Is the Bohr model of the atom still valid in quantum mechanics?

[mendocino]

Why does in QM the electron does not fall toward the nucleus? After all, the only force between nucleus and electron is attractive (- electron and + nucleus). Is the same reason that justifies the moon does not fall to the earth?

 

[ZapperZ]

Why does in QM the electron does not fall toward the nucleus? After all, the only force between nucleus and electron is attractive (- electron and + nucleus). Is the same reason that justifies the moon does not fall to the earth?

You may want to start by reading an entry in our FAQ in the General Physics forum.

Zz.

 

[mendocino]

You may want to start by reading an entry in our FAQ in the General Physics forum.

Zz.

Why don't you just answer my question?

 

[ZapperZ]

I gave you a place where you can find the answer and maybe even other possible questions that you might have (i.e. teach you how to fish rather than give you the fish), especially when the explicit answer to this every exact question has been addressed.

So why won't you go read it?

Zz.

 

[mendocino]

understandable physical picture

====================
WHY DON’T ELECTRONS CRASH INTO THE NUCLEUS IN ATOMS?

Contributed by Marlon and edited by ZapperZ

If one describes atoms using only the Coulomb forces, the electron and the nucleus will attract each other and no stable atoms could exist. Obviously this is not the case. Niels Bohr was the first (1913) to propose a better model, which consisted of electrons moving around the nucleus in circular orbits. Each orbit corresponds to a certain discrete energy level. This model is based upon the quantisation of the angular momentum.

Unfortunately, electrons moving in a circular orbit have an acceleration due to the centripetal force. In classical electromagnetic theory, an accelerated charged particle must emit EM-radiation due to energy conservation. Hence, the electron would lose energy and spiral down towards the nucleus. Again stable atoms could not exist. What is wrong now?

It turns out that the picture of electrons moving in circular orbits around the nucleus isn’t correct either(*). The solution here is the implementation of Quantum Mechanics via the Schrödinger Equation and the concept of wavefunction. By applying such formalism, the “electron” occupies a volume of space simultaneously, so that it is “smeared” in a particular geometry around the nucleus. While there are no more “orbits”, we do use the term “orbitals” to indicate the shape of such geometry. However, this term should not be confused to mean an orbiting electron similar to our planets in the solar system. By describing the system in terms of the QM wavefunction, it creates stable states for the nucleus+electrons system that matches very well with experimental observation of standard atomic spectra.

Since there are no more “orbits” in the conventional sense, the problem of electrons radiating due to an accelerated motion is no longer meaningful. It explains why we have stable atoms.
Zz.

Basically, I think what you said is this
we solve Schrödinger Equation and the math shows that the wavefunction of electron is “smeared” in a particular geometry around the nucleus.

I think that's well-known, but it does not explain anything
What I'm looking for is an understandable physical picture
why electron does not fall toward the nucleus

 

[malawi_glenn]

Why don't it gives us the physical picture? Why does people think that physics "must" be understood in pictures and be intuitive in the "common" (i.e. every day life) sense?

What if nature itself is weird? (As Bohr told Einstein)

You can't think of electrons beeing small balls, following sharp trajectories. One have to abandon the classical picture of nature and instead see how nature works, and what "laws" it follows? Nature happens to be quantum mechanical in the so called microscopic scale. And as I said, we just have to deal with it.

 

[dst]

Basically, I think what you said is this
we solve Schrödinger Equation and the math shows that the wavefunction of electron is “smeared” in a particular geometry around the nucleus.

I think that's well-known, but it does not explain anything
What I'm looking for is an understandable physical picture
why electron does not fall toward the nucleus

The electron doesn't "orbit" the nucleus, what we see is a probability of the electron being in any of those regions. Now, think back to time before any of that was known - if you repeated an experiment to test the existence of electrons, you would more or less always get the same results, then you would either conclude that it's a "plum pudding" or that there's a mini solar system. But, we're talking about a certain scale where the particles stop looking like particles (there's a certain limit for that, but I can't remember) and are better described by waves.

I'm sure someone with an actual knowledge of what goes in in dwarf-land can clear it up.

 

[mendocino]

Why don't it gives us the physical picture? Why does people think that physics "must" be understood in pictures and be intuitive in the "common" (i.e. every day life) sense?

What if nature itself is weird? (As Bohr told Einstein)

You can't think of electrons beeing small balls, following sharp trajectories. One have to abandon the classical picture of nature and instead see how nature works, and what "laws" it follows? Nature happens to be quantum mechanical in the so called microscopic scale. And as I said, we just have to deal with it.

How come we do see electron tracks in bubble chamber?
How can that be explained away using Quantum Mechanics of wave?

 

[f95toli]

How come we do see electron tracks in bubble chamber?
How can that be explained away using Quantum Mechanics of wave?

You shouldn't take the "wave-nature" of the electron too litteraly. When we talk about particles and waves in QM we are really referring to classical analogies that are often convenient since they help us understand what is going on, it doesn't mean that an electron is a "wave" in the classical sense (waves in water etc); it simply means that electrons (and everything else) has wave-like (and at the same time particle-like) properties.
In the case of the bubble chamber it is probably more conventient to think of the electron as a particle since its particle-like properties "dominates" (i.e. it behaves more or less like a classical particle).

This can be quite confusing. However, it is important to remember that this confusion only arises because we are trying to describe QM phenomena- and the math that is needed to describe these phenomena- using analogies from our "classical" world.

 

[lightarrow]

Why does in QM the electron does not fall toward the nucleus?

Who told you it's not already "fallen"? What exactly is an electron in an atom? Did you know that, at least for the fundamental state of hydrogen atom, the electron has a non zero probability to be located in the nucleus? Teachers at school, as well as school books, don't always explain things correctly.

 

[ZapperZ]

Who told you it's not already "fallen"? What exactly is an electron in an atom? Did you know that, at least for the fundamental state of hydrogen atom, the electron has a non zero probability to be located in the nucleus?

Er.. you may want to check this. There is a difference between <psi|psi> and <psi|r|psi>. Look at the distribution of the latter function, which is a more accurate description of the "location" of the electron.

Zz.

 

[jtbell]

How come we do see electron tracks in bubble chamber?
How can that be explained away using Quantum Mechanics of wave?

We don't actually see the electrons, we see bubbles at locations where the electron has interacted with the surrounding fluid. The bubbles are much much larger than the quantum mechanical uncertainty in the position of the electron which comes from their wavelike behavior by way of the Heisenberg Uncertainty Principle.

 

[malawi_glenn]

Very good call here f95toli. My teacher said "The electron is neither a ball (particle) or a wave, an electron is an electron".

lightarrow: yes the electron in ground state (and other states) have non zero probability to be "inside" the nucleus, therefore we can have internal conversion.

 

[lightarrow]

Er.. you may want to check this. There is a difference between <psi|psi> and <psi|r|psi>. Look at the distribution of the latter function, which is a more accurate description of the "location" of the electron.
Zz.

Yes, but when I wrote <<Who told you it's not already "fallen"?>> I didn't mean to refer it to the radial distribution of probability, but to the fact that "fallen" refers to the classic description; how can we reason in those terms for an object which is described by a wavefunction? Since the electron is, in that example, in its fundamental state, in a sense we could then say that it has already "fallen" to the "lowest possible location".

When then I talked about <psi|psi>, I didn't mean to "prove" the previous assertion but simply add another fact difficult to explain with the orbiting particle paradigm.

 

[Xeinstein]

Yes, but when I wrote <<Who told you it's not already "fallen"?>> I didn't mean to refer it to the radial distribution of probability, but to the fact that "fallen" refers to the classic description; how can we reason in those terms for an object which is described by a wavefunction? Since the electron is, in that example, in its fundamental state, in a sense we could then say that it has already "fallen" to the "lowest possible location".

When then I talked about <psi|psi>, I didn't mean to "prove" the previous assertion but simply add another fact difficult to explain with the orbiting particle paradigm.

The question is if there is a force field that counteract the attractive force between the electron and the proton in nucleus
If yes, what's the force field and how it works?

 

[akhmeteli]

Er.. you may want to check this. There is a difference between <psi|psi> and <psi|r|psi>. Look at the distribution of the latter function, which is a more accurate description of the "location" of the electron.

Zz.

As the nucleus has finite dimensions, it seems the difference is quantitative, and technically the lightarrow's statement probably holds: "at least for the fundamental state of hydrogen atom, the electron has a non zero probability to be located in the nucleus".

 

[ZapperZ]

The question is if there is a force field that counteract the attractive force between the electron and the proton in nucleus
If yes, what's the force field and how it works?

Why do you even need such a force? What is the shortcomming of the coulombic force that so far can accurately fit with available experimental observation?

Zz.

 

[ZapperZ]

As the nucleus has finite dimensions, it seems the difference is quantitative, and technically the lightarrow's statement probably holds: "at least for the fundamental state of hydrogen atom, the electron has a non zero probability to be located in the nucleus".

Well, what is <psi|r|psi> for the s orbital at r=0?

Zz.

 

[akhmeteli]

Well, what is <psi|r|psi> for the s orbital at r=0?

Zz.

I agree, this value is zero. However, I am not sure this is fully relevant: as far as I remember, the radius of the nucleus is of the order of 10^(-13) cm (it was actually first measured in Rutherford's experiments). Therefore, the integral of |\psi|^2 over the volume of the nucleus (and this is exactly the probability for the electron to be inside the nucleus) is technically not zero.

 

[ZapperZ]

I agree, this value is zero. However, I am not sure this is fully relevant: as far as I remember, the radius of the nucleus is of the order of 10^(-13) cm (it was actually first measured in Rutherford's experiments). Therefore, the integral of |\psi|^2 over the volume of the nucleus (and this is exactly the probability for the electron to be inside the nucleus) is technically not zero.

That's why I asked for which of the two is more relevant in the quantity we want as the "position" of the electron. Note that |psi|^2 is simply the probability density. It has no associated physical quantity by itself. I could find <p>, or <r>, or <E>, etc.. by using the appropriate hermitian operator, but until I do that, |psi|^2 itself is not associated with any observable. So I'm not sure how you could use that as a basis to tell anything about the electron's position.

<r> is more relevant to a quantity that is associated with the electron's position. That's why, when we teach undergrad quantum mechanics, one of the function that the student has to know was u(r) =r*R(r), where R(r) is the radial solution of the wavefunction. There is a reason for this, because in most cases, u(r) is a function that will be needed when dealing with position distribution.

Zz.

 

[akhmeteli]

That's why I asked for which of the two is more relevant in the quantity we want as the "position" of the electron. Note that |psi|^2 is simply the probability density. It has no associated physical quantity by itself. I could find <p>, or <r>, or <E>, etc.. by using the appropriate hermitian operator, but until I do that, |psi|^2 itself is not associated with any observable. So I'm not sure how you could use that as a basis to tell anything about the electron's position.

<r> is more relevant to a quantity that is associated with the electron's position. That's why, when we teach undergrad quantum mechanics, one of the function that the student has to know was u(r) =r*R(r), where R(r) is the radial solution of the wavefunction. There is a reason for this, because in most cases, u(r) is a function that will be needed when dealing with position distribution.

Zz.

I am afraid you've lost me. Or maybe I was not attentive enough when reading your post. Your question was "what is <psi|r|psi> for the s orbital at r=0?" I was distracted by this "r=0" and somehow thought you asked about r|\psi|^2 at r=0. However, you are certainly right, and <psi|r|psi>=<r> is the average radius, but then the phrase "<psi|r|psi> for the s orbital at r=0" seems nonsensical. To the best of my knowledge, you cannot both average something and demand that it equals zero. Furthermore, I don't know how <r> is relevant to lightarrow's statement, as the probability for the electron to be inside the nucleus is, almost by definition, an integral of |\psi^2| over the volume of the nucleus. Otherwise, what's the meaning of your phrase "|psi|^2 is ... the probability density"? Let me remind you, lightarrow's statement was "at least for the fundamental state of hydrogen atom, the electron has a non zero probability to be located in the nucleus", and I still tend to believe it is technically correct.

 

[ZapperZ]

I am afraid you've lost me. Or maybe I was not attentive enough when reading your post. Your question was "what is <psi|r|psi> for the s orbital at r=0?" I was distracted by this "r=0" and somehow thought you asked about r|\psi|^2 at r=0. However, you are certainly right, and <psi|r|psi>=<r> is the average radius, but then the phrase "<psi|r|psi> for the s orbital at r=0" seems nonsensical. To the best of my knowledge, you cannot both average something and demand that it equals zero. Furthermore, I don't know how <r> is relevant to lightarrow's statement, as the probability for the electron to be inside the nucleus is, almost by definition, an integral of |\psi^2| over the volume of the nucleus. Otherwise, what's the meaning of your phrase "|psi|^2 is ... the probability density"? Let me remind you, lightarrow's statement was "at least for the fundamental state of hydrogen atom, the electron has a non zero probability to be located in the nucleus", and I still tend to believe it is technically correct.

But a probability density has no associated physical meaning. What quantity does that represent? To me, it simply defines the region of space that is covered by the wavefunction. It tells you nothing on the position of the object since the wavefunction tells you nothing until you associate it with a particular observable. While you can say that there's some probability for an electron to be within the confined of where the wavefunction is (which is obvious because of the wavefuction doesn't extend to other regions, you get no electron there), you cannot equate that as indicating an electron will be located at a particular position. rR(r) is zero at r=0.

Zz.

 

[akhmeteli]

But a probability density has no associated physical meaning. What quantity does that represent? To me, it simply defines the region of space that is covered by the wavefunction. It tells you nothing on the position of the object since the wavefunction tells you nothing until you associate it with a particular observable. While you can say that there's some probability for an electron to be within the confined of where the wavefunction is (which is obvious because of the wavefuction doesn't extend to other regions, you get no electron there), you cannot equate that as indicating an electron will be located at a particular position. rR(r) is zero at r=0.

Zz.

OK, let us consider the observable \theta(r_0-r), where r_0 is the radius of the nucleus, and \theta function is the integral of \delta function (\theta(x)=0 if x<0 and \theta(x)=1 if x>0). I guess this is a legitimate observable. If you disagree, please advise. Its average over the state is the probability for the electron to be within a sphere of radius r_0, and it is equal to the integral of |\psi^2| over this sphere.

Or I could just refer to the mathematical definition of probability density. If its integral over an area is not equal to the probability for the value of the variable (in our case it's the coordinate) to be in the area, what on Earth does one mean saying that |\psi^2| is the probability density? In general, what does Born's rule actually mean?

Another thing. I agree, rR(r) is zero at r=0. but it is not zero at, say, r=r_0/2, that is, inside the nucleus.

 

[ZapperZ]

OK, let us consider the observable \theta(r_0-r), where r_0 is the radius of the nucleus, and \theta function is the integral of \delta function (\theta(x)=0 if x<0 and \theta(x)=1 if x>0). I guess this is a legitimate observable. If you disagree, please advise. Its average over the state is the probability for the electron to be within a sphere of radius r_0, and it is equal to the integral of |\psi^2| over this sphere.

Or I could just refer to the mathematical definition of probability density. If its integral over an area is not equal to the probability for the value of the variable (in our case it's the coordinate) to be in the area, what on Earth does one mean saying that |\psi^2| is the probability density? In general, what does Born's rule actually mean?

Another thing. I agree, rR(r) is zero at r=0. but it is not zero at, say, r=r_0/2, that is, inside the nucleus.

Er.. there's a problem here. The solution you obtained from the Schrödinger equation is for a central potential OUTSIDE the nucleus. This is similar to solving the Gauss's law and getting the field dependence outside of the spherical charge source. Once you go inside the source, then you have to resolve for the field. So you can't use the same wavefunction and extrapolate it inside the nucleus. This isn't kosher.

The issue here isn't the "spread" of the wavefunction. The issue here is the confusion that the wave function, or specifically, R(r) is being confused as to also indicate the identical profile of the position. It isn't! R(r) for n=1, l=0 orbital strongly peaked at r=0, yet, rR(r) is zero there, which is completely contradictory to each other. So by simply using the wavefunction to deduce the position "distribution" leads to some severely misleading answers!

Zz.

 

[akhmeteli]

Er.. there's a problem here. The solution you obtained from the Schrödinger equation is for a central potential OUTSIDE the nucleus. This is similar to solving the Gauss's law and getting the field dependence outside of the spherical charge source. Once you go inside the source, then you have to resolve for the field. So you can't use the same wavefunction and extrapolate it inside the nucleus. This isn't kosher..

I fully agree with this remark. However, I guess you don't think that due to this circumstance the actual wavefunction inside the nucleus is identically zero. If you do, please advise.

The issue here isn't the "spread" of the wavefunction. The issue here is the confusion that the wave function, or specifically, R(r) is being confused as to also indicate the identical profile of the position. It isn't! R(r) for n=1, l=0 orbital strongly peaked at r=0, yet, rR(r) is zero there, which is completely contradictory to each other. So by simply using the wavefunction to deduce the position "distribution" leads to some severely misleading answers!

Zz.

I have no problems with this statement. However, I believe the issue was lightarrow's statement, which I quoted a couple of times. I still believe that technically it holds.

 

[lightarrow]

Er.. there's a problem here. The solution you obtained from the Schrödinger equation is for a central potential OUTSIDE the nucleus. This is similar to solving the Gauss's law and getting the field dependence outside of the spherical charge source. Once you go inside the source, then you have to resolve for the field. So you can't use the same wavefunction and extrapolate it inside the nucleus. This isn't kosher.

The issue here isn't the "spread" of the wavefunction. The issue here is the confusion that the wave function, or specifically, R(r) is being confused as to also indicate the identical profile of the position. It isn't! R(r) for n=1, l=0 orbital strongly peaked at r=0, yet, rR(r) is zero there, which is completely contradictory to each other. So by simply using the wavefunction to deduce the position "distribution" leads to some severely misleading answers!

Zz.

Sorry but are you saying that it's meaningless to talk about the probability to find the electron inside a finite little volume around r = 0? That would sound very strange to me. I agree that the fact <psi|psi> =/= 0 for r =/= 0 doesn't prove a non zero probability inside the nucleus, but it certainly implies a non zero probability for
0 < r < a where a is much less than <psi|r|psi>.
Otherwise, we should criticize the statement that the integral of <psi|psi> in a finite volume of space has a physical meaning = the probability to find the particle in that volume of space. Isn't it the very definition of the physical meaning of psi?

 

[ZapperZ]

Sorry but are you saying that it's meaningless to talk about the probability to find the electron inside a finite little volume around r = 0? That would sound very strange to me. I agree that this don't prove a non zero probability inside the nucleus, but it certainly implies a non zero probability for 0 < r < a where a is much less than <psi|r|psi>.

No, I said using the wave function itself to deduce the probability of such a thing is faulty just based on the profile of the wavefunction itself! I have said nothing about finding the electron within the nucleus (assuming you know the size of the nucleus). Again, R(r) peaked at r=0, yet, rR(r) goes to zero there! Don't you see how different this is? That is what I've been trying to point out! You can't get two functions more diverging from each other than having something peaked at one value while the other goes to zero!

So how can you speak with any degree of confidence that your usage of that function is roughly the same and justifiable when it varies that big?

Zz.

 

[lightarrow]

No, I said using the wave function itself to deduce the probability of such a thing is faulty just based on the profile of the wavefunction itself! I have said nothing about finding the electron within the nucleus (assuming you know the size of the nucleus). Again, R(r) peaked at r=0, yet, rR(r) goes to zero there! Don't you see how different this is? That is what I've been trying to point out! You can't get two functions more diverging from each other than having something peaked at one value while the other goes to zero!

That's perfectly clear and I certainly didn't mean to confuse one with the other.

So how can you speak with any degree of confidence that your usage of that function is roughly the same and justifiable when it varies that big?
Zz.

I only would like to know if it has a physical meaning or not to evaluate the integral of <psi|psi> dV for 0 < r < a and if that integral is = 0 or not.

 

[ZapperZ]

I only would like to know if it has a physical meaning or not to evaluate the integral of <psi|psi> dV for 0 < r < a and if that integral is = 0 or not.

I wouldn't know. I've never solved for R(r) inside the nucleus. Have you?

Zz.

 

[reilly]

Reverse beta decay does indeed occur, not in hydrogen, but in heavy neucleii, which typically have large radii --relatively speaking. The 1S state for hydrogen has an average radius of 3a0/2, while the probability density = 4 r*exp(-2r) = R(r)^^2
R, the wave function, is normalized so that the probability of finding the electron between r and r+dr is just R(r)^^2. So the most likely place to find electron where R's maximum is. that is, at r0 = a0, where a0 is the Bohr radius.

The radius of the proton is estimated to be roughly 10^^ -13 cm, while a0 is roughly 10^^-8. If, for some reason, you wanted to compute the reverse beta decay matrix element, p + [e] --> n + v(neutrino), a reaction that cannot happen for a bound electron,[e].

As a first approximation, simply use the standard 1S wave function to compute the probability of finding the electron within the proton's radius, or whatever multiple thereof you wish. You find, ballpark, this probability to be roughly 10^^-15. So, even if it could happen, it won't. But, if you wanted to compute the reverse beta decay matrix element, you could legitimately consider the combined wave functions of the electron and proton as constant over a very small volume, and so forth.

You could even go to models of the proton charge distribution or potential within the proton, like Coulomb outside the proton, and constant inside, or potential = oscillator potential , or ...

ZapperZ:If there's a problem here, it's avoided discovery for almost a century of intense work in both nuclear and atomic physics.

What is the problem?
Regards, Reilly Atkinson