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Natural Science Forum / Physics / Research / January 2008Atomic electrons
Atomic electrons spr The currently accepted atomic model (nucleus + electron cloud) assumes that every electron that can be ejected from a ground state atom or atomic ion in any low energy collision with an atom, ion or particle exists as an electron in that ground state. There would seem to be no way in which this assumption can ever be tested by experiment. Has anyone ever investigated an alternative model in which it is assumed that: (a) Any ground state atom or atomic ion is a single particle that contains no electrons but can, if a few electron-volts of energy are transferred to it in a collision, become unstable and decay into an ion and one more electrons. (b) When the energy transferred is in the range 80% - 99% of the ionisation energy the excited atom or ion that results will, most of the time (always in the case of the hydrogen atom), be a two body system - ion plus one electron. Phil Gardner Phil Gardnerwrote: >Atomic electrons spr > [quoted text clipped - 14 lines] >the time (always in the case of the hydrogen atom), be a two body >system - ion plus one electron. I'm not sure what you need for a direct proof of electrons "existing" in a ground-state atom, but there is plenty of indirect proof. I, for one, spent a great deal of laboratory-time scattering very-low-energy electrons off atoms; energies below 10 eV, and saw evidence that they "existed." These experimental results, and those of others, are well explained by unexcited "particle" electrons. Lots of work was done on this in the 60s and 70s, and there is a list of literature available, both theoretical and experimental. This area is referred to as elastic-electron-atom scattering. I suggest you hie to your local physics library and see what is available. > Atomic electrons spr > [quoted text clipped - 3 lines] > exists as an electron in that ground state. There would seem to be no > way in which this assumption can ever be tested by experiment. How about using x-ray scattering to determine the charge distribution of an atom at any given level of ionization? The spread of the charge distribution of a nucleus is known to be much smaller (about 10^5 times smaller) than the spread of the charge distribution of an atomic electron. So, if I understand correctly the assumption you are talking about, it's certainly experimentally testable. Google "x-ray scattering" or "x-ray spectroscopy". Computational chemistry is a major industry, with electronic structure calculation as one of its main goals. They wouldn't get very far if they couldn't compare their calculations to experiments. Igor >>Atomic electrons spr >> [quoted text clipped - 17 lines] > > Igor We have to be clear here that X-ray Mev scattering will establish the positions of atomic nuclei (Bragg Peaks) within some type of crystalline or polymeric structure but will not provide much if any information on electron distribution (ev energies) around any particular atom. Lower energy ev electrons are required as per the work of Murray Arnow. Determining atomic electron cloud structure with X-rays would be like determining a cumulus cloud structure with a M50 rifle. Richard > >>Atomic electrons spr > [quoted text clipped - 30 lines] > > - Show quoted text - But isn't that what X-Ray diffraction is all about? X-Rays are scattered primarily from the electrons. It is the cloud of electrons around the nuclei that diffract the X-Rays. Although the obvious structure of an X-Ray diffraction image is due to the periodic structure of the crystal, I am under the impression (but I'm not that knowledgable about it) that there is information in the brightness of the individual dots that gives information about the distribution of electrons in the clouds. Maybe it is like studying clouds of sand with an M50 rifle... Rich L. >>>>Atomic electrons spr >> [quoted text clipped - 43 lines] > > Rich L. e^2/r ~ 14 ev where r is 1 Angstrom (indicative of interatomic distance) and corresponding wave length hcr/e^2 ~ 8.61E-06 cm (ultraviolet) Of course higher energies would be warranted in the context of higher atomic weight (Z) materials Z e^2/r Of course to study arrays of atoms Bragg reflection requires incident radiation of < r/2 Higher frequencies towards x-rays would result in Bragg peaks dictated by atomic nuclei positions, but electron cloud resolution would decrease as this incident energy (E) (h nu) increases E ~ 1/r. Keep increasing the incident radiation energy toward gamma ray and you will blow the nuclei away too. Richard > How about using x-ray scattering to determine the charge distribution > of an atom at any given level of ionization? I don't know if this helps, really. But it's interesting. In Chapter 5 of Ballentine's text on quantum mechanics there is a nice (albeit brief) discussion of experiments that measure the probability distribution for the bound electron in atomic hydrogen. The technique involves ionizing the atom by an electron beam and measuring the momenta of the scattered and ionized electrons. The result depends directly on the momentum space probability distribution of the initially bound electron. Ballentine's discussion comes from here: B. Lohman and E. Weigold, "Direct measurement of the electron momentum probability distribution in atomic hydrogen" Physics Letters, 86A. 139-141 (1981) -charlie torre > Atomic electrons spr > [quoted text clipped - 3 lines] > exists as an electron in that ground state. There would seem to be no > way in which this assumption can ever be tested by experiment. Spectroscopy; Auger emission including muonic atoms and electron capture radioactive decay. Transition metal magnetic moments, especially inner shell lathanides and actinides. > Has anyone ever investigated an alternative model in which it is > assumed that: > (a) Any ground state atom or atomic ion is a single particle that > contains no electrons but can, if a few electron-volts of energy are > transferred to it in a collision, become unstable and decay into an > ion and one more electrons. Electron paramagnetic resonance says that is not even wrong. Ditto optical absorption, chemical bonding, Gouy balances, permanent magnets, semiconductors, Peltier heaters/coolers, BCS theory, ESCA/EDAX... the hydrogen 21-cm line, and molecular oxygen. > (b) When the energy transferred is in the range 80% - 99% of the > ionisation energy the excited atom or ion that results will, most of > the time (always in the case of the hydrogen atom), be a two body > system - ion plus one electron. Horrible.  SignatureUncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/lajos.htm#a2 [snip] > Has anyone ever investigated an alternative model in which it is > assumed that: > (a) Any ground state atom or atomic ion is a single particle that > contains no electrons but can, if a few electron-volts of energy are > transferred to it in a collision, become unstable and decay into an > ion and one more electrons. [snip] Oh yeah... you uncreated x-ray diffraction, too - especially difference maps. SignatureUncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/lajos.htm#a2 [Moderator's note: I've reformatted a bit; this post, in contrast to most of your posts, arrived rather garbled. -P.H.] > We have to be clear here that X-ray Mev scattering will establish the > positions of atomic nuclei (Bragg Peaks) within some type of crystalline > or polymeric structure but will not provide much if any information on > electron distribution (ev energies) around any particular atom. Lower > energy ev electrons are required as per the work of Murray Arnow. X-ray scattering does provide information about electron distribution. The Bragg peaks are there because of the periodic electron density inside the crystal (which happens to have the same periodicity as the nuclear positions). The intensity of the Bragg peaks gives information about the electron density within each crystal cell, as Rich L. has already pointed out in a parallel post. Both these features are explained by the fact that the X-ray diffraction pattern is basically the Fourier transform of the electron density distribution in the crystal (see for instance the Wikipedia article on X-ray Scattering). But you are right that lower energy probes are needed to work with single atoms, as opposed to crystals. Unfortunately, lower energy photons have the deficiency of also having poor spatial resolution. But it can still be useful. For example, visible light mostly couples to the electric dipole moment of atoms. While the full atomic charge density contains much more information, the electron dipole moment suffices for the purposes of counting atomic electrons, which should be of direct interest to the OP. In fact, there is a well known sum rule in atomic physics, the Thomas-Reiche-Kuhn sum rule aka the Oscillator Strength sum rule, which directly counts the number of atomic electrons. See for instance the corresponding Wikipedia page [1] or Bethe and Salpeter's classic book [2]. The sum rule has the form: Sum_n' f_{n,n'} = N, where N is the number of atomic electrons, n is some reference atomic energy eigenstate, and n' ranges over all other energy eigenstates to which dipole transitions are allowed. The f_{n,n'} factors are called "oscillator strengths" and are proportional to the intensity of the corresponding transition spectrum line and to a power of the line's frequency. Simple and direct. Bethe and Salpeter even give some numerical tables which show how the oscillator strengths add up to 1.000... for the Hydrogen atom. [1] http://en.wikipedia.org/wiki/Oscillator_strength [2] Bethe & Salpeter, _Quantum Mechanics of One- and Two-Electron Atoms_, (Plenum, 1977). Igor |
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