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Electron orbital

  1. Jun 19, 2011 #1
    The orbits of planets have some ellipticity. Is there any corresponding ellipticity in the orbital of electrons?
    Are we able to see some ellipticity at least for a many electron atom, if not for the Hydrogen like atoms? I expect to see some ellipticity.

    We solve the Hydrogen atom problem invoking the reduced mass that is equivalent to reducing the problem from the one of two bodies to the one of single body (the electron). How/what will be the solution if we solve the H-atom problem as a two body problem, that is, what will be the orbital for the nucleus as well as that of the electron that we get as the result?
     
    Last edited: Jun 19, 2011
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  3. Jun 19, 2011 #2

    ZapperZ

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    Er... look at the p, d, f, etc. orbitals. These orbitals cause the electrons to do more gymnastics than what you see in classical celestial orbits.

    Zz.
     
  4. Jun 19, 2011 #3

    jtbell

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    In Bohr's original model, electron orbits are circular. Sommerfeld extended the model to allow for elliptical orbits. Modern quantum mechanics (Schrödinger etc.) doesn't use classical, well-defined paths like planetary orbits. Instead it has orbitals which are three-dimensional probability distributions. Some orbitals are spherically symmetric, others are not, depending on the angular momentum.

    Pictures: http://chemlinks.beloit.edu/Stars/pages/orbitals.html
     
  5. Jun 19, 2011 #4

    SpectraCat

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    That is not really correct .. what we do is transform to the center of mass frame, so that we can treat the internal degrees of freedom related to the relative motion of the electron and nucleus separately from the translational motion of the center of mass of the atom. So solutions to the H-atom Schrodinger equation already account for the "motion" of the nucleus. The reason we tend to ignore the nuclear component is because the large mass discrepancy between the electron and proton means that the nuclear "motion" is tiny compared to that of the electron. However, you do not get correct agreement with experiment unless you take the finite mass of the nucleus into account.

    Note the quotation marks around "motion" in the above description ... that is because it is not clear that electrons and nuclei in atoms "move" in the way we are used to thinking about for classical particles. We know that they have non-zero kinetic energies, which imply motion for a classical system, but there is no clear way to translate the information about kinetic energies of electrons in atoms into velocities or trajectories, as would be possible for an equivalent classical problem. Rather, we have to satisfy ourselves with a probabilistic description of the relative positions of the electron and nucleus.
     
  6. Jun 25, 2011 #5
    Its just an electron standing wave which lies on the "orbit" if the circumference of the orbit is integral multiple of the period of electron wave oscillation.

    In the simplest case the probability is uniform round the "orbit". spherical s orbital.

    In other cases the probability has some bulges. p, d, f etc.

    Just wondering why don't we exclude classical theory while dealing with particles like electrons, protons etc. Upto atoms level classical theory is ok. But in electronic level it was wrong and it confuses a lot.
     
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