# Tag Info

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The number of photons from a light source is not necessarily inversely proportional to the frequency of the light emitted. It appears that this consideration followed from an artificial requirement of fixed power emitted by the source (perhaps this is what you meant by "constant intensity") across all frequencies, which is not necessarily true for a real ...

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I think a simple view is this: The solar cell must have a PN junction, which is a junction between p-type (many holes, no electrons) and n-type (many electrons, no holes) materials. Right where they meet there is actually a "depletion width" within which there is hardly any of either. Within this region, as photons come in they generate electron-hole pairs, ...

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It is the energy of the incident photon. At low Energy (Long wavelengths as in Radio waves and Microwaves, visible light) Rayleigh scattering would dominant, in which the scattered photon would have the same wavelength as the incident photon. At higher energy (shorter wavelengths as in Ultraviolet spectrum), the electron would absorb the energy of the ...

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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.

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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 ...

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Some notes: this Kinetic Energy that you use is the Maximum amount of kinetic energy among the electrons. it means maybe just a few of them has this amount of energy, other photo electrons with lower kinetic energies (therefor lower Velocity) just accelerate back to the plate that they been shot from. the Kinetic energy actually gets to Zero at the next ...

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The difference you are looking for is in the particle energy. The photon energy for visible light is about $3\,\mathrm{eV}$ (electron-volt), just enough to kick one electron from photocathode. In the photomultiplier, this electron is accelerated towards dynodes with high voltage. If it is $1000\,\mathrm{V}$, then the electron gains kinetic energy of ...

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The original problem can be seen in terms of energy and momentum conservation. Before scatter, there are two particles in the center of mass and the center of mass has an invariant mass larger than the mass of the electron. For total absorption of the photon there would be only the electron left. As the electron has a fixed mass and at the center of mass ...

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As the comments indicate, the answer truly is that the electrons in the solid are not really free. But wait, I hear you say -- the free electron model approximates the electrons in the solid as a free gas of electrons. It certainly isn't perfect, but it can't be that poor of a description. Yes it can, and I'll explain why. Consider what it means to say ...

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The negative potential works against the $K$ of the ejected electrons. So at very high negative voltages there is no photocurrent. As the voltage moves closer to zero, we hit the point where $f_2$ produces a current. If it's able to produce a current, this means that the electrons are ejected with a high kinetic energy, and thus the work function that binds ...

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When you say the stopping voltage determines the kinetic energy of the electrons you have that the wrong way round. For a given kinetic energy of photoelectron you must apply a potential to bring it to rest. So it is true that increasing the intensity of the incident light will increase the photocurrent as you have more electrons ejected from the metal, ...

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A photon that strikes the metal either has enough energy to release a valence electron from its bond with the atoms in the metal, from a certain frequency and up, or its energy is just not enough to eject the electron from the metal. The surplus of energy over the energy needed to eject the electron is used as kinetic energy of the electron and determines ...

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Higher intensity can contribute through secondary effects. As the beam intensity increases, then the likelihood of two photons impinging on an atom (almost) simultaneously also increases. If the two photon energies together are greater than the work function, then ionization can occur even though any single photon would be insufficient to do so.

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The photoelectric effect occurs when an atom (or ion) absorbs the energy of a photon and an electron is emitted. Some of the photon energy is used to liberate the electron, the rest goes into the electron kinetic energy. The same thing cannot happen when light interacts with a free electron. A simple proof shows that a free electron cannot absorb all the ...

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The incoming photon produces a photoelectron moving in the same direction as the photon i.e. down into the metal. Photoemission requires the electron to ricochet around inside the metal and either bounce back so it heads out of the metal, or transfer its energy to other electrons so they have enough energy to escape the metal. This is a highly inefficient ...

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The photoelectric effect - ejection of electrons from matter due to light - is a phenomenon that occurs when light shines on a piece of metal. I do not think such phenomenon was observed with free electrons in vacuo. ( Compton scattering is often described as if electrons interacting with the light were free, but it actually refers to an experiment where ...

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