# Tag Info

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You have to ask yourself if the electron can absorb the energy. For it to do so, there must be another energy level available to the electron inside the material. If there is, the photon is absorbed. Otherwise, it will be reflected. You need to look at the band structure of the material to decide what actually occurs.

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I would say it is both! Because of the abundance of electrons, the electric field at the battery pole/boundary, at the instant of turning on the switch (t=t0), is quickly (within a few Debye lengths) screened and cannot possibly reach the electrons further down the wire. However, the electrons at the vicinity of the pole that do feel the effect of electric ...

3

The momentum of an electron, which is not travelling at very high velocity will not have any relativistic effects. So, its momentum is given by $$p=m_0v$$ where $m_0$ is the rest mass of electron ($9.1\times 10^{-31}~\rm kg$) and $v$ is the velocity. But to observe phenomena like diffraction (which observed with radiations like X-rays), the energy ...

8

Often, when dealing with high-energy (relativistic) particles the rest mass of the particle can be neglected when performing calculations. Use your expression for $p$ from relativistic considerations, plug in the numbers and see the negligible change when you include and neglect to include the mass of the electron. A good tip for when you enter into higher ...

3

The 4-momentum vector is given by ${\bf p}=(\frac{E}{c},p^{1},p^{2},p^{3})$. Now taking the scalar product with itself we have, $${\bf{p.p}}=E^2-(pc)^2=m_{0}^2c^4$$ Now for extremely relativistic case , we can use the condition that $E\gg m_0c^2$, thus this yields $p=\frac{E}{c}$.

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As the energy of the electrons in that case is much greater than their mass, you can consider the approximation $E \sim pc$. So the formulas are equivalent.

2

The color of the photon is related to its frequency $f$, which can be related to the energy of the photon by the expression $E = hf$, where $h$ is Planck's constant. Thus the different colors of the emitted photons describes their different energies. The next step is to determine why specific elements emit certain colors. This has to do with the different ...

2

By definition, Thomson scattering is the elastic scattering of light by a free charged particle. Atoms cannot be described as such, but the electrons in an atom may approximate to free electrons if their binding energy is much lower than the photon energy. This might be true for X-ray wavelengths, although if the photon energy gets too high then elastic ...

0

The Thompson scattering is the low energy limit of the Compton scattering, which is given at tree level by the diagram (source): The vertex for this diagram is $ie\gamma^\mu$, as usual. Loop corrections to this vertex are given by the electromagnetic form factors, which in turn carry the information of, for example, the electron g-factor. In the article ...

2

You are basically asking a circular question of nomenclature. The Dirac quantum field is is a bispinor compactly packaging several degrees of freedom, such as the left- and right-handed Weyl spinors you wrote down the Lorentz transformation properties of. We call both left- and right-handed electrons "the electron", collectively, but of course they are ...

3

The Stokes law equation for the drag on the oil droplet is: $$F_d = 6 \pi \eta r v$$ wher $\eta$ is the viscosity of the air, $r$ is the radius of the oil drop and $v$ is the velocity of the oil drop. The trouble is that when the oil drop is very small its radius is comparable to the mean free path of the air molecules. That means the air no longer ...

1

The way mobility depends on average scattering time of the carriers is given here: A simple model gives the approximate relation between scattering time (average time between scattering events) and mobility. It is assumed that after each scattering event, the carrier's motion is randomized, so it has zero average velocity. After that, it accelerates ...

2

In short, you must consider the total elements of the system for conservation of momentum. In this case, nearly all of the momentum is exchanged between the electron and a photon that is absorbed or radiated away (the light). Momentum is conserved, and is largely balanced by this electron-photon interaction, although smaller amounts may be exchanged with ...

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Angular momentum is conserved only if there's no external forces, in this case the electron gains energy by light or by heat wich is kinetic energy. They are both external forces so the conservation of angular moment does not apply.

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To buid up a bit on Tamasger's awnser, BCS can indeed be very non-intuite at first glance. The picture where the electrons locally disturbs the lattice and in return attract another electron is a good one to understand the basic concept of pairing, but is far from being able to capture all the physics. Remember that phonons (the disturbance of the lattice) ...

1

No, and certainly not by the mechanism you describe. The "orbits" of electrons around nuclei are structures created by the electromagnetic force. This force is mediated by photons, which cannot pass out of the event horizon by definition of the event horizon. So even in the highly implausible scenario in which an atomic structure existed within a black ...

2

Courtesy of its spin the electron has a magnetic dipole moment. That means if we place it in a magnetic field the two states aligned with and against the magnetic field have different energies. The magnitude of the energy difference depends on the strength of the field and the size of the magnetic dipole moment, which in turn depends on the spin. So by ...

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The reason is that all experiments known can be explained by having two types of electric charge. To distinguish between the two types of charge them it is necessary to introduce labels, conventionally the labels were taken be "positive" and "negative". Because of history, electrons are given the label "negative".

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The battery is an energy source that supplies the electrical energy to the electrons in the conductors. There is no actual flow of electrons. It's the energy that is transferred. A conductor contains large no. of atoms tightly packed with plenty of availability of valence electrons that are ready to move out from the atom if you supply a little bit of ...

2

There is no structure of electrons as far as we know. It's a point entity. So it cannot be seen as something that has further structure or said to be having "parts". It's a fundamental particle

0

A filament of a light bulb can be thought of as being composed of a lattice of positive metal ions which are vibrating about fixed positions and a sea of mobile electrons which are responsible for the metal being an electrical conductor. With no external circuit present a chemical process within a battery moves mobile electrons within the battery to produce ...

0

I try to think of everything in terms of chemical potential. Many batteries utilize lithium ions to create a chemical gradient. This creates a driving force across the circuit, called voltage. Keep in mind that the electrons are not moving that quickly - it acts more as a wave travelling through. If you are interested, you can learn more about the actual ...

0

Okay, it's partially wrong. A battery is a device that maintains a constant potential difference between its terminals (generally through chemical stuff). Circuits are conductors. Conductors have electrons that can easily flow across their atoms. It's like conductors have a "sea" of electrons. The potential difference causes the circuit's electrons to move ...

2

There are two types of radiation that are emitted from an accelerating charged particle: synchrotron radiation (if the acceleration causes circular motion) and bremsstrahlung radiation (if the acceleration is from speeding up or slowing down). The total power emitted from synchrotron radiation is given by: $$P = \frac{q^2\gamma^4a^2}{6\pi\epsilon_0c^3}.$$ ...

1

The company explanation is wrong, really except for the first sentence. The correct hand waving explanation would be much longer. Materials are composed of atoms which contain electrons, and electrons have intrinsic magnetic moments and specific orbitals around atoms that are not affected by temperature. The molecular bonds are affected at high ...

1

It's possible that I do not understand this theory correctly, but it seems to me that it was disproved by experiment. Indeed, it is possible to strip an atom of its electron cloud. If the nucleus was purely an effect of the electron field, nothing would be left once the electrons are removed. This is not the case. See e.g. this post. Note that you can never ...

3

The way you combine quantum systems is not by summing their wavefunctions, but by taking the tensor product, see this question. In particular, if the systems you are composing are just single electrons described by square-integrable wavefunctions in $L^2(\mathbb{R}^3,\mathrm{d}x)$, then a system of $N$ electrons is described by a square-integrable function ...

0

Because you may imagine that the electrons "pile up" in a queue to get through the smallest resistor, and when the queue is too damn long, this resistor is not anymore the easiest way to the other side for the lastcoming electrons. A step by step thinking is this: The first electron arriving goes towards any resistor, since at this moment it doesn't know ...

0

Your confusion arises because you take the "visual model" too seriously. According to BCS theory, there is a certain kind of electron-phonon interaction, which can be shown to be an effective attraction between two electrons (that need to be on the surface of the Fermi-sea). This coupling generates a binded state in the right conditions, and electrons stay ...

0

What do you mean by positive electrons? There are no positive electrons. Once an electron is lost from an atom, it's an ion, not a positive electron. The ions cannot move because they are tightly packed. It's the electrons that are free to move. Now the flow of electron from positive terminal to negative terminal is related to conservation of energy. The ...

0

The photoelectric work function is primarily a surface effect, and for a given metal will vary significantly by crystal face. Note the variations given for silver, with the lowest, 4.26 eV, being from the polycrystalline form. Modelling of efficiency is complicated; from a macroscopic viewpoint one has the skin depth of the metal by wavelength, which ...

5

Before addressing your question, there is a point where I kind of disagree with Orca's answer that I'd like to discuss: I will begin with part 2 of your question about plane waves. The use of this Ansatz is the first clue that you are actually treating the situation quantum-mechanically, but ending up with a result that exactly matches the classical ...

3

We can treat this system classically because it is one of those nice situations in which the quantum-mechanical treatment produces the same results! I will begin with part 2 of your question about plane waves. The use of this Ansatz is the first clue that you are actually treating the situation quantum-mechanically, but ending up with a result that exactly ...

0

When light falls on the photoelectric surface, photoelectric radiations are emitted, when the energy is more than the work function i.e when more electrons are made to fall on the photoelectric surface,it will go beyond the work function and electrons will be emitted normally. No,there will be not the same effect because, $f=\frac1\lambda$,where wavelength ...

2

The diffraction pattern is due to elastic scattering from the "ion core", which is the stationary net charge of the atomic nucleus and it's bound electrons; these elastically scattered electrons don't lose any energy. The electrons which interact with the free electrons are inelastically scattered, and contribute a foggy background to the diffraction ...

0

What the scenario here is that you are comparing the photoelectric effect with X-ray diffraction. See a wave is a carrier of energy. IT could transmit energy from one point to another. But it cannot impart it's energy to another particle as a wave, but only by quanta of energy. That's what photoelectric effect tells us. An electron will be excited only if a ...

2

(1) If the wave is induced by and propagation from the voltage source (battery), then it should take the vector path of the magnetic field created by the battery, instead of the circuit path. Yes, and the magnetic field follows the circuit path, but remains outside the wires. The electric current in a wire is causing a magnetic field which is located ...

0

Actually the flow of electrons is not that cause the light to glow. The electrons are just carriers of energy. I will make it clear. Consider a bulb connected to a battery. The wire is a conductive metal which means that it is in solid state. So the inter atomic spaces is very less. So the electrons can't move that much freely. There arises a no. of ...

1

The battery sets up an electric field in the external circuit which all the mobile electrons feel in all the external circuit. This means that there is a force on all the mobile electrons in the external circuit. So these mobile electrons are accelerated by the electric field and gain kinetic energy from the electric field which is maintained by the battery. ...

2

The electron gains potential energy in the battery; this is transformed to kinetic energy, which from time to time gets dissipated in the inelastic collisions with the lattice. It's quite analogous to a (semi-elastic) ball jumping down some stairs (and then, in the battery, taking the elevator to get up again). This might be more intuitive.

1

Batteries use a type of reaction called a redox reaction that involves the transport of electrons. Rather then the carbon zinc battery, which is a bit complicated consider the simpler example of a zinc copper battery as taught in school science lessons across the world. The reaction is: $$Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$$ So the reaction dissolves ...

1

Via a series of chemical reactions a battery sets up a surplus of electrons on the zinc (negative) plate and a deficit of electrons (positive charges) on the carbon (positive) plate because it is energetically favourable to do that. You can think of the reaction as a zinc atom producing a zinc ion and two electrons with the release of energy. Assume that ...

2

A battery is both a sink and a source of electrons. It provides no net contribution of electrons to the external circuit, however. In the below schematic of an alkaline battery, which is a representative battery configuration: #3 is the metallic zinc anode #4 is a separator that conducts ions, but not electrons #5 is the nonmetallic manganese oxide ...

2

I went through unanswered questions, and stumbled over this... Did you find the original books? The mistake should be in your formula for the $\mu$ of a hollow sphere; the value with $1/5$ you gave is that of a solid sphere... The problem gets more simple I think, if you compare the two things directly: You get both, the angular momentum and $\mu$, from ...

-1

The classic cyclotron with constant acceleration frequency is not well suited for relativistic velocities because the particle's cyclotron resonance frequency decreases while relativistic mass increases 1. This implies that particles accelerated in a classic cyclotron could be treated classically.

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I've been collecting wave theory explanations of QM phenomena for years. The photo-electric effect is the easiest one. The much-talked about frequency dependence is an obvious consequence of the Schroedinger equation. The Compton effect is different: here, the coupling between the e-m field and the electron states is not controlled by the shared frequency of ...

0

No, it would violate both Conservation of Energy and Conversation of Momentum, this is without using Quantum Theory. A third law is Conversation of Angular Momentum, an electron has spin of 1/2 while a photon has spin of 1. Electron and positron can annhilate producing 2 photons if their spins cancel, or 3 if they are parallel. In a Centre of Mass frame ...

3

Accelerating and decelerating charges produce electromagnetic radiation. Bremsstrahlung : Bremsstrahlung (German pronunciation: [ˈbʁɛmsˌʃtʁaːlʊŋ] ( listen), from bremsen "to brake" and Strahlung "radiation", i.e. "braking radiation" or "deceleration radiation") is electromagnetic radiation produced by the deceleration of a charged particle when ...

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