View Full Version : Atomic partons?
Bob_for_short
Oct16-09, 11:47 AM
Let us consider an atomic orbital, for example n=2, l=1, m=0 in Hydrogen. The probability plot is given in the figure attached. This picture can be obtained experimentally: via X-ray and elastic electron scattering. Each negative sub-cloud carries a fractional charge. Fortunately, we have quantum mechanics to explain it but if the scattering experiments had been carried out before QM establishing, would we have advanced a parton-like hypothesis for atoms?
I do not know if it would be correct to consider each sub-cloud as an "elementary" particle with a fractional charge. Do you?
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Vanadium 50
Oct16-09, 06:37 PM
It's no more correct than to say that the left half of the 1s distribution has charge 1/2 and the right half also has charge 1/2.
Bob_for_short
Oct16-09, 06:52 PM
I did not understand your phrase.
I wanted to say that considering each sub-cloud as originating from some elementary particles (atomic partons) would be misleading.
Vanadium 50
Oct16-09, 08:22 PM
Dividing the orbital into "subclouds" is imposing your sense of pattern recognition on it. There's nothing whatsoever like a particle with fractional charge.
Bob_for_short
Oct17-09, 05:13 AM
This pattern can be observed experimentally, can't it?
Why we advance a parton model for protons? Following the experimental pattern.
Vanadium 50
Oct17-09, 05:39 AM
Bob, this is silly. An electron has charge -1. The fact that it has a probability distribution doesn't mean anything has a factional charge.
Bob_for_short
Oct17-09, 05:44 AM
Bob, this is silly. An electron has charge -1. The fact that it has a probability distribution doesn't mean anything has a factional charge.
I completely agree with you. Concerning atoms, I speak of a purely hypothetical situation. But it is similar to the deep inelastic scattering interpretation, isn't it?
Vanadium 50
Oct17-09, 07:24 AM
No Bob, it has nothing whatsoever to do with fractional charge and DIS.
I have answered your question four times now. This thread is done.
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