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Photoelectric effect and light polarization in Wikipedia

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lightarrow
#1
Jun17-09, 02:17 PM
P: 1,521
The wikipedia page on photoelectric effect
http://en.wikipedia.org/wiki/Photoelectric_effect

talks of a 5th experimental result, which I hadn't heard before:
"The direction distribution of emitted electrons peaks in the direction of polarization (the direction of the electric field) of the incident light, if it is linearly polarized."

Is that true?
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Bob S
#2
Jun17-09, 09:30 PM
P: 4,663
In Evans, "The Atomic Nucleus", it states that at low energies, the photoelectron direction is peaked along the electric field vector, which is orthogonal to the photon's direction of propagation.
lightarrow
#3
Jun18-09, 02:33 AM
P: 1,521
Quote Quote by Bob S View Post
In Evans, "The Atomic Nucleus", it states that at low energies, the photoelectron direction is peaked along the electric field vector, which is orthogonal to the photon's direction of propagation.
Don't have that book. Does he refer to real photoelectric effect (light on a solid surface) or to photoionization (light through an atomic gas)?

ZapperZ
#4
Jun18-09, 05:27 AM
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Photoelectric effect and light polarization in Wikipedia

Quote Quote by lightarrow View Post
The wikipedia page on photoelectric effect
http://en.wikipedia.org/wiki/Photoelectric_effect

talks of a 5th experimental result, which I hadn't heard before:
"The direction distribution of emitted electrons peaks in the direction of polarization (the direction of the electric field) of the incident light, if it is linearly polarized."

Is that true?
That passage is a bit vague. Is the experiment simply collecting ALL of the photoelectrons as a function of angle of emission? Or are they doing something similar to a momentum distribution curve?

If we assume that the linear polarization is parallel to the surface of the photocathode, then there are several issues when dealing with this. First is the question on whether we're dealing with a single-crystal photocathode, or a polycrystal. If that is the case, then the most naive answer to your question is "yes". One can then consider that there are electrons in the material moving in all directions. So when light with a certain direction of the E-field vector hits the surface, electrons in that direction would tend to be given the energy more "efficiently" than those already moving in other directions. Since photoemission preserves the in-plane momentum of the photoelectrons, these photoelectrons will be emitted more favorably, and therefore, have roughly the same direction as the polarization. The EXACT momentum (i.e. in-plane plus out of plane) depends on the photon energy.

Now, if this is a single-crystal, then it depends very much on the band structure of the material. If a particular direction is allowed by symmetry, then it will be the same as above. But if it isn't, or if that direction isn't that favorable for that band transition (i.e. there's an energy gap in that direction or if only an indirect transition is allowed), then what is written above doesn't hold true anymore.

Photoemission is solids is a very complex and rich process, so much so that we can probe the properties of the material (the photocathode). The standard photoelectric effect is an extremely simple and naive scenario of the photoemission process. Once one tries to peel away at the simplicity, then many other factors are involved in determining the characteristics of the photoelectron spectrum.

Zz.
lightarrow
#5
Jun18-09, 01:59 PM
P: 1,521
Thanks, ZapperZ.
Bob S
#6
Jun18-09, 03:04 PM
P: 4,663
From Bob S
In Evans, "The Atomic Nucleus", it states that at low energies, the photoelectron direction is peaked along the electric field vector, which is orthogonal to the photon's direction of propagation.
Quote Quote by lightarrow View Post
Don't have that book. Does he refer to real photoelectric effect (light on a solid surface) or to photoionization (light through an atomic gas)?
R. D. Evans has a plot of electron angular distribution relative to photon direction at 20 KeV photon energy showing a Gaussian-like curve with a max at about 80 degress wrt direction of photon, with half max points at about 40 degrees and 120 degrees. He states that the photons "tend to be ejected in the direction of the electric field". He refers to his paper with C. M. Davisson in Reviews of Modern Physics vol 24, page 79 (1952). See abstract and access to article at
http://prola.aps.org/abstract/RMP/v24/i2/p79_1
He also refers to (and plots) Fischer's nonrelativistic formula from Ann. Physik vol 8, page 821 (1931)
lightarrow
#7
Jun19-09, 06:50 AM
P: 1,521
Quote Quote by Bob S View Post
From Bob S
In Evans, "The Atomic Nucleus", it states that at low energies, the photoelectron direction is peaked along the electric field vector, which is orthogonal to the photon's direction of propagation.

R. D. Evans has a plot of electron angular distribution relative to photon direction at 20 KeV photon energy showing a Gaussian-like curve with a max at about 80 degress wrt direction of photon, with half max points at about 40 degrees and 120 degrees. He states that the photons "tend to be ejected in the direction of the electric field". He refers to his paper with C. M. Davisson in Reviews of Modern Physics vol 24, page 79 (1952). See abstract and access to article at
http://prola.aps.org/abstract/RMP/v24/i2/p79_1
He also refers to (and plots) Fischer's nonrelativistic formula from Ann. Physik vol 8, page 821 (1931)
Thank you, Bob S.


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