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I Physical Significance of 2s Orbital Peaks?

  1. Apr 30, 2016 #1
    The volumetric probability density function for the 1s orbital of the hydrogen atom has 1 peak, and it occurs at the Bohr radius.

    http://img1.mnimgs.com/img/shared/content_ck_images/images/probability%20density.png [Broken]

    For the 2s orbital, it has 2 peaks, neither of which coincides with an integer multiple of the Bohr radius.

    Is there any additional physical significance for the distance at which these 2 peaks occur?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Apr 30, 2016 #2
    the physical significance is that one has to throw out the Bohr's model of atom and as the probability density cloud is interpenetrating the electrons can not be thought of as confined to some circular orbital regions around the nucleus.
    therefore those effects which are due to exchange of electrons from one orbital to another or sharing can be accounted for.
    e.g. in He calculations for energy states we encounter overlap integrals or exchange integrals.
    though we are looking at s-states here - if one goes to p and other orbitals there shapes get further modified by the angular momentum considerations .
     
  4. Apr 30, 2016 #3
    So as far the as the Bohr model is concerned there is no further relation?

    By the way, you should return to my thread on blackbody peaks
     
  5. Apr 30, 2016 #4
    well, i could not follow the relation of the thread with black body radiation.
    i know that there are very many limitation of Bohr's theory of atomic structure.
     
  6. May 8, 2016 #5
    As I understand it, an electron in the H atom in the excited n = 2, state is most likely at the r corresponding to the peak in the square of the wavefunction. This will not be at an integral multiple of the Bohr radius. This idea is a total non - sequitur.
     
  7. Sep 2, 2016 #6
    so the fact that the 1S peak matches the Bohr radius is just a coincidence?
     
  8. Sep 2, 2016 #7

    Vanadium 50

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    The fact that it's proportional to the Bohr radius is not coincidence - that's required to get the right dimensionality and behavior. The fact that the constant of proportionality of is 1 and not 3/2 is because you chose to look at the distribution predicted by QM and ask "what is the mode of this distribution?" and not "what is the mean of this distribution?"
     
    Last edited: Sep 19, 2016
  9. Sep 2, 2016 #8
    sorry, is the mean related to the Bohr radii then?
     
  10. Sep 19, 2016 #9
    The expected value is related to the Bohr radii. The Bohr radii scale as ##n^2##. So, for the n=2 state, ##r=4a_0##, where ##a_0## is the Bohr radius.

    If you calculate an expected value for ##\frac{1}{r}##, it does give you ##\frac{1}{n^2 a_0}##. This follows from the virial theorem.

    You calculate the expected value by
    ##\int \frac{1}{r} (r R_{nl})^2##
    where R is the radial wavefunction.

    If you calculate the expected value for ##r## instead of ##1/r##, you get ##\frac{3}{2} a_0 n^2##
     
  11. Oct 2, 2016 #10
    the 1s peak coincides with the Bohr radius. But the 2s peaks do not coincide with the 2nd, or any of the Bohr radii. Does this indicate one of the failures of the Bohr model?
     
  12. Oct 2, 2016 #11

    Vanadium 50

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    You keep asking this question in different forms and not accepting the answer you get. See message #7.
     
  13. Oct 2, 2016 #12
    No, that is not true.
    First of all, I was replying to Khashishi, he sounds like he's on a different track. Secondly, with more info comes more questions. Its not the exact same question each time.

    Anyway, I did post #8. Can you elaborate on the mean? I'm sorry but I don't quite get it.
     
  14. Oct 2, 2016 #13

    Vanadium 50

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    As I said, the functional form is driven by the physics. The number "1" is your choice, and corresponds to the mode of the distribution. If you picked another reasonable choice, like the mean, you'd have a different number, like 3/2.
     
  15. Oct 4, 2016 #14
    I've read this thread from the top again, and I've realized that we were on entirely different wavelengths (no physics pun intended). I do understand what you mean now.

    I failed to state my question clearly.

    What I should've asked instead is: Is there any Bohrian (or Bohr-Sommefeldian of the Old Quantum Theory) equivalent to the peaks of the 2s and 3s curves (for the Hydrogen atom)?


    For example, if I asked: Is there any Bohr-Sommefeldian equivalent to the peak of the 1s curve?

    The answer would be: Yes there is, the peak occurs at the smallest radius at which an electron orbits the proton in the Bohr model.


    Of course, here "equivalent" is a vague term. This question is not really scientific, its answers can be diverse, but hopefully you and other forum goers are aware of interesting equivalences.
     
  16. Oct 4, 2016 #15
    In Old Quantum Theory, there were no 2s or 3s states. There were just the n=2 and 3 states.
     
  17. Oct 5, 2016 #16
    the azimuthal numbers had already been introduced in a rudimentary form though.

    anyway, if you read #14 again, I'm looking for equivalences, not differences
     
  18. Oct 5, 2016 #17

    vanhees71

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    I don't understand this debate. The physical meaning of the wave function is that its modulus squared gives the probability distribution for position of the electron bound to a proton. Case closed, question answered!
     
  19. Oct 8, 2016 #18
    we really are on different wavelengths. please read #17 carefully.
     
  20. Oct 9, 2016 #19

    vanhees71

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    I don't understand again. #17 is my own posting, which I read carefully again. So?
     
  21. Oct 9, 2016 #20
    Oops, my bad. I meant #14.
     
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