- #1
mistertaylor9
- 2
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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.htmlThe 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 shouldn't 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 isn't the case.
My only clue for the answer to my question is that maybe angular nodes are not represented by x-intercepts. Help meh!
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.htmlThe 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 shouldn't 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 isn't the case.
My only clue for the answer to my question is that maybe angular nodes are not represented by x-intercepts. Help meh!
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