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|WikiProject Elements||(Rated Start-class, Low-importance)|
Image in French article
There is an image of Janet's table in the French article fr:Charles Janet which would greatly improve this article by showing what his table looked like. However the image is generated by the French template fr:Modèle:Tableau périodique de Charles Janet which has a rather lengthy source code. Would importing the image into this English article require translating the whole source code and finding filenames equivalent to the French ones? Can someone figure out how to do this efficiently? Dirac66 (talk) 19:58, 31 March 2010 (UTC)
It is interesting to note that he was an Engineer. So even Engineers can learn something by thinking about the details of the periodic table.WFPM (talk) 17:44, 18 April 2010 (UTC) Thank You! Merci beaucoup!
Janet table as memory aid and predictor
There is an interesting way in helping to memorize the sequence of the elements of the Janet Periodic table. It involves the following format of the 8 series divisions: It goes:
2 = 2 2 = 2 2 4 2 = 8 2 4 2 = 8 2 4 4 2 4 2 = 18 2 4 4 2 4 2 = 18 2 4 4 4 2 4 4 2 4 2 = 32 2 4 4 4 2 4 4 2 4 2 = 32
Now does anybody want to argue against the idea that the next 2 (9th and 10th) series will go:
2 4 4 4 4 2 4 4 4 2 4 4 2 4 2 = 50 2 4 4 4 4 2 4 4 4 2 4 4 2 4 2 = 50
Comments as requested:
1. I agree that the Janet table is a useful memory aid for the known atoms.
2. The prediction of 50 for the next two rows is basically an extrapolation of the quantum energy levels from known (1-118) to unknown atoms. I think it should be valid, although relativity is known to play a role in very heavy atoms and could lead to some changes (which I don't know enough about to discuss in detail).
3. Verifying this prediction will require discovering the next 100 elements, so unfortunately neither you nor I is likely to live long enough to learn the experimental answer. Best wishes. Dirac66 (talk) 02:31, 7 August 2010 (UTC)
- I appreciate your quick reply and comment. However, I think you're being very diplomatic in neglecting to mention that the Janet Table format's first advantage is in its solving the problem of where to locate elements 119 and 120, which are within the present day scope of potential occurrence. Then, of course, the occurrence of 50 elements in each of the next 2 levels is probable in both circumstances, and, of course, the physical limitations of the nucleon accumulation process become the practical accumulation factor in either case. However, I like this set of ideas better as a geometric concept as indicated in my models, rather than as a mathematically based concept.WFPM (talk) 15:19, 8 August 2010 (UTC)See Talk:Nuclear model.WFPM (talk) 15:43, 20 January 2011 (UTC)
- Right!! But that implies that that they're the last part of the 8th structural series of the atom, as per the indications of the Janet table, rather than being the beginning of a new series, as implied by the Standard and IUPAC tables. And thus the controversy.WFPM (talk) 16:21, 18 January 2011 (UTC)
- I guess it depends upon whether you want to be able to visualize the nucleus as a 3-dimensional structural entity, or else just an accumulated cloud of nucleons.WFPM (talk) 01:03, 19 January 2011 (UTC)
- As a structural concept, you can visualize the first series as the creation on the nucleus of a planar atom of EE2He4, and then the second series as the creation of a second planar nucleus on top of the first. But the third series requires that the previous 2 planar nuclei be first surrounded by 6 side bonded deuterons before the third planar alpha nucleus is again added to the top. And so on, until the eighth series ends with the eighth planar nucleus being added to the top to add up to a total of 120 accumulated deuterons, with 16 of them having been converted into the planar alpha configuration. See Talk:Nuclear model
- Knowledge of the atom's atomic structure is important, of course, for all the elements, but is particularly important for certain key elements, like that of EE6C12 so that the process of atomic nucleus creation by nucleon accumulation can better be understood and detailed.WFPM (talk) 15:37, 20 January 2011 (UTC)
- Since you don't mention about the 2 4 -- subdivision logic re the groups of the table, I will note that as the model structure is created, it was noted that each group had at first 2 equal value locations for their accumulation locations on the structure. Then the next 4 elements likewise involved a greater, but equal distance location (from the center location), as did each of the succeeding 4 element additions. This geometric fact also adds to the mental ability to visualize and thus remember these sequences of additions. This might be considered to result in the additional subdivisions to have an increase in retained free energy. It also might result in a decrease in the stability factor in accordance with your argument about the increase in instability due to the electrostatic charge accumulation concept. Anyhow, I certainly thank you for your reply.WFPM (talk) 15:19, 8 August 2010 (UTC) See Alternative periodic tables
- Another feature of the format of the Janet table is to note that since the first 2 rows of the table feature the first two alpha particles of the first two rows, it is easy to imagine that the last 2 of each row also features the creation of an alpha particle within the nucleus. And therefor the the atom consists of an accumulation of nucleons around a core of up to 8 alpha particles. And that the remaining nucleons are those needed after the creation of each alpha particle to stabilize the expanding nuclear structure plus provide a base for the creation of the next successive alpha particle. In this context, the indication of the table would be that the first 2 plus 4 elements of row 3 would be accumulated around the first 2 alpha particles, and then the 3rd alpha particle would be created by the accumulation of the 2 deuterons of 11Na22 and then 12Mg24 to create the third alpha particle. This calls into question the ability of the triple-alpha fusion process to actually create a 3rd alpha particle with in the atomic nucleus, since the 3rd accumulated alpha particle would be assimilated within the nucleus as 2 alpha particles, with each helping bond the first 2 alpha particles together. Do you have any comments about this idea? Regards.WFPM (talk) 01:52, 1 December 2010 (UTC)
It is true, than the first period has structure: H, He, Li, Be
Physical and chemical properties of elements are more important parameters, than an external electronic configuration of atom of elements!
Clarification: Hydrogen is the likeness of halogens; Helium is certainly a noble gas; Lithium is an alkali metal; Beryllium is an alkaline earth metalloid. Alex makeyev (talk) 10:37, 25 September 2015 (UTC)
I would argue that the elements 1 H and 2 He make up the first period, and that the elements 3 Li and 4 Be make up a second period, with each period resulting in the creation of an alpha particle within the nucleus.WFPM (talk) 15:02, 15 September 2011 (UTC)
Note that that's the beginning of the proposed triple alpha accumulation process, with a very low probability of occurrence, and with the alternative possibility that the first two alpha particles are bound by accumulated deuterons (up to 6 at 10 Ne. and then the 3rd alpha particle is created by 11 Na and 12 Mg.WFPM (talk) 15:12, 15 September 2011 (UTC)
Hydrogen is not metal, chemically active gas. Its electronic structure is 1s1. But all subsequent elements with electronic structure ns1 are chemically active alkaline metals.
Hydrogen stands before helium, chemically inert gas, having electronic structure 1s2. But all subsequent elements with electronic structure ns2 are chemically active alkaline Earth metals.
All subsequent elements facing noble gases are chemically active gases or flying substance halogen. A typical Halogen have electronic structure np5. A typical Noble gas have electronic structure np6.
Helium is allowed to be noble gas which helium is in a reality.
It means also that the hydrogen is necessary to be allowed to be chemically active gas, not metal, especially not alkaline metal. As hydrogen is chemically active gas, not metal in a reality. Alex makeyev (talk) 18:04, 15 September 2011 (UTC)
I think that the important point is whether the growing size atomic nucleus is acquiring a physical structure or not and what its physical properties and associated physical activities are rather than its chemical activities, and that is the importance of the indications of the Janet Periodic table.WFPM (talk) 00:35, 16 September 2011 (UTC)
Force hydrogen to become alkaline metal, as all the others ns1 elements. Force helium to become alkaline Earth metal, as all the others ns2 elements. Alex makeyev (talk) 11:54, 16 September 2011 (UTC)
- Not at all, Alex. In the standard Periodic Table, Hydrogen, although not an alkaline earth metal, is already grouped as an ns1 element above Lithium! The only controversial change is grouping Helium as an ns2 element above Beryllium. As long as Helium is colored the same as Argon and the other nobel gases this should not be a catastropic issue. Let's face it, Element 118 (Ununoctium or Uuo) is grouped as an nd6 element under Radon in both tables despite not resembling any of the other noble gases - at least so far. — Glenn L (talk) 13:01, 16 September 2011 (UTC)
Type of complex physical and chemical properties, which exhibit collectives of atoms of the same element have a more important parameter than the structure of the External area of the electron cloud of atoms of the element. Alex makeyev (talk) 10:37, 25 September 2015 (UTC)
Yeah! so 118 is the last of the noble gasses and the question is what the atom will make after that? And the answer is more alpha particles, with 119 and 120, which made up only the 8th alpha particle within the nucleus. So each series line in the Janet table ends with the creation of an alpha particle and requires 4 more elements (every other series) to get the structure set up continue to do that.WFPM (talk) 18:48, 16 September 2011 (UTC)
- And the main difference between the 2 tables is as whether the original series begins with the sequence 2, 8, 8, (IUPAC table) or 2, 2, 8, 8, (Janet table). And on a structural basis, beginning with the atom EE4Be8, The whole thing boils down into whether the next 2 deuterons go on either individually, to make firstly OE5Be10 and secondly EE6C12, or else both together to make the same end process atom EE6C12, but with a different configuration. And it should be noted that to construct a 6 element deuteron wrap around the EE4Be8 structure, the minimum energy addition level is to first install a deuteron on one side to make an unbalanced OE nucleus, and then a second deuteron on the other side (same axial direction) to make a rebalanced EE Nucleus. Thus, a 6 (deuteron) addition to the nucleus is involved with a 2 + 4 = 6 deuteron addition as indicated in the Janet table. and the original organization of series additions is better understood as being the sequence 2, 2, 8, 8, in the Janet table than as being the sequence 2, 8, 8, of the IUPAC table.WFPM (talk) 19:37, 24 April 2012 (UTC)
If you desire to build a structural model, with incremental additions similar to that of the Janet Periodic table, you need to use a checkerboard, which is a nested set of increasing square spaces, and which is a central 2 x 2 = 4 square followed by a 4 x 4 = 16 second square and then a 6 x 6 = 36 third square and finally followed by a 8 x 8 = 64 outer square. Then the squares are filled in the sequence inside out, with 2 sets of 4 units in the center squares,= (2 x (2=2)), and then an additional 2 layers of 16,= (2 x (2+4+2)) and then 36, (2 X (2+4+4+2+4+2))and finally 64,= (2 x (2+4+4+4+2+4+4+2+4+2)) units, which adds up to an addition of 120 pairs of unit additions as needed to simulate the 120 pairs of nucleons needed to be accumulated by the simulation model. This will result in a model of all the possible A = 2Z isotopes, (up to Z=120), And then the other isotopes are created by adding "extra neutrons" to the neutron locations as needed to keep the model in a state of maximum rotational stability.WFPM (talk) 00:33, 21 August 2012 (UTC)
- So the deuterons of the atom are accumulated in a series of layers of subsequent accumulation on top of the previous, and the relationship of the side by side positioning of the subsequent layers to the previous layers is controlled by the compatibility requirements of both the direction of rotation (side by side rotation has to be opposite to be compatible), and also by the interaction of the magnetic force properties of the nucleons, which results in the positioning of a new outside layer addition to an atomic structure at a level such as to straddle the midpoint of the level of conjunction of the previous two layers. This results in the accumulation process being that of the creation of an increasing size octahedral (8 sided) structure.WFPM (talk) 02:45, 27 August 2012 (UTC)
I posted here The system of natural cycles automatisms vacuum and atomic levels of matter, vertical form by Julius Lothar von Meyer (1862, 1864, 1870); Dmitri Mendeleev (1869, 1870, 1906); Ernest Rutherford (1913); Niels Bohr (1913); Henry Moseley (1913); Charles Janet (1928); Alexander Makeyev (1999, 2010, 2012). And The System of a Natural Cycle of automatisms vacuum and atomic levels of matter, the spiral wave-particle form, by Alexander Makeyev (2000, 2010, 2012).
Here is clearly shows that every correct period (Natural Cycle) of the natural sequence of elements of atomic levels of matter (img. 1) is half the circumference of spiral elements (img. 2).
Shell numbers in the Janet periodic table
The article says, “Janet's table differs from the standard table in placing the s-block elements on the right, so that the subshells of the periodic table are arranged in the order (n-3)s, (n-2)p, (n-1)d, nf from left to right.” Are the shell numbers n–3, n–2, n–1, and n correct? Please explain with an example, referring to element names, too. Solo Owl 01:59, 22 December 2014 (UTC)