New answers tagged

2

Let the magnetic field be in the $\hat z$ direction. If you calculate the expectation values of $S_x$ and $S_y$, you find that they have time dependence like $\cos(\omega t),\sin(\omega t)$ while the expectation value of $S_z$ is constant. Explicitly, the Hamiltonian is $H = -\omega \sigma_z$. Using the Heisenberg equation of motion, \begin{align} \dot ...


0

As far as I know, the external magnetic field that produces a torque on the magnetic moment is not necessarily the reason for the precession of it around the direction of the magnetic field. The magnetic moment of an electron is proportional to it's spin and its revolving motion around the nucleus. so when an external field acts on it, it tends to align in ...


0

I will stick my neck out without researching this. I think this is a matter of the Pauli exclusion principle.


0

Dipole fields fall as $1/r^3$. The magnetic field due to a long linear current falls as $1/r$ near the wire. So you can ask yourself "Could I get the long wire behavior by integrating a bunch of dipole in a line?" If you try it \begin{align} E(r) &\propto \int_{-\infty}^{\infty} \frac{\mathrm{d}x}{\left(x^2 + r^2\right)^{3/2}} \\ &= \left. ...


4

@GerryHarp's answer contains the gist of the main idea, but there is a point that doesn't quite make sense: There is no such thing as bare ions crystal in solid state physics. To answer your last question: The "bare phonon frequencies" definitely do not refer to phonon frequencies in the absence of electrons. In fact, an ion crystal without electrons is ...


4

Lets start with the bare ions. You set up a linear lattice of protons where the boundaries are held in place by some means. The protons want to get as far apart from each other as possible. So the minimum energy state has the protons spaced on the line with equal distances between them. So yes, the ions form a regular lattice without the electrons. Now ...


1

It is possible to perform the Double-Slit Experiment using electrons. The resulting diffraction pattern matches that predicted by quantum mechanics, with characteristics defined by the deBroglie wavelength of the electrons. When I was in college, I watched a Professor perform this experiment in class. No other theory ever devised, aside from QM, explains ...


1

Photo-emission is not a simple one step process. The incident photon excites an electron, but the momentum of the initial photoelectron is in the same direction as the incident photon i.e. down into the bulk of the metal. For the electron to be emitted as a photo-electron it has either to backscatter off another electron in the metal or it has to transfer ...


1

There are several factors which determine the energy acquired by a photo-electron: the structure of the crystal surface determines the work function, which means that (111) surface is different from (100); only a perfect crystal has a single surface structure. There is always some bandwidth to the optical source, though it may be quite small. One should ...


1

The electron energy (the portion that changes at least) will be mainly due to kinetic energy (translational) and potential energy due to the potential difference between the cathode and the anode. The electron does have a "spin", but this spin isn't like that of a spinning sphere. The reason for the name spin is simply that the electron spin describes the ...


2

Okay, so just to be clear I am going to consider processes in which a photon and an atom at some energy level go in, and the photon and atom exchange energy (and momentum) such that a photon with a shifted (either higher or lower) energy comes out, while the atom ends up in a different internal electronic state than it started in. A general diagram looks ...


0

I found the answer I was looking for in Chemistry S.E. As you can see it is not that impossible to have a picture of what you calculate. http://chemistry.stackexchange.com/questions/51568/what-is-the-reason-why-protons-and-electrons-do-not-collide/51576#51576


1

This answer is not rigorous, just an afterthought: Remember that a photon has to be fully absorbed first, it will not absorb and emit simultaneously. This would leave us with an electron that is momentarily in a forbidden energy state. From that state it could emit a photon to jump to what should be its correct (i.e., allowed) energy level. Uncertainty will ...


-1

I believe if an atom receives too much energy it can be ionized. There are allowed energy levels but above these levels is the region for free electrons. The energy levels there are not quantized and can receive any energy. Photons with higher energy can put the total energy above the allowed energy levels therefore ionizing the atom. Another way to look at ...


1

You are unlucky, because the microworld of electrons nuclei, atoms and molecules has been studied with mathematical models for over a hundred years and it is not open to hand waving hypothesis of the type: I would say electrons are very tiny containers of energy, which can contain between a minimum and a maximum of such energy, depending on how much energy ...


0

Oxygen and Nitrogen do absorb 'light' but only in the ultra-violet region of the spectrum below approx 200nm, an area invisible to our eyes but easily observed by photomultipliers and similar detectors. This absorption is caused when an electron from a molecule's ground state is promoted to one of several electronically excited states. These states have such ...


1

This blog post is a good one to get a feeling of the Higgs mechanism. One should definitely separate the Higgs field from the Higgs boson. The Higgs boson is an elementary particle , attendant to the existence of the Higgs field as its excitation, and acquires its mass as all the other particles in the standard model table. The Higgs field that gives the ...


0

2) Does it include energy of electron due to electric field too? We do not know. If the electron was composed of charged elementary parts, its apparent inertial mass would be greater than sum of inertial masses of the parts. This is because such parts will act on each other with electromagnetic forces and in accelerated motion, sum of these internal ...


1

There is the concept of the self energy of the electron that can be described by its electric field. The model, which goes back to classical physics, can be described as a spherically distributed charges (that equal the charge of the electron) at infinity and the work required to bring the charges to the radius of the electron. The calculation would show the ...


0

To begin with, I can't think of any reason why an increase in the mass of an electrically charged object should affect the magnitude of the electric field it produces. You can easily check that $E=kq/r^2$ doesn't depend on $m$. Moving charges do produce magnetic fields, but this is a different matter (and the magnitude of the magnetic field has nothing to do ...


0

Electron bombardment of neutral atoms produces X-rays by ionizing the atoms, not by removing the outermost "valence" electrons like you do when you rub a balloon on your hair, but by removing the innermost electrons. The ionized atoms neutralize by picking up charge from the environment into their valence shells. However the hole is the inner shell is ...


11

The most direct answer I found is that N2 and O2 are very simple molecules. 2 atoms, tightly bound, no angles. Source From Source: Because the nitrogen gas molecule is so simple, it cannot do very much with the light energy that it absorbs. It can spin or vibrate only a little bit by stretching and pulling. Oxygen acts pretty much the same ...


5

Quantum mechanics says that only very specific wavelengths are acceptable for exciting an atom, wavelengths that are not those (most of the spectrum) passes by the atom without interacting much (just some scattering in the case of visible light). The allowed wavelengths are the ones that have energy extremely close to the energy difference between two ...


2

This is well explained on the basis of sub shell electronic configuration. But first, let's look it by the concept of shells alone. See in the example of calcium, it has $20$ electrons. Of course the outermost shell can accommodate $18$ electrons. If it goes like $2,8,10$ then the outermost shell contains 10 electrons. The stable state is either having an ...


2

In a degenerate gas of fermions, the fermions fully occupy momentum states from zero up to a momentum corresponding to the Fermi energy. It is the momentum of the fermions that leads to degeneracy pressure. As long as the kinetic energy of particles at the Fermi energy is much less than $kT$, then the fermions can be considered completely degenerate, so ...


2

So the entire electromagnetic force can be described as having these objects which interact by exchanging virtual photons. Photons -- light -- in some sense are the electromagnetic force. It is therefore unsurprising that light can be absorbed by a particle -- an electron, say -- as a sudden "push" which launches the electron in some new direction with ...


2

A very obvious answer is just to heat the air enough (flame). You can also accelerate electrons with electric field and inject them into air (plasma needles and a lot of commercial and lab plasma setups). Microwaves can be used to create and sustain ionization (think of microwave plasma experiments). The same goes for other wavelengths that resonate with ...


0

More energy is released from an electron dropping two levels compared to dropping one but every element has a unique arrangement of electrons. Some atoms can release more energy in one drop than another atom can do in two drops. It all depends on the energy levels of the valence electrons and their unique arrangement.


0

You are not the first to try think that one more undelying onion level ( or matriuska) lies within what are considered fundamental particles at present. Back in the late 1970s when the quark model was established , preons were the next hypothesis as the subcomponents of quarks A number of physicists have attempted to develop a theory of "pre-quarks" ...


0

Electrons and other leptons are, as far as we know, fundamental. They are not made out of quarks, they are not made out of anything! Neither of your assertions has any empirical evidence going for it. However, it seems you are really asking for reasons for charge discretization, since you seem to get the idea of your assertions from the charges of electrons ...


1

For some detailed analysis of the limits for electron microscopy, see: Viewpoint: What Are the Resolution Limits in Electron Microscopes? This is a brief review of the technology, and they summarize the recent improvements in resolution with: "The authors estimate that the resulting resolution limit is in the range $0.50–0.8Å$, which is consistent with the ...


1

Electrons can not loose their charge. It is not currently known to be made up of any other elementary particles, as discussed in the other postings. What makes it impossible are the conservation laws of charge, energy, and lepton number The one for charge would say that if it looses its charge something else has to appear with the same charge. That would ...


0

The concept of electric charge is introduced to explain experiments (originally from static electricity). It is found that only two types of charges are necessary and to distinguish them and to distinguish between they are given labels. The most convenient label is positive and negative (that has some mathematical advantages). It is pure convention that ...


-2

If the electron is said to be a charge carrier seems to be a little bit misleading. Electrons have the intrinsic property of electric charge, they are a charge. You can’t take away the charge from the electron, the electron is the charge. And as long as there are not found constituents of the electron it makes no sense to talk about a carrier property. ...


-1

All particles seems to be grouped under two distinct polarities based on the manner of attraction or repulsion. Those particles repelling one another are said to have like charges. Those that attract one another have different charges. Being a positive or negative charge, is a matter of convention already accepted by world scientific community. A Charge is ...


0

Electrons have charge, and that's not going to change, as you said. There are interactions that involve the charge going one way and something else going another way. One example would be reverse beta decay - Electron and a proton come in, neutron and neutrino go out (other particles will get involved, depending on the details). The neutrino carries the ...


-1

I don't know validity of my answer but still I would liked to propose it. What is the maximum energy you can extract from an electron here In above question we see their is a limit to energy we can extract from an electric field due to finite charge. If you somehow extract all that energy, you can eliminate its electric field and hence it is equivalent to ...


0

You can define velocity as $v=\frac{\sqrt{\langle \hat{p}^2 \rangle}}{m_e} $. For ground state $v=Z\alpha$.


2

It seems that some people liked this question so I shall post my thoughts so far. I don't have a definitive answer, but I did get some interesting results. Let $\rho_m(\boldsymbol r)$ and $\rho_e(\boldsymbol r)$ be the mass and charge densities of the electron. The $g$ factor is given by $$ g=\frac{m}{e}\frac{\int\mathrm d\boldsymbol r\ r^2\sin\theta\ ...


0

The work function doesn't depend on time because we're assuming a $\textbf{steady state solution}$. This means the differential equations which describe the total charge on the surface of the metal are at an equilibrium, where all the time derivatives are $0$. When you shine light on the metal, the system actually reaches this steady state almost ...


0

The more electrons we take out from the metal, the more ionized it becomes Looking at an experimental setup description (such as this one or this one taken at random from google), you should find that the target is not electrically isolated. Indeed the potential of the target can be directly controlled to change the behavior of the experiment. ...


2

If voltage is all that you know, then the answer is No. If you know how much charge $Q$ in Coulombs is added, you only have to divide by the charge $e$ on each electron (in Coulombs). Otherwise, if you have a parallel plate capacitor and you know the capacitance $C$, you can work it out from $Q = CV$. Your suggestion that the extra electrons 'push more' ...


-1

www.physicspages.com/2013/04/11/magnetic-dipole-moment-of-spinning-spherical-shell/ My search gives $\mu = \frac{e\omega R^2}{3}$ This gives $g = 5/3 = 1.667$ Did not you provided link given below? https://en.wikipedia.org/wiki/Electron_magnetic_moment#The_classical_theory_of_the_g-factor Which explains that non-uniform charge distribution can explain ...


0

An electron being a ball of uniform mass and charge is not consistent with its observed gyromagnetic ratio. The charge must be pushed out and the mass must be pushed comparatively inwards to satisfy the existing ratio of about 2. See Classical proof of the gyromagnetic ratio $g=2$


0

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.


2

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 ...



Top 50 recent answers are included