WHAT IS A “NEUTRON”

 

By: Clarence L. Dulaney

       2226 Fairgreen Drice

       Missouri City, TX 77489

        cldtx1@sbcglobal.net

 

A.    INTRODUCTION

 

The problem with the identity of “neutrons” starts with Rutherford and associates’ experiment on the size of nuclei     

  ]1.2]   In my paper, it is explained why nuclei are actually on the order of 10-8 cm  in size, nuch larger than Rutherford’s estimate of about 10-12 cm

 

Heisenberg Uncertainty Principle [3] showed that there could not be any electrons in Rutherford’s  tiny nuclei.  Thus, when it was discovered that there were “neutrons” in all nuclei, it was reasoned that “neutrons”  could not consist of a proton plus an electron, even though “neutrons” decay to a proton and an electron with a half life of about 10 minutes. “Neutrons” were supposed to be a “particle” which upon liberation from a nucleus decayed to a proton and an electron and to satisfy “quantum mechanics” both of these plus an antineutrino had to be created from the “particle” at the time of the decay.  Of course, the antineutrino is even harder to detect than the “neutron”.  (It seems like creation of the three particles would require much more energy than the simple ionization of the hydrogen atom which requires only 13.6 ev [4].)

 

 

B.    MASS OF THE “NEUTRON”

 

 

Chadwick [5] is given credit for discovery of the ‘”neutron” in 1932.  He bombarded 4Be with an a from 84Po.  After this, “neutrons” were produced from many atoms, and the “neutron” was shown thereby to be a constituent of all atomic nuclei excepting 1H.

 

H. Semat[6] gives a Table of 10 reactions that produce “neutrons”. He does not calculate possible masses.  It is interesting to do so.  The masses are from Lilley’s Table [7].  (Since there are known additional products, the masses given are maxima.)

 

                                          Mass of n in AMU

1)  2Li7 + a =  5B10 + n        1.005706

 

2)   4Be9 +a =  6C12 + n       1.014785

 

3)   5B11 + a  = 7N14 +n        1.008601

 

4)   7N14 +a   = 9F17 + n        1.003582

 

5)   9F19  +a   = 11Na22 +n      1.006569

 

6)   11Na23 +a = 13Al26 + n      1.005481

 

7)    12Mg24 +a = 14Si27 + n     1.000938

 

8)    13Al27  +a  =  15P30  +n     1.005826

 

9)    15P31  +a   =   17Cl34  +n    1.002603

 

10)   18Ar20  +a   =  20Ca23  +n  1.006223

 

 

 

 

 

 

 

 

The accepted value for the “neutron” mass was obtained was obtained by disintegration of deuterium with a 2.62 MeV g-particle from Thorium”C”.  This was done by Chadwick and Goldhaber in 1935 [8].  They did the experiment in an ionization chamber, and estimated the kinetic energy of the proton produced by the number of “kicks” registered by the oscilloscope.  They found this energy to be 0.25 MeV, and assumed the “neutron” to have the same kinetic energy, since its mass should be about that of the proton.  This means that 2.12 MeV was given to the deuterium atom.  It is known that the deuterium has a mass of 2.014102 AMU, and the proton 1.007825AMU.  [7].  The 2.12 MeV is equivalent to ,0.002276 AMU, so this gives 1.008553for the mass of the “neutron”.  Actually, this has been increased to 1.008665 AMU. 

 

It should be noted that the proton is not the only thing causing “kicks”  in Chadwick and Goldhaber’s  experiment.  The experiment was repeated by the Joliot-Curies[9] who found ovidence of “pair” formation, indicating that the the 2.62 MeV g-particle split into smaller g’s,  Thus,  the “neutron” mass may be lower than the accepted value. 

 

Looking at the data from the Table above, it will be noted that there is only one “mass” that is higher than the accepted value, and that all are questionable because of unknown additional particles or energies.

 

Thus,in summary, it seems certain that the currently accepted “neutron” mass is too high.  Note that it is probably lower than the sum of the masses accepted for the proton (1008725 AMU) plus an electron (.0.0000573AMU) which gives a total of 1.008398 AMU.

 

Let us consider the combining weight of “neutrons” for heavy elements [10],  As an example consider the addition of a “neutron” to 32Ge73 to form 32Ge74 The masses are [7]   73,921178 and 72.923459 AMU for a difference of 0.997719 AMU. Generally the mass difference for addition of a “”neutron” to a heavy element is very close to 1.0 AMU.  Remember there is no addition of energy on addition of a “neutron” to a heavy element [10]. (Thus there is no “relativistic” dimunition of mass) 

 

The  stable “heavy” elements run from 30Ni60 to 83Bi211 .  The masses concerned are 210.980383   and 59.930781. with a difference of 15104692 AMU.  The average is very nearly 1.0 as stated earlier.  Suppose there is a very small “binding energy” equal to the quotient of 0.0195021/151 or 0.003085 AMU or 0.306MeV per particle   (Something has to hold the “neutrons” in place.) 

 

 

C.    PROPERTIES OF “NEUTRONS”

 

 

Now it can be noted wny neutrons are called “neutrons” in my papers.  “Neutrons” are strongly believed to be equivalent to Hydrogen atoms, that is, a proton plus an electron.  As mentioned in an earlier paper[11], the proton in the stellar atmosphere must lose enough neutrinos to attain a mass of about 1.0 AMU to be stable in the high temperature conditions.  The “neutron” is a low mass proton plus an electron.  Once it is formed, it removes a  proton from  the fusion reaction process.  The neutron then awaits contact with a “heavy” element[10[. 

 

Where do the electrons come from not only that form “neutrons”, but that eventually become the “valence” electrons of atoms?  A paper by G. Fishman and D. Hartmann [12] explains how electrons (and positrons) are made from neutrinos (and antineutrinos).  There must be an extremely high temperature, probably 107 K or above, and there must be a tremendous number of neutrinos.  Both these conditions exist in the core of the sun [13].  Note that the neutrinos have spins, and are thus polar particles, and not rapidly expelled from the core, even though they have low masses (about 1/3 of an electron mass.)   Once the electron and the positron are formed at least two things can happen to each of them.  The electron can react with a proton to form a “neutron” and thus remove each from the “fusion” situation, or the electron can react with the positron to form a g-particle.,  The positron is a strong attractor and can attract a large number of neutrinos to become a proton.

 

We are thus left with coming up with some explanation of how the “neutrons” and “heavy elements” can be brought together in temperature regimes that are conducive to further reaction.  In other words, how are they removed from the core?

 

 

D.   CORONAL LOOPS, SUNSPOTS AND MAGNETISM

 

 

Heavy elements, particularly Fe, Co and Ni are made by fusion reactions in the solar core [10]. Consider particularly 26Fe56,3+ this is a triply charged ion that has 5 electrons with unpaired spins.

 

Possession of unpaired spins by an particle is one of the two ways that a magnetic field is produced,  The other method is in conjunction with an electrical current.  These two methods come together in coronal loops. 

 

Coronal loops are tubes of plasma that are consolidated by rotation of magnetic elements about the tube.  All components of the solar core are taken up by the loop, including “neutrons”. The rotating tube is projected outward and eventually makes its way, intact, to the solar surface.  Throughout the outward movement, the temperature decreases.  At the solar surface, the temperature is low enough that the “neutrons” and “heavy elements” can react.  Some of the reactions are slightly endothermic, which causes the local temperature to be lower than that of the photosphere, and thus appear as “sunspots”. Examination of the spectrum of the sunspots gives indication of, and abundances  of, most of the elements from 1H to 92U.[13.46ff].

 

(In the Appendix, it will be explained how  experiments may have  been designed and carried out to prove questionable theories.)

 

 

REFERENCES

 

1       E. Rutherford, Phil. Mag B, 21, 609. (1911)

 

2.  C. Dulaney, “Nuclear Size” *

 

3.A. Biesen, “Concepts of Modern Physucs”, McGraw-Hill, NY, (1969), 294ff

 

4.  C. Dulaney, “The Stationary Hydrogen Atom”

 

5. J. Chadwick, Proc. Roy. Soc. A, 136. 692 (1932)

 

6.  H.Semat, “Introduction to Atomic Physics”, Peinhart & Co,. NY, (1946), p308

 

7.  J. Lilley, “Nuclear Physics, Principles and Applications”J Wiley & Sons, NY,( 2001), 343ff

 

8..  J. Chadwick and M. Goldhaber, Proc. Roy. Soc. 181, 429, (1935)

 

9.  J.and F. Joliot-Curie, Comptes Rendu. 194, 703, (1935)

 

10 C.  Dulaney, “Heavy Elements”

 

11. C. Dulaney, “Moon-Hecht I”

 

12 C. Fishnman abd D. Hartmann, Sci. Amer., 277, July, 1997. P99

 

13. E. Novotny, “Introduction to Stellar Atmospheres and Interiors”, Oxford University Press, NY, (1973)p280ff

 

14. L. de Broglie, Ann. Phys. 3, 20, (1926)

 

15.C. Davisson and L. Germer, Phys. Rev. 30. 206 (1927)

 

16. G, Thomson, and A, Reid, Nature, 119, 896, (1927)

 

17. F. Sears, “Optics” Addison-Wesley, NY (1949) p270ff

 

18. R. Wollan and G. Shull, Phys. Rev, 73, 820, (1948)

 

19. T. Osgood, et al, “Atoms, Radition and Nuclei”, John Wiley&Sons, NY, 1955, p330

 

* Note that all my papers may be found  at:     http://mywebpage.netsvcape.com/clarencedulaney

 

 

 

APPENDIX     PARTICLES AS WAVES

 

 

In 1924, L. de Broglie was awarded a PhD in Paris [14]for his rhesis that particles can act as waves.  This disclosure led directly to “Wave Mechanics” of the quantum theory.

 

The disclosure also led to a flurry of rxperimental work attempting to demonstrate wave properties of particles.  One of the first of these was run by Davisson and Germer[15], who reflected fairly slow electrons (50-200V) off an “interesting” Ni surface.  The glass apparatus containing the Ni specimen broke while at high temperature and vacuum, causing the Ni to oxidize.  The apparatus was rebuilt, and the oxide reduced by heating in Hydrogen.  They then reflected the electron streams off the surface, and claimed they got the same kind of results the would have by reflecting x-rays (of about 0.5 Å wavelength) off a Nickel surface, although they did not produce any x-ray data.

 

In other words, they set out to show that electrons acted as waves, and they did so, at least to the satisfaction of the Nobel Prize Committee of 1937.

 

Other experiments were run particularly by G. P. Thomson[ [16] who passed somewhat faster electron streams through thin metal films, particularly gold.  He shared the 1937 Nobel.

 

About the same time the electron microscope was being developed to take advantage of the wave nature.  Given in bold type is a quotation from Sears [17.224].

 

“It should be pointed out that the ability of the electron microscope to form an image does not depend on the wave properties of electrons.  Their trajectories can be computed by treating them as charged particles, drflected by the electric and magnetic fields through which they move.  It is only when consideration of the limit of resolution arise that the electron wavelengths come into the picture,”

 

Sears[17.229ff] gives a formula for calculation of resolution of a (light) microscope which boils down to   r= Al,, where l is the wavelength,  the resolution decreases directly with the wavelength.

For electrons, the de Broglie wavelength is proportional to V-½,  Thus the “wavelength” of a 900 V electron is 0.408 Å.  This is smaller than the radius of the electron [4],  Thus, resolution would not be increased any more than if the electron were a simple charged particle.

 

The same is true for the diffraction experiments mentioned above.  All the effects could be calculated by assuming the elevtron is a simple, small, charged particle.

 

“NEUTRON” DIFFRACTION

 

G. O. Shull was awarded the Nobel Prize in 1994 (for work done in the 1940’s) for his work on “Neutron’ Diffraction”.  See a typical Laue type scintillation pattern he obtained in Osgood [19].

 

I have two comments about this work. 

 

1.    Remember that “neutrons” decay to a proton an electron and an antineutrino with a half life pf about 10 minutes.  If it takes say 20 minutes to set up the experiment, the “neutron” stream is essentially a proton and electron stream.

2.    A neutral particle such as a :neutron” cannot be detected with a scintillation screen.

 

I do not like being hoodwinked by establishment physicists, particularly by the Nobel Committee

 

 

 

© September 24, 2007, Clarence L. Dulaney