# Tag Info

4

Even if the laser had perfectly reflecting, i.e. lossless, mirrors at either end of the cavity, and both ends were sealed so no light could escape it would still require a continual power input. That's because excited atoms/molecules can decay by mechanisms that don't involve a photon e.g. collisional de-excitation. The lost energy goes into heating up the ...

0

It all depends on the amount of radiation to which one is exposed due to the day-to-day things used. Radium releases alpha particles which have very low penetrating power. Hence we are not in danger of those radiations.

10

One has to make clear that the watches we are using now are no longer using radium , because of radiation danger awareness. Radium dials are watch, clock and other instrument dials painted with radioluminescent paint containing radium-226. The 1900s (decade) were the peak of radium dial production, as radiation poisoning was then unknown; subsequently, ...

1

By far the most common isotope of Radium is 226Ra, which decays by emitting an alpha particle. Alpha particles have almost no penetrating ability, and in general externally- occurring alpha particles are absorbed by the outer layers of skin which are naturally sloughed off, so no permanent damage occurs. If you swallow it, that's a whole other (very sad) ...

-4

Humans can tolerate a certain amount of radiation. The watch contributes less radiation to our bodies than the soil. Radium emits x-rays, so yes, they can excite an electron.

2

It turns out there is a good way to estimate this. In PET (positron emission tomography), the patient is injected with a radioactive tracer that emits positrons (hence, the name). These positrons annihilate with electrons, and emit two 511 keV photons. Detecting these photons, one can estimate the location of the radiation. Doing this many times over, you ...

1

OK, let's take you questions one by one. Theoretically, the answer is yes. If you manage to observe the people on the planet from an area not in the vicinity of any significant mass, you would see the people moving much slower, clocks running slower, etc. Although the speed of light is a constant in a vacuum, the frequency of the light will be different ...

2

The most fundamental description of the electron we have at the moment is using quantum field theory. This describes the electron as an excitation of a single quantum field that spans all of time and space. It also neatly explains how electrons can be created and destroyed: add a quantum of energy to the quantum field and it appears as a newly created ...

3

In physics (as in all natural sciences) there are never any proofs. We write theories and believe them as long as there are no experimental results contradicting them. You can prove that a theory is wrong, but we can never be sure that it is correct. However, your question can be answered in a convincing way by looking at Fermi-Dirac and Bose-Einstein ...

0

As I understand it, to be able to distinguish between two things, is to be able, by some comparison between them , to find a property that is different between them. In macroscopical classical physics we are able to say a body 1 to be different than a body 2, even if they are identical in shape and all properties of their form because, if we have them in ...

0

Here is an explanation of electrical flow that might illuminate this matter: http://amasci.com/amateur/elecdir.html

1

That there are two distinct types of electric charge is a metaphysical fact. But nature is indifferent to what we choose to label these charges; up / down, left / right, positive / negative, black / white, etc. Electrons will still flow to the plate in a CRT regardless of how we choose to label the polarity of the charge on the electron and plate. ...

1

Benjamin Franklin proposed electric fluid theory and considered electric current to be flow of a charged fluid. He meant to use positive to denote a surplus of the fluid, negative as a deficit of it. No one knows how he came up with the choice, but it became the convention and as a result lead also to the labeling of charge. I know of no fact that could ...

0

As Zeldredge said the name is arbitrary and does not matter electron could have been positive and protron negative just the name.

2

Now, from Coulomb's Law we can find a vector for the electric field due to this electron at all points in this space. When you read about Coulomb's law, you can see that it describes the force between two charged particles. When you set them down such that they have no initial relative motion, they move together or apart in a linear fashion. But note that ...

2

The electron is an elementary particle in the underlying building blocks of matter organized in the elementary particles table of the standard model of particle physics. Elementary particles are point particles. The standard model is a precis of a very large number of measurements (data) fitted by mathematical models of theoretical physics. A point ...

2

A cathode ray tube where the electrons are randomly ejected and are not constrained in space the Pauli exclusion seems irrelevant.. In structured beams though, things may be different: The Pauli Exclusion Principle places a fundamental limit on the brightness of an electron beam. Developing a cathode which can reach this limit is useful for achieving ...

3

If by the exclusion principle you mean that the total wave function is anti-symmetric under particle exchange, then yes.

2

The Universe is indeed electrically neutral at the cosmological length scales which means that the total charge of the positively charged particles is equal to (minus) the total charge of the negatively charged particles. However, one must be more careful what these particles are. Electrons and protons are two dominant charged particle species. However, ...

2

Theoretically yes, the laser principle does not consume any material. There is a light source that excites the electrons in the material to higher levels, they deexcite to some intermediate one, here the avalanche of photons appears producing the laser light and leaving the electrons in the ground state. And you can repeat the process without a loss.

1

theoretically if its components never wore out then yes. however in practice things do wear out eventually and so no it could not be done in the same way that a perpetual motion machine can work in theory but not in practice.

0

For a series RL circuit with DC source and switch, there is a problem with opening the switch after it has been closed for some time. In the context of ideal circuit theory, the current through an inductor must be continuous since the voltage across is proportional the time derivative of the current through. Put less rigorously, if the inductor current is ...

2

This is a tough one. I'm going to take the liberty of rewording the question, and then use analogies to hopefully give some kind of answer that gets us part of the way there with this arguably obscure but worthy issue. How did the electron change from simply flying past the proton to adopting a quantised orbit around it? Because it was attracted towards ...

1

What makes you think that the question has an answer? The Bohr model has limited validity and this was realized from the start. In essence, you're describing a transition from an unbound state of the electron (with positive total energy) to a bound state (with negative total energy). This cannot happen all by itself, as the extra energy needs to go ...

1

It all comes down to the band structure. In a solid the energy levels are not the same as an isolated atom, there are bands instead of levels. So if you are thinking about an electron jumping between discrete energy levels then that is probably not going to happen. Particularly if you are thinking that an electron will de-excite emitting a photon, then that ...

1

The filling order of the shells is 1s, 2s, 2p, 3s, 3p, 4s, 3d, ... Silicon, with 14 electrons, has only filled 1s, 2s, 2p, 3s and half of 3p. It hasn't any electrons in 4s or 3d (in the ground state). Although the 3rd orbital can have a maximum of 18 electrons, the shell is considered full with 8 electrons if the 4s is not filled.

1

Boron can conceivably fit a maximum of 8 electrons in its outer shell. This could be achieved through boron covalently bonding with a non-metal (as boron is a metalloid). It is in the 2nd period (row) of the periodic table, hence has 2 'shells', following the 2n2 pattern for maximum amount of outer shell electrons (where 'n' is the amount of 'shells' the ...

3

The short, simple, and intuitive explanation is that in a superconductor state, electrons are paired (BCS case) because there is an effective interaction between them. To destroy such a pair and produce free electrons you need to invest a minimum of energy, which is this energy gap $\Delta$. This produces an excitation (2 free electrons), remember that SC ...

0

Proof time: Each level can have $n-1$ transitions This gives us $S = n-1 + n-2 + n-3+...+1$ Lets take $S$ and do this: $S = n-1 + n-2 + n-3+...+1$ $+S = 1 + 2 + 3 +...+ n-1$ $= n(n-1)/2$ (because there are as we can tell from $S = 1 + 2 + 3 +...+ n-1$, there are $n-1$ elements in the series) QED

1

You can derive it simply by noting that each level can have $n-1$ transitions,so we have $n-1+n-2+...+1=n(n-1)/2$

2

We know that solar cells generate electricity by utilizing the energy of the photon, This is an every day language, electricity. It means things electrical in general every day language. but how does it generate electricity forever? What is generated when the photons hit any material, is heat, and the sun's energy is at maximum 1300Watts per ...

0

This is a well known and kenned estimate. The Wikipedia article Observable Universe covers this well; look especially at the sections under headings "Mass of ordinary matter" and "Matter content — number of atoms" for how it is derived. In summary: Cosmological models, especially the Lambda CDM-Model refined from the FLRW Metric, imply a relationship ...

2

The angle is the same as long as you consider a free electron. Then they are parallel: $\vec{\mu}_\mathrm{elec}=-g_\mathrm{elec}\mu_\mathrm{Bohr}\frac{\vec{S}}{\hbar}$ with $g_\mathrm{elec}\approx 2$ (neclecting effects from quantum electro dynamics). But when dealing with bound electrons (e.g. in an atom), where the electron also has some orbital angular ...

2

Quantum mechanic predicts, that the allowed directions of the spins are quantized. This is one of the main findings of the Stern–Gerlach experiment. In a thermal beam I suppose the the spins to be equally in up and down. (There is no reason why they should not.) But "up" and "down" only correspond to a specific direction in space if there is an external ...

-2

The big questions with this experiment is something else. It is about of recording of where The particule goes. On The Left ore right. Or Both. When is recording The particule is moving as a particule , no interference. But When it is no recording by sensors it has a wave behaviour. Moreover other experiments confirm That The light is Smart. It know That it ...

4

An atom can be positively charged as well as negatively charged. The charged form of an atom (or molecule), which has either more or fewer electrons than protons, is called an ion. It cannot be both at the same time, though, and that is because you are viewing positive and negative charge as relative things ("atom A is negatively charged in atom C's view"), ...

-4

This may follow Einstein's equation and may appear to fit into classical picture, but this is taking place between two particles obeying Fermi Dirac statistics. Hence, it is a quantum phenomenon, and the direction of photon emission is arbitrary, as required by the fundamental assumption of quantum mechanics.

20

It is a standard exercise in quantum electrodynamics to find the angular dependence of the differential cross section. Which more or less means how probable it is for the photons to scatter at a certain angle, given the energy of the incident particles. So assuming the spins of the electron-positron pair is averaged, and that you don't care about the photon ...

0

In a lit candle, when gaseous candle wax reacts with the oxygen in the air, the atoms will be unstably excited. To be stable, the excited electrons will relax to the ground state by emitting photons with energy equal to the energy difference between the 2 states. The photons’ energy doesn’t change much, so the wavelength doesn’t change much. The chemical ...

0

To say I am skeptical would be an understatement. I briefly worked on this around 1997 and what I see here, although it matches his description of what he wanted, in no ways matches the effects he claimed for his "original" which exhibited massive positive feedback, ionization, extreme cooling and massive antigravity effects. The over-unity claims were just ...

0

Even if the electron is acted upon by a torque $\vec{\tau} = \vec{\mu} \times \vec{B}$, how does it deviate the path of the incoming electrons? To do such a task, you need a force. Where does that force come from? The force comes from this equation: $\vec{F} = -\vec{ \nabla }U$ where $U = - \vec{\mu} \cdot \vec{B }$. If the magnetic field \$ ...

0

Electron orbitals are more than just orbitals... it's really best to think of it as a shell, rather than an orbit. The Heisenberg uncertainty principle comes to play, stating the the electron doesn't orbit, but rather it is positioned at the nucleus with an uncertainty amounting to the size of the shell. The size of that shell can change if the electron ...

1

It will help if you study this diagram of what a vacuum tube is If a cathode is heated, it is found that electrons from the cathode become increasingly active and as the temperature increases they can actually leave the cathode and enter the surrounding space. When an electron leaves the cathode it leaves behind a positive charge, equal but ...

0

No, they don't. Only some of negative particles (electrons) on the top shell could do that. Atomic nuclei together with inner electronic shells do not move (almost). In metals they form kind of crystalline cubic grid.

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