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P-subshell orbital of Hydrogen

  1. Feb 12, 2013 #1
    I'm trying to understand electron orbitals. I'm a beginner. I'm looking at the shapes of p orbitals in a chemistry book, and it's talking about Hydrogen. I'm surprised to see that the orbitals are not spherically symmetric. How can that be when the nucleus is spherically symmetric? Can a hydrogen atom with one electron be excited so that its electron is in one of these orbitals that are not spherically symmetric? If so, how? That would seem to suggest there is some preferred rectangular coordinate system on the hydrogen atom.
     
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  3. Feb 12, 2013 #2

    DrClaude

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    It can depend on what caused the excitation. For instance, a photon has a direction of propagation and a polarization, which in effect break the spherical symmetry.

    In a case where the spherical symmetry should be preserved, you have to consider that the electron will not be in a definite p orbital, but in an equal superposition of all three p orbitals. A bit of math should convince you that the sum of spherical harmonics
    [tex] Y_{1,-1} (\theta, \phi) + Y_{1,0} (\theta, \phi) + Y_{1,1} (\theta, \phi)[/tex] leads to a probability distribution for the electron that is independent of [itex](\theta, \phi)[/itex].
     
  4. Feb 12, 2013 #3

    mfb

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    Space has the same laws of physics in all directions, but this does not mean that it particles in space have to be the same everywhere - on a larger scale, humans are not spherical, for example.

    Not if you fill all orbitals with electrons, or have superpositions of those orbitals with less electrons. Those shapes are mathematical solutions - it does not mean that you have to have an electron orbital in that shape.
     
  5. Feb 12, 2013 #4

    DrClaude

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    It can depend on what caused the excitation. For instance, a photon has a direction of propagation and a polarization, which in effect break the spherical symmetry.

    In a case where the spherical symmetry should be preserved, you have to consider that the electron will not be in a definite p orbital, but in an equal superposition of all three p orbitals. A bit of math should convince you that the sum of spherical harmonics
    [tex] Y_{1,-1} (\theta, \phi) + Y_{1,0} (\theta, \phi) + Y_{1,1} (\theta, \phi)[/tex] leads to a probability distribution for the electron that is independent of [itex](\theta, \phi)[/itex].
     
  6. Feb 12, 2013 #5
    Just to clarify, [tex] Y_{1,-1} (\theta, \phi)^2 + Y_{1,0} (\theta, \phi)^2 + Y_{1,1} (\theta, \phi)^2[/tex] leads to a probability distribution for that independent of [itex](\theta, \phi)[/itex].
     
  7. Feb 12, 2013 #6
    @mfb I was taking it as axiomatic that a proton is spherically symmetric. I think it follows that a proton cannot have properties that make some direction special. The drawings in my chemistry book made it looks like these p orbitals designate special directions.


    I think the comments will help my understanding, but I'll give it some time to digest. In particular, I had not considered a quantum superposition. (As I said, I'm only a beginner.)

    A hypothesis to test whether I've understood you: an electron could be in a quantum superposition of these orbitals. If we want to observe it in one of them, that forces it to be in just one of them (if this is possible---I don't know if that can or cannot be done in a lab), but the measurement we did would introduce the asymmetry in the problem that accounts for the orbital not being spherically symmetric.

    Thanks for your help.
     
  8. Feb 12, 2013 #7

    mfb

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    Neglecting its spin (where you get the same thing, just smaller), this is true.
    Well, that is a mathematical solution.
    Right
     
  9. Feb 13, 2013 #8

    DrDu

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    I don't understand what's so surprising, as this is also the case in classical mechanics.
    The motion of the planets around the sun takes place in the ecliptic although the potential of the sun is spherically symmetric.
     
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