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

In summary, the number of nodes in an orbital is not determined by the potential, but rather by the curvature of the wavefunction. Higher curvature results in more nodes, regardless of the potential. This explains why there are more nodes for higher values of n.
  • #1
applestrudle
64
0
... 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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  • #2
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.
 

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

Question 1: What does L = 0 (s) orbital mean?

L = 0 refers to the orbital angular momentum quantum number, which describes the shape of an electron's orbital. The s orbital, or "sharp" orbital, has a spherical shape and L = 0 means that it has no orbital angular momentum.

Question 2: How does increasing n affect the potential energy of the s orbital?

As the principal quantum number, n, increases, the s orbital's potential energy decreases. This is because the electron is located further from the nucleus, resulting in a weaker attraction and therefore a lower potential energy.

Question 3: Why are s orbital potentials V(r) more stable as n increases?

S orbitals become more stable as n increases because the electron is located further from the nucleus, resulting in a weaker attraction and a lower potential energy. This lower potential energy makes the electron less likely to be ionized or excited to a higher energy state.

Question 4: How does the shape of the s orbital change as n increases?

The shape of the s orbital does not change as n increases. It remains spherical, but the size of the orbital increases with higher values of n. This means that the probability of finding the electron further from the nucleus increases as n increases.

Question 5: Why do s orbitals with higher n values have a larger radius?

The radius of an s orbital increases with higher n values because the electron is located further from the nucleus. As the distance between the electron and nucleus increases, the orbital becomes larger and has a higher probability of containing the electron at any given time.

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