This question is directed at resident theoretical electron physicists in PF Regarding suborbitals L 0 _{I} → L 0 _{VII} corresponding to Periods 1 - 7 There are a total of 14 electron positions within Azimuthal Quantum Energy State 0 in Periods 1 through 7 Additionally, given that Azimuthal Quantum State 0 represents a 0° orbital plane angle along an x-z axis, and a limit of 1 electron pair per L 0 suborbital; This would then require 7 discrete coaxial suborbital paths for L 0 without violating the Pauli exclusion principle. Figure 1 illustrates 7 coaxial L 0 suborbital paths and 14 electron standing wave positions, numbered L 0 _{I} → L 0 _{VII}, with 4 electron pairs aligned along the z axis and 3 electron pairs aligned along the x axis. Question 1 : Are the x-z axis suborbital paths shown in Figure 1 in agreement with the current standard model? Question 2 : Are the electron standing wave positions shown in Figure 1 in agreement with the current standard model? (assuming the answer to Question 1 is Yes). Thanks in advance ~ E:S
2 obvious errors : Firstly, x - z axis should be x - y axis Secondly, The electron positions in Figure 1 (above) should be aligned with the y axis since the electron positions are actually (+) and (-) nodes in standing waves, rather than independent particles as illustrated in the correct version of Figure 1 (below)
L=1ħ, 2ħ, 3ħ +/- nodal point charges This question is directed at resident theoretical electron physicists in PF Regarding Azimuthal Quantum Energy States : L+1ħ, +2ħ, +3ħ → L -1ħ, -2ħ, -3ħ The point charges for each (+) and (-) node are presumably counted as discrete electrons which would result in : 3 suborbital shells (x 2) for L +1ħ and -1ħ respectively, with 6 nodes (electron point charges) per suborbital shell a total of 36 electron positions for Periods 2 → 7 L +1ħ and -1ħ illustrated in Figure 2 and Figure 3 2 suborbital shells (x 2) for L +2ħ and -2ħ respectively, with 10 nodes (electron point charges) per suborbital shell, a total of 40 electron positions for Periods 4 → 7 L +2ħ and -2ħ illustrated in Figure 4 and Figure 5 and 1 orbital v (x 2) for L +3ħ and -3ħ respectively, with 14 nodes (electron point charges) per orbital shell a total of 28 electron positions for Periods 6 → 7 L +3ħ and -3ħ illustrated in Figure 6 and Figure 7 Question : Are the point charges at each standing wave node counted as discrete electrons, or are the number of suborbital paths double the number illustrated in Figures 2 → 7 ? Thanks in advance E:S
L=0 Suborbital Configuration This question is directed at resident theoretical electron physicists in PF Regarding suborbitals L 0 _{I} → L] 0 _{VII} corresponding to Periods 1 - 7 A total of 14 electron positions exist within Azimuthal Quantum Energy State 0 in Periods 1 through 7 Additionally, Azimuthal Quantum State 0 represents a 0° orbital plane angle along an x-y axis, and a limit of 1 electron pair per L 0 suborbital; This suggests 7 discrete coaxial suborbital shells for L 0 without violating the Pauli exclusion principle, although there is no mention of these 7 discrete coaxial suborbital shells in any physics textbook known to me. Figure 1 illustrates 7 coaxial L 0 suborbital shells and 14 electron positions, numbered L 0_{ I} → L 0 _{VII}, with 7 electron pairs aligned along the y axis, each electron representing 1 standing wave nodal loop, and the crest of each node being 1 negative point charge. Question : Are the 7 suborbital shells illustrated in Figure 1 in agreement with the current standard model? Thanks in advance ~ E:S
What do you mean by suborbitals ? And what does [itex]L0[/itex] mean? the n=2 shell with 0 angular momenta? Also what does Azimuthial Quantum Energy State mean (I haven't heard of azimuthial energy before- the energy is not direction dependent)? Please fix your terminology... Please give a link if you want to ask something in particular... The standard model does not study the atomic structure, so in any case the question about the SM has little (if not nothing) to do with this topic.
L 0 refers to Quantum Number 0 typically known as s shell (ground state) (See attached image - vectors.png) I use Quantum Number rather than the outdated and misconceived sharp, diffuse, principle and fundamental convention. There are 14 electron positions in Periods 1 → 7 These 14 positions require 7 L 0 sub-shells This is more clearly explained through the shell structure of element 118 Ununoctium 0^{2}→0^{2} 1^{6} → 0^{2} 1^{6} 2^{10} → 0^{2} 1^{6} 2^{10} 3^{14} → 0^{2} 1^{6} 2^{10} 3^{14} → 0^{2} 1^{6} 2^{10} → 0^{2} 1^{6}
and how exactly are you assigning 14 electrons to an s? The s is characterized by L=0 (zero orbital angular momentum, so Lz=0 too). Then for the electrons you have their spins being 1/2, so the most you can put to an s-shell is 2. Except for if you mean different s shells. In that case I still don't get your question. Also the Lz=0 doesn't mean that your electron wavefunction is on a plane. The s=0 is better seen as a sphere (so there is no angular dependence -Lz is the generator of rotations- on the probability wavefunction)
This is quantum physics, I moved the thread and merged it with the others, the problem seems to be the same in both cases. There are no "paths" or "positions" for electrons in an atom (or for any other quantum-mechanical object), and no "electron pairs aligned along [something]". The questions assume the existence of something that does not exist, so it is impossible to answer them.
Simply by following the Aufbau principle : ending at the shell structure of element 118 Ununoctium 0^{2}→0^{2} 1^{6} → 0^{2} 1^{6} 2^{10} → 0^{2} 1^{6} 2^{10} 3^{14} → 0^{2} 1^{6} 2^{10} 3^{14} → 0^{2} 1^{6} 2^{10} → 0^{2} 1^{6} The arrows (→) indicate Aufbau per Period
Those 14 electrons correspond to the principal quantum numbers n=1, n=2 and so on, with 2 electrons of opposite spin in each orbital. They have different radial distributions of their orbitals. Also see my previous post if you missed it in the merging process.
I am sorry if this is going to appear more than once, but for some reason I couldn't find my answer... Then your question is answered by: Lz=0 doesn't mean that your electron wavefunction is on a plane. The s=0 is better seen as a sphere (so there is no angular dependence -Lz is the generator of rotations- on the probability wavefunction)
Actually, the question has completely changed please revert the topic to it's original form and delete the previous topic The Topic has become a complete mess the images and questions have been revised since the original topics posted days ago the original topics from the Atomic Forum should have been deleted. Or.. better yet - Please delete the whole topic and I can repost Thanks
The question has changed, but I think the problem is the same in both cases. Do you know how atomic orbitals look like? There is no such thing as a "path", or "position". It is a 3-dimensional probability distribution.
Sorry, but the Topic has become so distorted and jumbled it no longer makes any sense the fundamental questions have changed and the images have changed. So the Topic in it's jumbled form is useless to me and impossible to reply to for óthers
The problem is that the original 2 Topics posted in the Atomic Forum were reposted in the Quantum Physics Forum in a completely new form with revised questions and images And now the reposted (corrected) Topics have been deleted and replaced with the 2 previous (obsolete) Topics which have been merged together into 1 useless topic, and the result is complete confusion. I have reposted the new (lost) topics as a single Topic : https://www.physicsforums.com/showthread.php?p=4799033&posted=1#post4799033 This topic can now be deleted, and the thread restarted in the new topic.