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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 ...


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 ...


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.


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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 ...


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 ...


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

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 ...


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 $p=\frac{E}{c}$.


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

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\ ...


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

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 ...


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' ...


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 ...


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 ...


1

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 ...


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 ...


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 ...


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 ...


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 ...


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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 ...



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