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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 ...


2

Of course. This is how antennas work. The Maxwell equations make this possible. The metal as a conductor has electrons in what is called the conduction band, and these are free from the ions in the metal lattice. The oscillating electric field $\vec E$ or $\vec D = \epsilon \vec E$ given by $$ \nabla\times {\vec H} = \frac{4\pi}{c} {\vec J} - ...


0

I think this question does not actually address the photoelectric effect because this would be strongly dependent on the frequency of the light source. To me this seems to be a question regarding the relationship between intensity, power and work. While this is technically not how an electron is emitted in real life, one could calculate the power incident ...


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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 ...


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Millikan was the first to experimentally discover the photoelectric effect. The 1923 Physics Nobel Prize was awarded to Millikan "for his work on the elementary charge of electricity and on the photoelectric effect". Einstein did no experimental work regarding the photoelectric effect. Also, the 1918 Nobel Prize was awarded to Max Plank "in recognition of ...


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What Einstein added to the discussion was the idea that electromagnetic energy comes in little particle-like packets. That was a very radical concept at the time (and frankly still is).


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Your confusion is justified since the value of the work function depends on the configuration of the atoms in the material, and therefore can vary significantly. The work function values can be found in this Wikipedia article. I also have the 1975/1976 edition of the Handbook of Chemistry and Physics which contains a table of work functions, and separately ...


0

The work function doesn't depend on time because we're assuming a $\textbf{steady state solution}$. This means the differential equations which describe the total charge on the surface of the metal are at an equilibrium, where all the time derivatives are $0$. When you shine light on the metal, the system actually reaches this steady state almost ...


0

With a classical wave model for light and a classical mechanics model for matter, the energy which is absorbed by one particle will be a function of the wave amplitude $E$. If we reduce the wave amplitude the absorbed energy $W$ will smoothly go to zero. In mathematical terms: $$ W(E\rightarrow0,\omega) \rightarrow 0 $$ If an electron requires a minimum ...


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The more electrons we take out from the metal, the more ionized it becomes Looking at an experimental setup description (such as this one or this one taken at random from google), you should find that the target is not electrically isolated. Indeed the potential of the target can be directly controlled to change the behavior of the experiment. ...


0

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.


1

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 ...


0

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 strength or amplitude of a light wave is proportional to its brightness, suggesting that a bright light should be strong enough to generate a large ...


1

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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