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If electron radiates photons towards proton, then due to 3rd law of Newton electron will move away from proton. But since it's not happening, this means electron and proton are emitting photons in same direction and thus the whole atom moves(or revolve) as the electron spins. Thus atoms are unstable until they complete there incomplete shell/shells.

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It's actually quite simple. If the electromagnetic field (i.e. photons) were charged, that would imply non-linear self-interactions such as those occurring for the gluon octet that mediates the strong force. The gluonic Lagrangian takes the form $$\mathcal{L}_\text{SU(3)} = -\frac{1}{2} \, \mathrm{tr}(F^2) = -\frac{1}{4} \, \sum_{a = 1}^8 F_{\mu\nu}^a \, F_a^... 0 Most people believe that you need mass to transmit a force, or even to be able make one, but the more fundamental concept happens not the force but the momentum. The momentum is the capacity to interact with another entity and transfer it some change in speed. In the case of the photon, it is massless but however has momentum and energy. The energy and ... 0 This is newton second law:$$ \mathbf F = \frac{\Delta\mathbf p}{\Delta t} $$As you can see, its variation of momentum that brings force. So, its the transfer of momentum. Granted, photons are massless, but they do have momentum. There are two ways of a photon transfer momentum to a solar sail: The photon is absorbed (also heats up the sail), or, the ... 2 Okay, so: Unfortunately, the word "mass" has been used in two different ways in physics. One was the way Einstein used it in E=mc^2, where mass is really just the same thing as energy (E) but measured in different units. This is the same "m" that you multiply velocity by to find momentum (p), and thus is sometimes called the inertial mass. It's also the ... 2 This shows a schematic illustration of the Pound-Rebka experiment: At the bottom we have ^{57}Fe source that emits gamma rays upwards with an energy of 14.4 keV. The frequency of the gamma ray is 3.48 \times 10^{18} Hz, but let's just call this \nu_0 to avoid messing around with figures. The gamma ray travels upwards, and as it travels it is red ... 1 In principle any acceleration of an electron causes some radiation, and an electron has to accelerate in order to leak from one plate to the other. However: the velocities, and therefore the accelerations, of electrons in electrical circuits are small. Calculating the electron drift velocity is an exercise routinely given to students and the results tend ... 0 If you are interested in the physics of solar cells this series of lectures is great. It may at times be over your head but you should be able to get a general idea. But to answer your question: Yes, the electron is excited by the photon and will then travel through the circuit, retaining some of the extra energy that was given to it by the photon. To go ... 1 So, lets step back a bit. First lets look at avalanche breakdown. Electrons are constantly scattering, off atoms and other electrons, with some average scattering rate under given conditions. In a semiconductor, avalanche breakdown occurs when the field is strong enough that a free conduction electron gains (through accelerating in the field) a threshold ... 1 I don't know about sources that emit polarization-entangled photon pairs, but polarization-entangled pairs can be obtained from a polarized source in many ways. One example is Spontaneous Parametric Down Conversion (SPDC). Quoting from Wikipedia: In a commonly used SPDC apparatus design, a strong laser beam, termed the "pump" beam, is directed at a BBO ... 2 The diagrams that typically figure in textbooks and pop-sci books have the transverse scale massively exaggerated. The slits for a optical wavelength version of the experiment are typically less than a millimeter apart and on order of a tenth of a millimeter or less in width. This means that they are separated by less than the natural width of the incident ... 0 I like the following explanation. Although mostly mathematical, it illustrates a point Feynman tried to convey in his lectures. Energy is always conserved, and whenever it seems like it isn't you just need to look harder. Let's recall that in the context of classical particle mechanics one defines the energy of the system of n particles as the sum of the ... 5 The photons are not radiated only in perpendicular (normal) direction from the star surface. In fact, they are radiated into all available directions (180 degrees) from every point on a star surface. The same applies for galaxies as well. For this reason there are photons sent into all directions (including the Earth direction) from each point of a star / ... 9 This is what you seem to be imagining as happening: Whereas this is more accurately what's happening: 0 Now that discussion of this topic has ceased, may I offer my own simplified answer based on what I have learned? Firstly, no source of light can produce identical photons with exactly the same energies and direction. If the spread of energies is ΔE then the spread of wavelengths Δλ = ΔE λ2/hc. The maximum number of wavelengths of path difference before ... 0 You cannot immobilize light, but you can absorb it, convert it to heat. A first step in measuring light intensity (watts/square meter) is to make a black body to absorb the light. Then, by conservation of energy, the heating of that black body tells you how much light is being absorbed, in watts. The illuminated area of the black body is the "square ... -2 The amplitude of the wave components change, but the photon itself has no size. -3 Well, a photon does not have a size. Though it does have relativistic mass which depends on the energy/wavelength:$$ E =mc^2 \leftrightarrow m = \frac{E}{c^2} = \frac{h}{\lambda c} 

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A light wave is a traveling disturbance in the electric and magnetic fields. In the far-field these two components are coupled and in-phase so that one can define the amplitude entirely in terms of either the electric field strength or the magnetic field strength; by convention we give the electric field strength. So, in SI, the amplitude of a light wave is ...

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Your reasoning is quite correct, and you can see exactly this phenomenon in a photomultiplier tube. The photomutiplier tube uses very thin metal sheets, and when a photon strikes the sheet the primary photoelectron is emitted in the same direction as the incident photon and escapes from the far side of the sheet: The quantum yield for this process is ...

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If I may simplify this a bit.........I will compare to a gong. When you stike that gong sound propagates though air. Photons are the end result of gamma radiation. When you 'bang a gong' that energy is released. 'It' must radiate out. Same as with light energy or photons.

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So an amplitude in terms of an electromagnetic wave is related to its intensity. So technically, the more intense a light wave is at a point, the more its amplitude. What is the intensity Intensity is inversely proportional to the A^2 of a wave. If you know the intensity, then the amplitude can be found using this formula I=1/2*pvw*A^2 where p is density, v,...

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In the experiments for photoelectric emission, the light is incident on one face of the emitting plate, for example the anode, when determining the stopping potential. The electrons are emitted by this face of the plate. Why are the electrons emitted in the direction of the incident light, and not opposite to it? In this answer the energy momentum balance ...

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There isn't a straight answer to this question, which sheds light onto the meaning of the subtle word coherence, because what we tend to call "decoherence" can have two main roots. A practical experimental meaning of the word coherence is "ability to show interference", and there are two ways whereby observable interference can disappear: (1) (energy) ...

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Assuming that with "oscillation of the wave function" you mean that the state evolves in time with a relative phase $E/\hbar$ as $\alpha\lvert 1 \rangle +\mathrm{e}^{\mathrm{i}E/\hbar t}\lvert 2 \rangle$, that's really an "accident" because given an energy, how else are you going to get a frequency in quantum mechanics out of it if not by dividing by $\hbar$?...

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CuriousOne makes a good point in comments and I will elaborate on that. While you are asking about the coherence length of a single photon, in the experiment that you describe you will have to detect many photons to judge if they are detected in both arms with equal probability. Where do this photons come from? If you assume that the photons are identical, ...

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If you go to the wiki page for coherence length , the following is found as introduction: In physics, coherence length is the propagation distance over which a coherent wave (e.g. an electromagnetic wave) maintains a specified degree of coherence. Wave interference is strong when the paths taken by all of the interfering waves differ by less than the ...

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By length maybe you mean wavelength. A single photon traveling at the speed of light and oscillating at a certain frequency will oscillate through one cycle every wavelength or say 500 nm. As you increase the length of one arm of the experiment the interference will go in and out of phase every one half cycle or every 250 nm.

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The speed of light is always 299,792,458 meters per second, regardless of the speed of the source and the speed of the receiver. According to Newton's corpuscular theory, both speeds of the source and the receivers were being added or subtracted from $c$. According to an aether theory of the electromagnetic field, the speed only depended on the speed of the ...

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Einstein's theory of relativity. Seriously. That's the answer to your question. The purpose of Einstein's theory was to describe what the universe would be like if the speed of light was an absolute constant. You can read all about it on Wikipedia: https://en.wikipedia.org/wiki/Theory_of_relativity

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The speed of light is the same in all frames. If you are moving relative to Earth at 99% the speed of light and you measure the speed of light you find that the speed of light is the same. Assume you shine light in the direction of your motion. From the perspective of an observer on Earth the photons creep forwards at 1% c relative to your frame. However ...

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Both the electron and the photon are subject to quantum effects and your intuition that the answer will be a probibility distribution rather than a number (obtained from evaluating a Feynman diagram)is correct. There is indeed a formula for this process and the answer will also depend on the details of the accelerating force.

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I will address: My question - Does not photon, which is supposed to be quantum of electro-magnetic field, interact with an electron "electromagnetically"? A photon and an electron are elementary particles, quantum mecanical entities. Probabilities of interaction in quantum mechanics are calculated from the wave functions of the system in QED, using ...

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