# 'Looking' at atoms

Tail
Can anyone explain to me just how we can 'look' at an atom and its, um, components? And how the process of 'looking' changes the object we're looking at?

arcnets
Hi Tail,
are you asking how to isolate a single atom, or how to probe atomic properties in general?

Greetings !
Originally posted by Tail
Can anyone explain to me just how we can 'look' at an atom and its, um, components? And how the process of 'looking' changes the object we're looking at?
We look at atoms the same way we look at larger objects -
by detecting the EM waves - light that bounces off them.
Occasionaly, like in particle colliders, we can also observe
the path and effect of a particle in different layers of a material
and thus deduce it's properties.

The process changes the object because every EM wave also
has mommentum. So, if you detect a particle it no longer
goes in the same direction and/or the same speed as before -
a change in velocity.

Live long and prosper.

Tail
arcnets-
I'm guessing the second.

drag-
thanks! This was about what I wanted. The thing is, I remember reading about it, but don't recall what exactly was said.

How does the frequency and energy of the EM waves affect the precision of measurements and how much it changes the subject? Also, do particles have a real color?

E8
Also, do particles have a real color?

Yes and no I guess (in a loose philosophical sense since we can argue all day about what is 'real'). What we, as humans, see of an atom or collection of atoms is a very narrow range of the spectrum and the sensors in your eyes assign a particular color to a particular wavelength. What we as humans see as red might be seen as green to a martian. It depends on how we're 'wired' in particular.

So to you a rose is red and the color is 'there' but if you're standing next to a bee, bees only see black, white, and yellow, the red color is not 'there.' You and the bee see the same EM wave differently and each of your sight receptors have assigned a different color to the same wavelength.

I must mention that atoms emit a variety of different wavelengths and not just one. Here's a great link to the absorption/emission spectra of all the elements known to man:

http://javalab.uoregon.edu/dcaley/elements/Elements.html [Broken]

You should click on 'emission,' and click on the different elements to see what colors are sent out by each element that OUR particular senses can sense.

http://www.lbl.gov/MicroWorlds/ALSTool/EMSpec/EMSpec2.html
(The EM spectrum, notice how little the visible region is compared to everything else)

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Tail
Thank you, E8! That was very interesting. And does an electron, for example, have a colour?

And I'd like to repeat my previous question: How does the frequency and energy of the EM waves affect the precision of measurements and how much it changes the subject?

E8
Do you mean color or colour? If you mean color then the same answer applies as before only the difference is how the electron relates to the EM wave. The electronic vibration or excitation/relaxation is what sends out an EM wave(*kind of the same thing as vibration but more energetic, to the point where the electron jumps into higher orbitals). Every outer (or valence) electron vibrates differently w/ each element so that a particular element has it's own EM wave fingerprint it is this fingerprint that physicists, inorganic and organic chemists, Doctors, microbiologists, etc etc., use to determine unknown substances. That's how scientists were able to determine the gas composition of other planets and stars without physically being there.

I'm not sure if the nucleon emits anything we can percieve as light even though I'm sure they send out photons called gluons, so w/ that caveat I'll say that the electron, and not the proton or neutron, is what is responsible for visible light.

I'm not really sure how to answer your other question. Somebody else who works in a physics lab may. My only exposure to light is a spectraphotometer I use to measure bacterial density.

Tail
Is there a difference between 'color' and 'colour'?

Anyway, can an electron reflect light? If so, what frequency is it?

Gold Member
No thee's do difference between color and colour, color is the US spelling colour is the UK spelling. Yes, electrons can scatter (or reflect if you prefer) photons, the frequency is related to the angle of scattering.

arcnets
Originally posted by Tail
Anyway, can an electron reflect light? If so, what frequency is it?

Yes, it can. The process is called Thompson scattering.

Gold Member
I was thinking of Compton scattering but Thompson scattering does the job too (in my post above).

Tail
So electrons don't reflect light of certain frequency?

arcnets
Originally posted by Tail
So electrons don't reflect light of certain frequency?
They scatter electromagnetic radiation of any frequency, but the cross section will of course heavily depend on frequeny. The natural guess is, it will work best when the photon energy is in the range of electron rest mass (511 keV, meaning hard X-ray). But what happens above that, I'm not sure. Unfortunately, the math gets real complicated for the 'relativistic domain'. Interesting question - I'll do some more research. Hang on...

Tail
Interesting question - I'll do some more research. Hang on... [/B]

Do electrons actually absorb any light?

arcnets
I must correct myself. The total cross-section for Thompson scattering seems to be independent of frequency, and seems to equal [pi]r^2, where r is the 'classical electron radius'. However, the differential cross-section seems to depend heavily on frequency. In the sense that you get almost isotropic scattering for low frequencies, but mostly forward scattering for high frequencies.

This is easy to understand, since a long wave hitting a small target will cause spherical waves centered at the target, while in the high-energy domain, a photon will not be deflected much by the target, and will keep most of its original momentum.

Gold Member
Yes electrons absorb photons too, that's what happens for lower energy photons. Classical Thompson scattering is elastic whereas quantum mechanical Compton scattering is inelastic.

arcnets
Originally posted by Tail
Do electrons actually absorb any light?
No. A free electron can't absorb any photon. Being an elementary particle, it doesn't have the internal degrees of freedom to absorb any energy.

Now you could ask: Why can an electron annihilate with a positron (matter + antimatter), resulting in 2 photons of 511 keV (electron rest mass) each?

The standard answer is: e- and e+ together form a virtual particle (the Z boson WRONG! - see below) which, being virtual, is not stable and decays into 2 photons.

Now, draw the Feynman graph of this, rotate it by 90 degrees, and you get the graph of Thompson scattering (remember e+ changes to e- as you apply time reversal. The photon does not change, since it's its own antiparticle). The Z WRONG! - see below is off-shell (virtual) in the old graph, so it's off-shell in the new graph.

See: e- and photon form a virtual Z WRONG! - see below, but this is not stable. And since it's not stable, you can't call it absorption.

Tail, I realize I might be losing you somewhere, but I like your way of asking very simple questions that demand quite complicated answers.

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Gold Member
sorry, yes arcnets right electrons only absorb light when they can move up an energy level.

arcnets
Don't be sorry, jcsd. And remember, it's not only the bound electron which absorbs the energy, but the system nucleus-shell, including e.m. fields between them...

Tail
So what HAPPENS when light touches a free electron? Does the electron reflect it? Or does the light push it?

Gold Member
In Compton Scattering the collision is inelastic so some of the photons momentum is imparted to the electron (this is why the photons wavelength changes), Thompson scatteribng is elestic so no energy is imparted to the system by the photon and it maintains it's orginal wavelength.

arcnets
Agree.

Originally posted by Tail
So what HAPPENS when light touches a free electron? Does the electron reflect it? Or does the light push it?

Concerning particles it's IMO always problematic asking 'what happens'. The answer depends on what model you use.

For low frequencies, I think it's best to use the model of electromagnetic fields. You know that electromagnic radiation can be described as oscillating electric and magnetic fields. Now if you place a single electron in such a field, it will feel an electrostatic force proportional to the electric field vector. This will accelerate the electron. Once in motion, the electron will also feel a Lorentz force, since it's also in a magnetic field. The overall effect will be an oscillating motion of the electron. Now an oscillating charge produces its own electromagnetic wave, much like a transmission antenna. This is the scattered radiation which we can detect.

For high frequencies, I think it's better to use the photon model. In this model, a photon collides with the electron, and the result is a virtual particle called a Z boson WRONG! - see below. Since energy and momentum can not both be conserved in such a collision, the Z boson WRONG! - see below is virtual, or unstable. It thus decays into an electron and a photon. These final particles have the same energies as the initial ones, but may have different momenta. The scattered radiation which we can detect consists of the final photons.

There must, of course, be an intermediate domain where aspects of both models apply, and my guess is this happens when the photon energy is in the range of electron rest mass, which is 511 keV, meaning hard X-ray. Any good theory of the process should cover both domains, and should show that the transition between the two domains is continuous. I guess the theory that we have, does this, but I'm not sure. Any comments on this?

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arcnets
Ooooops...!
I must correct myself. The virtual particle involved in the process is of course *NOT* a Z boson. It's a virtual electron. How could I be so stupid...

arcnets
Here's the graph. Look at graph (b):

http://www.physics.sfsu.edu/~bland/courses/490/labs/b4/4diag.html [Broken]

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