Problems with De Broglie's Wavelength

In summary, the De Broglie wavelength fails to explain why the electron does not crash into the nucleus in the Bohr model of the hydrogen atom. Despite De Broglie's attempts to create a wave-matter version of the model, it is not a proper quantum-mechanical description. Modern quantum mechanics and the historical models are more accurate and useful in understanding the behavior of particles. De Broglie's equations were based on analogies and were later combined with Schroedinger's wave equation to form a more comprehensive theory. However, even this theory has flaws and is considered a stop-gap solution. It is important to fully understand quantum mechanics before attempting to explain it to others.
  • #1
Dick Niggle
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I always used the matter-wave as an explanation for why the electron does not crash into the nucleus. As a standing wave, how could it? The wave would have to increase in frequency, almost infinitely, until it was the size of a point, the nucleus, which seemed like an impossible task.

However, using De Broglie's wavelength to explain this failed miserably. My goal was to simplify the situation as much as possible. So, we will look at just a lone Bohr's model of the hydrogen atom in an empty void of space.

I computed the wavelength of the electron orbiting at the Bohr's radius after finding it's velocity using a few formulas and simple algebra and obtained about 0.5 nm. The electron's wavelength is pretty small.

When trying to compute the wavelength of the proton, I expected that I could somehow arrive at 0 or a very small number that I could explain as insignificant, but the proton's wavelength is larger than the electron's wavelength and approaches infinity.

The electron is bound to a very high speed around the proton, so no matter what, it's wavelength becomes very small. The proton, however, is stationary, even to the perspective of the electron. So, somehow it has an infinite wavelength and shouldn't even be there.

If I decide to give the atom some velocity, both the proton and the electron waves decrease in length, because this extra velocity is being added to the electron, too. So, it seems, no matter what, the proton has a considerably greater wavelength than the electron.

Why is this so? Can you not describe more than one object at a time with De Broglie's wavelength? Are the proton and electron wave function somehow intertwined? It appears that the matter-wave no longer explains why the electron does not crash into the nucleus for me. I have looked and looked for an answer to that, but none sufficed. I suppose I will never understand. How do I fix this problem so that there can be a standing wave orbiting a solid nucleus like my textbook described.
 
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  • #2
The Bohr model is wrong. You shouldn't be surprised to get wrong results if you apply it to anything apart from electrons in hydrogen(-like) atoms.
The de-Broglie wavelength can be useful sometimes (double-slit experiments for example), but it is not the proper quantum-mechanical description either.
 
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  • #3
So, the De Broglie wavelength really does fail in this scenario? Did not De Broglie create a wave-matter version of Bohr's model? Maybe I should read his doctoral thesis.
 
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  • #4
I suggest reading more modern quantum mechanics.
The historical models are interesting in the historic context only.
 
  • #5
I'm just trying to make sense of his model, and when using his equation, it isn't coming together for me.
 
  • #6
It does not make sense for hydrogen atoms, unless you apply it to the relative motion of electron<->proton and the motion of the whole atom separately.
 
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  • #7
Dick Niggle said:
I'm just trying to make sense of his model, and when using his equation, it isn't coming together for me.

It has several flaws that are easily seen when jumping to a frame where the particle is at rest.

Its purely for historical interest.

The history is:
1. De Broglie, based on the idea that waves can act like particles, thought the converse may be true, and by various analogies came up with some equations.
2. Someone commented to Schroedinger, if waves were involved then there should be a wave equation. Based on that he came up with one - although his reasoning was wrong:
http://arxiv.org/pdf/1204.0653.pdf
3. Experience with the equation showed it couldn't really be waves in any usual sense. Because of that Schroedinger wasn't happy with ever being inolved in it.
4. There was another equally successful theory at the time called Matrix Mechanics. It sort of smelled right that they were really the same and in 1926 Dirac came up with a theory that satisfied physicists in combining them, called the transformation theory. It contained this thing called the Dirac Delta function that the math at that time was unable to handle rigorously, and it wasn't until Von Neumann published his mathematical Foundations Of QM, that it was fully combined in a theory satisfactory to both mathematicians and physicists. These days, with Rigged Hilbert Spaces, Dirac's methods can be done rigorously:
http://www.lajpe.org/may08/09_Carlos_Madrid.pdf

That's the historical version, and how its usually explained in most textbooks. However IMHO a better starting point is the following:
http://www.scottaaronson.com/democritus/lec9.html

Thanks
Bill
 
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  • #8
Thanks. I will be taking quantum next semester. It's kind of embarrassing, but I run an SI class and I would explain this wave thing in relation to De Broglie to tell them why the electron does not fall into the nucleus, but after reviewing it myself. I don't know why the electron doesn't fall into the nucleus. De Broglie offers kind of a really simple explanation, but when looking into it in more depth, it no longer made much sense. So, idk. I doubt quantum will tell me exactly why, either.

I am looking more into it right now and actually reading his thesis before I move on to the other stuff. There's no way you can win a nobel prize and your entire research be this bad. I believe that I am not getting the full picture.
 
  • #9
Dick Niggle said:
There's no way you can win a nobel prize and your entire research be this bad. I believe that I am not getting the full picture.

You are getting the full picture. It is this bad. It was recognised from the start it was a stop gap.

If you want to talk about it to a class try something like:


But I have to say if you don't understand what's going on its probably not a good idea giving a talk about it.

Thanks
Bill
 
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  • #10
You might be interested in our most recent Insights article.

Edit: Hmm, didn't read it before I posted it, does not fit to the topic, sorry.
 
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  • #11
is this something anyone is suppose to understand? I thought it was just a concept, like what really is an electron, anyway, and how is it charged?

Who knows.

You know, there might be a small center of mass that the proton would revolve about, that may give a better result than the original problem.

The above article doesn't really do it for me. All it is really saying is that the electron doesn't fall into the nucleus because the electron doesn't fall into the nucleus. That's all I really got out of it. It doesn't explain why the force should be zero below the Bohr radius. I'm pretty sure that should require an explanation.

To me, De Broglie offers the simplest and best explanation I've come across. If only I can fix this one tiny inconsistency.
 
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  • #12
Not sure matter waves make explanations clearer.

λ = h / p
frequency f is mc^2 / h
wave speed is λ * f
speed of the matter wave is c^2 / v (what is now called the phase velocity) always exceeding c.

The matter wave of a particle must always exceed c while neither its mass nor energy do so, only the governance of certain interference effects do so.
This may be more difficult to explain than the original question...
 
  • #13
HA! With the center of mass calculated and the speed of the proton there in that tiny little radius created, the wavelength of the proton becomes smaller than the electron's.

Actually, their difference doesn't seem like a whole lot. The electron has a wavelength of 0.3 nm, which is about one wavelength around the circumference of its orbit at n=1, and the proton's is 0.18 pm, which is also about one wavelength around the circumference of its orbit, maybe at n=1 of some nuclear shell (idk), if I computed it correctly.

After reviewing some links, perhaps I am going about this concept the wrong way.

The SI class I run is just for general chemistry I. We don't usually go over anything other than the wave properties of the electron, and then, sometimes, I'll say when viewed as a standing wave, it doesn't easily collapse. We don't go in depth into much of anything, but half of the class ends up failing, anyway. The retention rate is like 40%. It was so disappointing to find out that everyone taking general chemistry hates physics and chemistry, and trying to help people who hate the subject and resent you is very frustrating. I just try to make things more interesting by asking them questions. Like, why doesn't the electron fall into the nucleus?

I believe the textbook explains it by this standing wave concept. We, obviously, look at quantum numbers, but not the Schrodinger Equation. It is very basic.
 
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  • #14
I'm not sure why it's not letting me edit, but I did the math wrong, again. I believe the proton should have a wavelength of 10 pm, which corresponds to n=0. So, the proton can not exist as a circular, standing wave. It is probably appropriate in the model to take the proton as a point mass.

I think my original concept was actually right. At n=1, the circular wave contains only one wavelength. It can not exist as a standing wave if it were to approach any closer to the nucleus. Furthermore, if you could somehow increase the waves' frequency and number of wavelengths and try to squeeze it into the nucleus, it would take an almost infinite amount of energy to condense the wave enough.
 
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  • #15
Dick Niggle:

I'm at about the same knowledge level as you are, and also was wondering about the De Broglie wavelength of the proton in the hydrogen atom; posing that question in this thread: https://www.physicsforums.com/threads/debroglie-wavelength-of-proton-in-h-atom.799937/ However, I quickly realized that both the proton and electron would have the same De Broglie wavelength.

But I don't want to add confusion to this issue, and will defer to the far more knowledgeable people on this thread, and the other thread, where it was stated that the De Broglie concept was a heuristic model that has long been superseded by a much more rigorous, and complete, theory. I'll check out the links provided by bhobba.
 
  • #16
Dick Niggle said:
I don't know why the electron doesn't fall into the nucleus.

Actually the electron does sometimes "fall into the nucleus," in some sense. See the graphs for the electron probability density for the ground state of hydrogen in figures 3.4 and 3.6 on this page:

http://www.chemistry.mcmaster.ca/esam/Chapter_3/section_2.html

With hydrogen, nothing actually happens as a result of the non-zero probability of the electron being "in the nucleus", because there's no outcome that satisfies energy conservation. The proton can't simply absorb the electron and convert into a neutron, because an isolated neutron has more mass than an isolated proton, and there's no place for the required additional energy to come from.

However, certain heavier nuclei can capture the electron, convert a proton into a neutron, and change the total nuclear binding energy in such a way that the new isotope has less mass than the original one, releasing energy in the process.

https://en.wikipedia.org/wiki/Electron_capture
 
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  • #17
jtbell:

That was a great explanation as to why the electron can't "fall into the nucleus" in a hydrogen atom, since it's energetically not favored.

I looked at the links that were provided by bhobba, and the one most useful for a novice like me was http://www.scottaaronson.com/democritus/lec9.html It looks like an entire course in QM is available on line from that site.
 
  • #18
I just had a funny thought. This is all over the simplest atom not being "all-inclusively" and therefore not accurately described. With quantum physics clenched tightly in one hand and general relativity warping the other you'd think someone would have looked for the answers in the most energetically neutral atom of Iron. I am certainly curious of how well the de Broglie wavelength applies to that atom.
 
  • #19
jerromyjon said:
most energetically neutral atom
What is that supposed to mean?
jerromyjon said:
I am certainly curious of how well the de Broglie wavelength applies to that atom.
No difference to other atoms.

General relativity is completely negligible for atoms.
 
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  • #20
mfb said:
What is that supposed to mean?
Nuclear fusion.
 
  • #21
The value of the binding energy per nucleon (the thing where nickel is special, and iron is not so far away) has nothing to do with the discussion here.
 
  • #22
mfb said:
(the thing where nickel is special, and iron is not so far away)
What is that supposed to mean?
 
  • #23
Nickel-62 (not iron) has the highest binding energy per nucleon.
The binding energy of some iron isotopes is still quite large, so stable fusion does not work with iron.
 
  • #24
I wasn't shooting for highest anything. I'm thinking about a nirvana atom. All at 1:1's with itself.
 
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  • #25
Then why did you ask about iron in particular, and why do you expect something special there?
 
  • #26
I don't know, I thought I was on to something but when I try to put the pieces together it's not adding up. From what I understand about nuclear fusion energy is released while nucleons are fused up to iron, then above iron more energy has to be added to fuse larger atoms. I know it's "special" but I'm not sure how or why.
 
  • #27
jtbell said:
Actually the electron does sometimes "fall into the nucleus," in some sense.
Can we say: "The electron always falls into the nucleus, but it's also constantly "tunneling out" and showing up at arbitrary places all around the nucleus" ...?
 
  • #28
Swamp Thing said:
Can we say: "The electron always falls into the nucleus, but it's also constantly "tunneling out" and showing up at arbitrary places all around the nucleus" ...?

No.

Electrons do not have properties like fall etc when not observed.

What's going on when not observed the theory is silent about.

Thanks
Bill
 
  • #29
In addition, there is no tunneling involved. The potential in the nucleus is much lower than the electron energy.
 
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1. What is De Broglie's Wavelength?

De Broglie's Wavelength is a concept in quantum mechanics which states that particles, such as electrons and photons, can also exhibit wave-like behavior and have a wavelength associated with them. This wavelength is inversely proportional to the particle's momentum.

2. What are some problems with De Broglie's Wavelength?

One of the main problems with De Broglie's Wavelength is that it only applies to objects with mass, meaning that it cannot be used to describe the behavior of massless particles like photons. Additionally, it does not fully explain the wave-particle duality of particles, leading to the development of more complex theories.

3. How does De Broglie's Wavelength relate to the uncertainty principle?

The uncertainty principle, proposed by Heisenberg, states that it is impossible to simultaneously know the exact position and momentum of a particle. De Broglie's Wavelength is related to this principle as it shows that the more precisely we know a particle's momentum, the less we know about its position.

4. Can De Broglie's Wavelength be applied to macroscopic objects?

No, De Broglie's Wavelength is only applicable to microscopic particles. This is because the wavelength becomes extremely small for larger objects, making it impossible to observe any wave-like behavior.

5. How has De Broglie's Wavelength contributed to our understanding of quantum mechanics?

De Broglie's Wavelength was a groundbreaking concept that helped bridge the gap between classical mechanics and quantum mechanics. It helped explain the wave-like behavior of particles and opened the door for further research and development of quantum theories.

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