THE STATIONARY HYDROGEN ATOM

By: Clarence L. Dulaney

2226 Fairgreen Drive

Missouri City, TX 77489

e-mail cldtx1@sbcglobal.net

 

Abstract:- The Bohr "solar atom" theory is shown to be incorrect in two  respects, electrical and mechanical. A model is developed in which the electron, in the ground state, is stationary, separated from the proton by one or more layers of small, neutral particles. This model readily explains how the spectrum arises and how the dark line spectra and fluorescence occurs. Chemical bonds and spatial isomerism are also readily explained.

An explanation of how electromagnetic radiation arises is given.

This paper was rewritten to correct errors pointed out by J. Alford in personal conversations in November 200111.  Any resulting errors are the fault of the author alone.

 

THE HYDROGEN SPECTRUM

One of the interesting things about the  spectra of atoms is that they are made up of relatively distinct lines.  The spectrum of Hydrogen is particularly simple because there is only one electron per atom.  It soon became evident that spectra were connected with electrons.

The study of the hydrogen atom spectrum in the late 1800’s and early 1900’s led to several empirical relations which are cited by Osgood,etal1. It was noted in 1885 by Balmer that the wavelengths of the visible spectrum of hydrogen could be represented by: l = 3645.6 n2/(n2-4) where n is an integer 3 or greater and l is the wavelength..

In 1889, Rydberg showed that the Balmer formula could be represented in terms of wavenumbers as: v = R ((1/22) - (1/n2)) where v = f/c, R = 109,678 cm-1 and n is an integer 3 or greater..

Lyman in 1906 showed that the lines of the ultraviolet spectrum of hydrogen can be expressed in terms of wavenumbers as: v = R (1 - (1/n2) where n is an integer 2 or greater..

In 1908, Paschen showed that for H lines in the infrared, v = R((1/32) - (1/n2)) where n is an integer 4 or larger.

In 1908, Ritz propounded his "Combination Principle" in which he noted that the wavenumbers (and thus the frequencies) of all H atom spectral lines could be expressed as the sum or difference between wavenumbers (or frequencies) of other H lines.

THE HYDROGEN ATOM

In 1911, Rutherford and co-workers2 irradiated gold films with a particles. Based on large scattering angles, they calculated that the gold nucleus containing the positive charge is about 10 -12 cm in diameter. (However, see my paper on "Nuclear Size"3.). This led to speculation that the electrons rotated about the tiny nuclei like planets about the sun.

Niels Bohr, then a student of Rutherford, in 19134 proposed a "solar" hydrogen atom with the electron rotating about the central proton.

According to basic electrical theory, two unlike charges that are free to move cannot be in equilibrium unless they are touching, (or separated by a barrier). Also, a charge that is rotating about a central force is accelerated continuously and as such would continuously radiate energy and rapidly fall into the center.

To overcome these objections, Bohr postulated that the electron would not radiate if it were in an orbit in which its angular momentum is an integral multiple of h/2p, where h is Planck’s constant.

However, the above is not the only problem with the "solar" hydrogen atom. Such a system is obviously one involving a central force. It is well known from basic principles of mechanics that r2w is a constant for bodies rotating in different orbits about the central force5. This does not hold true for Bohr’s hydrogen atom. For Bohr’s atom, r = n2(h2/4p2me2) and w = (1/n3)(8p3e4/h3). The two terms in parentheses not involving n do not change with orbit. Call them A and B respectively. For orbit 1, r12 = A2 and w = B, so the product is A2B. For n = 2, the product is 2A2B which is obviously different from the value for orbit 1. Any model with a continuously rotating electron is thus impossible. 

ENERGETICAL CONSIDERATIONS

Paraphrasing Osgood, page 95ff,   “The stationary state of the Hydrogen atom finds the electron in “orbit” n=1.  By various processes that add energy, the electron may find itself in any of the other states of higher energy.  Eventually the electron will fall toward the nucleus until it reaches the ground state, or some intermediate “orbit”.  In so doing, the total energy of the atom is reduced.  At each step, some energy is radiated as a photon of some particular frequency.  Eventually the electron falls back to orbit 1” 

This explanation seems to be incomplete, since it does not really explain the absorption spectrum, or particularly fluorescence, which is emission of a single line by an atom caused by absorption of energy and emission of a single,slightly lower energy line.  There does not appear to be any way for  the electron to have any “ground state”position than “orbit” 1.   

It is proposed that the “ground state for Hydrogen consists of a numbers of atoms with the electron in any possible “orbit”.  For  the emission spectrum to be generated, it is only necessary for enough energy to be added to the assemblage  to ionize the electron from a particular “orbit”.  For example, 1.8 electron volts would ionize electrons from “orbit” 3 (and from all those higher numbered).  Once the energy source is removed, the electrons start back toward the proton.  None can go back further than “orbit” 3.  

THE STATIONARY ATOM

To solve the problems of the Bohr atom, let us assume that the electron is stationary. and separated from the proton by layers of a small, neutral particle.  We will call layer one orbit one just from force of habit.  The second orbit is two particles away and so forth.  In our example above, the electron is ionized from orbit 3,  It then begins to return to higher orbits. At each layer, part of the energy is emitted as a ray of light.  This continues until level 3 at which point  the last of the energy is emitted.

If the electron from orbit 1 is ionized, this requires 13.6 ev.  The same sort of thing is repeated, and the entire emission spectrum of Hydrogen is emitted.

It is proposed that the small neutral particle is a neutrino.  Assuming this is so, what is the diameter of the neutrino?  It is reasonable to assume that the electron in orbit 1 requires 13.6 ev for ionization.   Thus 13.6 ev = -e2/d   and  d = (4.8 x 10-10 esu)2/(13.6 ev x 1.602 x 10-12 erg/ev) = 1.058Ε.  Orbit 2 is 2d away from the proton, orbit 3 is 3d away and so on.

What energy is required to start the process, and what is the radius of the first "orbit"? It seems reasonable that the Lyman series describes the first orbit. From above, the wave number v of the Lyman series is  v = R(1-(1/n2)) = f/c.  Looking at the first “orbit”, f = cR= 3.2903 x 1015 Hz.    The energy corresponding to this frequency is hf or  2.1792 x 10-11ergs. This amounts to 13.6ev. This is interesting, since 13.6 eV is the ionization energy for the Hydrogen atom.

After the ionization energy is removed from the atoms, they settle back down into an equilibrium situation in which the electron may be separated from the proton by any number of layers of neutrinos. The Balmer lines are generated when the electron in the second level is excited, the Paschen lines when the electron in the third level is activated, etc. See Table I for comparison of calculated and actual results.

The "Fraunhofer Dark Lines" that occur when an elementary spectrum is passed through the gaseous element are readily explained by this model. The energy of the elementary spectrum line excites an electron with the same energy up to the next level, with the energy of the line being absorbed in the process.

                       

TABLE I

SELECTED SPECTRAL DATA FOR HYDROGEN

…….."ORBITS"..  F x1014Hz……….l,CALC. , Ε…….l, ACTUAL8

……..1-2…………29.275……………1216……………...1215.7

……..1-3…………30.848……………1023.7……………1025.83

……..1-5…………31.990…………….949.7……………..949.76

……..2-3………….6.1742…………...6562.8……………6562.79

……..2-4………….6.9117…………..4861.3……………4861.33

……..2-6………….7.3140…………..4101.8……………4101.74

……..3-4………….1.5999………….18751…………….18751.1

……..3-5………….2.3405………….12818…………….12818.1

……..4-5………….0.74054…………40511…………….40500

……..5-6………….0.40227…………74577…………….74000                      

Fluorescence is also explained in that the energy of the exciting line does not have to be exactly the same as an electron level, just higher. The quiescent electron acquires enough energy to be raised by at least one level, and then falls back to a lower level, radiating a lower level wavelet compared to the exciting line. If any energy remains it is dissipated as heat.

Much simpler explanation of chemical covalent bonds, and of the geometrical isomerism of molecules may be given by the stationary atom model. For example, carbon atoms will have their four "valence" electrons at the vertices of a regular tetrahedron as the most stable configuration, although other configurations with slightly less stability may also be possible.

Covalent bonds can consist of two stationary electrons that actually are shared by the two atoms, just as pictured for years by chemists.

Since the electrons do not move about the bonded atoms, the bonds may be considerably more “fixed”, so that spatial isomers such as “boat” and “chair” forms of certain compounds may be physically separated, even though they only vary in physical arrangement of the atoms.

WHAT CAUSES RADIATION OF ELECTRODYNAMIC WAVES ?

It is noted by Pierce10 that when an electron is accelerated toward a positive charge, an electric wave traveling at speed c is emitted.  (See also my paper11, “C….and All That” .).  In all the cases discussed above, the excited electron ends up being accelerated toward the proton. 

The electron’s energy is the difference between the starting level, and that of the next inner neutrino level.  The speed the electron attains is determined by the kinetic energy.  Even at the highest possible energy, 13.6ev, the velocity is only a negligible fraction of c11, 

The electron is brought to a complete, at least momentary, stop by collision with the next neutrino layer.  A neutrino is knocked out of this layer, and is attached to the electromagnetic wave, the assemblage going off at speed c, with the energy the electron had when moving.  Half of the “rays” have a spin of +½, and the other half a spin of -½, both caused by the neutrinos.  These spins account for the  polarizability of the light.

The electron, if it has any residual energy, is once again accelerated toward the proton, and the process repeats itself, except for a different energy.  Because of the huge number of atoms involved, the entire spectrum is accounted for in this fashion.

.

 

SUMMARY

The Bohr model of the hydrogen atom is shown to be impossible from electrical, and mechanical considerations, as is any "solar" atom.

A model with a stationary electron is proposed. In this model, the electron is separated from the proton by one or more layers of small, neutral particles, proposed to be neutrinos. The spectrum of the atom arises from jumps of the excited electron between various "orbits", with the energy of the various lines equal to the differences of energies between the starting and ending orbits..

This model removes the main objection to decrease of charge with speed6, since the  atom can have zero residual charge with no “solar”electron.

This model readily explains the "Fraunhofer Dark Lines", fluorescence, spatial isomerism and other chemical properties of molecules.

An explanation of how electrodynamic radiation arises is given.

BIBLIOGRAPHY

[1] T. Osgood, et. al., "Atoms, Radiation and Nuclei" , John Wiley & Sons, NY, (1959), p 93-98.

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

[3] C. Dulaney, “Nuclear Size"

[4] N. Bohr, Phil Mag. 26, 1, 857 (1913)

[5] M. Spiegel, "Theoretical Mechanics" , Schaum’s Outline Series, McGraw-Hill, NY (1967) p122.

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

[7] C. Dulaney, “Planck’s Constant and the Energy of Radiation"

[8] G. Herzberg, "Atomic Spectra and Atomic Structure" , Dover, NY (1945),, p24

[9] S. Attwood, " Electric and Magnetic Fields". , Dover, NY, (1967), p462

[10] C. Dulaney, "Why ‘C’"

[11] C. Dulaney, “C, Charged Particles, Electrodynamics,  Radiation and all That” 

[12] Personal Conversation with J. Alford, Nov. 10, 2001, e-mail jephysics@aol.com

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

© 8/8/2006 by Clarence L. Dulaney