BY

Until 1911, the most famous theory of the structure of the nucleus was the "currant bun" (or "plum pudding") theory of J. J. Thomson, in which electrons were dispersed among the positive charges like the currants in a bun.

In 1911, Rutherford and co-workers¹ irradiated gold foil with a particles. They found by observing the tracks of the particles as indicated by scintillations on a fluorescent screen that about 1 in 20,000 particles was deflected by over 90° from its original path. A relatively simple calculation based on Coulomb’s Law indicated that the atoms appeared to have positive nuclei with diameters on the order of 3 x 10^-12 cm. The extra-nuclear electrons were supposed to rotate about this positive nucleus like planets in a miniature solar system.

Several reasons were propounded that electrons could not exist in nuclei. The main objection was based on an extension of the Heisenberg Uncertainty Principle². This theorem states that if a particle were confined within a space of radius x, the momentum thereof must be at least (h/4p)x. For a nucleus of radius 10^-12cm, the momentum of an electron would be about 5 x 10^-16gcm/sec, and the kinetic energy calculated by Einsteinean relativity would be at least 9 MeV. If the electron remained in the nucleus, the acceleration would be so great that radiation would be quite intense. Of course, such radiation is not noted in normal atoms. Discovery of the neutron in 1932 further obviated the need for electrons in the nucleus.

Another reason that there are not supposed to be electrons in nuclei is the hyperfine spectrum discovered by A. A Michelson^2,3 at the turn of the century. Spectral lines are broadened by hyperfine splitting due to nuclear "spin". In 1926 G. Uhlenbeck and S. Goudsmit⁴ introduced the concept of spin to explain the spectra of alkali metals. It appears that all atomic particles have spins, and that nuclei have resultant spins that are the vector sum of those of the nuclear particles. Those nuclei that have non - integral spins give rise to hyperfine spectra. One nucleus, ₇N¹⁴, if it had electrons in the nucleus would have 14 protons and 7 electrons. Since protons are supposed to have +½ spin and electrons -½ spin, the resultant spin of the nitrogen nucleus should be +3½. Actually, the nuclear spin for nitrogen is 1, indicating an even number of nuclear particles.

Consider the force in a gold nucleus of 3 x 10^-12 cm diameter. Each gold nucleus has 79 protons, each with a charge of 4.8 x 10^-10 esu. The force caused by like charge repulsion would be F = q²/r² or 7.1 x 10⁶ dynes, an immense force for such a small particle. H. Yukawa in 1935⁵ proposed a particle to moderate this force. His particle (which was later called the meson) was to have a charge equal to that of the electron. and a mass about 200 times that of the electron, so that it could stay in the nucleus. It was also supposed to pass rapidly between the protons and neutrons, disappearing after every encounter. The particle was required to have a very short half-life, about 10^-8 seconds, but this was supposed to be long enough for its moderating activity.

When the small gold nucleus with "solar" electrons was proposed, and particularly after Bohr’s explanation of the hydrogen spectrum, the idea was immediately and enthusiastically received. Nobody stopped to calculate what the density of the gold nucleus would be, or for that matter what the force between the protons would be, because, until 1932, there was still the possibility that there might be electrons in the nucleus. What would be the density of the 3 x 10^-12cm nucleus? With a volume of 1.44 x 10^-35cm³ and a mass of 3.27 x 10^-22 gram, the density would be 2.27 x 10¹³g/cm³. This should be large enough to cause fusion of the nuclear particles.

What Rutherford et al measured in 1911 was probably not scintillations caused by deflected a particles, but by particles that were produced by transmutation reactions. It was not until 1919 that Rutherford’s group⁷ found that when nitrogen was irradiated by a’s the product was oxygen isotope 17 plus a proton. By this time the solar atom was so entrenched that nobody thought about the a-deflection experiments.

Suppose the following occurred: ₇₉Au¹⁹⁶ + a = ₈₁Tl²⁰⁰ = ₈₂Pb²⁰⁰ + e^- This alchemist’s nightmare is only one possibility of the kind of transmutation reactions that could have occurred. The electron produced in this and other reactions first of all could not pass through the gold foil, and secondly would make just as big a scintillation as an a. As a result, the size of the nucleus may be considerably bigger than the 1911 estimate.

Suppose the diameter of the gold nucleus is actually 8.5 x 10^-9 cm. This is slightly smaller than the kinetic theory atomic size of about 10^-8cm. This diameter gives a density of 1000 g/cm³ for the gold nucleus, and would make the "currant bun" model possible once again.

What would it do for the force between the protons? With such a large size, the force would be 1.01 dynes, or a million times less than for the Rutherford nucleus, even neglec- ting the possible presence of electrons. What about the possibility of electrons? The calculated Heisenberg momentum would be 1.24 x 10^-19 gcm/sec compared to 5 x 10^-16 gcm/sec for the Rutherford nucleus, and a kinetic energy which we can calculate from the Newtonian formula of 5.28 eV, a quite reasonable value that would not preclude electrons in nuclei.

Let us also consider the hyperfine situation for ₇N¹⁴ , both with neutrons and with electrons. First, with 7 each of protons and neutrons, and a spin of 1, six of the neutrons would have a spin of -½ and the seventh would require a spin of +½. Just how does the one neutron know that its spin is to be +½, and isn’t such a neutron an anti-particle. What keeps it from annihilating a normal neutron?

With electrons present in the nucleus there would be 14 protons and 7 electrons for a spin of 3½. Thus it appears that there must be association with 5 more extranuclear electrons to make the spin come out 1. This would not only make the spin right, and would make the intra-nuclear force virtually zero. This concept is reasonable, because there would be substantial attraction between the 7 un-neutralized protons and the extranuclear electrons. This would make the intra-nuclear force virtually zero.

Because of the reasons cited, the diameters of nuclei are proposed to be much nearer the kinetic theory estimate (of about 10^-8 cm) than the value calculated by Rutherford. It is also proposed that all of the extranuclear electrons save the "valence" electrons are associated with the nucleus, thus making the proton, proton force almost zero.

What about neutrons? A paper in progress⁸ will show that what we call neutrons are hydrogen atoms, and/or high speed, zero charge protons. Also, what we call negative mesons (from cosmic rays) are electrons moving at light speed⁹, and there are no mesons needed or found in nuclei.

Bibliography

[1] E. Rutherford, Phil Mag., 21, 669 (1911)

[2] "Atoms, Radiation and Nuclei" by T. H. Osgood et al, John Wiley & Sons, NY (1964), p-321ff

[3] "Introduction to Atomic Physics", by H. Semat, Rinehart and Co. NY (1946), p368ff

[4] G. Uhlenbeck and S. Goudsmit, Nature, 117, 264 (1926)

[5] H. Yukawa in "Fundamentals of Nuclear Physics" Edited by R. Beyer, Dover, NY (1949). p139-148

[6] T. H. Osgood, op. cit., p-375ff

[7] E. Rutherford, Phil. Mag.. 31, 581, (1919)

[8] "What is an Atom" by Clarence L. Dulaney, Unpublished Paper (1997)

[9] "A New Look at Relativity" by Clarence L. Dulaney, paper delivered at the San Luis Obispo, CA meeting of the Natural Philosophy Alliance, July, 1997

By Clarence L. Dulaney

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Missouri City, TX 77489

First Submitted: December 22, 1997