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8

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.


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 ...


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

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 ...


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 ...


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 ...


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 ...


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, \begin{equation} {\bf{p.p}}=E^2-(pc)^2=m_{0}^2c^4 \end{equation} Now for extremely relativistic case , we can use the condition that $E\gg m_0c^2$, thus this yields $p=\frac{E}{c}$.


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 ...


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 ...


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 ...


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

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.


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 ...


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 ...


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

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.


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 ...


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 ...


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: $$P = \frac{q^2\gamma^4a^2}{6\pi\epsilon_0c^3}.$$ ...


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


2

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

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".


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 ...


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 ...


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 ...


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

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. ...


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 ...


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 ...



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