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Orbital Nodes (Diagrams of Psi/Psi^2)

  1. Sep 17, 2011 #1
    I am having a hard time understanding the nodes on the graphs of Ψ for atomic orbitals. Take for example, the graph of Ψ for a 2s orbital as seen on the left here:

    http://winter.group.shef.ac.uk/orbitron/AOs/2s/wave-fn.html

    The graph intersects at two points, but these two intercepts represent the same spherical node.

    Now consider the graph of Ψ for a 5d orbital.

    http://winter.group.shef.ac.uk/orbitron/AOs/5d/wave-fn.html


    The number of of radial nodes is 2 (from equation n-l-1) and the number of angular nodes is l=2. If one radial node causes the graph to cause the x-axis twice (on each side of nucleus, as in the case of the 2s orbital) then shouldnt there be 2 nodes * 2 intercepts/node = 4 intercepts just for the angular nodes? And then 2 more intercepts for the angular nodes? But the graph hits the x-axis at 5 (instead of 6) separate points which would imply 5 nodes but this isnt the case.

    My only clue for the answer to my question is that maybe angular nodes are not represented by x-intercepts. Help meh!!!
     
    Last edited: Sep 17, 2011
  2. jcsd
  3. Sep 17, 2011 #2

    diazona

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    Homework Helper

    What they're plotting on the (2D) graph in those figures is a cross section of the wavefunction. The cross section is taken along a radial line, i.e. a line that passes through the origin. So what you're seeing in the 2D graph is just the radial part of the wavefunction. The angular part is just a scaling factor.

    In other words, you know that the wavefunction can be decomposed as
    [tex]\psi(r,\theta,\phi) = R(r)Y(\theta,\phi)[/tex]
    where R and Y are the radial and angular parts, respectively. For a radial line, [itex]\theta[/itex] and [itex]\phi[/itex] are fixed, so [itex]Y(\theta,\phi)[/itex] is just a constant for any given radial cross section. If you switch to a different radial cross section, though, it would have a different value. For example, in the 5d orbital you showed, the cross section was taken along the line [itex]x = y[/itex]. If they had used [itex]x = 3y[/itex] instead, the 2D graph would be smaller, but it would still have the same shape.

    There are some particular angles at which the scaling factor [itex]Y(\theta,\phi)[/itex] is zero; in this case, [itex]\phi = \{0,1,2,3\}\pi/2[/itex]. If they had plotted a cross-section corresponding to one of those angles, the entire graph would be zero. Those angles are the angular nodes. You're correct that the angular nodes are not represented by x-intercepts.

    By the way, since the cross sections they show in those graphs are taken along a full line, not a ray emanating from the origin, they actually show the radial wavefunction twice. So I'd actually suggest that you look only at the right half of the graph; the left half is just the mirror image. The center of the graph, of course, corresponds to the center of the atom, which is why a node at the center isn't part of a pair.
     
  4. Sep 17, 2011 #3
    You have cleared a lot up for me. Thank you!
     
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