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Bohr model & Atom Stability

  1. Dec 6, 2007 #1
    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?
     
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  3. Dec 6, 2007 #2

    ZapperZ

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

    Zz.
     
  4. Dec 6, 2007 #3
    Why don't you just answer my question?
     
  5. Dec 6, 2007 #4

    ZapperZ

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    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.
     
  6. Dec 11, 2007 #5
    understandable physical picture

    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
     
    Last edited: Dec 11, 2007
  7. Dec 11, 2007 #6

    malawi_glenn

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    Why dont 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 wierd? (As Bohr told Einstein)

    You cant 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.
     
  8. Dec 11, 2007 #7

    dst

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    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 existance 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.
     
    Last edited: Dec 11, 2007
  9. Dec 11, 2007 #8
    How come we do see electron tracks in bubble chamber?
    How can that be explained away using Quantum Mechanics of wave?
     
  10. Dec 11, 2007 #9

    f95toli

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    You shouldn't take the "wave-nature" of the electron too litteraly. When we talk about particles and waves in QM we are really refering 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.
     
  11. Dec 12, 2007 #10
    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.
     
  12. Dec 12, 2007 #11

    ZapperZ

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    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.
     
  13. Dec 12, 2007 #12

    jtbell

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    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.
     
  14. Dec 12, 2007 #13

    malawi_glenn

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    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 probabilty to be "inside" the nucleus, therefore we can have internal conversion.
     
  15. Dec 12, 2007 #14
    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.
     
  16. Dec 18, 2007 #15
    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?
     
    Last edited: Dec 18, 2007
  17. Dec 18, 2007 #16
    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".
     
  18. Dec 18, 2007 #17

    ZapperZ

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    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.
     
  19. Dec 18, 2007 #18

    ZapperZ

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

    Zz.
     
  20. Dec 18, 2007 #19
    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.
     
  21. Dec 18, 2007 #20

    ZapperZ

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    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.
     
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