Why doesn't the electron fall into the nucleus?

[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...

 

[malawi_glenn]

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.

i) you should not in one sentence disprove QM and in another one praise feynman and yukawa

ii) The atoms has NOT definite size, its "size" is a statistical statement. Like the root mean radius square and mean radius etc, only "mean" no "definite".

can you stop with this circus? It is clear to me that you have never studied QM from a textbook used in college...

 

[granpa]

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 doesn't the Earth fall into the sun?

 

[malawi_glenn]

why doesn't the Earth fall into the sun?

not a perfect analogy, since a radial accelerating (non quantum) CHARGE will emit radiation and hence loose energy and decrease its radial distance.

 

[Dadface]

Take it back to basics and look at what an electron actually is.The electron has a rest mass and can display a relativistic mass increase.This fact alone appeals to intuition and leads us to imagine the electron as being a particle.The particle model may be over simplistic but it still has a tendency to persist in the mind even,I suspect, occasionally in the minds of some of the most able practitioners of QM.The particle model has its successes but also has its limitations as evidenced by the discussion here.
Let's stick with the particle model for a moment and pick up on granpas "non perfect" but neverthless relevant analogy..."why doesn't the Earth fall into the sun?"and let us now apply this to the electron and ask again "why does the electron not fall into the nucleus" Let the nucleus and the electron be at a position of momentary rest and about to approach under the influence of the Coulomb force.Using just the particle model we might predict that the electron and nucleus eventually collide and make actual contact.Good, but this is not what we observe the reason being that the particle model breaks down and we have to resort to the more powerful wave model and QM.QM,QED and the like may not appeal to intuition but nevertheless they work.Whether or not there will come a greater reconciliation between particle and wave remains to be seen.

 

[alxm]

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.

I don't have a problem with saying that. The kinetic-energy operator then, operating on the electronic wave function, increases as r \rightarrow 0. If it didn't, the wavefunction would diverge, since the potential diverges there.

Now, unlike a classical particle, knowing the kinetic energy at any point in space requires knowing its kinetic energy at every point in space. And it the electron doesn't have a definite position and so on and so forth.

But as long as you recognize that you're not dealing with a classical particles and so on, I don't see a problem. I doubt most chemical physicists would take issue with a statement like "electrons move faster near the nucleus", although they'd no doubt prefer the more relativistically-correct "have higher momentum" to "move faster".

 

[Halcyon-on]

The interpretation given by Bohr is that the electron must follow orbit with an integer number of periods, that is only periodic orbits are allowed.

 

[Count Iblis]

The interpretation given by Bohr is that the electron must follow orbit with an integer number of periods, that is only periodic orbits are allowed.

No, according to Bohr Sommerfeld theory (a.k.a. "Old Quantum Theory"), the integral of P dQ integrated over a periodic orbit where P is the conjugate mometum to Q is a multiple fo Planc's constant. So, if you take Q to be the angle then the conjugate momentum is the angular momentum. Then, since for a circular orbit, the angular mometum is conserved we have:

L* 2pi = n h -------->

L = n h/(2pi) = n hbar.

f you then consider a classical orbit and the you only allow angular momenta that satisfy the above quantization rule, you find the equations for the Bohr orbits.

Since this is already too complicated to explain in the dumbed down high school physics classes, what they do is they use de-Broglie matter wave hypothesis (which came after Bohr Sommerfeld theory) and then they derive the result obtained by Bohr (who frst simply posulated that L is quantized and later with Sommerfeld came up with Old Quantum Theory).

But then it is incorrect to say that this is the derivation pesented by Bohr. So, I guess that http://insti.physics.sunysb.edu/~siegel/plan.html" , as these textbooks apperently don't even get the history correct.

 

[Halcyon-on]

Good to know! I have many other item to add to the Siegler´ s list!

Concerning the atomic orbits, the Bohr Sommerfeld condition can be formulated in energy-time rather that in momentum-space, in fact the waves are classic and on-shell. The intagral along the orbits of duration t is simplified being the energy constant. The resultin condition is

E t = n h

in this way you avoid the assumption of circular orbits. Morover from de Broglie E = h v (v=1/T is the frequence and T the period of the n-th orbital) you get
v t = n -> t = n T
and finally you see that the duration of the orbits is such that it contains a integer number of orbits. So the fundamental idea came from de Broglie, but there is an additional periodicity condition.

Here n is the atomic number. The quantization of the angular momenta is a consequence of the periodicity coming from the spherical symmetry of the problem (theta = theta + 2 pi ) whereas the spin can be interpreted as antiperiodicity of the electron wave. In this semi-classical way many other non-relativisitc quantum problems can be solved. (examples http://people.ccmr.cornell.edu/~muchomas/8.04/Lecs/lec_bohr-sommerfeld/notes.pdf )

 

[alxm]

In this semi-classical way many other non-relativistic quantum problems can be solved.

Well, semiclassical models have their uses, but they fail in general pretty miserably at describing atoms and molecules, with the sole exception of the Bohr model, really.

Perhaps you'd be interested in reading about some recent toyings with a semiclassical http://chaosbook.org/projects/bertelsen.ps.gz" model. They 'cheat' in a few ways (as does the Bohr model), but it's surprisingly good given the model.

But in general, semiclassical attempts at atoms/molecules are a dead end. E.g. the Thomas-Fermi model cannot form stable molecules. I doubt any semiclassical theory can, given the importance of exchange energy in chemical bonding.

 

[feynmann]

i) you should not in one sentence disprove QM and in another one praise feynman and yukawa

ii) The atoms has NOT definite size, its "size" is a statistical statement. Like the root mean radius square and mean radius etc, only "mean" no "definite".

can you stop with this circus? It is clear to me that you have never studied QM from a textbook used in college...

No one is saying in one sentence disprove QM and in another one praise feynman and yukawa. What was disproved is "Copenhagen Interpretation", not Quantum mechanics.
The probability of wavefunction is "Copenhagen Interpretation", Not "Quantum Mechanics" itself. There is No probability of wavefunction in Bohm's version of Quantum Mechanics

It's clear your knowledge of quantum mechanics is full of nonsense, since you don't even know the difference between quantum mechanics and "Copenhagen Interpretation", what a "Science Advisor". Would you stop calling yourself "Science Advisor"? I certainly don't need your "advice".

 

[alxm]

The probability of wavefunction is "Copenhagen Interpretation", Not "Quantum Mechanics" itself. There is No probability of wavefunction in Bohm's version of Quantum Mechanics

Wrong. |\psi(\mathbf{x},t)|^2 is a particle's spatial probability density in quantum mechanics. Regardless of the interpretation.

It's clear your knowledge of quantum mechanics is no better than high-school level, since you don't even know the difference between quantum mechanics and "Copenhagen Interpretation".
What a Science Advisor

Maybe you should actually study the basis of the http://en.wikipedia.org/wiki/Bohm_interpretation" of which you speak before making such claims, in particular since I already explained this once.

The Copenhagen interpretation amounts to the claim, that that probability is truly random, i.e. indeterministic, rather than a result of incomplete knowledge of a deterministic system. i.e. hidden variables - such as the Bohm interpretation.

If you have some issue with how you can describe a deterministic system can be described in a probabilistic/statistical way, then your misunderstanding is with statistical mechanics, not QM.

 

[ajw1]

Wrong. |\psi(\mathbf{x},t)|^2 is a particle's spatial probability density in quantum mechanics. Regardless of the interpretation.

But when using the probability wave as answer for the question of the TS it seems to me that one needs the Copenhagen interpretation of this wave, being that the particle really is 'spreadout' in it's wave (in superposition).

When one images the electron being like a fly circling around in a room one can calculate a probability wave for the fly being somewhere, but there is a great chance he finaly will be caught by the sticky strip in the middle.

 

[Modman]

This is similar to the question to the question: "Why doesn't the Earth fall into the Sun?". In both, the answer is that the revolving object has enough inertia to prevent it.

Also, there is a confusion over quantum physics. While we hold Heisenberg's uncertainty principle to be true, this does not mean we do not know what they do in there. Think of the uncertainty principle as a veil over a fitting room: we know what goes on inside the fitting room, but we cannot see it without violating a law (physics or US) .

 

[ytuab]

Quantum physics replaced Bohr-Sommerfeld theory in 1920's.
Since then many phsicists such as Pauli, Heisenberg, Dirac... have given up explaining the
motion of an electron concretely.

because by equating the angular momentum of the spinning sphere of the electron to 1/2
h-bar, the sphere speed leads to about one hundred times the speed of light.

The orbital angular momentum of the electron in the ground state of the hydrogen atom is zero, so the Coulomb potential is infinitely negative when the electron is close to the nucleus.

 

[thoughtgaze]

This is similar to the question to the question: "Why doesn't the Earth fall into the Sun?". In both, the answer is that the revolving object has enough inertia to prevent it.

Also, there is a confusion over quantum physics. While we hold Heisenberg's uncertainty principle to be true, this does not mean we do not know what they do in there. Think of the uncertainty principle as a veil over a fitting room: we know what goes on inside the fitting room, but we cannot see it without violating a law (physics or US) .

I don't know why people keep using this reasoning. Clearly the earth/sun analogy does not work at this level ELSE IT WOULD DEFINITELY FALL INTO THE NUCLEUS!

 

[malawi_glenn]

No one is saying in one sentence disprove QM and in another one praise feynman and yukawa. What was disproved is "Copenhagen Interpretation", not Quantum mechanics.
The probability of wavefunction is "Copenhagen Interpretation", Not "Quantum Mechanics" itself. There is No probability of wavefunction in Bohm's version of Quantum Mechanics

It's clear your knowledge of quantum mechanics is full of nonsense, since you don't even know the difference between quantum mechanics and "Copenhagen Interpretation", what a "Science Advisor". Would you stop calling yourself "Science Advisor"? I certainly don't need your "advice".

I perfectly know the difference, it is custom to drop to the "copenhagen interpretation", since that is the standard in physics today. As we had in another thread "only amateur worries about different interpretations". So QM and Copenhagen interpretation of QM are interchangeable to many (almost all) physicists.

So let us go back to the atom again, it has no definite size, this is an experimental fact as well. If you claim that the atom has a definite size, you should have it backed up with articles. Claim was "hydrogen atoms has definite size. It's size is about the Bohr's radius", now prove it. Also you used the word "superposition" for continuous variable, position, I don't believe that makes sense in Real Analysis...

So let us go back to Yukawa again. You said that Yukawa used the Heisenberg uncertainty relation to show that the strong force is mediated by pions. That is an insult to Yukawa, we actually went through Yukawa's theory in my quantum field class recently, and there is no Heisenberg principle used what so ever, just pure and nice Quantum Field Theory. Many popular science book uses Heisenberg to explain alot, even yukawa's theory, so one will, as you just proved, get the impression that it is used. But now you encountered an Science Advisor, who knows the things we are discussion in detail, since he has worked with these things, not just read them on wikipedia.

One also uses the Heisenberg principle in discussion about Feynman diagrams and virtual particles, but these "explanations" are never used in REAL textbooks. In REAL textbooks, one presents the REAL arguments. The reason for why Heisenberg principle is so applicable to explanations of quantum phenomenon is that is really easy to do so, it is really "the ultimate probabilistic" entity. It is almost like the good ol "God of the gasps", whenever one couldn't find an explanation due to lack of knowledge, one "blaimed" God. Today, people who does not know quantum mechanics uses Heisenberg Uncertainty principle for their explanations...

 

[Modman]

I only use the analogy to explain the effects of inertia on the electron. Only a loss of energy ( in the form of heat ) can bring the electron closer to the nucleus. There is obviously no friction in subatomic space, so there is no other reason that the electron would lose energy.

 

[thoughtgaze]

I only use the analogy to explain the effects of inertia on the electron. Only a loss of energy ( in the form of heat ) can bring the electron closer to the nucleus. There is obviously no friction in subatomic space, so there is no other reason that the electron would lose energy.

Well in your analogy there IS reason the electron would lose energy if it were "orbiting" around the nucleus. It can be shown that an accelerating electron radiates energy. If it's in "orbit" in the usual sense, there is some sort of centripetal acceleration at all times. The stability of the energy state has nothing to do with it being frictionless system.

 

[edguy99]

Another way to think of this is what happens once the electon is within the bohr radius (53 picometers) of the hydrogen proton? The max energy that holds the electron to the proton is 13.6evolts, that does not increase, ie you never need more then 13.6 evolts of energy to pull an electron away from a hydrogen proton.

If one views the bohr radius as about the size of the first s orbital of hydrogen, it appears that it does not matter where the electon is, it does not require more energy to pull it out even if it is very very close to the proton.

Does this not lead to the conclussion that the hydrogen electron does not feel the pull of the proton (ie, does not gain energy beyond 13.6 evolts) once inside the first s orbital?

 

[granpa]

Well in your analogy there IS reason the electron would lose energy if it were "orbiting" around the nucleus. It can be shown that an accelerating electron radiates energy. If it's in "orbit" in the usual sense, there is some sort of centripetal acceleration at all times. The stability of the energy state has nothing to do with it being frictionless system.

the only possible explanation is that the electron is spread out over the whole atom due to the uncertainty principle and it is the whole 'electron cloud' that is spinning around the nucleus. hence the electric and magnetic fields arent changing so there is no radiation of energy.

Quantum physics replaced Bohr-Sommerfeld theory in 1920's.
Since then many phsicists such as Pauli, Heisenberg, Dirac... have given up explaining the
motion of an electron concretely.

because by equating the angular momentum of the spinning sphere of the electron to 1/2
h-bar, the sphere speed leads to about one hundred times the speed of light.

the whole electron cloud would not be spinning faster than the speed of light.

The orbital angular momentum of the electron in the ground state of the hydrogen atom is zero, so the Coulomb potential is infinitely negative when the electron is close to the nucleus.

but the electron still has spin

 

[Dadface]

Another way to think of this is what happens once the electon is within the bohr radius (53 picometers) of the hydrogen proton? The max energy that holds the electron to the proton is 13.6evolts, that does not increase, ie you never need more then 13.6 evolts of energy to pull an electron away from a hydrogen proton.

If one views the bohr radius as about the size of the first s orbital of hydrogen, it appears that it does not matter where the electon is, it does not require more energy to pull it out even if it is very very close to the proton.

Does this not lead to the conclussion that the hydrogen electron does not feel the pull of the proton (ie, does not gain energy beyond 13.6 evolts) once inside the first s orbital?

If the first orbital is a sort of stable state then the electron can feel an even greater force if it got even closer to the nucleus for example if a collision pushed it momentarily inwards before it moved outwards..This would not necessarily change the excitation or ionisation energies because energy lost on the inward journey between levels might be regained on the outward journey.It's just a thought and I will give it some more thought.

 

[alxm]

The orbital angular momentum of the electron in the ground state of the hydrogen atom is zero, so the Coulomb potential is infinitely negative when the electron is close to the nucleus.

The Coulomb potential is infinitely negative as long as the nucleus is modeled as a point charge. Angular momentum has nothing to do with it at all.

The divergence of the Coulomb term is canceled out by the divergence of the kinetic-energy term. The wave function is not divergent at the nucleus, just non-smooth. It forms a cusp.

 

[alxm]

Another way to think of this is what happens once the electon is within the bohr radius (53 picometers) of the hydrogen proton?

A hydrogen electron in the 1s state spends almost 1/3 of its time within the Bohr radius. So?

The max energy that holds the electron to the proton is 13.6evolts, that does not increase, ie you never need more then 13.6 evolts of energy to pull an electron away from a hydrogen proton.

That'd be conservation of energy.

If one views the bohr radius as about the size of the first s orbital of hydrogen, it appears that it does not matter where the electon is, it does not require more energy to pull it out even if it is very very close to the proton.

That doesn't make sense. Why do you need to assume the Bohr radius is the 'size' to then conclude that it does not matter where the electron is?

An orbital is an energetic eigenstate of the Schrödinger equation, and with the exception of the occasional infinitesimally-small node, they have nonzero values for the location-probability over all space. So knowing the location of an electron at any given moment tells us nothing about its energy. Which wouldn't necessarily be the case if it could be described classically.

Does this not lead to the conclussion that the hydrogen electron does not feel the pull of the proton (ie, does not gain energy beyond 13.6 evolts) once inside the first s orbital?

Again, that's just conservation of energy. That's never been an issue. The issue is why this energy couldn't be radiated away.

 

[edguy99]

Again, that's just conservation of energy. That's never been an issue. The issue is why this energy couldn't be radiated away.

If an electron inside of the bohr radius (53pm) no longer feels an attraction to the proton, then presumbably there is no need to radiate energy. Doesn't it basically just sit around somewhere inside "about" this radius "most" of the time?

 

[ytuab]

the only possible explanation is that the electron is spread out over the whole atom due to the uncertainty principle and it is the whole 'electron cloud' that is spinning around the nucleus. hence the electric and magnetic fields arent changing so there is no radiation of energy.

the whole electron cloud would not be spinning faster than the speed of light.
but the electron still has spin

If the whole electron cloud is spinning, this means the orbital angular momentum in the ground state of hydrogen is not zero. And it also radiates energy outside.

The Coulomb potential is infinitely negative as long as the nucleus is modeled as a point charge. Angular momentum has nothing to do with it at all.

The divergence of the Coulomb term is canceled out by the divergence of the kinetic-energy term. The wave function is not divergent at the nucleus, just non-smooth. It forms a cusp.

If the nucleus is not a point charge, the electron penetrates into the nucleus many times a day. The electron is probably scattered by the nucleus.
Experimentally the scattering of electrons from nuclei has given us the most precise information about nuclear size and charge distribution.