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Why do electrons orbit nucleus...

by Carbonoid
Tags: electrons, nucleus, orbit
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Carbonoid
#1
May15-04, 06:07 PM
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If oppositely charged particles attract, then why do electrons orbit the nucleus and not stick to it as the nucleus is positively charged and the electrons being negatively charged?
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mathman
#2
May15-04, 06:18 PM
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There is no simple answer. Quantum mechanics says this is the way it is. For example, hydrogen atom ground state (electron at its lowest energy) is at a lower energy level than a neutron. The net result is a free neutron splits into a proton plus electron (plus anti-neutrino), while the hydrogen atom does not collapse into a neutron (plus neutrino).
arivero
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May16-04, 11:13 AM
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If masses attract between them, why does the moon orbit the earth instead of falling towards us? (or towards U.S., if Hollywood prefers ;)

marcus
#4
May16-04, 11:44 AM
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Why do electrons orbit nucleus...

the moon is prevented from falling to earth by angels who push it
sideways so as to avoid collision
this was already known by Kepler
however if the angels should ever stop pushing
it has been found that
the moon would fall directly onto Zaragoza
quartodeciman
#5
May16-04, 02:07 PM
P: 383
What happens when a beam of sub-relativistic electrons (NEGATIVELY-charged) is fired point blank into proton targets (electronically-stripped Hydrogen)?

There is not much discussion of this online. Most discussions of targetting nuclei are of the Rutherford experiment type, with + charged probe particles.

The best I can glean about - charge experiments is that (1)ordinary electron capture takes place or (2)scattering. Electrons can emit photons and perhaps set their angular momenta to correspond to requirements of Hydrogen electron states. If not, then the electron probe "loses" the location of the proton, though obviously still subject to its attraction. This follows from the DeBroglie wave of the electron. At low momentum, the wavelength of an electron is much larger than the size of the proton. So the proton itself apparently vanishes as far as providing a target for the electron is concerned. Finally, there is a strong force barrier if an electron gets too close. The negative d quark get pulled toward the electron while the two positive u quarks are pushed away in the opposite direction. This strains the "strings" connecting them and causes maximum resistance to that coulomb attraction. The electron just bounces off, thereby scattering.

At higher momenta, the deBroglie wavelength of an electron would be small enough to sense the electrical substructure of the proton. This was crucial in verifying the quark idea experimentally.

At extreme high momenta, like modern electron-proton collider experiments, then the electron probe evidently can disrupt the proton substructure and produce spectacular events. My guess is that these occur outside the proton proper and would involve weak interactions.

If an electron were actually to make it into a proton, then the proton would deform to surround the invader with uniformly isotropic positive charge (same at every angle), and thereby allow the electron to pass right through. What doesn't happen is that the electron joins the strongly bound proton, because electrons don't feel the strong force (they are colorblind).

maybe maybe
arivero
#6
May17-04, 07:18 AM
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Quote Quote by marcus
the moon is prevented from falling to earth by angels who push it sideways so as to avoid collision this was already known by Kepler
Pesky angels, they are even unable to trace a circle rightly. Ellipses around poor Kepler's head
however if the angels should ever stop pushing it has been found that
the moon would fall directly onto Zaragoza
I could even guess the street... well, in any case, happy to know that the catastrophe filmmakers will lost another possibility of destroying Los Angeles or some similar crowded town in the States.

Anecdotically, Zaragoza was (is?) the emergency airport for the Space Shuttle if something does not work during launching. Never used, even if it should :-(
ZapperZ
#7
May17-04, 07:45 AM
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Quote Quote by quartodeciman
What happens when a beam of sub-relativistic electrons (NEGATIVELY-charged) is fired point blank into proton targets (electronically-stripped Hydrogen)?

There is not much discussion of this online. Most discussions of targetting nuclei are of the Rutherford experiment type, with + charged probe particles.
I'm not sure why you would want to consider sub-relativistic electrons here. There isn't much "discussion" about this because this is uninteresting. All you get is H atom if the electron gets captured at all.

At extreme high momenta, like modern electron-proton collider experiments, then the electron probe evidently can disrupt the proton substructure and produce spectacular events. My guess is that these occur outside the proton proper and would involve weak interactions.
"Outside the proton proper"? Such collision can result in a weak interaction that involved a change in one of the QUARKS inside the proton. So I would hardly call that "outside the proton proper".

Zz.

http://www.anl.gov/OPA/factsheets/r10-02.htm
quartodeciman
#8
May17-04, 11:51 AM
P: 383
ZZ,

Quote Quote by ZapperZ
I'm not sure why you would want to consider sub-relativistic electrons here. There isn't much "discussion" about this because this is uninteresting. All you get is H atom if the electron gets captured at all.
It is interesting inasmuch as it is part of the historical record. Even as the Rutherford experiment is interesting, so would negative electron scattering be interesting in the same context. It is an amplification of Carbonoid's original question about electrons not falling to the nucleus by electrical attraction. Now we would deliberately invite something to happen by deliberately shooting negative probes at the positive nucleus. Of course it is more spectacular when the electrons have higher energies and momenta. We have learned since that time to expect fireworks when the speeds are high. It took Rutherford about 10 years after his classic alpha-scatter experiment to realize that something else (a nuclear barrier) was present.

Quote Quote by ZapperZ
"Outside the proton proper"? Such collision can result in a weak interaction that involved a change in one of the QUARKS inside the proton. So I would hardly call that "outside the proton proper".
I was thinking about experiments like the following:

Deep Inelastic Scattering of electrons from protons --->
http://www.physics.nmt.edu/~raymond/...k/node188.html

note the sentence: "A sufficiently energetic photon is able to knock a single one of these particles out of the proton, ..."
ZapperZ
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May17-04, 01:29 PM
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Quote Quote by quartodeciman
ZZ,

It is interesting inasmuch as it is part of the historical record. Even as the Rutherford experiment is interesting, so would negative electron scattering be interesting in the same context. It is an amplification of Carbonoid's original question about electrons not falling to the nucleus by electrical attraction. Now we would deliberately invite something to happen by deliberately shooting negative probes at the positive nucleus. Of course it is more spectacular when the electrons have higher energies and momenta. We have learned since that time to expect fireworks when the speeds are high. It took Rutherford about 10 years after his classic alpha-scatter experiment to realize that something else (a nuclear barrier) was present.
I shoot electrons at stuff all the time. That's part of my work. However, we were talking about electron-proton colliders. Some of the things you cited were experiments shooting electrons at FIXED targets, or a piece of material. The physics is very different between those two. Even keV electrons being shot at a material can produce a range of interesting stuff, from producing secondary electrons, to x-rays, to Auger processes. You need to be clear and narrow down just exactly what you want to discuss.

I was thinking about experiments like the following:

Deep Inelastic Scattering of electrons from protons --->
http://www.physics.nmt.edu/~raymond/...k/node188.html

note the sentence: "A sufficiently energetic photon is able to knock a single one of these particles out of the proton, ..."
I have no problem with that quote. I merely want to point out that this is certainly NOT a process that is "outside the protron proper". A "parton" is part of the proton. It isn't "outside" the proton.

Zz.
quartodeciman
#10
May17-04, 03:12 PM
P: 383
ZZ,

Quote Quote by ZapperZ
However, we were talking about electron-proton colliders
Sorry! I was talking to the original topic question.

Quote Quote by ZapperZ
You need to be clear and narrow down just exactly what you want to discuss.
Thanks for mentioning those other phenomena in the kev range. My main point was that one doesn't seem to find electrons just tumbling down to the proton and sticking close to it.

Quote Quote by ZapperZ
A "parton" is part of the proton. It isn't "outside" the proton.
I must be unable to read and understand this account. What does "knock ... out of the proton" mean? The accompanying illustration doesn't help. I will agree that this stuff seems to take off from the immediate vicinity of the proton.
ZapperZ
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May17-04, 04:56 PM
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Quote Quote by quartodeciman
I must be unable to read and understand this account. What does "knock ... out of the proton" mean? The accompanying illustration doesn't help. I will agree that this stuff seems to take off from the immediate vicinity of the proton.
It appears that you have great difficulty in accepting that a proton is made up of "stuff" and that an electron can cause a change in that stuff. You continue to tow the line that things appear in the "vicinity" or "outside" of the proton. I have no other means of making myself clear even after saying this repeatedly, so I will end my effort here.

Zz.
quartodeciman
#12
May17-04, 10:25 PM
P: 383
Too bad.

I don't have trouble with the proton as a parton stream. A boson (photon?) from that high momentum electron knocks one out, say a u quark. But does the proton disappear as a result? Is the yield (minus the scattered electron) all mesonic? I couldn't tell from this and other descriptions. I figured the ousted u quark couldn't be left naked, so a u/u-bar pair must have been raised, with the new u replacing the ousted u and leaving the proton intact (there are still two unaffected quarks). Meanwhile, the new u-bar would join the ousted u quark and make a meson. Wrong? What happens to the baryon number?

Oh, I forgot! ZZ is done with me. :)


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