Element Formation By H, D, He, C And O Building Blocks Hydrogen .
Johann Marinsek XII 2015 foundationsofphysics.org Cab contact@foundationsofphysics.org Element formation by H, D, He, C and O building blocks Hydrogen: magnetic coupling of proton and electron Bonding of atoms due to magnetic coupling Neither nucleus nor electron orbital! Empirical evidence for low energy fusion Periodic table due to element formation and configuration of atomic building blocks Refutation of quantum mechanics based periodic table Periodic pattern is not due to electron valence configuration. Extra nuclear electrons don’t exist. Hydrogen consists of a proton and an electron that are magnetically coupled. Elements are fusion products of hydrogen, deuterium, helium, carbon and oxygen building blocks. Helium consists of four hydrogen atoms. Neutrons are not elementary nuclear particles. “Neutrons” are excited hydrogen atoms. During the fission of an element, hydrogen atoms can be decay products. They are excited and decay: H* p e-. The atom is not a composition of a nucleus and an electron shell. Nuclei don’t consist of protons and neutrons. The compact atom is a structure of hydrogens. Moseley’s experimental data of X-rays spectra are not a proof for the existence of Z protons in nuclei where Z is the atomic number in the periodic table. How were the various elements formed? The most common production line by fusion is the assembling of an element given and helium to create a new element. For example: 52 Cr 4He 56Fe ; 56 Fe 4He 60Ni Superscripts indicate the number of hydrogens that constitute the element. The new periodic table is based on the aufbau process by consecutive fusions of additional helium atoms. Periods, physically and chemically similar elements are due to the succession of fusions and the resulting atomic configurations of the elements. Moseley ignored that not all isotopes (35Cl and 37Cl for example) of an element can exhibit the same X-ray spectrum because that is due to the unique atomic number Z for the element (Z 17 for Cl). Obviously, the number of atomic building blocks (H’s) and their architecture determine frequencies and not the hypothesized proton or electron number Z! 1
The alkali elements have an increasing mass number but the chemical behaviour remains unchanged because their surface remains the same. So it is assumed that this mass increase is nothing but an increase in oxygen building blocks that are chemically inactive. There is a core under a chemically active surface. 35 Cl and 35Cl as well as the alkalis Na, K, Rb, Cs are chemically equivalent. For example the combinations NaCl, KCl, CsCl are possible. When 35Cl and 35Cl are treated as isotopes and not as elements then one can also the alkalis treat as isotopes. Only their surface is chemically equivalent, their core comprises different numbers of oxygens as building blocks. If the alkalis are treated as elements, then also 35Cl and 35Cl can be treated as elements as well as 63Cu and 65Cu. Moseley’s claim that atomic numbers determine the number of protons is pointless. It is the number of hydrogens (or mass number A) that determines the atoms Element formation due to quantum mechanics untenable In the chapter Origin of the Elements produced by the Nuclear Science Division ---- Lawrence Berkeley National Laboratory, the graph depicted above shows the formation of the elements according to current physics. l How were the numerous elements of our cosmos formed? Current QM theory claims that the infernal heat of stars delivers the energy necessary to form nuclei by fusion. For our sun Nuclear Science Division (chapter 10) claims: The basic reaction chain (86% of the time) is the fusion sequence: 1 H 1 H 2 H e υe 2 H 1H 3He γ 3 He 3He 4He 1H 1H These fusion reactions occur only at the centre of the Sun where the high temperature ( 107 K) gives the hydrogen and helium isotopes enough kinetic energy to overcome the long-range repulsive Coulomb force and come within the short-range of the attractive strong nuclear force. The processes described above produce only nuclei of the elements: In the infernal heat of about 107 K atoms cannot have electron shells, neutrons would decay into protons and electrons. The elements cannot pull on their electron jacket Therefore, Lawrence Berkeley National Laboratory physicists claim that the origin of the elements takes place at temperatures of “only” 4000 K, 400.000 years after the big bang! At this time it is assumed that 2
the average energies of nuclei and electrons were low enough to form stable hydrogen and helium atoms. As we can see, this claim is a black box: At the surface of our sun the temperature is about 5800 K. The temperature at which the formation of the elements allegedly functions is 4000 K, not essentially lower than at the surface of the sun. Think for instance at the formation of caesium. The first ionization energy for caesium is only 3,89 eV, i. e. with only 3,89 eV the outermost electron can be removed! The second ionization energy of Cs is 25,1 eV, and so on. Thermal ionization works at temperatures 4000 K. Note how thermal ion sources work: Samples are deposited on specially treated filaments (usually rhenium or tantalum), then carefully dried. The filaments are heated slowly, leading to evaporation and vaporization of the sample. The resulting positively ionized atoms are accelerated from the ion source by an electric field, then mass-separated by an electromagnet. Source: hods/ionization.html Recall that the melting point of a tantalum filament is about 3000 C . Of course, the ionization of the sample takes place earlier at lower temperatures. For instance caesium has the boiling point at 944 K and the critical point at 1938 K. www.wikipedia.org/wiki/Thermal ionization thermal ionization specifies: The hot filament reaches a temperature of less than 2500 degrees Celsius, leading to the inability to create atomic ions of species with a high ionization energy, such as Osmium (Os), and Tungsten (H f -W). It follows from these facts that at 4000 K the orbitals of the elements cannot be formed! Nuclei are not able to bind electrons. Then one must ask the question: How could nuclei arrange their orbitals? Current physics offers no comment on this topic For example, Cs has 55 protons. According to Bohr, the 1s electron has the velocity v Z 2π e2/nh. For hydrogen (Z 1) the velocity is about 3000 km/sec, for Cs (Z 55): v 55 3000 km/sec! How does the nucleus capture the electron? Forget Bohr! According to current theory the 1s electron is not an orbiting one. Why does it not crash into the nucleus? Concerning electron orbital formation there is a black box According to QM, nuclear reactions are described by particle exchanges: An element can lose or receive protons, neutrons and electrons. Consider the often-cited reactions: 2 H 14N 3He 13C Even if we assume that atoms possess their electron jacket in the infernal heat, particle exchange would destroy the electron jackets and the aufbau of new electron jackets according to QM rules is not explainable: N loses a p to get C, H receives a p to get He. These processes must change or destroy the electron shells. Then the new elements possess new shell structures that form themselves according to QM rules? What is really going on is not explainable in terms of QM. 23 Then consider the element formation Na 16O 39K. Neutrons and protons of O join the Na nucleus, 8 electrons join the Na electron orbital according to the rules of QM (Pauli, Hund). The graph from invsee.asu.edu shows for example p-orbitals, they are impossible in the assumed heat. The enormous heat of about 4000 K is not the right environment for the production of K. There must be low energy fusion on Earth that produces K! 3
For example: 23Na 16O electrical energy 39K. Nuclear fusion sequences: the triple alpha process, carbon burning. http://wapedia.mobi/en/Silicon burning process claims: stars with normal mass (no greater than about 3 solar masses) run out of fuel after the hydrogen in their cores has been consumed and fused into helium. Stars with an intermediate mass (greater than 3 but less than about 8-11 solar masses) can go on to "burn" (fuse) helium into carbon by means of the triple-alpha process. α α α 3α . As massive stars contract, their cores heat up to 600 MK and carbon burning begins which creates new elements as follows: C He O, O He Ne (names of the elements stand for their nuclei!) Comment: The cited fusion sequences correspond to nuclear fusions only! For example there is no triple alpha process α α α 3α but He He He C This theory does not and cannot describe how for example the oxygen nucleus is capable of arranging its well defined orbital structure! Of course during the burning time electrons cannot be arranged into orbitals. Recall that the first ionzation energy of O is only 13,6 eV. But even if the nucleus of O is not in a hot environnement, the question remains: How is the arrangement of the electrons possible? Why „cold“ fusion is alledgedly impossible: the assumed positively charged nuclei repel each other Wikipedia claims that Because nuclei are all positively charged, they strongly repel one another. Normally, in the absence of a catalyst such as a muon, very high kinetic energies are required to overcome this repulsion Extrapolating from known rates at high energies down to energies available in cold fusion experiments, the rate for uncatalyzed fusion at room-temperature energy would be 50 orders of magnitude lower than needed to account for the reported excess heat. The assembling of elements that posses neither nuclei nor electron shells If there are no extra-nuclear electrons and if there are no neutrons, then the fusion of elements can be thought of as assembling of building blocks to new elements. All elements are specific configurations of hydrogen atoms. (William Prout’s thesis in 1819.) Hydrogen atoms can be combined to building blocks. Building blocks are D, He, C, O, These building blocks are combined to form the various elements. The elements don’t possess electron orbitals, they are compact atoms. 4
The bonding of the compact atom is due to magnetic coupling. Hydrogen comprises a proton and an electron. Protons and electrons are tiny magnets. Hydrogens as the basic building blocks dont’t repel one another, they attract each other when they come near in the right position. The problem of repelling protons exists only for the untenable nuclear model of QM! During decomposition of an atom, hydrogen atoms are destabilized and decay: H H* p eWhat’s the matter with neutrons? . there are no such things as a neutron in nuclei. (Dulaney) What we recognize as neutrons after fission of an atom are exited hydrogen atoms that decay: H* p e-. The neutron is not an elemental particle. It consists of a proton and an electron. See below under References: 1: Monti and his references to Borghi 2: Clarence Dulaney . there is no such thing as a homepage 3: Marinsek, this website: Neutrons Are Decaying Excited Hydrogen Atoms Orbital is not the domicile of electrons The QM atom contains Z electrons in the shells, Z is the atomic number. The atom of QM does not contain more electrons because the neutrons in the nuclei are treated as elementary particles that do not contain electrons. Only after an intermediating process the end product of the decaying neutron are proton and electron. This theory is wrong. So-called neutrons are in reality stable hydrogens that consist of a proton and an electron: H p eOnly in the case of fission some hydrogens are excited and decay into their composite parts electron and proton: H* p eSee the article on neutron. The QM atomic model contains protons, neutrons and electrons but not positrons! This model is untenable because it cannot explain the occurrence of positrons when unstable atoms decay. Element assembling obeys mass conservation Assembling by magnetic coupling If there are no extra-nuclear electrons and if there are no neutrons, then the fusion of elements can be thought of as assembling of building blocks to new elements. Building blocks are H, D, He, C, O, 4 H’s unite to 4He. 4He itself is also a building block for other elements: 3 4He s unite to 12C, 5 4He s unite to Ne. It is assumed that 3 4He s, and 4 4He s can form stable configurations. Four 4He s form the oxygen building block 16O. 5
The most abundant elements of the cosmos are H and He. The most abundant elements of our globe are H, He, C, O, Si, Al, Fe. Oxygen plus carbon unite to silicon: 12 C 16O 28Si Fe can be formed this way: 52Cr 4He 56Fe Note therefore that only the atomic weight must be conserved for possible fusions. The number of protons (or vice versa the number of neutrons) is pointless, they don’t exist, see the article Moseley’s law refuted. The assembling 52Cr 4He 56Fe shows that only a helium brick can be assembled to 52Cr in order to build 56Fe! The example 12C 16O 28Si shows that here the right building block is oxygen. Oxygen and sodium can be fused to potassium this way: 23Na 16O 39K and 7Li 16O 23Na. Assemblings are possible because the surface of Na and Li are equal. Different are the atomic cores that comprise oxygen atoms. Note very well that not all configurations of hydrogen atoms are stable. Oxygen is stable. The formula is 16O, which means that the total number of hydrogens (H’s) is 16. 16 O consists of four 4He. 4He is two 2D on top of each other, see the picture. A He atom that would consist of two 2D side by side is instable. Because this configuration is unstable also two of them on top of each other are instable. This atom would be 8Be that is unknown QM cannot explain why there is no element 8X. Some remarks regarding molecular and chemical bonding Bonding is due to magnetic coupling because each hydrogen atom is a tiny magnet; presumption: hydrogen itself is composed at least of 4 elementary ring magnets that are charged — —. But there can be more neutral parts — inside, we don’t know. Metals or molecules are atoms conjoined in some way. Bonding is due to magnetic coupling. Bonding agents are hydrogen atoms. There are 3 varieties of H-bonds: A single hydrogen: —; 2 hydrogen atoms in series: —— or 2 hydrogen bonds parallel: Obviously, these are first attempts to explain the crystal structures by atom nuclei that have bonding hydrogen atoms attached at their surface. In a group there is a periodic increase of nuclear mass, for example 16, 32, 80, 128, 176. But the number of bonding or co-ordinating hydrogen atoms is the same for every atom in the group, so that chemical similarities seem to be explainable by the same surface of these atoms. Some considerations about metal lattices are also in the article Molar heat capacities of metals depend on crystal structures. See also The Nature of Chemical Bond 6
Basic assembling’s to make possible atomic configurations Remark: The number of neutral electron –positron pairs is unknown proton 1 electron /positron electron - ortho 2D H para-2D para 4He Element formation The graph shows some formations of elements: Two 1H can be combined to 2D. , 1 H 1H ortho- 2D The alternate assembling gets 1 H 1H para- 2D Then ortho- 2D ortho- 2D parallel para- 4He A helium atom that would consist of two para- 2D side by side is unstable: para- 2D para- 2D 4He* heat Regarding this process wikipedia http://en.wikipedia.org/wiki/Cold fusion#CITEREFSaeta1999 shows the lack of understanding of current physics: Deuteron fusion is a two-step process, in which an unstable high energy intermediary is formed: D D 4He* 24 MeV Here the varieties of deuterium and helium are not distinguished! As explained above there are para- and ortho- varieties of both helium and deuterium. Because the configuration 4He* is unstable, also two of them on top of each other are unstable: 8 Be* is unstable. It decays in 7 10-17 sec into 2α. However, 9Be is stable. The requirements for stability are unknown. The synthetic isotope 8Be* has a half-life time of only 6,722 · 10 17 s. QM cannot explain why there is no element with the mass number A 8: 8Be. Only the stable 9Be is known. The synthetic isotope 8Be has a half-life time of only 6,72 · 10 17 s. This is the proof that 8Be is not stable. Regarding the formation of carbon, it is possible that even the unstable 8Be* can be a partner as a short time intermediate in the process of carbon formation: two helium 4He unite to ßBe* and ß Be* unites immediately with 4He to carbon: 2 4He 8Be* 4He 12C 7
Empirical evidence for Low Energy Atomic Reactions (LEAR) Yasuhiro Iwamura et. al. presented some examples of element formations at the ICCF11conference, in Marseilles, France, on October 31-November 5, 2004. These japanese physicists observed for instance the process: 88 Sr 8? 96Mo Iwamura et. al. reported that the process mentioned works with D2 but not with H2. Some preliminary remarks to this process: A stable element 8Be for the unknown reactant 8? does not exist. As explained above the short living unstable 8Be* can work as an intermediate catalyst. But instead of the addition of 8Be* we can explain the formation of 96Mo also as a fast on-going combination of 88Sr with 2D: 88 Sr 2D 2D 2D 2D 96Mo 53.5 MeV. Note that all reactants do not possess extra-nuclear electrons in orbitals. The assembling of the compact modules functions due to magnetic forces. Recall that this process functions only when the sum of the mass numbers of the deuterium reactants is 8 and when the sum of their atomic number is 4. Iwamura et. al. did not mention the nature of D2. The mass number of D2 is 4, but what is the matter with its atomic number? According to current theory the D2 molecule possesses two “orbiting” electrons and therefore for the reaction D2 should be written as 4D. But there is no causal explanation how the nuclei of two D2 molecules can be combined with the nucleus of Mo without the total destruction of the electron orbitals of Mo and D! Above I proposed an atomic model that is comprised of hydrogen atoms only. The mass number of an element is the number of H atoms. Neutrons are not elemental particles but decay products: H* p e-. During fission some H-atoms are excited: H H*, and decay. According to that model two 1H can be combined to ortho-D: 1 H 1H 2D, or vice versa, 2D 1H 1H Further, 2D 2D 4D; this is not a molecule but an atom. 4D does not have to necessarily identical with helium 4He. Perhaps there is an assembling of protium possible that has a different shape but the same mass numbe. Note also that there are ortho- and para-helium atoms! According to the proposed “natural” series of helium fusion (see the periodic table below) 96 Mo is the result of two steps: 88 92 Sr 4He 92Zr; Zr 4He 96Mo Perhaps the experiment of Iwanura et.al. also works with helium Perhaps the experiment of Iwanura et.al. also works with neon in this manner: 86 Sr 20Ne 106Cd Papers of Iwamura et.al. are easily available at the website of New Energy Times: ura.shtml Archives for LEAR (Low Energy Atomic Reactions) are available at x.shtml 8
The periodic aufbau of the elements Vertical columns show the building blocks Oxygen, Helium and Deuterium. Horizontal listed elements of the second table (below) are generated by the successive addition of 4He. ê Columns: element formation with D, He, C and O building blocks ê 2 1 D ê He 6 Li ê 4 7 Li ê 2D 9 Be ê 16O 25 Mg 24 Mgê 16O 40 Ca H 1 7 Li ê 16O 23 Na 16Oê 39 K 7 Bê 16O 12 Cê 16O 16 Oê 16O Li ê 12C 19 F ê 16O 27 28 32 35 11 Al Si S Cl H ê 1H 2D ê 2D 4 He ê 16O 20 Ne ê 16O 36 Ar The helium building block 4He Add 4He to an element and generate the following elements Example first row: Li 4He 11B; 10B 4He 14N; 11B 4He 15N; 15N 4He 19F; 19 F 4He 23Na; 23Na 4He 27Al; 27Al 4He 31P 31 P 4He 35Cl; 37Cl 4He 41K Proposals for combinations of elements, example Vanadium. 7 a) Helium addition chain: 45Sc 4He 49V (49V: half life 0,9 y) Or b) 51 V 11B 40Ar, or 51 V 7Li 44Ca; 103 Example Rh: 103Rh 23Na 80Se; or Rh 99Tc 4He Example Au Tl: 197Au 4He 201Tl (201Tl: half life 3,04 d, for half lifes see www.priodictable.com. There are also decay chains leading to this isotope.) Example I Cs: 129 I 4He 133Cs (129I has a half life of 1,57 107 y). Break of the 4He series and continuation by an isotope, examples: 31 P 4He 35Cl; 35Cl 4He 39K; 41K 4He 45Sc 9
Periodic table Series of element formations by fusion with 4He helium building blocks Series Li - B - N - F - Na - Al - P - Cl - K - Sc - V - Mn - Co - Cu - Ga - As - Br - Series He - Be - C - O - Ne - Mg - Si - S - Ar - Ca - Ti - Cr - Fe - Ni - Zn - Ge - Se - Lathanoids He I II 6 Li 7 Li 9 Be 23 24 10 Na B 11 B 13 C 12 C 27 13 Al Mg 14 39 K K 40 Ca 44 Ca 85 Rb 45 41 86 88 Sr Sr 133 138 48 28 Si Sc Ti Y 89 90 92 Zr Zr Cs 177 Ba 142 III IV V VI VII �———— 4He �——— 4 VIII 14 N 15 N O 31 P �——— 4He 32 S 75 As 51 52 93 96 98 V V 55 Cr Nb 56 Mo Mo Hf 181 Ce 186 Ta W Mn Fe Tc 99 Tc 102 Ru 97 59 Co 60 Ni Rh 106 Pd 108 Pd 191 187 193 Re Re 190 Os 192 Os 69 65 71 Cu Cu 64 Zn 68 Zn 107 Ag 109 Ag 103 185 63 112 Ir Ir 196 Pt Cd 197 Au Hg 202 Hg 200 19 20 16 �——— He �——— 4He 49 IX Ga Ga 72 Ge 74 Ge 113 In 115 In 116 Sn 120 Sn 119 201 209 205 Bi 78 Se Sb 123 Sb 124 Te 130 Te Tl Tl 206 Pb 210 208 Bi3207 Po Po F Ne 35 Cl Cl 36 Ar 40 Ar 79 35 Br 81 Br 37 82 36 Kr 127 I I 128 Xe 134 Xe 129 211 At 85 212 Rn Some remarks: there is not always a strong succession of elements with steps of 4He. Consider the step 9Be 4He 13C, there is no immediate succession 13C 4He 17X because 17X is a non-existing element. The way of element evolution is now developing: 16 O by : 3 4He 12C; 12C 4He 16O! econd example: 6Li 4He 10B stop! Then 11B 4He 19F; Then 23Na 4He 27Al or alternatively 11 19 F 4He 23Na. B 16O 27Al 3th example 45Sc: 3ways are conceivable: 1: 15N 22Ne 37Cl; 37Cl 4He 41K; 2: 27Al 18O 45Sc. 3: 23Na 22Ne 45Sc. 41 K 4He 45Sc We don’t know all naturally occurring combinations In the article Moseley’s law refuted it is shown that from Moseley’s investigations it is impossible to conclude that atoms have Z protons in a kernel that is surrounded by Z shellelectrons. 10
Helium Spiral Lanthanoids Pr, Pm, Eu, Tb, Ho, Tm, Lu La Ta Y Nb Sc Cs Re V Rb Tc Al K Mn Na B Li Co Rh Ir F N Cl 24He Cu Ag Br Au P I Ga Element Formation Li He B B He N Sb N He F .!!!!!!!!!!!!!!!!!!!!!!!! As In Tl Bi The following table shows some elements with He-4 building blocks Mg 6 He C-12 3 He O-16 4 He Ne-20 5 He Si-28 7 He S-32 8He 2O Ar-36 9 He Trends or breaks of trends for electron affinity, ionization energy and electrical conductivity (see below) are explainable due to changes of the building block configurations of the elements. So changes in the configuration of the elements may explain the changes of physical properties. The trend of physical properties for the alkalis Li-Na-K can be explained in that way because of K Na O and Na Li O the closed oxygen building block does not change significantly physical properties like ionization energies, crystal structure and electrical properties. Neon 20Ne being a building block: 7 Li 20 27Al 8 31P 12 C 20 Ne 35Cl 16 O 20 Ne 39K 20 Ne 11 B 20 15 N 20 19 F 20 Ne Ne 28Si 82 Kr 20 Ne 32S 86 Sr 20 Ne 36Ar 87 Ne 20Ne 40Ca 170 Be 20 Ne 102Pd Ne 106Cd Rb 20Ne 107Ag Er 20 Ne 190Pt 11
Atomic properties related to atomic structures Electron affinity in kJ/mol (see Wikipedia and periodic table.com) By definition, electron affinity is the energy given off when a gaseous atom adds an extra electron to become a negatively charged ion. For example F e- F- energy released. But some elements such as He, Be, N, incorporate an electron without energy released. This might be interpreted so that their atomic surface is very stable. Ia Li 60 Na 53 K 48 Rb 47 Cs 46 II a B 27 Al 42 Sc 18 Y 30 Hf 0 Ib Be 0 Mg 0 Ca 2 Sr 5 Ba 14 II b C 122 Si 134 Ti 8 Zr 41 Ce 92 III a IV a Va VI a VII a V 51 Nb 86 Ta 31 Mn 0 Tc 53 Re 15 Co 64 Rh 110 Ir 151 Cu 119 Ag 126 Au 223 Ga 41 In 39 Tl 36 III b IV b V b VI b VII b Cr 65 Mo 72 W 79 Fe 15 Ru 101 Os 104 Ni 112 Pd 54 Pt 205 Zn 0 Cd 0 Hg 0 Ge 119 Sn 107 Pb 35 VIII a N 7 P 72 As 79 Sb 101 Bi 91 IX a F 328 Cl 349 Br 324 I 295 VIII b O 141 S 200 Se 195 Te 190 IX b Ne 0 Ar 0 Kr 0 Xe 0 1 3 5 7 9 2 4 6 8 10 Remarks: electron affinity defines some vertical physical groups (some of them shown coloured). Electronegativity shows columns of the known chemical groups. Elements shown red are appended. Li 0.98 Na 0.93 K 0.82 Rb 0.82 Cs 0.79 Be 1.57 Mg 1.31 Ca 1 Sr 0.89 Ba 0.89 Al 1.61 Sc 0.95 Y 1,22 La 1.1 C 2.55 Si 1.9 Ti 1.54 Zr 1.33 Ce 1.12 Ge 2.01 Sn 1.96 Pb 2.33 N 3.04 P 2.19 As 2.18 Sb 2.05 Bi 2.02 O 3.44 S 2.58 Se 2.55 Te 2.1 Po 2.0 F 3.98 Cl 3.16 Br 2.96 I 2.66 He 0 Ne 0 Ar 0 Kr 0 Xe 2.6 Ionization energy eV Ia Li Na K Rb Cs 5,4 5,2 4,3 4,2 3,9 II a B 9,3 Al 6 Sc 6,5 Y 6,4 Hf 6,6 Ib Be 9,3 Mg 7,6 Ca 6,1 Sr 5,7 Ba 5,2 II b C 11,3 Si 8,2 Ti 6,8 Zr 6,8 Ce 5,5 III a IV a Va VI a VII a V 6,7 Nb 6,9 Ta 7,9 Mn 7,4 Tc 7,3 Re 7,9 Co 7,9 Rh 7,5 Ir 9,1 Cu 7,7 Ag 7,6 Au 9,2 Ga 6 In 5,8 Tl 6,1 III b IV b V b VI b VII b Cr 6,8 Mo 7,1 W 8 Fe 7,9 Ru 7,4 Os 8,7 Ni 7,6 Pd 8,3 Pt 9 Zn 9,4 Cd 9 Hg 10,4 Ge 7,9 Sn 7,3 Pb 7,4 VIII a N 14,5 P 10,5 As 9,8 Sb 8,6 Bi 7,3 IX a F 17,4 Cl 13 Br 11,8 I 10,5 VIII b O 13,6 S 10,4 Se 9.7 Te 9 IX b Ne 21,6 Ar 15,8 Kr 14 Xe 12,1 1 3 5 7 9 2 4 6 8 10 12
Remarks: At a first glance all columns and all rows show the existence of corresponding physical groups. Going down the columns, ionization energy decreases. Going in a row, ionization energy increases. Only after column VII there is a break of these trends. Explanation: successively adding helium atoms may at one point change the main configuration of the atom and change therefore its physical behaviour. But direct measuring of ionization energy is not trivial. We get nearly exact values for the resonance frequencies of the elements that are the cause of ionization. These frequencies can be obtained by extrapolation of spectral lines of a series. Due to ν c/λ the frequency can be calculated. Molecular properties Magnetic type, crystal structure b bcc f fcc t: trigonal h:hexagonal tc: triclinic te: tetragonal Ia b Li b Na bK b Rb b Cs II a t B d f Al p h Sc h Y h Hf p p p p p Ib h Be d h Mg p f Ca p f Sr p b Ba p III a II b h C d th Si d h Ti p h Zr p h bCe p IV a or: orthorhom. Va VI a f Cu d f Ag d f Au d p paramagnetic d diamagnetic VII a b V b Nb h Ta b Mn p h Tc h Re h Co f f Rh p f Ir p or Ga d te In d h Tl d III b IV b V b VI b VII b b Cr af b Mo p bW p b Fe f h Ru p h Os p f Ni f f Pd p f Pt p h Zn d h Cd d t Hg d f Ge d te Sn p f Pb d f ferromagn. af anti-f. VIII a h N d tc P d t As d t Sb d mcl Bi d IX a mcl F Cl d Br d or I VIII b O d or S d Se d t Te d IX b f Ne d f Ar d f Kr d f Xe d Remarkable is a nearly strict demarcation line between paramagnetic and diamagnetic elements! Crystal structures show coincidences with columns: Elements of groups Ia and Ib have bcc structure, except Be. Group IX b is fcc and so on Electrical conductivity 106 S/m Ia Li 11 Na 21 K 14 Rb 8,3 Cs 5 II a Be 25 Al 38 Sc 1,8 Y 1,8 Hf 3,3 Ib II b Mg 23 Ca 29 Sr 7.7 Ba 2,9 Ti 2,5 Zr 2,4 Ce 1,4 III a IV a Va VI a VII a V 5 Nb 6,7 Ta 7,7 Mn 0,6 Tc 5 Re 5.6 Co 17 Rh 23 Ir 21 Cu 59 Ag 62 Au 45 Ga 7,1 In 12 Tl 6,7 III b IV b V b VI b VII b Cr 7.9 Mo 20 W 20 Fe 10 Ru 14 Os 12 Ni 14 Pd 10 Pt 9,4 Zn 17 Cd 14 Hg 1 Sn 9,1 Pb 4,8 VIII a N P 10 As 3,3 Sb 2,5 Bi 0,8 13
Remarks: the tables for electrical conductivity show vertical groups, whose members possess the same order of magnitude. There are also horizontal groups, for example Mo-Ru-Pd-Cd. Remarks to group formation The Ti-Cr-Fe-Ni-Zn- and the V-Mn-Co-Cu-Ga-As horizontal groups According to Gmelin the metals Mn-Co-Ni-Fe and V-Cr-Mo form groups. Criteria are electronegativity and ionization energies. Here is the proposal that uses He building blocks to explain periodic patterns: Example: 55Mn 4He 59Co Possible big jumps are: 51V 12C 63Cu and 48Ti 12C 60Ni The alkali elements Let us discuss the atoms of the alkali metals Li, Na, K, Rb, Cs and Fr at the beginning. In terms of current theory they have one valence electron in its outermost shell in common. This is the reason for low ionisation energies and on the other side for a great capability of chemical reactions. All alkali metals ( molecules) have a bbc crystal structure. Now let us explain the alkali family in terms of the hydrogen structure model of the atom. There is no explanation for this fact in terms of the Rutherford-Bohr model. The alkali elements have an increasing mass number but the chemical behaviour remains unchanged because the surface remains the same. So it is assumed that this mass increase is nothing but an increase in oxygen building blocks that are chemically inactive. There is a kernel under a chemically active surface. For it forms a kernel under a chemically active surface. The alkalis are chemically equivalent. Therefore they can form molecules with 35 Cl and 35Cl. For example the combinations NaCl, KCl, CsCl are possible. When 35Cl and 35Cl are treated as isotopes and not as elements then one can also the alkalis treat as iso
Elements are fusion products of hydrogen, deuterium, helium, carbon and oxygen building blocks. Helium consists of four hydrogen atoms. Neutrons are not elementary nuclear particles. "Neutrons" are excited hydrogen atoms. During the fission of an element, hydrogen atoms can be decay products. They are excited and decay: H* p e-.
Mila Formation 130 Kalshaneh Formation 130 Derenjal Formation 130 Ilebeyk Formation 130 The Tippecanoe Sequence: 133 Shirgesht Formation 133 Niur Formation 133 . Khaneh Kat Formation 217 The End of the Absaroka Sequence in the Central and Northern Arabian Gulf
Generalized coordiDate finite element lDodels ·11 17 'c. IT,I .f: 20 IS a) compatible element mesh; 2 constant stress a 1000 N/cm in each element. YY b) incompatible element mesh; node 17 belongs to element 4, nodes 19 and 20 belong to element 5, and node 18 belongs to element 6. F
bliss element glass tile anatolia tile stone anatoliatile.com bliss - element glass tile 01 3" x 9" element ice beveled glass tile 38-018 3" x 12" element ice glass tile 38-000 3" x 12" element ice artisan glass tile 38-010 1.5" x 6" element ice stacked glass mosaics 35-114 element ice chevron glass mosaics 35-122
CHEMISTRY — RELEASED ITEMS 6 Go to the next page. 15 The table below shows the electron configurations of three elements. Element Electron Configuration Element 1 1s22s22p2 Element 2 1s22s22p5 Element 3 1s22s22p4 What is the order of the elements from smallest to largest atomic radius? A Element 1, Element 2, Element 3
nite element method for elliptic boundary value problems in the displacement formulation, and refer the readers to The p-version of the Finite Element Method and Mixed Finite Element Methods for the theory of the p-version of the nite element method and the theory of mixed nite element methods. This chapter is organized as follows.
Finite element analysis DNV GL AS 1.7 Finite element types All calculation methods described in this class guideline are based on linear finite element analysis of three dimensional structural models. The general types of finite elements to be used in the finite element analysis are given in Table 2. Table 2 Types of finite element Type of .
map element are to the values for a CEDS element. CEDS Element Data Model ID This is the version-specific CEDS identifier. Use this identifier when mapping to CEDS. CEDS Element Global ID The CEDS global ID is the numeric CEDS ID assigned to an element. This remains the same across all versions of CEDS. Figure 7 Individual Element Details tab.
The Element How finding your passion changes everything By Sir Ken Robinson Summary by Kim Hartman This is a summary of what I think is the most important and insightful parts of the book. I can’t speak for anyone else and I strongly recommend you to read the book in File Size: 600KBPage Count: 15Explore further(PDF) The Element - Ken Robinson Simona Ana - inson-epub.pdf - Google Docsdocs.google.comThe Element by Ken Robinson. How finding your passion ry - A summary of the .www.coursehero.comRecommended to you based on what's popular Feedback
