Why doesn't the electron fall into the nucleus?

[thoughtgaze]

why doesn't the electron fall into the nucleus!?

yes I know this question must have been asked before right? tried looking for it on here but I couldn't find it. the only answer I have gotten to this was just that particles are quantized... yes yes, fine... but WHY? for instance, why does an electron and a positron collide so readily? but an electron and a proton don't? someone explain please and thanks.

P.S.

I have heard an argument that deals with the uncertainty relation but it doesn't make any distinction between a proton and a positron... i don't think... any HELP!?

 

[ZapperZ]

Please read an entry in the FAQ thread in the General Physics forum.

Zz.

 

[feynmann]

yes I know this question must have been asked before right? tried looking for it on here but I couldn't find it. the only answer I have gotten to this was just that particles are quantized... yes yes, fine... but WHY? for instance, why does an electron and a positron collide so readily? but an electron and a proton don't? someone explain please and thanks.

P.S.

I have heard an argument that deals with the uncertainty relation but it doesn't make any distinction between a proton and a positron... i don't think... any HELP!?

The proton itself is not important. The important thing is its charge, that will create an electrical field in which the electron can feel and it must do work to gain its potential energy.

 

[thoughtgaze]

Please read an entry in the FAQ thread in the General Physics forum.

Zz.

Thanks ZapperZ... checked it out. Makes sense, but it seems like i could make that same argument with an electron and positron, yet they annihilate quite readily no? Or am I mistaken?

 

[jtbell]

Electrons sometimes do "crash into the nucleus," because an atomic electron's Schrödinger wave function and probability density are generally not zero at the nucleus. In fact, for the lowest energy state (1s), the probability density is maximum at the nucleus! In certain types of nuclei, this causes a decay mode called "electron capture," in which the electron is "converted" to a neutrino, and a proton is converted to a neutron, via the weak interaction.

But the weak interaction is a lot weaker than the electromagnetic interaction which causes positron-electron annihilation, so it doesn't happen very often, relatively speaking.

 

[alxm]

See also https://www.physicsforums.com/showthread.php?t=304488" thread. Just to list the last two times it got asked around here.

 

[feynmann]

Thanks ZapperZ... checked it out. Makes sense, but it seems like i could make that same argument with an electron and positron, yet they annihilate quite readily no? Or am I mistaken?

If it makes sense to you, can you explain "why doesn't the electron fall into the nucleus!?" in simple term?

 

[malawi_glenn]

If it makes sense to you, can you explain "why doesn't the electron fall into the nucleus!?" in simple term?

what simple terms means is subjective ;-)

 

[feynmann]

what simple terms means is subjective ;-)

Richard P. Feynman: If you can't explain it to a high school student you probably don't understand it.

 

[malawi_glenn]

Richard P. Feynman: If you can't explain it to a high school student you probably don't understand it.

here we go again, haven't we told you that it is not applicable? R. Feynman also said "shut up and calculate".

If I can't explain Baker-Cambell-Hausdroff formula for a high-school student, then I have not understood it? What authority is Feynman when it comes to epistemology??!

The language of physics is not intuitive daily language, but math math math...

The terms "fall into the nucleus" is not even well defined in quantum mechanics, it is not even a stringent mathematical formulated statement.

 

[malawi_glenn]

Thanks ZapperZ... checked it out. Makes sense, but it seems like i could make that same argument with an electron and positron, yet they annihilate quite readily no? Or am I mistaken?

You have to differ the bound states and interactions in general.

Electrons and protons will interact if you shoot electrons against proton target (i.e free scattering)

Electrons and protons will form bound state with a certain probabilty (scattering length is negative).

Same with positrons, there are bound states of electrons and positrons too, with a certain 'lifetime'.

Everything is clear and free from "paradoxes" if one does the real calculations, now i just stated the results for you.

 

[kexue]

yes I know this question must have been asked before right? tried looking for it on here but I couldn't find it. the only answer I have gotten to this was just that particles are quantized... yes yes, fine... but WHY? for instance, why does an electron and a positron collide so readily? but an electron and a proton don't? someone explain please and thanks.

P.S.

I have heard an argument that deals with the uncertainty relation but it doesn't make any distinction between a proton and a positron... i don't think... any HELP!?

Why some annihilation or creation processes occur and others do not? I think the easiest answer: because of conservation laws (momentum, energy, lepton number, baryon number, charge). If they are respected, annihilation or creation processes can happen.

Positron and electron are antiparticles, so obviously they can annihilate and become an photon.

I do not know exactly what you get when you crash an electron and proton at very high energies
(pions), but they will then annihilate. High energy because E=mc².

 

[malawi_glenn]

Why some annihilation or creation processes occur and others do not? I think the easiest answer: because of conservation laws (momentum, energy, lepton number, baryon number, charge). If they are respected, annihilation or creation processes can happen.

Positron and electron are antiparticles, so obviously they can annihilate and become an photon.

I do not exactly, what you get when you crash an electron and proton at very high energies
(pions), but the annihilate. High energy because E=mc².

I just gave the answer, one has to distinguish from scattering and scattering which leads to bound states.

 

[ytuab]

No one can answer this question correctly.

Quantum mechanics has given up explainig the motion of electrons clearly, as Pauli and Dirac do since 1920's.

 

[kashiark]

No one can answer this question correctly.

Quantum mechanics has given up explainig the motion of electrons clearly, as Pauli and Dirac do since 1920's.

what are you talking about ever heard of quantum electrodynamics?

 

[alxm]

No one can answer this question correctly.

Quantum mechanics has given up explainig the motion of electrons clearly, as Pauli and Dirac do since 1920's.

What a load of nonsense. Quantum physics does an excellent job of explaining the motion of electrons.
The proof of that is sitting right in front of you in the form of lasers, semiconductors and other bits of technology that directly followed from quantum theory of atoms, molecules and solids.

Not to mention providing a theory that reproduced and explained all of chemistry.

 

[feynmann]

yes I know this question must have been asked before right? tried looking for it on here but I couldn't find it. the only answer I have gotten to this was just that particles are quantized... yes yes, fine... but WHY? for instance, why does an electron and a positron collide so readily? but an electron and a proton don't? someone explain please and thanks.

The simple answer is that if the electron stay too far away from the the nucleus, the potential energy will increase, if it stay too close to the nucleus, the kinetic energy will increase. So it will stay where the sum of potential and kinetic energy is minimum.
Feynman gave his explanation in his famous book, "Feynman lecture on physics", volume 3, section 2-4, <The size of an atom>, He used the argument that deals with uncertainty relation.

 

[Count Iblis]

The electron actually has a nonzero probability to be inside the nucleus. When you think of something falling in classical physics, then what you implicitely assume is that when the macroscopic object hits the target (say the ground), it loses energy and then doesn't bounce back up.

It is this dissipation of energy that doesn't happen in quantum physics for the electron in the ground state. The electron does interact with the nucleus, not only via the Coulomb interaction but also via short range interactions (some QED corrections can be modeled approximately as a Dirac delta potential).

 

[feynmann]

The electron actually has a nonzero probability to be inside the nucleus. When you think of something falling in classical physics, then what you implicitely assume is that when the macroscopic object hits the target (say the ground), it loses energy and then doesn't bounce back up.

It is this dissipation of energy that doesn't happen in quantum physics for the electron in the ground state. The electron does interact with the nucleus, not only via the Coulomb interaction but also via short range interactions (some QED corrections can be modeled approximately as a Dirac delta potential).

What happen when the electron is inside the nucleus? Will they stick together or not, due to the attraction of their opposite charges?

 

[Count Iblis]

What happen when the electron is inside the nucleus? Will they stick together or not due to their opposite charges?

"Sticking together" corresponds to electron capture. Then an electron merges with proton, producing a neutron and a neutrino. But if this is energetically not possible such a transition cannot happen. Then the ground state is stable.

An electron in the ground state should be interpreted as a superposition of the electron being in all possible position (with appropriate amplitudes). These possible positions include the region inside the nucleus.

So, it isn't like the electron moving into the nucleus from the ground state, rather the electron in the ground state is, in a certain sense, always partially inside the nucleus.

 

[feynmann]

" An electron in the ground state should be interpreted as a superposition of the electron being in all possible position (with appropriate amplitudes). These possible positions include the region inside the nucleus.

That's just the "Copenhagen interpretation", but it is by no means the only interpretation.
The fundamental equation of "quantum mechanics" does not imply this at all.
They are not at equal footing, the fundamental equation vs Copenhagen interpretation

 

[thoughtgaze]

If it makes sense to you, can you explain "why doesn't the electron fall into the nucleus!?" in simple term?

Well, in very very simple terms, an electric charge will radiate energy if it is accelerated. Since, at these scales it no longer makes sense to think of them "orbiting" in the usual sense, it so too no longer makes sense to think that these electrons are accelerating, thus no energy is being radiated from the electron and there you hAVE IT LADIES AND GENTS! a stable bound state.

But I think the real issue comes from these other interactions which I guess I have to study. What was really puzzling me is why then is it less likely to have stable POSITRON ELECTRON atoms let's say. Opposite charge? Taking into account electric fields and what not, seems like you could make the same argument. Malawai Glenn and JTBELL gave answers that imply the story is a little deeper than merely taking into account the uncertainty relation. Taking into account the uncertainty relation implies an electron might want to stay a Bohr radius away from the positron but this is less likely according to some who have replied.

 

[Count Iblis]

That's just the "Copenhagen interpretation", but it is by no means the only interpretation.
The fundamental equation of "quantum mechanics" does not imply this at all.
They are not at equal footing, the fundamental equation vs Copenhagen interpretation

Huh? The Copenhagen interpreation has nothing to do with this. That deals with observers measuring a system and then asking about what happened to the pother possible values that could have been found insted of the avue that was actually found.

In this case the point is simply that the state |psi> of the electron can be spoecified in terms of any basis. If you use the states |x> corresponding to the electron being at position x as your basis for the Hilbert space then the components of the |psi> in the "direction" |x>, given by the inner products:

<x|psi>

for all possible |x>

completely define |psi> as you ave specified all the componens of the vector |psi> in the iunfinite domnensional Hilbert space. Now, <x|psi> is a function of the position x, and we call it the "wavefunction". You can equally well consider other basis vecotrs, like the states |p> corresponding the electron havng a well defined mometum p. Then specifying <p|psi> for all momenta p fixes |psi>.

 

[alxm]

That's just the "Copenhagen interpretation", but it is by no means the only interpretation.
The fundamental equation of "quantum mechanics" does not imply this at all.

Yes it does. $$|\psi|^2$$ is the location-probability density, in straight-up QM formalism, regardless of the interpretation.

Interpretations concern whether or not that probability is a true-non-deterministic probability (e.g. Copenhagen interpretation) or a result of deterministic-but-unknown variables (e.g. Bohm interpretation).

Well, in very very simple terms, an electric charge will radiate energy if it is accelerated. Since, at these scales it no longer makes sense to think of them "orbiting" in the usual sense, it so too no longer makes sense to think that these electrons are accelerating, thus no energy is being radiated from the electron and there you hAVE IT LADIES AND GENTS! a stable bound state.

You posit that the electron cannot accelerate, meaning change its overall energy, and hence is bound, meaning it cannot change its overall energy. That amounts to a tautology.

The problem of 'why doesn't the electron fall into the nucleus' was formulated about a century ago knowing full well that charged particles moving in a circular 'orbit' would radiate energy and fall in. That's why they posed the question. If not a circular orbit or similar (and they knew that was probably the case), then what, and why?.

Malawai Glenn and JTBELL gave answers that imply the story is a little deeper than merely taking into account the uncertainty relation.

No, the uncertainty relation is enough to explain why the electron does not simply sit in a stationary position inside the nucleus. An electron in an s-type state will have a non-zero probability of being at r=0, as per the solution to the Schrödinger equation, which implicitly takes into account the uncertainty principle. And at that point, the electron will have a kinetic energy maximum. (as it is lowest in potential energy)

The rare (for most atoms: never) event of K-capture doesn't relate to the 'why don't electrons fall in' question, because it has no bearing on issue of why electrons spend most of their time so (relatively) far away from the nucleus.

 

[ytuab]

I meant Quantum physics has given up the measurement of position and momentum of
an electron at the same time.

To answer this question of thoughtgaze clearly, we must explain the motion of an electron
concretely. But it's impossible.

 

[thoughtgaze]

You posit that the electron cannot accelerate, meaning change its overall energy, and hence is bound, meaning it cannot change its overall energy. That amounts to a tautology.

I was just summing up the explanation given to me on this forum. Which instead of explaining why to full extent explains how it is possible to be in a stable equilibrium around the nucleus, simply because the picture of electrons around the nucleus of an atom does not fit with the standard orbital motion of planets and hence doesn't need to radiate energy.

No, the uncertainty relation is enough to explain why the electron does not simply sit in a stationary position inside the nucleus. An electron in an s-type state will have a non-zero probability of being at r=0, as per the solution to the Schrödinger equation, which implicitly takes into account the uncertainty principle. And at that point, the electron will have a kinetic energy maximum. (as it is lowest in potential energy)

Right, so as to what accounts for the bohr radius, or rather the radius of the overall "spread" of the electron you are merely taking into account the electric forces and the uncertainty relation, so an electron around a positron in its stable equilibrium should have the same bohr radius? What I'm driving at here is why the electron and positron should annihilate so readily instead of forming atoms as easily as a proton and electron. Malawi Glenn and jtbell gave me some answers to think about for a while.

 

[alxm]

I meant Quantum physics has given up the measurement of position and momentum of an electron at the same time.

To answer this question of thoughtgaze clearly, we must explain the motion of an electron
concretely. But it's impossible.

Nobody was ever really trying to measure the position and momentum of an electron at the same time. The HUP was found long before experimental accuracy got to that level.

The fact that the position and momentum cannot be simultaneously determined in experiment does not mean that these values are unknown or unknowable. They're entirely known, insofar we can predict every possible experimental result, assign probabilities to each possible result, and calculate the expectation value of multiple measurements. What I cannot do is predict the exact result of any single measurement. So what?

I think you've confused the fact that measured values are 'random' (using quotes to please Feynmann), with the idea that they're equally probable, which they're not. If all values were always equally probable, then yes, we would not be able to say anything at all about electron motion. But the fact is that all values are not equally probable means that yes, we can know something about electron motion. Most importantly we can know the expectation value, the result over multiple measurements, which is usually what you're measuring anyway. And we know that that expectation value corresponds exactly to the classical value, as you move from the quantum domain to the classical one.

I've said it before but will say it again: Knowing momentum and position simultaneously and exactly is not a prerequisite for motion, or to understanding and describing motion.

 

[alxm]

Right, so as to what accounts for the bohr radius, or rather the radius of the overall "spread" of the electron you are merely taking into account the electric forces and the uncertainty relation, so an electron around a positron in its stable equilibrium should have the same bohr radius?

The uncertainty relation includes momentum, hence mass. The (exact, proper) derivation of the Bohr radius is done assuming the proton has infinite mass (the Born-Oppenheimer approximation), i.e. neglecting 'recoil' of the proton from the motion of the electron. It's stationary. It's therefore also not an entirely exact result. (then there's relativistic effects, and then there's QED effects). So it's off by 0.1% or something. You can't directly get the Bohr radius from the uncertainty principle - Feynman made some reasonable but hand-waving assumptions when he did so. But the uncertainty principle does dictate the correct 'ballpark figure'.

With a positron-electron pair you have two particles of near-equal mass, the Born-Oppenheimer approximation is as invalid as it gets, and you have to solve the Schrödinger equation in an altogether different way, with altogether different results.

 

[feynmann]

Huh? The Copenhagen interpreation has nothing to do with this. That deals with observers measuring a system and then asking about what happened to the pother possible values that could have been found insted of the avue that was actually found.

<Copenhagen interpreation> http://en.wikipedia.org/wiki/Copenhagen_interpretation

<Principles>
1. A system is completely described by a wave function ψ, which represents an observer's knowledge of the system. (Heisenberg)
2. The description of nature is essentially probabilistic. The probability of an event is related to the square of the amplitude of the wave function related to it. (Born rule, due to Max Born)

 

[malawi_glenn]

What a load of nonsense. Quantum physics does an excellent job of explaining the motion of electrons.
The proof of that is sitting right in front of you in the form of lasers, semiconductors and other bits of technology that directly followed from quantum theory of atoms, molecules and solids.

Not to mention providing a theory that reproduced and explained all of chemistry.

what alxm mean is that one has to give up the INTUTIVE meaning of what motion is.

 

[Count Iblis]

<Copenhagen interpreation> http://en.wikipedia.org/wiki/Copenhagen_interpretation

<Principles>
1. A system is completely described by a wave function ψ, which represents an observer's knowledge of the system. (Heisenberg)
2. The description of nature is essentially probabilistic. The probability of an event is related to the square of the amplitude of the wave function related to it. (Born rule, due to Max Born)

And the collapse of the wavefunction which distinguishes Copenhagen from Many Worlds.

 

[malawi_glenn]

The simple answer is that if the electron stay too far away from the the nucleus, the potential energy will increase, if it stay too close to the nucleus, the kinetic energy will increase. So it will stay where the sum of potential and kinetic energy is minimum.
Feynman gave his explanation in his famous book, "Feynman lecture on physics", volume 3, section 2-4, <The size of an atom>, He used the argument that deals with uncertainty relation.

these are just semiclassical fairytales, which are 'false'. Every student of quantum mechanics knows that the position of the electron is probibalistic, one does not use the uncertainty realtion to derive properties of atoms. The electron will not stay in one place, the explanation has just killed itself, since one invoked the uncertainty principle, it can not be in one definite position...

The kinetic energy of an electron in an atom is given by the kinetic energy operator and nothing else.

Explanations of quantum mechanics must be done with mathematical rigour, not just invoking the uncertainty principle all the time.

 

[feynmann]

these are just semiclassical fairytales, which are 'false'. Every student of quantum mechanics knows that the position of the electron is probibalistic, one does not use the uncertainty realtion to derive properties of atoms. The electron will not stay in one place, the explanation has just killed itself, since one invoked the uncertainty principle, it can not be in one definite position...

The kinetic energy of an electron in an atom is given by the kinetic energy operator and nothing else.

Explanations of quantum mechanics must be done with mathematical rigour, not just invoking the uncertainty principle all the time.

Feynman never say that his explanation is "semiclassical fairytales", I believe he intended to explain it in terms of "quantum mechanics", otherwise, He would say so. You are probably the first person to say that it's "semiclassical fairytales".

I guess what you are saying is since the position of the electron is probabilistic, so they do not have potential energy and kinetic energy. I don't think that's true. The Schrodinger Equation itself is based on potential and kinetic energy. Yukawa invoked the uncertainty principle to predict that the pions are the carriers of strong force and that was confirmed by experiment. He got Nobel prize for this work. I guess his work is also "semiclassical fairytales" to you.

 

[feynmann]

"Sticking together" corresponds to electron capture. Then an electron merges with proton, producing a neutron and a neutrino. But if this is energetically not possible such a transition cannot happen. Then the ground state is stable.

An electron in the ground state should be interpreted as a superposition of the electron being in all possible position (with appropriate amplitudes). These possible positions include the region inside the nucleus.

So, it isn't like the electron moving into the nucleus from the ground state, rather the electron in the ground state is, in a certain sense, always partially inside the nucleus.

The argument that the ground state should be interpreted as a superposition of the electron being in all possible position and the position of the electron is probabilistic fails to explain why hydrogen atoms have definite size. It's size is about the Bohr's radius.

 

[malawi_glenn]

Feynman never say that his explanation is "semiclassical fairytales", I believe he intended to explain it in terms of "quantum mechanics", otherwise, He would say so. You are probably the first person to say that it's "semiclassical fairytales".

I guess what you are saying is since the position of the electron is probabilistic, so they do not have potential energy and kinetic energy. I don't think that's true. The Schrodinger Equation itself is based on potential and kinetic energy. Yukawa invoked the uncertainty principle to predict that the pions are the carriers of strong force and that was confirmed by experiment. He got Nobel prize for this work. I guess his work is also "semiclassical fairytales" to you.

He didn't explain this in QM-terms, since the properties are found by solving the schrödinger equation - not by applying the HUP in this way. HUP is just a statistical statement of observables.

Yes, but one can not say that when the electron is closer to the nucleus, it has more KE, etc, since such statements of the electron is meaningless in QM.

No Yukawa did not use the uncertainty principle, he did advanced quantum field theoretical calculations. Which I actually went through this weekend...