Toonces

2005-Sep-16, 08:25 AM

If electrons are held in orbit by electromagnetic attraction to nuclei, shouldn't electrons crashing into the nucleus be a common event? If so, why have I never heard about this?

View Full Version : Crashing electrons

Toonces

2005-Sep-16, 08:25 AM

If electrons are held in orbit by electromagnetic attraction to nuclei, shouldn't electrons crashing into the nucleus be a common event? If so, why have I never heard about this?

Jens

2005-Sep-16, 08:59 AM

Good question. I can't say I totally understand the answer, but here one is:

http://www.madsci.org/posts/archives/feb99/918286257.Ch.r.html

http://www.madsci.org/posts/archives/feb99/918286257.Ch.r.html

Toonces

2005-Sep-16, 09:44 PM

Thanks for the link!

jkmccrann

2005-Dec-27, 04:08 PM

Today we have abandoned the solar system model of the atom in favor of the Schrödinger model: electrons are standing waves distributed through the spherical volume of the atom. Nevertheless, we still understand electronic energy levels as being quantized. Electrons are not permitted to fall into the nucleus because this would involve a violation of quantization.

Finding the electron in the nucleus would also be a violation of the Uncertainty Relation: we would know the electron's position and momentum simultaneously! In fact, the electron's velocity can be approximated by solving the expression for the uncertainty in velocity, using the electron's rest mass and the mean distance of the electron from the nucleus (which can be calculated from the expression for the atomic orbital -- a wave equation! -- in which the electron is located).

Electrons are inherently fuzzy particles, especially when they are in atoms -- and the "fuzz" is on the order of the size of the atom. This means that, while the electron has a certain probability of being "in" the nucleus, the probability can never be 100%.

So, in fact the electron can be in the nucleus except that it can't because..... Have to love that fuzzy logic.......

Finding the electron in the nucleus would also be a violation of the Uncertainty Relation: we would know the electron's position and momentum simultaneously! In fact, the electron's velocity can be approximated by solving the expression for the uncertainty in velocity, using the electron's rest mass and the mean distance of the electron from the nucleus (which can be calculated from the expression for the atomic orbital -- a wave equation! -- in which the electron is located).

Electrons are inherently fuzzy particles, especially when they are in atoms -- and the "fuzz" is on the order of the size of the atom. This means that, while the electron has a certain probability of being "in" the nucleus, the probability can never be 100%.

So, in fact the electron can be in the nucleus except that it can't because..... Have to love that fuzzy logic.......

papageno

2005-Dec-27, 07:42 PM

If electrons are held in orbit by electromagnetic attraction to nuclei, shouldn't electrons crashing into the nucleus be a common event? If so, why have I never heard about this?

It is not a common event, but something like that does happen: electron capture (http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/radact2.html#c3).

It is not a common event, but something like that does happen: electron capture (http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/radact2.html#c3).

snarkophilus

2005-Dec-28, 04:56 AM

So, in fact the electron can be in the nucleus except that it can't because..... Have to love that fuzzy logic.......

Not really. It would be allowed in the nucleus but for the Uncertainty Principle. However, because of Heisenberg, it isn't.

Actually, the way it's worded there, you could have an electron in the nucleus. All you need is to not know its momentum to some degree. Since the nucleus isn't a single point, there's some room for error. Quantum theory predicts a 0 probability of the electron being precisely in the centre of the nucleus, in the same way that probability theory predicts a 0 chance of throwing a dart and hitting the exact centre of a dart board. But if you have a dart with thickness, there's a non-zero chance of some part of the dart hitting the centre.

Not really. It would be allowed in the nucleus but for the Uncertainty Principle. However, because of Heisenberg, it isn't.

Actually, the way it's worded there, you could have an electron in the nucleus. All you need is to not know its momentum to some degree. Since the nucleus isn't a single point, there's some room for error. Quantum theory predicts a 0 probability of the electron being precisely in the centre of the nucleus, in the same way that probability theory predicts a 0 chance of throwing a dart and hitting the exact centre of a dart board. But if you have a dart with thickness, there's a non-zero chance of some part of the dart hitting the centre.

papageno

2005-Dec-28, 03:22 PM

Quantum theory predicts a 0 probability of the electron being precisely in the centre of the nucleus, in the same way that probability theory predicts a 0 chance of throwing a dart and hitting the exact centre of a dart board. Actually, that is not entirely correct.

See here (http://www.chemistry.mcmaster.ca/esam/Chapter_3/section_2.html) for the radial part of the lowest-level electronic wavefunction in the Hydrogen atom: at r = 0 the probability is non-zero (fig. 3-4).

The effect of this can be observed in the hyperfine structure (http://www.pha.jhu.edu/%7Ert19/hydro/node9.html) of atomic spectra.

See here (http://www.chemistry.mcmaster.ca/esam/Chapter_3/section_2.html) for the radial part of the lowest-level electronic wavefunction in the Hydrogen atom: at r = 0 the probability is non-zero (fig. 3-4).

The effect of this can be observed in the hyperfine structure (http://www.pha.jhu.edu/%7Ert19/hydro/node9.html) of atomic spectra.

snarkophilus

2005-Dec-28, 05:31 PM

Actually, that is not entirely correct.

See here (http://www.chemistry.mcmaster.ca/esam/Chapter_3/section_2.html) for the radial part of the lowest-level electronic wavefunction in the Hydrogen atom: at r = 0 the probability is non-zero (fig. 3-4).

The effect of this can be observed in the hyperfine structure (http://www.pha.jhu.edu/%7Ert19/hydro/node9.html) of atomic spectra.

I was thinking in terms of Fig 3.5, where if you look for a particle at a certain point you get a zero probability of finding it there. You can't determine the position exactly. But to find an electron very close to the centre of the nucleus you have a non-zero chance (as is evident from the non-zero value of the wave function at that point in Fig 3.4).

On a side note, I think the explanation below that diagram

Since the curve is unsymmetrical, the average value of r, denoted by rbar, is not equal to rmax.

isn't strictly true. (I know, I'm picky.)

I just read over the math to make sure that what I'm saying here is right. It is beautiful. Whoever invented spherical harmonics deserves a medal or something. And don't you love looking at pictures of orbitals?

Now comes a question (or two) that I don't know enough to answer. Are there concerns akin to the Pauli principle when it comes to protons and electrons (and neutrons, I suppose)? Is there a reason beyond simple energetic concerns why a proton and a neutron can't overlap? Does that then apply to electrons and protons? Can we use that as an argument for saying that an electron can't penetrate a proton/neutron (and therefore the non-free space portions of a nucleus)? Or is a proton in real life more of an infinitesimal point with associated forces that make it seem like it has a size, so that there's always a non-zero probability of finding an electron within its radius?

I'll have to get around to getting a physics degree, too, just so I can answer these kinds of questions for myself. :)

See here (http://www.chemistry.mcmaster.ca/esam/Chapter_3/section_2.html) for the radial part of the lowest-level electronic wavefunction in the Hydrogen atom: at r = 0 the probability is non-zero (fig. 3-4).

The effect of this can be observed in the hyperfine structure (http://www.pha.jhu.edu/%7Ert19/hydro/node9.html) of atomic spectra.

I was thinking in terms of Fig 3.5, where if you look for a particle at a certain point you get a zero probability of finding it there. You can't determine the position exactly. But to find an electron very close to the centre of the nucleus you have a non-zero chance (as is evident from the non-zero value of the wave function at that point in Fig 3.4).

On a side note, I think the explanation below that diagram

Since the curve is unsymmetrical, the average value of r, denoted by rbar, is not equal to rmax.

isn't strictly true. (I know, I'm picky.)

I just read over the math to make sure that what I'm saying here is right. It is beautiful. Whoever invented spherical harmonics deserves a medal or something. And don't you love looking at pictures of orbitals?

Now comes a question (or two) that I don't know enough to answer. Are there concerns akin to the Pauli principle when it comes to protons and electrons (and neutrons, I suppose)? Is there a reason beyond simple energetic concerns why a proton and a neutron can't overlap? Does that then apply to electrons and protons? Can we use that as an argument for saying that an electron can't penetrate a proton/neutron (and therefore the non-free space portions of a nucleus)? Or is a proton in real life more of an infinitesimal point with associated forces that make it seem like it has a size, so that there's always a non-zero probability of finding an electron within its radius?

I'll have to get around to getting a physics degree, too, just so I can answer these kinds of questions for myself. :)

papageno

2005-Dec-28, 09:46 PM

Actually, that is not entirely correct.

See here (http://www.chemistry.mcmaster.ca/esam/Chapter_3/section_2.html) for the radial part of the lowest-level electronic wavefunction in the Hydrogen atom: at r = 0 the probability is non-zero (fig. 3-4).

The effect of this can be observed in the hyperfine structure (http://www.pha.jhu.edu/%7Ert19/hydro/node9.html) of atomic spectra.

I was thinking in terms of Fig 3.5, where if you look for a particle at a certain point you get a zero probability of

finding it there. That is a distribution function: it does not give directly the probability of finding an electron in a point.

The function in that figure gives "the total amount of electronic charge lying between the spheres of radius r and r + Dr".

You can't determine the position exactly. But to find an electron very close to the centre of the nucleus you have a non-zero chance (as is evident from the non-zero value of the wave function at that point in Fig 3.4). Considering that the size of a nucleus is about four orders of magnitude smaller than the typical size of an orbital, "very close" is like saying r = 0.

On a side note, I think the explanation below that diagram

Since the curve is unsymmetrical, the average value of r, denoted by rbar, is not equal to rmax. isn't strictly true. (I know, I'm picky.)

I just read over the math to make sure that what I'm saying here is right. It is beautiful. Whoever invented spherical harmonics deserves a medal or something. And don't you love looking at pictures of orbitals? But we were talking about the radial part of the wave function, not the angular part (where the spherical harmonics come into play).

Now comes a question (or two) that I don't know enough to answer. Are there concerns akin to the Pauli principle when it comes to protons and electrons (and neutrons, I suppose)? Yes.

For example, the Hund rules for the level occupancy, are explained partly with the Pauli principle.

Is there a reason beyond simple energetic concerns why a proton and a neutron can't overlap? In what sense?

I think that this is the realm of Quantum Chromodynamics (quarks and stuff).

Does that then apply to electrons and protons? Can we use that as an argument for saying that an electron can't penetrate a proton/neutron (and therefore the non-free space portions of a nucleus)? No.

Pauli's principle works for identical particles.

Or is a proton in real life more of an infinitesimal point with associated forces that make it seem like it has a size, so that there's always a non-zero probability of finding an electron within its radius? Nucleons (proton and neutron) are composed of quarks, hence their size comes from the structure.

The electron, on the other hand, seems to be point-like.

I'll have to get around to getting a physics degree, too, just so I can answer these kinds of questions for myself. :) For starters, you can keep reading this board. :clap:

See here (http://www.chemistry.mcmaster.ca/esam/Chapter_3/section_2.html) for the radial part of the lowest-level electronic wavefunction in the Hydrogen atom: at r = 0 the probability is non-zero (fig. 3-4).

The effect of this can be observed in the hyperfine structure (http://www.pha.jhu.edu/%7Ert19/hydro/node9.html) of atomic spectra.

I was thinking in terms of Fig 3.5, where if you look for a particle at a certain point you get a zero probability of

finding it there. That is a distribution function: it does not give directly the probability of finding an electron in a point.

The function in that figure gives "the total amount of electronic charge lying between the spheres of radius r and r + Dr".

You can't determine the position exactly. But to find an electron very close to the centre of the nucleus you have a non-zero chance (as is evident from the non-zero value of the wave function at that point in Fig 3.4). Considering that the size of a nucleus is about four orders of magnitude smaller than the typical size of an orbital, "very close" is like saying r = 0.

On a side note, I think the explanation below that diagram

Since the curve is unsymmetrical, the average value of r, denoted by rbar, is not equal to rmax. isn't strictly true. (I know, I'm picky.)

I just read over the math to make sure that what I'm saying here is right. It is beautiful. Whoever invented spherical harmonics deserves a medal or something. And don't you love looking at pictures of orbitals? But we were talking about the radial part of the wave function, not the angular part (where the spherical harmonics come into play).

Now comes a question (or two) that I don't know enough to answer. Are there concerns akin to the Pauli principle when it comes to protons and electrons (and neutrons, I suppose)? Yes.

For example, the Hund rules for the level occupancy, are explained partly with the Pauli principle.

Is there a reason beyond simple energetic concerns why a proton and a neutron can't overlap? In what sense?

I think that this is the realm of Quantum Chromodynamics (quarks and stuff).

Does that then apply to electrons and protons? Can we use that as an argument for saying that an electron can't penetrate a proton/neutron (and therefore the non-free space portions of a nucleus)? No.

Pauli's principle works for identical particles.

Or is a proton in real life more of an infinitesimal point with associated forces that make it seem like it has a size, so that there's always a non-zero probability of finding an electron within its radius? Nucleons (proton and neutron) are composed of quarks, hence their size comes from the structure.

The electron, on the other hand, seems to be point-like.

I'll have to get around to getting a physics degree, too, just so I can answer these kinds of questions for myself. :) For starters, you can keep reading this board. :clap:

snarkophilus

2005-Dec-29, 07:58 AM

That is a distribution function: it does not give directly the probability of finding an electron in a point.

The function in that figure gives "the total amount of electronic charge lying between the spheres of radius r and r + Dr".

Well, yes, because the probability of finding an electron at a precise point is 0. You can't know position exactly. That's the basic premise of quantum theory. That's all I was really saying. Certainly for an s orbital, you're most likely to find the electron very close to the nucleus. When you do the detections to get Fig 3.4 experimentally, you don't really measure any electrons exactly at r=0. You can get ones pretty darn close, to within your instrument precision, but you can never be certain that it was really at the dead centre.

And that's not true just for points, but for planes (and infinitesimally thin things like spherical shells). 4*pi*r^2*dr = 0 if dr = 0, and similar equations hold when you integrate over different spatial components. For the point in the dead centre, both r=0 and dr = 0.

Anyway, that's moot given the information you've given me below, that a proton has an actual size and isn't just a point, because now there's a window for error where you can say "this electron has probability p of being within the region defined by the proton's boundary X, and an error in momentum measurement of dp >= hbar / 2*dX."

On that note, to answer the original question, maybe electrons regularly hit the nucleus, but just don't usually react. Or maybe there's an aspect of the weak force (or the strong force) that we haven't accounted for.

But we were talking about the radial part of the wave function, not the angular part (where the spherical harmonics come into play).

Um... that has nothing to do with it. I'm just saying you can have a curve that is not symmetric but still have rbar = rmax. It'd be a weird curve, perhaps, but it can certainly be done. It just seemed like a hand-waving argument to me: better would have been to say something like, "when you do the math, you find that rbar does not equal rmax."

Maybe this actually comes up with more complex systems. It would be interesting to try to find one, I think.

Is there a reason beyond simple energetic concerns why a proton and a neutron can't overlap?

In what sense?

I think that this is the realm of Quantum Chromodynamics (quarks and stuff).

Yes, it is. I haven't gotten around to learning that yet, though I'm reading Feynman's QED, which is a nice book. QCD is the other end of things, but the ideas behind each theory seem similar at a first glance. Fortunately, I know most of the group theory stuff already, so it shouldn't be too hard to pick up.

I suppose that I mean it in the sense of "does QCD tell me that you can't pack as many quarks as you want into as small a volume as you want?" That's assuming that the quarks themselves have a shape and size, and it's not just the hadron that they form that has those properties.

Nucleons (proton and neutron) are composed of quarks, hence their size comes from the structure.

The electron, on the other hand, seems to be point-like.

Thank you! That is terribly interesting to me, and relevant to this discussion (as I mentioned above). I'll have to hunt down a particle physics book some time and find out why, exactly (if it's even known why, exactly).

For starters, you can keep reading this board. :clap:

With good responses like that, you can bet I will! :D

The function in that figure gives "the total amount of electronic charge lying between the spheres of radius r and r + Dr".

Well, yes, because the probability of finding an electron at a precise point is 0. You can't know position exactly. That's the basic premise of quantum theory. That's all I was really saying. Certainly for an s orbital, you're most likely to find the electron very close to the nucleus. When you do the detections to get Fig 3.4 experimentally, you don't really measure any electrons exactly at r=0. You can get ones pretty darn close, to within your instrument precision, but you can never be certain that it was really at the dead centre.

And that's not true just for points, but for planes (and infinitesimally thin things like spherical shells). 4*pi*r^2*dr = 0 if dr = 0, and similar equations hold when you integrate over different spatial components. For the point in the dead centre, both r=0 and dr = 0.

Anyway, that's moot given the information you've given me below, that a proton has an actual size and isn't just a point, because now there's a window for error where you can say "this electron has probability p of being within the region defined by the proton's boundary X, and an error in momentum measurement of dp >= hbar / 2*dX."

On that note, to answer the original question, maybe electrons regularly hit the nucleus, but just don't usually react. Or maybe there's an aspect of the weak force (or the strong force) that we haven't accounted for.

But we were talking about the radial part of the wave function, not the angular part (where the spherical harmonics come into play).

Um... that has nothing to do with it. I'm just saying you can have a curve that is not symmetric but still have rbar = rmax. It'd be a weird curve, perhaps, but it can certainly be done. It just seemed like a hand-waving argument to me: better would have been to say something like, "when you do the math, you find that rbar does not equal rmax."

Maybe this actually comes up with more complex systems. It would be interesting to try to find one, I think.

Is there a reason beyond simple energetic concerns why a proton and a neutron can't overlap?

In what sense?

I think that this is the realm of Quantum Chromodynamics (quarks and stuff).

Yes, it is. I haven't gotten around to learning that yet, though I'm reading Feynman's QED, which is a nice book. QCD is the other end of things, but the ideas behind each theory seem similar at a first glance. Fortunately, I know most of the group theory stuff already, so it shouldn't be too hard to pick up.

I suppose that I mean it in the sense of "does QCD tell me that you can't pack as many quarks as you want into as small a volume as you want?" That's assuming that the quarks themselves have a shape and size, and it's not just the hadron that they form that has those properties.

Nucleons (proton and neutron) are composed of quarks, hence their size comes from the structure.

The electron, on the other hand, seems to be point-like.

Thank you! That is terribly interesting to me, and relevant to this discussion (as I mentioned above). I'll have to hunt down a particle physics book some time and find out why, exactly (if it's even known why, exactly).

For starters, you can keep reading this board. :clap:

With good responses like that, you can bet I will! :D

papageno

2005-Dec-29, 01:54 PM

That is a distribution function: it does not give directly the probability of finding an electron in a point.

The function in that figure gives "the total amount of electronic charge lying between the spheres of radius r and r + Dr". Well, yes, because the probability of finding an electron at a precise point is 0. You can't know position exactly. That's the basic premise of quantum theory. That's all I was really saying. When dealing with wave functions, textbooks always consider the probability to find the particle within a small volume about a certain postion, never to find the particle at a position. (But this is mathematical rigour.)

Anyway, do not confuse the distribution function with the wave function.

If you want to know the probability of finding the electron in a small volume around r = 0 (where the nucleus is), you have to look at the wave function.

And you can see from the graph the wavefunction (and corresponding probability P) is not zero.

The distribution function is given by 4\pi r^2 P, and at r = 0 it is zero (because of the r^2 factor).

Certainly for an s orbital, you're most likely to find the electron very close to the nucleus. When you do the detections to get Fig 3.4 experimentally, you don't really measure any electrons exactly at r=0. You can get ones pretty darn close, to within your instrument precision, but you can never be certain that it was really at the dead centre. Well, we know if the nucleus captures the electron...

And that's not true just for points, but for planes (and infinitesimally thin things like spherical shells). 4*pi*r^2*dr = 0 if dr = 0, and similar equations hold when you integrate over different spatial components. For the point in the dead centre, both r=0 and dr = 0.

Anyway, that's moot given the information you've given me below, that a proton has an actual size and isn't just a point, because now there's a window for error where you can say "this electron has probability p of being within the region defined by the proton's boundary X, and an error in momentum measurement of dp >= hbar / 2*dX."

On that note, to answer the original question, maybe electrons regularly hit the nucleus, but just don't usually react. Or maybe there's an aspect of the weak force (or the strong force) that we haven't accounted for. It is possible, since hyperfine structure is ususally observed, and this depends on interaction betweeen nucleus and electrons at r = 0.

But we were talking about the radial part of the wave function, not the angular part (where the spherical harmonics come into play). Um... that has nothing to do with it. I meant that the spherical harmonics have nothing to do with the radial distribution function, because the spherical harmonics describe the angular part of the wave function.

As a function of r, the distribution function is not symmetric,(but this is consequence of the fact that there is no "hard" boundary to the distance electron-nucleus.

One of the first things in textbooks, is the problem of a quantum particle in an infinite potential well (hard boundaries left and right). In this case the system is symmetric, and the average of the position is where the probability has a maximum.

I'm just saying you can have a curve that is not symmetric but still have rbar = rmax. It'd be a weird curve, perhaps, but it can certainly be done. I do not think so, but I cannot prove it.

It just seemed like a hand-waving argument to me: better would have been to say something like, "when you do the math, you find that rbar does not equal rmax."

Maybe this actually comes up with more complex systems. It would be interesting to try to find one, I think.

(Sorry for the bad math typing, but I am used to LaTeX.)

The function in that figure gives "the total amount of electronic charge lying between the spheres of radius r and r + Dr". Well, yes, because the probability of finding an electron at a precise point is 0. You can't know position exactly. That's the basic premise of quantum theory. That's all I was really saying. When dealing with wave functions, textbooks always consider the probability to find the particle within a small volume about a certain postion, never to find the particle at a position. (But this is mathematical rigour.)

Anyway, do not confuse the distribution function with the wave function.

If you want to know the probability of finding the electron in a small volume around r = 0 (where the nucleus is), you have to look at the wave function.

And you can see from the graph the wavefunction (and corresponding probability P) is not zero.

The distribution function is given by 4\pi r^2 P, and at r = 0 it is zero (because of the r^2 factor).

Certainly for an s orbital, you're most likely to find the electron very close to the nucleus. When you do the detections to get Fig 3.4 experimentally, you don't really measure any electrons exactly at r=0. You can get ones pretty darn close, to within your instrument precision, but you can never be certain that it was really at the dead centre. Well, we know if the nucleus captures the electron...

And that's not true just for points, but for planes (and infinitesimally thin things like spherical shells). 4*pi*r^2*dr = 0 if dr = 0, and similar equations hold when you integrate over different spatial components. For the point in the dead centre, both r=0 and dr = 0.

Anyway, that's moot given the information you've given me below, that a proton has an actual size and isn't just a point, because now there's a window for error where you can say "this electron has probability p of being within the region defined by the proton's boundary X, and an error in momentum measurement of dp >= hbar / 2*dX."

On that note, to answer the original question, maybe electrons regularly hit the nucleus, but just don't usually react. Or maybe there's an aspect of the weak force (or the strong force) that we haven't accounted for. It is possible, since hyperfine structure is ususally observed, and this depends on interaction betweeen nucleus and electrons at r = 0.

But we were talking about the radial part of the wave function, not the angular part (where the spherical harmonics come into play). Um... that has nothing to do with it. I meant that the spherical harmonics have nothing to do with the radial distribution function, because the spherical harmonics describe the angular part of the wave function.

As a function of r, the distribution function is not symmetric,(but this is consequence of the fact that there is no "hard" boundary to the distance electron-nucleus.

One of the first things in textbooks, is the problem of a quantum particle in an infinite potential well (hard boundaries left and right). In this case the system is symmetric, and the average of the position is where the probability has a maximum.

I'm just saying you can have a curve that is not symmetric but still have rbar = rmax. It'd be a weird curve, perhaps, but it can certainly be done. I do not think so, but I cannot prove it.

It just seemed like a hand-waving argument to me: better would have been to say something like, "when you do the math, you find that rbar does not equal rmax."

Maybe this actually comes up with more complex systems. It would be interesting to try to find one, I think.

(Sorry for the bad math typing, but I am used to LaTeX.)

Wolverine

2005-Dec-31, 05:16 AM

Welcome back, papageno. Nice to see you posting again.

papageno

2005-Dec-31, 08:23 PM

Welcome back, papageno. Nice to see you posting again. Thanks! :)

Bob Angstrom

2006-Jan-01, 08:46 AM

If electrons are held in orbit by electromagnetic attraction to nuclei, shouldn't electrons crashing into the nucleus be a common event? If so, why have I never heard about this?

Could the formation of degenerate matter, such as the formation of neutron stars, be considered an example of electrons crashing into the nucleus?

Could the formation of degenerate matter, such as the formation of neutron stars, be considered an example of electrons crashing into the nucleus?

papageno

2006-Jan-01, 01:27 PM

Could the formation of degenerate matter, such as the formation of neutron stars, be considered an example of electrons crashing into the nucleus? In this case, gravity is the major factor: electrons and nuclei are "pushed" together by the mass around them.

The OP was referring to the electromagnetic attraction between nucleus and electrons, which is not enough to crash the electrons in the nucleus.

The OP was referring to the electromagnetic attraction between nucleus and electrons, which is not enough to crash the electrons in the nucleus.

Spherical

2006-Jan-04, 04:21 PM

It is not a common event, but something like that does happen: electron capture (http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/radact2.html#c3).

Speaking neutrinos, the ethereal cousins of electrons:

http://www.ps.uci.edu/~superk/announce.html

http://wwwlapp.in2p3.fr/neutrinos/anhistory.html

If the blooming things have mass, and no one of the three flavors has the same mass, where does the mass go when they change flavors?

Speaking neutrinos, the ethereal cousins of electrons:

http://www.ps.uci.edu/~superk/announce.html

http://wwwlapp.in2p3.fr/neutrinos/anhistory.html

If the blooming things have mass, and no one of the three flavors has the same mass, where does the mass go when they change flavors?

Grey

2006-Jan-04, 05:46 PM

I suppose that I mean it in the sense of "does QCD tell me that you can't pack as many quarks as you want into as small a volume as you want?" That's assuming that the quarks themselves have a shape and size, and it's not just the hadron that they form that has those properties.Yes. Quarks are fermions, so they'll obey the exclusion principle, too. Of course, that's not quite "can't be in the same spot", but rather "can't occupy the same energy level", but that's the case for electrons, too.

sarongsong

2006-Jan-06, 08:03 AM

How fast do electrons travel?

01101001

2006-Jan-06, 08:26 AM

How fast do electrons travel?

What's your energy budget? You can get them very close to the speed of light if you can afford it.

What's your energy budget? You can get them very close to the speed of light if you can afford it.

sarongsong

2006-Jan-06, 08:54 AM

$11---just curious about silver's (in particular), in its natural state. Ballpark figures OK.

HenrikOlsen

2006-Jan-06, 02:25 PM

$11---just curious about silver's (in particular), in its natural state. Ballpark figures OK.

Are you thinking orbital velocity, or the speed they move at when a current is running through it?

Are you thinking orbital velocity, or the speed they move at when a current is running through it?

Grey

2006-Jan-06, 03:11 PM

Are you thinking orbital velocity, or the speed they move at when a current is running through it?And if you're thinking about the former, do you mean a classical calculation of what the electron's speed would be, if it actually orbited like a planet around the nucleus, or do you want the actual velocity at which the wave representing the electron propagates? I'm not sure what meaning the former really has, and the latter, of course, is probably best considered to be zero! That is, an electron at a given energy level has a time-independent probability distribution, so it's really not moving at all.

sarongsong

2006-Jan-06, 10:35 PM

Well, this is certainly one of the most educational threads I've encountered on BAUT and very much appreciate any assistance in grasping this topic. Previously, I thought electrons orbited their nucleus in planetary fashion, much as my avatar indicates, so my question was for orbital velocity. Now this concept of an electron being a wave instead---and not moving at all, is where I need more comprehension. References welcome, thanks.

Spherical

2006-Jan-06, 10:43 PM

Well, this is certainly one of the most educational threads I've encountered on BAUT and very much appreciate any assistance in grasping this topic. Previously, I thought electrons orbited their nucleus in planetary fashion, much as my avatar indicates, so my question was for orbital velocity. Now this concept of an electron being a wave instead---and not moving at all, is where I need more comprehension. References welcome, thanks.

Hang onto your hat! You're in for a wicked ride.

Hang onto your hat! You're in for a wicked ride.

peteshimmon

2006-Jan-06, 10:56 PM

I read it takes between 10 and 15 minutes for

a free neutron to dissociate to an electron and

proton. Such a human time interval, a tea break

or a fag break! Wonder if its an indication of

the strength of the zero point energy field.

a free neutron to dissociate to an electron and

proton. Such a human time interval, a tea break

or a fag break! Wonder if its an indication of

the strength of the zero point energy field.

01101001

2006-Jan-06, 11:38 PM

And if you're thinking about the former, do you mean a classical calculation of what the electron's speed would be, if it actually orbited like a planet around the nucleus, or do you want the actual velocity at which the wave representing the electron propagates?

Which reminds me of a chemistry class, when the professor said, "Well, that's it for the Bohr Model (http://en.wikipedia.org/wiki/Bohr_model) of atoms. Now we will move on to what is really happening."

Wiseacre me, sitting in the back of the large auditorium, flipped to a fresh page in my spiral-bound notebook -- I'd never do it to actual class notes -- and everyone heard:

Riiiiiiiiiiip! Crumple. Crumple. Crumple.

Laughter ensued.

Which reminds me of a chemistry class, when the professor said, "Well, that's it for the Bohr Model (http://en.wikipedia.org/wiki/Bohr_model) of atoms. Now we will move on to what is really happening."

Wiseacre me, sitting in the back of the large auditorium, flipped to a fresh page in my spiral-bound notebook -- I'd never do it to actual class notes -- and everyone heard:

Riiiiiiiiiiip! Crumple. Crumple. Crumple.

Laughter ensued.

snarkophilus

2006-Jan-07, 04:30 AM

And if you're thinking about the former, do you mean a classical calculation of what the electron's speed would be, if it actually orbited like a planet around the nucleus, or do you want the actual velocity at which the wave representing the electron propagates? I'm not sure what meaning the former really has, and the latter, of course, is probably best considered to be zero! That is, an electron at a given energy level has a time-independent probability distribution, so it's really not moving at all.

Are you saying that the electron really is a wave? Because although the probability distribution is static, the electron itself is considered to be (in most formulations) a particle with a definite energy at any given point, and therefore it has a definite speed (even if you can't find that position/speed exactly). It might not orbit like a planet, but it does move in some fashion about the nucleus.

Sarongsong, to find the average speed of an electron, you just need to know the energy of its orbital. There are books with tables of those things, or you can download a computer program to calculate them. It's a devilishly complicated thing once you get beyond hydrogen.

There's a cool little problem associated with electron speeds. Basically, you can't get more than 138 electrons around a nucleus with 138 protons, because an electron would have to move at light speed (actually, on average it would have to move faster than light speed).

Are you saying that the electron really is a wave? Because although the probability distribution is static, the electron itself is considered to be (in most formulations) a particle with a definite energy at any given point, and therefore it has a definite speed (even if you can't find that position/speed exactly). It might not orbit like a planet, but it does move in some fashion about the nucleus.

Sarongsong, to find the average speed of an electron, you just need to know the energy of its orbital. There are books with tables of those things, or you can download a computer program to calculate them. It's a devilishly complicated thing once you get beyond hydrogen.

There's a cool little problem associated with electron speeds. Basically, you can't get more than 138 electrons around a nucleus with 138 protons, because an electron would have to move at light speed (actually, on average it would have to move faster than light speed).

sarongsong

2006-Jan-07, 09:24 AM

...Sarongsong, to find the average speed of an electron, you just need to know the energy of its orbital. There are books with tables of those things, or you can download a computer program to calculate them. It's a devilishly complicated thing once you get beyond hydrogen...Ah---thanks! So far, have come across this site (http://www.webelements.com/webelements/elements/text/Ag/econ.html) and its accompanying The Orbitron (http://www.webelements.com/phpAdsNew/adclick.php?n=a04de058) link, for starters. Will check the library tomorrow.

swansont

2006-Jan-07, 01:48 PM

Are you saying that the electron really is a wave? Because although the probability distribution is static, the electron itself is considered to be (in most formulations) a particle with a definite energy at any given point, and therefore it has a definite speed (even if you can't find that position/speed exactly). It might not orbit like a planet, but it does move in some fashion about the nucleus.

Sarongsong, to find the average speed of an electron, you just need to know the energy of its orbital. There are books with tables of those things, or you can download a computer program to calculate them. It's a devilishly complicated thing once you get beyond hydrogen.

There's a cool little problem associated with electron speeds. Basically, you can't get more than 138 electrons around a nucleus with 138 protons, because an electron would have to move at light speed (actually, on average it would have to move faster than light speed).

There are times you can treat the electron as a particle, and there are times you can't.

In larger atoms, one must make relativistic correction to the electron speeds (relativistic QED), but I need a link supporting your contention that superluminal speeds are required. I strongly suspect that is a classical analysis.

Sarongsong, to find the average speed of an electron, you just need to know the energy of its orbital. There are books with tables of those things, or you can download a computer program to calculate them. It's a devilishly complicated thing once you get beyond hydrogen.

There's a cool little problem associated with electron speeds. Basically, you can't get more than 138 electrons around a nucleus with 138 protons, because an electron would have to move at light speed (actually, on average it would have to move faster than light speed).

There are times you can treat the electron as a particle, and there are times you can't.

In larger atoms, one must make relativistic correction to the electron speeds (relativistic QED), but I need a link supporting your contention that superluminal speeds are required. I strongly suspect that is a classical analysis.

Grey

2006-Jan-09, 09:08 AM

Are you saying that the electron really is a wave? Because although the probability distribution is static, the electron itself is considered to be (in most formulations) a particle with a definite energy at any given point, and therefore it has a definite speed (even if you can't find that position/speed exactly).I'm saying that the electron is a quantum entity, which behaves sometimes like we imagine a classical wave to behave, and sometimes like a classical particle behaves. An electron bound in an atom does have a definite energy (or, more precisely, the overall configuration of electron and nucleus and any other electrons present has a definite energy). However, to suggest that is has a definite speed even though we might not know what that speed is is missing the point of quantum mechanics. The uncertainty in an electron's position and momentum is not a limitation on what we know, it's a limitation on what the universe knows.

It might not orbit like a planet, but it does move in some fashion about the nucleus.That's true in the sense that if you were to make repeated position measurements, you'd find a pointlike electron in different spots. If you made repeated momentum measurements, you'd likewise get various values for the particle's momentum. But it's decidedly not true that the electron is moving in some kind of orbit that could be envisioned classically (whether planet-like or not), and we just don't know the details of what that orbit is like.

Sarongsong, to find the average speed of an electron, you just need to know the energy of its orbital.Except that just gives you the difference in potential energy of various orbitals. Even for an orbiting planet, its kinetic energy is not equal to the potential energy lost in getting to its present position from an unbound state.

It might not orbit like a planet, but it does move in some fashion about the nucleus.That's true in the sense that if you were to make repeated position measurements, you'd find a pointlike electron in different spots. If you made repeated momentum measurements, you'd likewise get various values for the particle's momentum. But it's decidedly not true that the electron is moving in some kind of orbit that could be envisioned classically (whether planet-like or not), and we just don't know the details of what that orbit is like.

Sarongsong, to find the average speed of an electron, you just need to know the energy of its orbital.Except that just gives you the difference in potential energy of various orbitals. Even for an orbiting planet, its kinetic energy is not equal to the potential energy lost in getting to its present position from an unbound state.

Powered by vBulletin® Version 4.2.3 Copyright © 2020 vBulletin Solutions, Inc. All rights reserved.