L = 0 (s) orbital potentials V(r) as n increases - why are...

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The discussion centers on the behavior of L = 0 (s) orbital potentials V(r) as the principal quantum number n increases. It is established that for l = 0, the effective potential is solely influenced by the Coulomb potential, which is expressed as ~ -k/r^2. The presence of nodes in the wavefunctions is explained by the relationship between kinetic energy and wavefunction curvature, where increased curvature results in more nodes. Specifically, for n = 2, there is 1 node; for n = 3, there are 2 nodes, illustrating a clear pattern as n increases.

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applestrudle
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... why are there more nodes/zeros?

If l = 0 then the angular momentum contribution to the effective potential is zero, and there is the coublomb potential only. So shouldn't it always go as ~ -k/r^2 (k = constant) like the n=1 s orbital?

Why is it that for n = 2 is there 1 zero, n= 3 there is 2 zeros, etc?

Thanks
 
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This has nothing to do with the potential. Even for a particle in a box where there is no potential, the excited wavefunctions have nodes. This relates to kinetic energy being the higher the more curved the wavefunction. Increasing curvature necessarily leads to the appearance of nodes.
 

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