THE SIZE OF NUCLEI

 

By: Clarence L. Dulaney

 

2226 Fairgreen Drive

 

Missouri City, TX 77489

 

e-mail cldtx1@sbcglobal.net

 

 

ABSTRACT;  This paper discusses the possibility that the so-called anomalous a-particle dispersion by gold foil (in 1911)could have been caused by smaller charged particles produced by disintegration of the a's, or by transmutation reactions.  The latter type of reactions were not known until 1919.

If nuclei are close to the kinetic theory size, nuclear electrons may be possible, moderating intranuclear forces.

 

INTRODUCTION

 

Until 1811, the most famous theory of nuclei was the "currant bun" theory of J. J. Thomson, in which the electrons were supposed to be interspersed among the positive charges  like 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 were 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 extranuclear electrons were assumed to rotate about this nucleus like planets in a miniature solar system.

 

NO ELECTRONS IN NUCLEI

 

Several reasons were propounded that electrons cannot 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 with radius x, the momentum thereof must be at least (h/4p)x.  For a nucleus of about 10-12 cm, the momentum of an electron would be about 5 x 10-16 gmcm/sec and the kinetic energy calculated by STR 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 for normal atoms.  Discovery of the neutron in 1932 further obviated necessity of electrons in nuclei.

 

NUCLEAR SPIN AND THE HYPERFINE SPECTRUM

 

Another reason that there are not supposed to be electrons in the nucleus is the hyperfine spectrum discovered by A. A. Michelson2.,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 spins5 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.  Consider 7N14.  If it had electrons in the nucleus, It would have 14 protons and 7 electrons according to the theorists. (why not 14 electrons ?)  Since the proton spins are each +˝ and those of the electron each -˝, the resultant would be 3˝.  Actually, the measured spin is 1, indicating an equal number of nuclear particles. 

 

NUCLEAR FORCE-YUKAWA AND THE MESON

 

Consider the Coulombic force in a gold nucleus of 2 x 10-12 cm in diameter.  This nucleus has 79 protons, each with a positive charge of 4.8 x 10-10 esu.  The force  F = (q/r)2 = 7.1 x 106 dynes, an immense force for such a small particle.  All kinds of fantastic schemes were devised to allow for such a force.  Finally, in 1935, H. Yukawa6 devised an exchange particle, which he called a meson.  This particle was supposed to have a mass of 200 times that of an electron. so it could stay in the nucleus.  Its charge was that of an electron, and it had a very high speed, but it was to disappear after each encounter with a proton.  The particle was to have a very short half-life, about 10-8 sec.  (Even though the acceleration was quite high, and the meson was destroyed with each encounter, nothing was said about radiation.)

 

A NEW LOOK AT NUCLEAR DIAMETERS

 

When the small gold nucleus, with the "solar" extranuclear electrons was proposed, and particularly after Bohr's explanation of the Hydrogen spectrum, it was enthusiastically received.  Nobody stopped to calculate what the density of the gold nucleus would be, nor for that matter what the force between the protons would be, because until 1932 there was still the possibility that electrons could be part of the nucleus. 

 

What would be the density of the Rutherford gold nucleus?  With a volume of 1.44 x 10-35 cm3 and a mass of  2.27 x 10-22 gm, the density would be 2,27 x 1013 gm/cm3.  This should be high enough to cause fusion of the nuclear particles.

 

The scintillations that Rutherford et al measured almost surely were not due to deflected a-particles, but from particles produced by disintegration of the a's, or from 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 experiment. 

 

However, note that the "solar" atom has been shown10 to be mechanically and electrically impossible.

 

The a is made up of 4 protons and two electrons.  It is unstable, unless it finds two more electrons somewhere and becomes He.  Otherwise, it breaks into its constituents.  The half-life for this disintegration is roughly 5 x 10-7 sec. (My calculation based on the increase of the rate of ionization of air (Osgood, op cit p297ff.))  The particles of the disintegration all make the same kind of scintillations as would the alpha.  It thus is postulated that the  original conclusions of the 1911 experiment were wrong, and that the nucleus probably is close to the kinetic theory size.

 

Suppose the gold nucleus is actually about 10-8 cm.  This diameter gives a specific gravity of about 1000, and would make the "currant bun" model once again possible.  With such a reasonable size, the force between the protons would be 101 dyne, a million times smaller than that for the Rutherford nucleus, even neglecting the possibility of electrons.

 

If electrons are present, the calculated Heisenberg momentum would be roughly one millionth that for the Rutherford nucleus, and the kinetic energy would be 5.3 eV,even supposing the electrons were moving.  Thus, the intranuclear force should be very nearly zero.

 

Let us now consider the hyperfine situation for 7N14, both with neutrons and with electrons.  With 7 neutrons and a resultant spin of 1, 6 of the neutrons would have spins of -˝, and the seventh would need to have a spin of +˝.  How would the seventh neutron know to have a + spin?  Wouldn’t it be an anti neutron?  Wouldn't it annihilate a - spin neutron?   With electrons, there could be 14 protons and 12 electrons to account for the +1 spin.  The other two electrons would be extranuclear or "valence" electrons. The intranuclear force would be virtually zero.  It is proposed that the nuclear particles would be stationary, positioned so that the energy would be a minimum.  To moderate the electron, proton forces, it is proposed that neutrinos and antineutrinos are present between them. 

 

SUMMARY

 

Because of the reasons cited, the diameters of muclei 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 atomic electrons save a small number of "valence" electrons are associated with the nucleus, thus making  the intranuclear force essentially zero. 

 

What about neutrons?  Se my paper8 "What is an Atom" which shows that what we call neutrons are hydrogen atoms or high speed, zero charge protons.  Also, what have been called negative mesons (from cosmic rays) are electrons moving at high speed9, and there are no mesons found or needed in nuclei.

 

Finally, see my paper10 "The Stationary Hydrogen Atom" for evidence that "valence" electrons do not rotate about nuclei, but are stationary, separated from the nucleus by layers of neutrinos.

 

REFERENCES

 

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

 

[2] T. Osgood, et al, "Atoms, Radiation and Nuclei", John Wiley and Sons, NY, (1964)p321ff

 

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

 

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

 

[5] C. Dulaney, "Why Spin"

 

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

 

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

 

]8] C. Dulaney, "What is an Atom?"

 

[9] C. Dulaney, "Charge vs Speed"

 

[10] c. Dulaney, "The Stationary Hydrogen Atom"

 

NOTE:  All my papers may be found at:  http://mywebpage.netscape.com/clarencedulaney/

 

© August 19, 2006.  Clarence. L. Dulaney