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Photon absorption

by touqra
Tags: absorption, photon
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touqra
#1
Jul11-07, 02:48 AM
P: 282
All the while, I thought that an atom only absorbs a photon that precisely corresponds to one of the transition level (resonance), but now I read that the atom will also absorb photon with wavelengths different from the transition level in question.
So, my question is, suppose I have two energy level = E, and the photon has energy 2E, how would the atom actually absorb E amount to the transition and what happens to the other leftover E ?
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olgranpappy
#2
Jul11-07, 03:24 AM
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well, some "leftover" energy could go into the recoil of the nucleus...

Or, if your atom has two electrons in the lower energy level (different spin states tho) they could each make a transition, thus absorbing 2E.

furthermore, energy levels in "real" systems made of atoms are not perfectly sharp, thus there is no one single frequency at which absorption is possible.
Cthugha
#3
Jul11-07, 03:35 AM
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Are you talking about single atoms?
Single atoms usually do just absorb photons with energies, which correspond to one of the transitions. The linewidth of the absorption is always a bit broadened due to the uncertainty principle (homogenous or natural broadening) and due to the motion of atoms(inhomogenous broadening), but these ranges are still narrow.

Maybe you are not talking about single atoms, but stoms in solids. Here the situation is different as there are collective effects of the solid like oscillations of the lattice (phonons), oscillations of the electrons (plasmons) and such stuff, which can carry away energy as well.

meopemuk
#4
Jul11-07, 03:41 AM
P: 1,746
Photon absorption

One common effect is two-photon absorption. In this case the photon energy should be E/2. In high intensity fields an atom can absorb two E/2 photons simultaneously and jump between levels separated by E.
lightarrow
#5
Jul12-07, 06:43 AM
P: 1,521
Still another possibility, if the photon energy is high enough, is the recoil of an electron by Compton effect.
wesclark
#6
Sep13-07, 10:26 AM
P: 1
I was reading Leonard Susskind's "The Cosmic Landscape," and he was describing how spectral absorption lines are created. I wondered if there is an exact difference in energies between two electron orbitals, then isn't a photon of exactly that energy required to be absorbed by the electron to make the quantum jump? An if so, where would such a special and exact photon come from? Reading this thread made me laugh at my "rookie" mistake. Electrons and photons interact is all kinds of ways per QED, of course, and as pointed out here, the uncertainty principle gives lie to any notion of exactness.
f95toli
#7
Sep13-07, 12:13 PM
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The linewidth of an energy level is simply given by 1/lifetime, this is just classical physics (the "classical uncertainty principle", the relation between Fourier transform pairs), it is essentially the same thing as the quality factor of a resonator.
Incidently, the lifetime of a photon in a high-quality resonator is also given by its frequency divided by Q; so in this case classical physics and QM give the same result.

In atomic clocks energy levels with long lifetimes are used for precisely this reason; long lifetimes give a small bandwidth.
Hans de Vries
#8
Sep13-07, 02:59 PM
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Quote Quote by touqra View Post
All the while, I thought that an atom only absorbs a photon that precisely corresponds to one of the transition level (resonance), but now I read that the atom will also absorb photon with wavelengths different from the transition level in question.
So, my question is, suppose I have two energy level = E, and the photon has energy 2E, how would the atom actually absorb E amount to the transition and what happens to the other leftover E ?
There is the Kramers' dispersion theory equation which led via for instance a joint paper
from Kramers and Heisenberg to Heisenbergs discovery of his Matrix Quantum mechanics.
This plays in 1924. kramers' dispersion relation between the atom and the radiation field:

[tex]{\cal M}\ =\ E\frac{e^2}{4\pi^2 m}\left(\sum_{abs}\frac{f_i}{\nu_i^2-\nu^2} - \sum_{emis}\frac{f_j}{\nu_j^2-\nu^2} \right)[/tex]

Where the [itex]f[/itex] are interaction coefficients, the [itex]\nu_i,\ \nu_j[/itex] are the characteristic frequencies
of the atom and [itex]\nu[/itex] is the frequency of the radiation field.

You see that the the chance for interaction is highest if the incoming frequency is
equal to the characteristic frequency but is not zero if they are not equal. The break-
through in this formula was the inclusion of the second (negative) term which describes
the emission stimulated by the incoming radiation.


Regards. Hans

P.S. The birth of matrix mechanics is described in van der Waerden's book: "The sources
of quantum mechanics" which includes all the relevant papers.
starlindisima
#9
Mar28-11, 03:21 AM
P: 2
Hans de Vries : I'm doing my thesis and i want to know if you can put the reference of your information please...your information its very good, it was what i was looking for.
thanks.
ZapperZ
#10
Mar28-11, 05:03 AM
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Quote Quote by starlindisima View Post
Hans de Vries : I'm doing my thesis and i want to know if you can put the reference of your information please...your information its very good, it was what i was looking for.
thanks.
This thread had its last activity in 2007. Nevertheless, I think Hans has provided a reference at the end of his last post:

Quote Quote by Hans de Vries
P.S. The birth of matrix mechanics is described in van der Waerden's book: "The sources of quantum mechanics" which includes all the relevant papers.
Zz.
starlindisima
#11
Mar28-11, 12:47 PM
P: 2
Quote Quote by ZapperZ View Post
This thread had its last activity in 2007. Nevertheless, I think Hans has provided a reference at the end of his last post:



Zz.
Thanks.


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