the proton itself, because the proton has a significantly higher mass than the electron. This is why electrons revolve "around" the proton, and hence form the outer layer of an atom. Quantum mechanically, electrons could never form a ...

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Is the working principle of light bulb reversible? No. Electric potential energy is converted into thermal energy in the light bulb filament. At a certain temperature range the filament will light up; that is, it will radiate with a wavelenght in the range of visible light. That this process is non-reversible might be clear if you consider some more ...

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Careful! The essence of your question is a good thought but I think you are having several misconceptions. First of all, it doesn't make sense to talk about a phonon being directed towards a single atom. Phonons are delocalized. Secondly, the "input of temperature required to eject an electron" is a dangerous idea. You need to input energy to eject an ...

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The hyperphysics site you mention states spin rate of some $10^{32}$ radian/s would be required to match the observed angular momentum. Classical angular momentum is calculated as $I\omega$, where $I = \frac{2}{5}mr^2$ for a sphere. The mass of an electron is $9.11\times10^{-31}$ kg and the site mentions an upper limit of $10^{-3}$ fermis or $10^{-18}$ ...

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Do anodes emit virtual photons representing their positive electrostatic potential? No. Virtual photons aren't real photons. There's this lie-to-children popscience myth that they're short-lived real photons that pop into existence like magic, but they aren't. They are abstract things used in QED to model interactions. If you think about an electron and ...

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There exist two frameworks in which to look at how fields and interactions appear: a) when talking of elementary particles and their interactions it is the quantum mechanical framework; this is the underlying framework from which emerges b) the classical mechanics and electrodynamics framework. The classical emerges smoothly in a computable way. An ...

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The metal plate is typically attached to a circuit which collects the ejected electrons making the system net neutral overall just with a current flowing. Also they are usually more easily ejected because the plate is at a potential attached to a battery. However, if the plate were suspended by an insulator in a vacuum and it continuously lost electrons ...

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In regard to the first, we have a 5v supply, which from my understanding means that if you were able to enclose a coulomb of charge eminating from the negative terminal, you'd find that it has 5 joules of energy. This isn't the typical understanding. A 5V (ideal) supply (source) maintains a 5V potential difference across the terminals independent ...

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First, a couple of asides: A. When circuits are drawn, the convention is that straight lines represent perfect conductors. Any real resistance is shown as a resistor, and hence, gets included in any IR drops. You don't go and say "well, there are some other resistances." If they are important, they are shown. Circuit analysis of diagrams is done with this ...

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I'd like to start with a slightly pedantic aside: I think you should try to avoid the phrasing of electrons "having energy"; what they have is potential energy, and potential energy is relative. This is important to remember. Question 2 is trivial, so I'll answer that first: there is no such thing as zero resistance, so the wire from R3 to the batter's ...

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Power in a circuit is defined by P= I x E P is power in watts or joules per second I is the flow of electrons in coulombs per second or amps E is the electormotive force or voltage the energy of the electrons is lost to heat your first question the power supply see only the circuit as a whole it dose not see the individual resistors for R1 the total ...

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Assume that the free space is vacuum and the battery is perfectly insulated as well as all the electrical contacts. In this case there are two possibilities for an electronic device to fail: 1. The induced emf in the inductive elements of electronics device because of the changing electrical field caused by the addition of electrons. 2. Most of the ...

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I would say (almost) yes... If you are familiar with QFT, then you probably know that the mass $m$ of a given field, say a scalar $\phi$, is introduced via the square term $\frac{m^2}{2} \phi^2$ in the Lagrangian. This however comes at the expense of some of the symmetry of the theory. In particular, if you naively apply this idea within the Standard Model, ...

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A classical "shell" of charge of radius $R$ will appear to have a field, outside the shell, corresponding to the charge all being at the center of the shell. If the charge on the shell is $q$ then a charge $dq$ being pulled in from infinity will cost an energy $k_e~q~dq / R$; without loss of generality this is the same if $dq$ is spread over an infinitely ...

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The mass of a fundamental particle turns out to be quite an elusive concept, because massless particles act as a source of gravity and they carry momentum. What then is special about mass? Where mass comes in is in explaining the relationship between the total energy of a particle and its momentum. For any particle we have the expression for the total ...

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It is not enough for it to be moving - it needs to accelerate (or decelerate). An accelerating charged particle will emit radiation and it will lose energy as a result. An excellent example would be the loss of energy of charged particle in synchrotron accelerators. They emit... synchrotron radiation. This is either a boon (e.g. the Diamond light facility ...

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