# Tag Info

## New answers tagged photoelectric-effect

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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.

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How photo electric effect happens? According to quantum theory (based upon which the photoelectric effect) is explained, the electrons in an atom are in quantized states. For a given photosensitive material the electrons in the valence shell are having unique quantized energy levels. When a light of suitable energy equal to the energy level of the electron ...

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The photoelectric effect is explained by a cool experiment which exactly displays why the wave model doesn't apply to the idea of photons. Here's a diagram of the experiment: The wave model insinuates that the energy of an electromagnetic varies by the intensity (square of the amplitude of the wave), i.e. $E \propto I \propto A^2$ Physically, this ...

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Let us first describe a relevant experiment: You have a photomultiplier tube, hooked to a loudspeaker for convenience. If you shine on the detector with light you hear noise, which is louder if the light source is brighter. But if you only take a very feeble light, you'll notice a peculiar thing: The loudspeaker does not make noise anymore but produces ...

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Classical electromagnetism has no way of linking energy of a wave to its frequency. In classical E-Mag energy density is usually given by $$u = \frac{1}{2}( \epsilon E\cdot E + \frac {B\cdot B}{\mu} )$$ As you can see this has no frequency dependence in it at all. And in photoelectric effect experiment suggests that frequency affects the energy carried by ...

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Theoretically speaking, you can use Doppler effect to change the frequency of light... But the change is so tiny that it is of practically no use at low speeds... I'm pretty sure there is no other direct way... Hope this helps!!!

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There is a general way to increase the frequency of light. It is Doppler Effect. You can keep the observer (target for photoelectrons) and move the source of light towards it. But it is not practical, because you will need very high relative velocity to have any considerable change in the frequency.

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Frequency of the emitted light depends on the nature of the source. Eg: Incandescent -- depends on the temperature of the material Discharge Tubes -- depends on the characteristic spectrum of the gas

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I am also pretty much unclear about it. But thing of fire, fire at the bottom or where the fire starts, the acceleration of charged particles is more, which means the frequency is also more and wavelength is short. As, $$E=hv$$ Where E is total energy, h is Planck's constant and v is frequency. The acceleration of charged particles is more as the ...

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The photoelectric work function is primarily a surface effect, and for a given metal will vary significantly by crystal face. Note the variations given for silver, with the lowest, 4.26 eV, being from the polycrystalline form. Modelling of efficiency is complicated; from a macroscopic viewpoint one has the skin depth of the metal by wavelength, which ...

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photoelectric effect can occur only in metals because in metals,electrons are freely moving. so when light fall on the surface of the metal, the electrons get sufficient energy to come out of the metal surface. But in non-metals,electrons are stationary and can't come out and no photoelectric effect is occured.

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we know that frequency is inversely proportional to the wavelength of light by the formula $f=\frac 1\lambda$' So frequency will increase if wavelength is decreased.

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When light falls on the photoelectric surface, photoelectric radiations are emitted, when the energy is more than the work function i.e when more electrons are made to fall on the photoelectric surface,it will go beyond the work function and electrons will be emitted normally. No,there will be not the same effect because, $f=\frac1\lambda$,where wavelength ...

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The diffraction pattern is due to elastic scattering from the "ion core", which is the stationary net charge of the atomic nucleus and it's bound electrons; these elastically scattered electrons don't lose any energy. The electrons which interact with the free electrons are inelastically scattered, and contribute a foggy background to the diffraction ...

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What the scenario here is that you are comparing the photoelectric effect with X-ray diffraction. See a wave is a carrier of energy. IT could transmit energy from one point to another. But it cannot impart it's energy to another particle as a wave, but only by quanta of energy. That's what photoelectric effect tells us. An electron will be excited only if a ...

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The saturation current is the same because the RATE of charge flow per unit time is the same. Shorter time, smaller charge flow. The charge is not fixed. The amount of photoelectrons emitted is maxed. So all that are emitted are being collected

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I've been collecting wave theory explanations of QM phenomena for years. The photo-electric effect is the easiest one. The much-talked about frequency dependence is an obvious consequence of the Schroedinger equation. The Compton effect is different: here, the coupling between the e-m field and the electron states is not controlled by the shared frequency of ...

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Well, it's better not to argue about words ... but I would say, that it makes more sense not to consider the photoelectric effect as ionisation. (There is no ion produced anyway, is there? :)) There is nothing special about charges in metals (and there is also nothing special about the slow free electron produced), and nothing "bad", whereas ionising ...

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