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How can light interact with atoms? |
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| Feb16-13, 05:24 PM | #1 |
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How can light interact with atoms?
Atoms are much, much smaller than the wavelength of visible light. Visible light is 4,000 - 7,000 angstroms in wavelength while atoms are ~1-7 angstroms.
How can something interact with a wavelength when that wavelength is 10^3 larger than it? |
| Feb16-13, 05:57 PM | #2 |
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Can you swim in water waves which are longer than 2m? Do they interact with you?
You have a similar situation with visible light and atoms. |
| Feb16-13, 06:01 PM | #3 |
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Water waves are different from electromagnetic waves.
If I send a EM wave with a wavelength of 5000m to a 1m antenna, there will little if any interaction. |
| Feb16-13, 06:37 PM | #4 |
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How can light interact with atoms?
Atoms are not metallic antennas. In a semi-classical picture, the electrons can follow the (slowly changing) electromagnetic field, and get in resonance with that field.
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| Feb17-13, 12:00 PM | #5 |
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| Feb17-13, 12:12 PM | #6 |
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Note that if you are referring to the fact that visible light interacts with bulk materials, you cannot think of it as interacting with single atoms at a time. The EM wave interacts with the material as a whole.
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| Feb17-13, 03:43 PM | #7 |
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Recognitions:
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| Feb17-13, 04:38 PM | #8 |
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So are you saying that atoms do the same thing? That would make sense to me. Unless the electron is the correct size, it shouldn't be able to get in resonance with a wave larger than itself, correct? If I hit a pendulum at a uniform rate which is lower than its resonant frequency, it cannot be in resonance. Of course, if we take the aforementioned effect then the atom could have a large electric field which then makes it *look* electrically bigger and thus, get in resonance the proper way. Just note, I'm not here to argue, I just it all to make sense. :) |
| Feb17-13, 04:54 PM | #9 |
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Mentor
Blog Entries: 27
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Zz. |
| Feb17-13, 06:40 PM | #10 |
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The wavelength as emitted or absorbed has more to do with energy levels (mass and velocity of electron) than just the raw wavelength of a released photon. λ = hc/ΔE Do a google search for emmission spectrum and Bohr model. |
| Feb17-13, 07:14 PM | #11 |
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Hi Menaus...
Based on your implied model, interactions would not be well defined....that's not the model of quantum theory. Note that when you shine a flashlight on a piece of wood for example, not much happens. More happens if the material happens to be photoelectric. And of course a LASER can set a piece of wood on fire, so there is definitely something of interest here between light and matter. One way to gain insights about wave-matter interactions is to read about the 'photoelectric effect'....you'll find its the energy of individual photons that controls the interaction...a valuable hint from Einstein!! Sub atomic particle interactions [even with 'continuous' electromagnetic waves] are best thought of as interactions of field [wave] quanta, that is photons. Quanta are localized 'particle' aspects of waves...localized energy packets. In other words, the Standard Model of particle physics, which describes particles and their interactions, is based on quantum theory. So if you check here: http://en.wikipedia.org/wiki/Standar...rticle_content you'll see everything is described in terms of particles....light is represented as photons for example. In quantum theory, particles are are described via the continuous wave function in a higher-dimensional space in which the wave function exists, what physicists refer to as ‘configuration space’. So while the evolution in time appears continuous and deterministic, photons, like all quantum particles, interact at a single point statistically, regardless of the size of their wave packet dimension. Quantum activity is largely a statistical phenomenon so it doesn't comport in general with everyday macroscopic observations and classical theory....in which everyday things SEEM continuous, like time and distance and lightwaves. To learn more about the wave=particle duality of light, check out 'double slit experiment'. Richard Feynman says if you understand that experiment you understand quantum mechanics... A related way to think about light and matter particle interactions is to note that both exhibit wave particle duality; in other words, even matter particles have a characteristic wave, a DeBroglie wave....reflecting the wave-particle (dual) nature of matter. So in classical theory there is nothing but a point particle, quantum theory unfolds a more complete picture revealing wave-particle duality, Heisenberg uncertainty,etc,etc. |
| Feb17-13, 07:25 PM | #12 |
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Resonance in such antenna occurs when the length of the antenna satisfies $$ d = m\frac{\lambda}{2},~~, m= 1,2,3,... (*) $$ so indeed such antenna has to be as long as the wave it is to be in resonance with. The atom, however, is modeled differently. It has resonance frequencies too, but these are not given by such formula. That formula is not something fundamental - there are even macroscopic antennae that do not obey it either. For example, you can receive long wave transmission with pocket radio. Antenna in such device has additional network attached to it, which influences the resulting resonance frequency of the antenna. Changing parameters of the network allows you to change this resonance frequency, so you can be in resonance with any transmission you want, within some range of course. For the atom, things are more complicated and the standard understanding of its resonances is based onm Schroedinger's equations rather than on the antenna theory (which is based in the macroscopic electromagnetic theory anyway). Physically, the differences are these: in the simple antenna, the resonance frequency is determined mainly by the simple boundary conditions for the current in a straight wire, since in metal there are no effective forces on the charge. In the atom, however, besides boundary conditions (which are very different by the way) there are long-range binding forces due to nucleus as well as repulsing forces between the electrons. These co-determine the resulting resonance frequencies of the atom, and it turns out that resonance light has two orders longer wavelength than the size of the atoms. The resonance frequencies of the atom can be determined by the time dependent Schroedinger's equation, which leads to formula for the resonance wavelength of the atom $$ \lambda = \frac{hc}{E_n -E_m},~~~n, \neq m = 1,2,3,... $$ where ##E_n,E_m## are solutions of the time-independent Schroedinger equation $$ \hat H \Phi_n = E_n\Phi_n. $$ To summarize, the atom is a microscopic object, modelled differently than simple antenna and that allows us to explain its different behaviour. |
| Feb17-13, 07:38 PM | #13 |
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Dave |
| Feb17-13, 10:45 PM | #14 |
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The electric nature of the electron and proton would definitely suffice as an antenna. In a way, the atom could be looked at as a LC circuit seeing as how the electron and proton are connected electrically (forming the 'capacitor'), and the electron itself is spinning (forming the 'inductor') Thank you, this helps. :) I was showing that the logic doesn't follow, so it doesn't then make any sense to me. You don't seem to be following what I've been saying otherwise you would understand that I didn't infer that the quoted sentence was true. Why are you reverting to an argumentative tone when I am trying to get the answer to a question that in a way that makes sense to me? Look forward two sentences and you have my actual postulate. |
| Feb17-13, 11:07 PM | #15 |
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how does a IR CCD camera work? lol the pixel elements are smaller than the wavelengths of the light they are interacting with! how does a solar cell work, optical wavelengths are far bigger than the size of the PN junction?
oh, another one: how do you produce optical frequencies? in LEDs the light emitting layer is thinner than the wavelength of of light! atoms emit light of wavelengths much larger than themselves all the time! the details of how light interacts with matter at the microscale is dealt with in molecular spectroscopy for molecular systems and condensed matter physics for periodic systems. it is not easy. |
| Feb18-13, 01:50 AM | #16 |
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| Feb18-13, 02:23 AM | #17 |
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Take also in mind that the first analytic treatment of absorption and emission was that of Heinrich Hertz who considered point dipoles. The low efficiency in terms of radiated power is due to the fact that the radiated power depends on the absolute value of the dipole moment. The shorter the dipole, the higher the charges and currents have to be. So the answer to your question is that the electric charges moved in electronic transitions in atoms and molecules are enormous as compared to that moved usual antennas. Try to calculate the transition dipole moment per mole of a substance and compare it to that of a metallic wire formed from 1 mole of the same substance of e.g. 1 m length and a voltage of 1V between its ends. |
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