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Author’s Note:  This model is an un-reviewed analysis of isotopes.  It is an original work thought up independently by the author because of his amazement at the descriptions that he has read.  This work may not be reproduced or transmitted without the author’s permission.  The author apologizes for any error or any lack of understanding within this work.

 Isotopic Stability

             Isotopes that have masses recorded to the millionth of a gram per mole were analyzed in this work to gain insight into nuclear structure with regard to predicting radioactive behavior.  728 such isotopes in total were analyzed.  The hope was to make predictions regarding the twenty-one undiscovered elements up to the maximum possible atomic number of 137 resulting from the Bohr Model of 1913 (see below).

            Before continuing, it is worth noting some facts regarding nuclei.  Individual neutrons are radioactive and individual protons have a positive electrical charge.  There are no known nuclei of just neutrons.  There are no known nuclei of just protons greater in number than one.  Obviously, within limits, neutrons squelch proton repulsion in nuclei while protons squelch neutron radioactivity in nuclei.

            Mass defects based upon hydrogen-1 mass times nuclear proton number plus neutron mass times nuclear neutron number subtracted by isotope mass were calculated and plotted to yield the result below.  The ordinate has units of grams per mole.

Click here for Graph 1

            Another plot was prepared of Mass Defect versus Mass Number.  The ordinate has units of grams per mole.

Click here for Graph 2

            The author assumed that the density of protons and neutrons are identical.  As a result, some of the mass defect is possibly a consequence of neutron shrinkage to proton size to facilitate a closest packed arrangement.  The proton is known to be less massive than the neutron by 0.0008398441gmol^-1.  The resulting corrected mass defect plot is shown below and indicates that the author’s approach could be correct.  If correct, there are at least two contributors to mass defect.  The ordinate has units of grams per mole.

Click here for Graph 3

            A plot was then prepared of corrected mass defect per nucleon versus atomic number.  The ordinate values were multiplied by 1000 to yield mg per mol per nucleon.  The curve is obvious for predictions.  Curiously, while ⁵⁶Fe followed by ⁶⁰Ni and ⁶²Ni are the most stable isotopes in the plot, three of the twenty-two most stable isotopes are radioactive!  They are, in stability order, ⁵⁹Ni, ⁵³Mn and ⁶³Ni.  ⁵⁹Ni being the fifteenth most stable isotope.  The ordinate has units of grams per mole per nucleon.

Click here for Graph 4

Amazingly, the previous two plots indicate that nuclear stability is independent of nucleon number as the patterns do not distinguish between radioactive and non-radioactive isotopes!

            One other plot was prepared of percent neutrons among nuclear nucleons versus atomic number as a predictive aid.  It demonstrates a beautiful pattern that nicely illustrates the quantized nature of neutron addition to nuclei.  This is to be expected due to the apparent fact that neutrons add in whole number units.  Note how the percentage curves grow closer together as added neutrons represent a smaller fraction of total nucleon number.

Click here for Graph 5

            Since nuclear stability is not apparently related to mass number or atomic number alone, there must be another explanation for why some nuclei are stable and others are not.  Protons and neutrons are composed of quarks and apparently are stable in many isotopic compositions.  The only explanation for these observations that immediately suggests itself is that nuclear stability is thermal in nature.  Therefore, all nuclei, no doubt, have radioactive character.  ‘Stable’ nuclei merely have immeasurably long half-lives.  This is to be expected as all radioactive nuclei follow a first order decay relationship of ln[A]_t = -kt + ln[A]_o, where k = ln2/t_1/2.  The rate constant, k, is known to be temperature dependent in non-nuclear reactions.  It is suggested in this work that all nuclei become detectably radioactive at an appropriately high temperature.  The residual radioactivity in ‘stable’ isotopes could, partially at least, account for phenomenon such as background radiation and ‘cold fusion.’  It is therefore imperative to study the temperature dependence of radioactivity in order to make use of the Arrhenius equation, ln(k₂/k₁) = E_a/R(1/T₁-1/T₂).  Room temperature radioactivity could be compared to liquid nitrogen temperature radioactivity to provide useful data.  A better comprehension of quark behavior should result.

It seems that nuclei should be made to undergo fission by photon bombardment of sufficient energy to over come fission activation energy.  This phenomenon is observed with the fission of atmospheric nuclei under cosmic ray bombardment.  Large nuclei are known to produce energy when they undergo fission.  When large, stable nuclei are identified that become radioactive under a ‘slight’ temperature increase, they may become a useful power source under appropriate illumination.  This will be true especially if the fission daughters are stable and the light source’s power requirement is relatively small compared to the fission’s power output.

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Bohr Model Result

Niels Bohr’s assumption that m_eVr = (nh)/(2π) yields equation (a), V = (nh)/(2πm_er).  Since centripetal force must sum to 0 with potential energy, i.e., F_c + V = 0, results in equation (b), (m_eV²)/r + (-ke²)/r² = 0.  If (a) is plugged into (b) and solved for r, the result is r = (n²h²)/(4π²m_eke²).  Z⁺e^- is -e² which accounts for the negative potential energy.  In any case, assuming point charges and solving for V after r is determined yields a maximum Z of 137 before c is exceeded.  V = 0.846c for the 1s electron of element 116, ₁₁₆El.  Eka-dubnium of ground state electronic configuration, 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰7p⁶8s²6f¹⁴7d³, will likely be the final element of the Periodic Table.

Data Source

(1)  CRC Handbook of Chemistry and Physics, 79th ed., David R. Lide, Ed. CRC Press LLC, 1998, pp. 11-42 thru 11-149 & the inside back flap.

© April 2006 by Dr. Thomas F. Tekut

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Section I

Author’s Note:  This model is an un-reviewed model of gravity.  It is an original work thought up independently by the author because of his amazement at the descriptions of gravity that he has read.  This work may not be reproduced or transmitted without the author’s permission.  The author apologizes for any error or any lack of understanding within this work.

Gravity as a Natural Consequence of Relativity

            The current best model of gravity invokes gravitons as gravitational force transmitters.  The graviton is a particle of no rest mass that travels at the speed of light.  To date, no such particle has been detected, nor have gravity waves.  The graviton is part of string theory.  It is apparent that gravitons must has a mass equivalent when traveling at the speed of light.  Isolated matter should then lose mass as they emit gravitons.  No such mass loss is known.  Actually, no interaction of gravitons with matter at all is known.  It is postulated here that gravity is of a different origin.

            Image a point particle of the mass of a hydrogen atom existing in a positive, one dimensional universe.  It’s an x-axis universe.  This ‘universe’ is at a temperature of exactly 25^oC.  At the origin is the source of a gravitational ‘field’ that ‘exerts’ an acceleration of exactly 9.80ms^-2 at the hydrogen atom’s position.

            F=m(-a) on the ‘atom.’  It is (1.00794X10^-3kgmol^-1/6.0221415X10²³mol^-1)

(-9.80ms^-2) = -1.6402X10^-26N toward the origin.

            Imagine the ‘atom’ now emits two photons.  One heads toward the origin and the other heads away.  At the hydrogen’s surface,’ the photons have the same energy calculated by E=hcλ^-1.  E=mc², so mc²=hcλ^-1 and, therefore, m=hc^-1λ^-1.  Gravity attracts light so F=ma for the origin headed photon and F=m(-a) for the other photon.  Therefore, ΔF= hc^-1λ^-1(-a) - hc^-1λ^-1(a) = -2hac^-1λ^-1 = -4.4572X10^-36N.  Note that gravitational force is directly proportional to local ‘field’ strength and inversely proportional to the maximum wavelength of emitted light.  It is also directly proportional to Kelvin temperature.

            9.7192X10^-6m wavelength photon pairs, from Wien’s Law (λ_max = Constant/T)(1), heading either toward the hydrogen or away will exhibit the same ΔF.  The total force experienced by the hydrogen is then equivalent to 3679934930 such pairs of 25^oC photons!  Gravity is then, from an alternate viewpoint, the result of a relativistic time dilation in a ‘gravitational field’ altering photon emission and absorption times.  Gravity apparently originates from a particle itself interacting within its environment.

            How does the photon ‘know’ it is in a field?  Background radiation means that the universe is nothing more than a volume of photons.  All matter absorbs photons.  Furthermore, all matter of a temperature above 0K emits photons.  The wave nature of photons causes interference with the immediate photon environment at the particle ‘surface’ (i.e., the time environment).  In the future, the temperature dependence of gravity should to be measured to determine the actual relativistic contribution to the total.

            As a concluding observation, the author cannot help but note that Newton’s Law of Gravity, F = G (m₁/r)(m₂/r)  appears similar to a one dimensional rate law,  rate = k[][].

Data

c = 299792458ms^{-1  (}2)

h = 6.6260693X10^{-34Js  (}2)

H_mass = 1.00794X10^-3kgmol^-1

N_A = 6.0221415X10²³mol^{-1  (}2)

Wien’s Law constant = 2.8977685X10^{-3mK  (}2)

Sources

(1)  Foundations of Astronomy, 1994 ed.,  Michael A. Seeds, Wadworth, Publishing Co., 1994.

(2)  http://physics.nist.gov/cuu/Constants/

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Section II

Author’s Note:  This model is an un-reviewed model of a neutron star.  It is an original work thought up independently by the author because of his amazement at the descriptions of pulsars that he has read.  This work may not be reproduced or transmitted without the author’s permission.  The author apologizes for any error or any lack of understanding within this work.  This model is limited by the paucity of M1 facts available to the author.

Radioactive Model of a Neutron Star Hypothesis

            The current model of neutron stars neglects some facts.  The prevailing model has a ‘crust’ capping a liquid neutron interior.  The interior supposedly contains protons and electrons to account for the strong magnetic field.  If the neutron star formed by matter collapse into neutrons, how could the molten interior harbor protons and electrons?  If they are produced by neutron decay, would not the star heat up and burst as a consequence?  Also, the interior volume would likely change from the radioactive decay which would further weaken the star’s structure.  How could neutrons decay when under neutron star interior pressure?  In any case, how could a giant atomic nucleus, which is what a pulsar is, have a molten interior?  States apply to atoms and compounds not to nuclei.  Are there any atoms that have such a molten inner nucleus covered by a nuclear crust?

The smallest neutron star, or pulsar, would have a mass of 1.44 suns and a radius of about 10.km according to theory.  The sun’s mass is 1.9891X10³⁰kg.  M1, the famous Crab Nebula is thought to be a neutron star of such size.  This object spun once every 0.033099324s on June 28, 1969.  It rotation slows by 36.5ns per day.  The change in kinetic energy can be calculated by the equation, ΔK.E. = 0.5Iω₁² – 0.5Iω₂², where I is the rotational inertia and ω₁ is the angular velocity in radians on the given date and ω₂ is that on the succeeding day.  The result is a daily loss in rotational energy of 5X10³⁶J.

            What is the source of the pulsar’s energy loss?  It’s is hypothesized that since both large atomic nuclei as well as free neutrons are radioactive, the radioactive decay of neutrons in the pulsar is ultimately the source for M1’s energy loss.  Free neutrons decay with a half-life of 10.3min.  All large atomic nuclei decay.  A pulsar should behave no differently.  From E=mc², the rotational energy loss daily of the Crab Nebula would be 5X10¹⁹kg per day.  This represents a trivial daily mass loss that would allow a 155 million, or so, year existence for the pulsar if that rate was constant.  In any case, M1, in total, emits about 4X10³⁶J per day.  The two numbers have led investigators to conclude that pulsars somehow convert rotational energy to electromagnetic energy.  It should be noted that the luminosity of M1 is due to both the cloud’s emission and the pulsar’s emission.  Unfortunately, the mass, radius and luminosity of the pulsar have not been directly measured.  The assumed values are likely in error.

            When a neutron decays, a proton and an electron form releasing 1.2535X10^-13J of energy.  An antineutrino, υ, also forms.  The observed daily pulsar rotational energy loss would therefore be equivalent to 4X10⁴⁹ neutron disintegrations.  This corresponds to an E=mc² mass loss of approximately 5X10¹⁹kg.  At that rate, all neutrons will decay to protons and electrons in a little over 100,000 years!  No doubt these numbers are in error but the calculations are amusing.  Presumably, the pulsar decays in its surface layer.  The loss of neutron surface volume should be non-uniform over the surface since the neutron star is likely not a perfect sphere due to rotation.  Also, the surface cannot be uniform anymore after a neutron decays.  The consequence would be an occasional ‘gravitational resurfacing.’  This would cause a shrinking of equatorial radius resulting in a rotational rate increase.  This is observed as pulsar ‘glitches.’  The ‘glitches’ also imply that the pulsar is a rigid object.  At least the surface layer, if not the entire object, must be rigid.

            A neutron star’s magnetic field appears offset from the axis of rotation.  What is the magnetic field’s origin?  No doubt none arises from a neutral neutron interior.  The thought that gravitational contraction somehow concentrates a star’s magnetic field upon pulsar formation is illogical given that a neutron star is composed of neutrons having no electrical charge.  The field must be generated above the surface.  One field source would be surface movement of the electron and proton plasma from neutron decay.  The other, is the rotation of the charged plasma from the pulsar’s rotation.  The high pulsar temperature would prevent hydrogen formation by the ions.  Rotation would tend to move the plasma toward the equator.  The proton and electron plasma would emit north and south magnetic polar jets that result due to high temperature plasma leaving the surface along magnetic field lines.  The escaping plasma would get pulled toward the equator by centrifugal force.  The result is the observed offset magnetic field.  The offset angle would depend mainly upon relative jet and surface plasma masses.  Basically, the field and associated plasma can be considered as a separate object encapsulating a rotating neutron star.  The field’s plasma will prevent a direct view of the pulsar’s surface.  Also, the tidal interaction between the plasma jets and the neutron star would account for the pulsar’s slowing rotation rate and provide frictional heating.  Also, the jets will distort from the tidal drag.  Ultimately, the pulsar will cease rotation and the rotational component of the magnetic field will dissipate.  The field would flare up temporarily from ‘glitches’ allowing a temporary plasma flow.

            A simple mathematical model of the above external field idea leads from

m_jetωr_jet = m_eqωr_eq to m_jet/m_eq = cosθ/sinθ, where θ is the polar jet offset angle from the rotational axis.  The radii are from the axis of rotation to the surface.  Subscript, eq, refers to the equatorial plasma bulge.  The conclusion is that the jets actually contain more of the plasma mass as the angle decreases.  At a 45^o angle, the leaving jets and equatorial bulge are of equal mass.  Since the angle is likely close to 45^o, the pulsar must be losing mass quickly.  Perhaps, a 100,000 year lifespan is reasonable after all.  With around 1500 known pulsars one would expect a supernova every about ninety years.  With three visible supernovae in the last millennium and many opaque objects in the Milky Way, that might be near the correct rate.

The external magnetic field concept would apply to black holes as well.  Their jets should be more cylindrical due to their source being from a ring of matter encircling, and encapsulating, a larger spherical hole.  Also, a black hole’s magnetic field would be aligned with the axis of rotation since the field would result from inflowing matter not from surface as with a neutron star.  Of course, whether a singularity can rotate at all is another thought entirely.  The consequence of the field origin is that a black hole jet will not rotate in a ‘lighthouse’ manner.

            It is of interest to determine the pulsar’s temperature contribution to energy output.  It is thought that a young pulsar is at a temperature of 1X10⁶ million Kelvin.  Wien’s Law allows the thermal emission to be calculated as peaking at about 3nm.  This is in the X-ray region.  The Stefan-Boltzmann Law allows a value of 6.16X10³⁰J per day to be calculated for the emission.  This represents a negligible contribution to energy emission for the Crab pulsar.

            The correctness of this model can be ascertained by detection of emitted antineutrinos from the Crab nebula.  The antineutrino emission rate determination will allow an excellent estimate of the pulsar’s lifespan.  This might be done using a proton detector making use of the reaction, υ + ¹H à n + e⁺.  Such a detector might be a tube of hydrogen with detectors along its length.  The tube could be insulated from noise by encapsulating the length within a tank of hydrogen.  Such a detector should be used at high altitude to minimize atmospheric proton interference.  Interestingly, pulsars should create antimatter within space.

Data                                                                            Equations

M1:      radius = 10.km  (1)                                           K.E._rotation = 1/2Iω²

            Mass = 1.44sun (2)                                          I_sphere = (2/5)mr²

            T = 1X10⁶K  (1)                                                          ω = 2π(rotations/s)

            Period = 0.033099324s                                   E=mc²

                           on 6/28/69  (3)                                              Vsphere = (4/3)πr³

            period decay = 3.95X10^-8s/day  (4)                  S.A._sphere = 4πr²

            luminosity = 5X10³⁸erg/s  (5)                            Wien’s Law:  λm = 2.898X10^-3mK/T

(1)

solar mass = 1.9891X10³⁰kg  (6)                                  Stefan-Boltzmann Law:

                                                                                    E(Jm^-2s^-1) = σT⁴

neutron decay:  nàp⁺ + ^o_-β + 0.78235MeV  (6)           σ = 5.67X10^-8Jm^-2s^-1K^-4  (1)

Sources

(1)  Foundations of Astronomy, 1994 ed.,  Michael A. Seeds, Wadworth, Publishing Co., 1994.

(2)  http://en.wikipedia.org/wiki/Neutron_Star and others

(3)  www.space.com/reference/brit/nebulae/structure.html and others

(4)  http://www.seds.org/messier/m/m001.html

(5)  hyyp://www.core.org.cn/NR/rdonlyres/Physics/8-282JSpring2003/OF44051C-C729-4F36-B4F9-6563732C3A6F/0/ps9.pdf

(6)  Handbook of Chemistry and Physics, 79^th ed., David R. Lide, Ed., CRC Press LLC, 1998.

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Section III