# Tag Info

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There is one, and only one, electron, neutron, proton. If they are in rest to the observer and fields are not involved, the electron always has the identical energy, electric field and magnetic dipole moment. The same we say about the proton and the neutron too. Photons are never in rest, they all have the same velocity in vacuum. Furthermore they have an ...

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No because according to particle nature of light there is a collision between an electron and photon. Thus one photon can emit only one electron the rest energy get converted to overcome collisions by other metal atoms and kinetic energy.

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Simply we say that when light ray hits the boundary of the 2 media at more than critical angle, it refracts so much { >90deg. } that it gets directed back into the same medium.

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The ideal perfectly smooth flat surface has translational invariance symmetry. That means that there is no mechanism for the scattering of a wave in the horizontal direction, and no mechanism for the change in the component of wave vector parallel to the boundary. That is, the horizontal component of wave vector is conserved. For light incident at angles ...

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Imagine the speed of light to be $1$ meter per second and the speed of light in the medium with a high refractive index to be $\frac{1}{2}$ meters per second. If you have a single peak of a wave in the slower medium, that peak must move forwards at speed $\frac{1}{2}$, no matter what angle it's facing. In the faster medium, that peak must move forwards at ...

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The radiation of an accelerated or a decelerated charge particle is called "Bremsstrahlung" radiation, which means (roughly) deceleration radiation. The radiation is a continuous spectra. i.e. it's not the emission of one photon. The charged particle which is accelerating emits continuously photons of different energy depending upon the rate of change of ...

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For the forces between elementary particles we have Feynman diagrams, where there exists a mediating particle for the interaction. In the simplest diagrams: for the strong it is the gluon, for the weak it is Zs and Ws and for the electromagnetic it is the photon. Here is Bhabha scattering, where the electron and the positron ( attractive force) are first ...

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Actually A.Zee's book on "QFT in a nutshell" has a very nice explanation on this on chapter I.5, I will breifly sketch it (this is a very rough skeptch), $$Z=\int DA e^{iS(A)} =e^{iW(J)}$$ where W(J) is given by, $$W(J)=-1/2 \int \int d^4xd^yJ(x)D(x-y)J(y)$$ where D(x-y) is the photon propogator and J(x) and J(y) refer to two lumps of matter Plugging in ...

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You have a misconception, photons are massless, but there is a momentum term that is usually left out of the energy equation. Since a photon has no mass, so it can travel at the speed of light with no problems. You're thinking of e = mc^2, and since a photon has non zero energy, obviously it must have mass? No! That equation is the simple version for the ...

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Its hard to imagine a photon with a frequency as low as one cycle per second considering green light with a 500nm wavelength has a frequency of about 600 trillion cycles per second. Considering your question from the point of view of a photon traveling at c with a low frequency of 1 cps helps to imagine the process in slow motion. If a photon has an ...

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Well I would argue a much simpler and shorter explanation: That the measurement of a photon will collapse the wave function of the photon and therefore the system will no longer be a quantum mechanical one but classical. Before the measurement there will be a error due to the uncertainty principle, see it like the photon wave is hitting the detector but the ...

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The simple answer is no, but that's because two separate laser beams cannot directly "com(ing) from same direction". Any two beams from separate sources which shine on a target with the centers aligned will have slightly different angles, and the distances from the emitter (and therefor the relative phase) will vary with location on the target. It's true ...

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For two perfect laser sources without noise (infinite narrow linewidth) and of same amplitude, you could achieve this when both sources have the same frequency and same polarization. You need to control the length of one pathway to control the relative phase to $\phi_0=\pi$: $E=E_1+E_2=E_0 (sin(\omega t)+sin(\omega t+\phi_0))$ The problem in general will ...

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It depends, if the frequency is identical and the waves are in the same phase, they do not cancel out. Only when the phase is exactly 1/2 out of phase, they will completely cancel out.(if you point them from the exact same location) !http://cns-alumni.bu.edu/~slehar/PhaseConjugate/PhaseConjugate_files/image009.jpg You can take a look here. This aren't two ...

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Ignore the diagram. It's complex because it includes all the various bits of electronics required for the experiment and that confuses the issue. The experiment is just a variant of Mössbauer spectroscopy. The experiment uses a $^{57}$Fe source that emits a gamma ray with an energy of 14.4 keV. Since this energy matches the spacing in the energy levels of ...

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You need to distinguish between conductors and non-conductors. If your material is non-conducting then the EM radiation can pass through any size of hole regardless of whether the hole is larger or smaller than the wavelength. The power transmitted is just the incident power per unit are multiplied by the hole area, exactly as you would expect. All very ...

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The problem may be that you you are thinking about a string. It may also be that you are mixing classical and quantum views of physics. Never the less, let's try some visual images. Let's move to 2 dimensions. A classical electric field at a point tells you which about the force on a test charge at that point. You can represent that as an arrow. If you put ...

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It seems logical that a photon has to have a cross-section. But since the QED is applied to everything and not only to inner nucleus interactions, the photon as an indivisible particle is "unfashionable". Now a photon is a excitation in a global electromagnetic field. This field is endless and so even from methodological side it is not possible to search for ...

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Time appears to slow down as light passes a strong object of gravity. Time is not really a physical property, time is simply a unit of measure invented by earthlings. Time appears to change because gravity bends light and it actually travels further to get to the observer so it takes longer to get there. Time is actually distance travelled. The light that ...

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After reading some fundamental mathematics and physics or better to say becoming a sophomore you can start reading these books but absolutely some topics need more than fundamental mathematics and physics. Fundamentals of Photonics (Bahaa Saleh, Malvin Teich): This book provides an introduction to the fundamentals of photonics. Fundamentals of Photonics ...

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According to Einstein and many others, light speed c0 in vacuum is universal and measures about 299,792,458 m/s. So it is never possible to change the light speed in vacuum, which is the absolute upper limit for everything. Gravity does not affect a light ray, but the space and time through which the ray travels. In a strong gravitational field, the time ...

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It has been shown experimentally that the formation of H2 can happen in the presence of free electrons: so a photon is emitted in this two step process, taking energy away . Note that this is for low densities. The three body process in the answer by John dominates with increasing density, as discussed in the link.

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Two isolated hydrogen atoms cannot form an $H_2$ molecule for the simple reason that they have too much energy. Any system formed from the two atoms will have an energy greater than the dissociation energy of $H_2$ so no bound state will be formed. Observation tells us that the process must happen because there is a lot of $H_2$ around. It happens when ...

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I would suppose the short answer is no, but photons are affected by gravity. What is happening is naturally quite relativ to the observer. Suppose you are sitting on the photon, travelling past the gravity source with the speed of light. As has been argued above you would experience the force of the gravitational pull as an acceleration toward the source. ...

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You don't feel acceleration. When onboard the ISS, you are accelerating towards the earth (down) due to gravity: if you didn't, you would just fly away from the planet. Because you and the ISS are accelerating exactly the same way, you don't feel a thing. You don't feel a force if it's accelerating you: you feel pressure caused by opposing forces. Here ...

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To talk about acceleration in space is a little bit dangerous without exact definition. One has to separate free fall and acceleration from an impulse. Imagine, you are inside the ISS during an orbit correction. The impulse from the rocket engine you could feel, you get some weight, and and this is an acceleration. In all other time you are weightless and ...

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Photons are blue-shifted when attracted by gravity (I mean - moving towards a mass, not moving at right angles to the gravitational field like in an orbit). They can't go faster, but their energy goes up.

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The type of interaction depends on the energy of the photon, based on the Klein-Nishina formula. From Wikipedia: Very low energy photons (visible light; as long as the photon energy is much less than the mass energy of the particle, i.e. Compton wavelength) yields Thompson scattering, which is elastic scattering with electrons. Low energy photon (a few ...

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Does a proton have a "bandgap"? If yes, what happens when a photon is absorbed by a proton? For single protons, as in a plasma , there exists Compton scattering . The photon transfers part of its energy to the proton and scatters off at a lower energy/frequency, the proton taking up the energy-momentum balance. This is a continuous spectrum, from very ...

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A system can absorb a photon if the energy of the photon matches an excitation in the system. So the hydrogen atom can absorb a photon if its energy matches one of the frequencies in the hydrogen spectral series. A proton is a composite object and it does have a spectral series. However the excited states of the proton involve rearrangements of the energy ...

1

In optics, you rather speak of intensity of light which is the energy per time and per surface. The brighter the light the more intensity it has. The energy transported by a light beam per minute is proportional to the squared amplitude of a wave or the number of photons times their energy. $E\propto |E_0|^2$ for waves and $E=n \cdot E_{photon}=n \cdot h ... 2 (1) it's the energy in the sense that the photon oscillates at a certain frequency. (2) i'm not sure you can physically explain a light wave. More light is just more photons, more energy is photons with higher frequencies. (3) when it comes to the particle nature of light the photon has a frequency. It frequently oscillates through positive and negative ... 1 (sorry, I couldn't write this in the comment section) Have you met the postulates of quantum mechanics? Here is a summary of them http://vergil.chemistry.gatech.edu/notes/quantrev/node20.html Postulate 3 says if an observable has associate a (hermitian) operator, the only values we would observe for one photon the spin-angular momentum are the eigenvalues ... 0 It is called Quantum Mechanics and to understand the processes a lot of graduate studies are necessary. A photon is a quantum mechanical entity/particle. The classical magnetic field , electric field and electromagnetic radiation emerge from the underlying quantum mechanical level by the very large number of the elementary underlying processes involved. ... 2 The concept of a photon comes from quantum field theory, but the radiation from your remote control is well in the classical regime, so we need to figure out how to go from the quantum field theory to the classical field theory. Now, the way we do that is by forming so called minimum uncertainty states. There's a Heisenberg uncertainty principle for fields ... 3 Most likely, your garage door opener operates at a frequency of 315 MHz. Multiplying by Planck's constant, that means each photon has energy of about$2\times 10^{-25}$joules. Most likely, your garage door opener operates at about$1/10$of a watt (or less, per comments below). So each second, it emits 1/10 of a joules of energy. That's$(2/10) ...

2

Yes, the requirement for at least two photons is because a single photon would violate conservation of momentum. See my answer to Particle anti-particle annihilation and photon production for a (very simple!) proof of this. Annihilation can produce more than three photons. In fact the decay of ortho-positronium to two photons is forbidden, and it (mostly) ...

2

Yes. It is much easier to think of this in terms of conservation of momentum: Because light (and electromagnetic radiation in general) has momentum, you will have to gain momentum in the opposite direction to conserve total momentum --- just like if you were to throw the flashlight. It is difficult to think of this in terms of forces because we tend to ...

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The only reason I could imagine a frequency shift is fluorescence (ok, this is cheating :-) ) ultra slight Doppler effect due to the thermal motion of scatterer/reflector atoms. (at macroscopic scale for many photons and atoms, it's more like an ultra slight frequency blur).

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Now generally, it is true that light incident upon an object is slightly shifted in frequency, but when we come to point of your question there are other factors weighing in. In these cases it is better to look at the phenomena in Maxwell's wave model of light. When light falls upon a body, it energizes the lattice to vibrate (this can be thought of as ...

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I agree with what John Rennie said, "A photon doesn't interact with a single electron, it interacts with the entire molecule." The 'probabilistic process' is a better way of stating 'Give it a shot, and see what happens.' The probability between relaxing and splitting, or whether the photon and the molecule reacts at all, sounds good to me. Please ...

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When a molecule absorbs a photon it reaches to an excited state and there are various mechanisms in which the molecule can relax. Dissociation of the molecule is just one of the possibilities. It is not necessary to ionize (to separate the electron from) the molecule for dissociation to occur. What is necessary is to excite a bonding electron, that is, an ...

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Consider this nomenclature for simplicity: Photon 1 is the photon emitted by the spaceship when it was at rest with respect to the outside observer and Photon 2 is the photon emitted by the spaceship when it is moving at a uniform speed with respect to the outside observer. Since the mechanism through which the photon is emitted from the spaceship is not ...

1

This isn't really how it works. A photon doesn't interact with a single electron, it interacts with the entire molecule. Suppose you take the example of ozone photolysis to $O_2$ and an oxygen atom. We can do a calculation for ozone and come up with a series of molecular orbitals, then put two electrons in each orbital. So far so good. But if you remove an ...

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The work function of the photocathode is such that it can eject no more than one electron for each photon. Well, it's possible that very energetic photons can eject two, but that process is rare, and can be suppressed with filters. So that's the sense that it's said that PMTs detect single photons. But only 30% to 70% of photons in the proper energy ...

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Unfortunately PMTs don't do that, at least not cleanly. One of the crucial performance characteristics for photon counting is the peak-to-valley ratio of the PMT, which you get by measuring a pulse-height distribution. PMTs produce dark noise, which means they indicate "photons" that aren't there and they can't cleanly tell the difference between one and two ...

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If your photon has not enough energy to excite the electron then it will just not be absorbed and will pass by, and if you have an electron with an excess energy, it will be absorbed and a photon with the excess energy will be automaticaly emitted and the electron will jump to an excited state. So yeah in your case, you might have a photon with $0.1 \space ... 1 Use the equation E=hc/λ. In case you don't know, h is Planck's constant and λ is the wavelength in nano meters. convert electron volts to joules (E) using 1 electron volt = 1.60217662 × 10^-19 joules. Planck's constant is 6.62607004 × 10^-34 m2 kg / s 0 How a classical electromagnetic wave emerges from innumerable photons can be seen in this blog entry. It is not simple, one needs quantum field theory to start with. One should get the interaction of a single photon with a crystal lattice , and one can get a quantum mechanical solution, which will give the probability of the photon to scatter or go through ... 2 As dmckee says in a comment, the proof is ridiculously simple. Suppose we work in the centre of momentum frame so the total momentum is zero. The particle comes in with some momentum$p$and the antiparticle comes in with the opposite momentum$-p$, and the two annihilate. Suppose the annihilation produced a single photon. The momentum of a photon is:$\$ p ...

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