# Wavefunction probability nodes within 2s electronic orbitals

1. Jun 26, 2013

### Salman2

I have some questions based on the figures provided on this helpful educational link:
http://chemed.chem.wisc.edu/chempaths/GenChem-Textbook/Orbitals-896.html

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For discussion, let us work with two isotopes, He-3 (ppn) and Li-6 (pppnnn). Each (p+) would have an electron (e-) associated with it.

For He-3, the two electrons would have singlet spin, thus {e-$\uparrow$}{e+$\downarrow$}. They are both in 1s orbital, and circular images of this 1s orbital are shown in the above internet link. For discussion, the two protons of He-3 would likewise have singlet spin within the first 1s nuclear energy shell {(p)$\uparrow$(p)$\downarrow$}, but these are not presented in the internet link.

For Li-6, a third (-e) is added to the 2s orbital and an image of the 2s orbital is provided in the internet link, which clearly shows a dense 1s central orbital surrounded by a diffuse electron cloud that I assume would be associated with the third (p) present in Li-6 that is not found in He-3. Within the nucleus of Li-6 the third (p$\uparrow$) likewise moves to a higher energy shell, the 1p, in a similar way that the third (e-) moves to a higher energy state, the 2p orbital.

Next, there is a comment in the text of the above internet link that within the electron 2s orbital of Li-6 there exists a 'node' where the quantum wavefunction of all three (e-) have 0.0 amplitude, thus 0% probability of existence at that specific state space. So, if true, within the Li-6 isotope there must exist a quantum state space (associated with the node) where none of the three (e-) are present. Likewise, none of the more central energy shells associated with the three (p) wavefunctions would be present at the location of the 2s electron orbital 'node', given that it is well removed from the central nuclear energy potential.

I think I have the above correct, but please let me know if I error. OK, now for my questions.

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Question #1: Does the above mean that there exists a state space at the 2s electron orbital 'node' within Li-6 where the HUP would not apply ? I mean, if the wavefunction probability for the three electrons = 0.0 amplitude, then would we not know with 100% probability the momentum and position simultaneously for all three (e-) within Li-6 at the 'node' state space ? Would this be a correct quantum interpretation of the 'node' within the 2s orbital of Li-6 isotope ?

Question #2: Is the existence of the 'node' within the 2s orbital of Li-6, where none of the three (e-) exist, caused by repulsive EM force between the two (e-) in the 1s orbital and the single (e-) in the 2s orbital ? And if yes, why is this repulsive EM force between electrons (e-) stronger than the attractive force between three protons (p) and electrons that would tend to bring the third (e-) in the 2s orbital closer to the nuclear center potential, and thus occupy the 'node' state space with at least a small probability amplitude ?

Question #3: My final question relates to the third (p) within Li-3, which moves to a higher energy shell, the 1p shell (as opposed to the first two p found in the 1s shell), in a similar way the third (e-) in Li-6 moves to a higher 2s energy orbital. Is it possible that the third (e-) found in the higher energy 2s orbital of Li-6 is directly associated via EM attraction to the third proton (p) in the 1p nuclear energy shell, and not the two protons associated with the lower energy 1s nuclear shell ?

Last edited: Jun 26, 2013