Timeline for Distance of Orbitals from Nucleus relation with Energy level and penetration power
Current License: CC BY-SA 4.0
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| Jun 8, 2023 at 16:50 | comment | added | bm27 | So considering all the factors including the spherical coordinates what would be the size comparison for the subshells of the 3rd shell? | |
| Jun 8, 2023 at 9:29 | comment | added | Stephen Elliott | @Banaj Mahajan : please see my updated answer to your question. The ionization-energy picture is not clear from just the radial-coordinate plot. Plots versus the 3 spherical coordinates are necessary to get a more complete picture about contributions to the energy aside from the radial-distribution contribution to the ionization-energy. I suspect that a good part of the periodic table can be calculated straight-forwardly, but I have not had the time yet to take on this expanded task versus just the Helium calculations that were very successful. | |
| Jun 8, 2023 at 9:20 | history | edited | Stephen Elliott | CC BY-SA 4.0 |
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| Jun 7, 2023 at 8:10 | history | edited | Stephen Elliott | CC BY-SA 4.0 |
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| Jun 7, 2023 at 8:03 | history | edited | Stephen Elliott | CC BY-SA 4.0 |
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| Jun 7, 2023 at 7:54 | history | edited | Stephen Elliott | CC BY-SA 4.0 |
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| Jun 7, 2023 at 4:17 | comment | added | bm27 | How would the energy of S-shell be lower if its radius is greater than that of D-shell, wouldn't it be greater as further from the nucleus the magnitude of Columbic attractive force would decrease and hence the energy would decrease, as closer to the nucleus more negative the energy is and hence smaller. I'm not sure the energy we are talking about is the same but I think they have the same magnitude, as the energy the electron has in an orbital should be the same as its binding/ionization energy. | |
| Jun 6, 2023 at 12:00 | comment | added | Stephen Elliott | If the S-shell radius is larger than the D-shell radius, that would put the S-shell-energy conceptually lower. It is the nucleus that binds the electron; so if the distance is expected to be further from the nucleus, then the nucleus's effect is weaker. The energy you are talking about is the binding energy to the nucleus. Far away from the nucleus, this energy is Hydrogen-like, corresponding to $e^{+1}$. Closer to the nucleus this binding energy is greater because of the lower shielding of the other electrons. Please keep on asking be questions if this is not clear. | |
| Jun 6, 2023 at 11:52 | history | edited | Stephen Elliott | CC BY-SA 4.0 |
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| Jun 1, 2023 at 17:08 | comment | added | bm27 | thanks again for your answer but I think you misunderstood my question (or maybe I haven't explained it well), I'm not bothered with the size comparison of orbitals of different shells (K,L,M...) , my confusion arises from the fact that according the average radius formula mentioned in my question and also the plots (even in the reference you shared) depict that the $s$ subshells is larger than $d$ for the same $n$ value, despite the fact that it has less energy and less penetration power. Is there a reason for that or the information provided to me is false? | |
| Jun 1, 2023 at 12:58 | history | edited | Stephen Elliott | CC BY-SA 4.0 |
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| Jun 1, 2023 at 12:56 | history | edited | Stephen Elliott | CC BY-SA 4.0 |
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| Jun 1, 2023 at 12:54 | history | edited | Stephen Elliott | CC BY-SA 4.0 |
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| S Jun 1, 2023 at 12:53 | review | First answers | |||
| Jun 1, 2023 at 17:55 | |||||
| S Jun 1, 2023 at 12:53 | history | answered | Stephen Elliott | CC BY-SA 4.0 |