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322) Rutherford-Bohr model questioned

Ludwik Kowalski; 4/11/2007
Department of Mathematical Sciences
Montclair State University, Upper Montclair, NJ, 07043

This morning I received an e-mail message from a stranger, P.J.J. He wrote: ". . . This is a long and clumsy prelude to my quest for your help in determining the validity if any in the posit of the hydreno (not hydrino) theory. . . ."

Well I never heard about hydrenos. But Google helped me. A book entitled "Lattice Nested Hydreno Atomic Model" by Mark Porringa, is available over the Internet:

Jumping from link to link I finally found something scientific at

It is an extract from Porringa’s book. What follows are my comments on his objections to the Bohr-Rutherford model of atoms. His words are in blue; my comments are in black.

“ . . . Aside from its undisputed accuracy in the prediction of the spectral lines of hydrogen, the Bohr-Rutherford atomic model appears to be in need of some serious rework . . . . Another objection is raised regarding Rutherford’s [29] calculation of the size of the nucleus, which could be off by a very wide margin, due to several invalid assumptions including the erroneous exchange of terminology used in his analysis that permitted the “surface” of the nucleus, to be considered equivalent to the “center” of the nucleus.  From the outset, Rutherford presumes a point like nature for the nucleus with a dimension conjectured to be less than 10-14 m. If the nuclear charge is in reality dispersed over a considerably larger volume than the presently conjectured 10-42 m3, the coulomb interaction with a high velocity charged particle would be much different than Rutherford’s presumption.  A larger nucleus, with a much reduced electric field intensity, would permit the high energy alpha particles used in the Gold foil experiment to come much closer to the nucleus, possibly striking it.”

Yes, this would happen if the nucleus were much larger than 10
-14 m. and the dependence of the differential cross section on the angle would be very different from what was actually measured in 1911. (Scattering would not be due to Coulmbe force only). I measured differential scattering cross sections many times (not only on gold, and not only for alpha particles) and I know how well the results confirm the point-like approximation. This approximation, by the way, is no longer valid when the energy of the projectile is sufficiently large. The size of a nucleus can easily be estimated from the energy at which a cross section, for example at 150 degrees, starts to deviate from what is predicted by the point-like approximation. I have done this, several times, on the basis of data collected during my postdoctoral work at Columbia university.

“ Rutherford also makes the mistake of presuming that the probability of large angle deflections is directly related to the thickness of the gold foil, ignoring the obvious tendency of the alpha particle to be channeled within the highly organized atomic lattice by the presence of the concentrated charge of the gold nuclei, and secondarily by the electron cloud surrounding it. “

I see no mistake in Rutherford's analysis of experimental data. He made an assumption (independent scattering of alpha particles by individual point-like nuclei) and showed that experimental data, for very thin foils, were in good agreement with this model. What is wrong with this kind of validation?

“The beam trajectory being normal with respect to the foil, and therefore parallel to the internal lattice structure, would effectively prevent large angle deflections beyond the first two staggered layers of the cubic close-packed crystal structure.  

A gold foil is not a monocrystal. Therefore the internal lattice structures are not always parallel to trajectories of alpha particles.

Barring a direct hit, or very close miss on a gold nucleus within these first two layers, the high energy alpha particle would simply continue on its merry way between the nuclei, effectively channeled through the entire foil, without any further prospects of high angle deflections. Only small deflections would be possible for the remainder of its travel, and these would be entirely insufficient to bump the alpha particle out of the defined channel provided by the lattice... “

But scattering at large angles does take place. The results, at low energies, are found to be consistent with the idea of independent scattering by individual point-like nuclei. That was the central point of Rutherford's discovery (1911). the only mistake he made, as far as I know, was in an argument (in the 1937 book) that practical applications of nuclear energy will not possible. He did not anticipate a possibility of chain reactions. But that is a different topic.

“The thickness of the foil used in the experiment is cited, near enough, as 10
-8 m.  Given the well-established atomic dimension of 10-10 m, the foil was less than 100 atoms thick, making it extremely unlikely that a single alpha particle could experience more than one large angle deflection.  The proportion of large angle deflections greater than 90o was found to be only one in 20,000. Following the reasoning above, about half of these deflections happen at the surface of the foil, the remainder occurring at the staggered layer of atoms immediately below the surface of the foil of the ccp structure. “

What evidence is available to support a statement that “about half of these deflections happen at the surface of the foil.” ? In my opinion it is a speculation that has nothing to do with reality. All experimental data with which I am familiar, including my own, are consistent with the idea that the above assumption has nothing to do with reality.

“In other words if the foil could have been reduced to only two layers thick the number of high angle deflections would have been essentially the same as that observed for the 100 atom thick foil.  The remaining 98 atomic layers would have virtually no effect with regard to high angle scattering.  Rutherford’s inclusion of a thickness factor in his probability analysis would therefore appear quite unjustified, and would by itself give rise to a 100 fold error in the size of the nucleus....  ”

When the target thickness is increased from 2 to 100 layers, the number of high angle deflections increases by the factor of 50. Once again, this is a well established experimental fact.

“Making no presumptions about the relative size of the nucleus, and the nature of the Coulomb interaction, a reasonable estimate could be asserted by considering only the high angle scattering at the surface of the foil, which clearly indicates that only about 1/10,000 of the surface area is occupied by something substantial (a heavy nucleus); the remainder being mostly empty space occupied by the orbital electron clouds. ”

Which experimental data are being used to justify this 1/10,000 factor?

“This clearly and simply implies that the diameter of the nucleus could conceivably be roughly 1/100 of the atoms diameter.  The result of this “ball park” analysis is that the nuclear radius may be about a thousand times larger than the presently conceded dimension of 10
-15 m, which results in an atomic diameter in the range of 10-10 m, with a nucleus of 10-12 m; a much different view from that presently accepted. Even at 10-12 the absolute Casimir pressure is still a very substantial 1018 kPa and certainly adequate to hold the nucleus together against the substantially reduced coulomb repulsion of the protons....”

That reminds me a well known Polish saying: "Gdyby baba wasy miala toby dziadem sie nazwala." And the Russian sayng "Yielsi-by-da dakaby da wortu rosli boby to polutchilsia by nie rot a celyi ogorod". The English proverb, I was told, is "If wishes were horses beggars might ride."

“A thousand fold increase in the nuclear diameter, also results in a million fold reduction in the intensity of the mutual coulomb repulsion of the protons of the rarified nucleus, assuming for simplicity a radial field attenuation, according to the normal 1/r
2 relation. 

Strong nuclear forces have short range. What would keep protons inside atomic nuclei?

Such a dramatic reduction greatly improves the prospects of fusion at low energies and calls into question the present understanding of thermonuclear explosions, which are  heavily reliant on the notion of an extremely intense coulomb repulsion within the nucleus....  

Yes, thermonuclear reactions would be taking place at lower temperatures if atomic nuclei were much larger than they really are.

Thermal neutron absorption cross-sections cast further doubt on the notion of an extremely small nucleus.  Gold 198 and Cadmium 113 for instance have cross-sections approaching 27,000 barns, which on a statistical basis, just so happens to equate to an apparent nuclear dimension of 10
-12 m, again a thousand times larger than the presently accepted nuclear dimension.   “

Very large cross sections (for 0.025 eV neutrons) do not conflict with small sizes of atomic nuclei. Cross sections become very small at 0.1 MeV and above.

“. . . Why is it that no stable isotopes exist for elements having an odd number of neutrons and protons?  Why is it that only one stable form exists for over 25 elements while others have up to 10?  What is behind the enigmatic chemistry of Nitrogen. Why does the hot fission of Uranium result in unequal fission fragments.  How does cold fission frequently occur without radioactivity or the release of the vast amounts of energy characteristic of thermonuclear reactions.  Why is it that no one seems to even bother anymore with such fundamental questions?”

Yes, there are many things we do not understand, including cold fusion. That is why being a scientist is so exciting. Rutherford and Bohr are giants on whose shoulders we are standing.

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