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No look at IsosortB there is no neutron bombardment! the decay mode is beta minus, which adds an extra shell electron at each decay stage.

As you can now read IsosortB I presume you have noticed that Uranium is produced from Thorium

I suppose it hasn't helped that all the access permissions were set incorrectly! Could someone please check that all the spreadsheets are accessible and downloadable? Thanks

This is the same data as you will find in IsosortB, column AE lines 483 to 493 What point are you trying to make?

Not quite 238U is the longest lived of the 238 sequence, which commences 238Th for b- chain, and 242Fm (alpha) to 238Cf and then b+ to 238U The further (shorter life) decays happen for all subsequent decays for all sequences above 208 The point of the "source" is there are no precursors, so these must be the primary products of all stellar processes including supernovas

Actually there are two possible decay paths from 238U see http://periodictable.com/Isotopes/092.238/index.full.dm.html The most probable (alpha decay) can ultimately lead to 206Pb or 208Pb. the other most unlikely mode is a 2b- decay to 238Pu which yields several of the rare earths 235U has four definite possible paths see http://periodictable.com/Isotopes/092.235/index.full.dm.html the most probable leading to 207Pb with a very low probability to 206Pb 238U is not a "source" isotope for this see http://periodictable.com/Isotopes/092.238/index2.full.dm.html

Apologies I set the permissions incorrectly I hope I have now changed it,could someone check please. You will find larger versions of the graphs in the spreadsheet

Some observations on abundance: The data for this is in spreadsheet Fe56-58.xls which can be downloaded from https://drive.google.com/file/d/0B3pdkE0Liyu2SFJlYVpWclFueVk/edit?usp=sharing From this the nuclear density curves can be drawn for the decay chains leading to the stable Isotopes. As can be seen the diagram for 56Fe is considerably more complex than those for 57 and 58. This complexity is mainly due to the multiple decay modes of some Isotopes. The data tables therefore also show the relative probabilities of the daughter Isotopes. It is clear from the data that the most probable decay mode also has the highest decay energy. This allows a reasonable estimate of the proportions even where a probability is not given in the source data. From the raw data, 56Fe has seven source isotopes; 57Fe has six source isotopes; and 58Fe has just four. But this simple view can be misleading. In 56 Fe the chain from 61AS has only a 0.023% probability in the decay from 60Ge to 56Ni. Similarly the decay from 57Ti is only 0.3% and from 57V only 0.4%. In the 57 chains only 58Sc has a full yield to 57Fe. The decay of 59Sc gives only 30% leading to 57Fe; similarly 58Ga allows 35% from 57Zn; 60As 35% also from 57Zn; and 59Ga only 3% at 58Zn. The remaining chain 57Ca is shared with 56Fe which takes 30% of the initial decay and another 33% at 57Sc. The net yield to 57Fe is 66% of 70% or only 46.2% For 58Fe the chain from 59Sc is 70% in favour, but at 59V less than 1% continues to become 58Fe. The chain from 59Ga (and 59Zn) effectively terminates at the extremely long lived 2 beta decay 58Ni. Leaving just the chain from 58Sc, and the OP2 58Co.

I was expecting to see the “orphan parent” isotopes as products, or by-products of the high energy fusion events. However I cannot find any reference as to how these isotopes are created. The only thing I can think of is that the pressure wave has two stages: the high pressure producing the fusion events, and the low pressure producing the oprphan parents.

Yes the 12 year 3H to 3He is the longest of all the steps but not in the lifetime of a star. However it could account for the comparitively much larger abundance of 4He. I add them together to show the total energy produced. The key point is that the fusion process generates the source isotpes as the starting point. the solar energy output is produced by the decay process

Some helpful pictures 3He.jpg shows the density curves for the decay chains from source isotopes to 3He 4He.jpg shows the density curves for the decay chains from source isotopes to 4He Helium energy.jpg shows the total energy release by the decay of each isotope to the helium end product The supporting data sheet can be downloaded from https://drive.google.com/file/d/0B3pdkE0Liyu2SmUyZGxmSUp6OUU/edit?usp=sharing Note no quarks anywhere!

Quite right, but these are alpha emitters 106Te has the lowest number of nucleons, but the vast majority of alpha emitters have over 200 nucleons. The “source" isotopes are quite different. they are not alpha emitters, instead these condense individually to become 4He atoms The data sheets are now available. They are all Excel spreadsheets: Isodata.xls is the main data capture document https://drive.google.com/file/d/0B3pdkE0Liyu2YTZ2bVFORFVzYTQ/edit?usp=sharing IsosortA.xls is the same data sorted by nuclide number, and then by Proton number. This contains the initial sorting codes and colour highlights https://drive.google.com/file/d/0B3pdkE0Liyu2QXdad3FReFZlaFk/edit?usp=sharing Stabanal.xls is the analysis of the stable isoropes (including the 2beta stable) https://drive.google.com/file/d/0B3pdkE0Liyu2eldSby04clRvTlU/edit?usp=sharing IsosortB.xls expands the sorting codes, and contains the mass calculations https://drive.google.com/file/d/0B3pdkE0Liyu2UGFaZ3VMSEhFcWM/edit?usp=sharing OPanal.xls analyses the “orphan Parents” https://drive.google.com/file/d/0B3pdkE0Liyu2ZzloUlZPdkJITEE/edit?usp=sharing DEanal.xls analyses decay energy patterns https://drive.google.com/file/d/0B3pdkE0Liyu2OWZiUDBqeEhGbGs/edit?usp=sharing headanal.xls looks at the isotoes which ar the “heads” of the nuclide number set https://drive.google.com/file/d/0B3pdkE0Liyu2RDVsY2t2YWFzSEk/edit?usp=sharing ANMandDensity.xls has the main curves for these values https://drive.google.com/file/d/0B3pdkE0Liyu2SGJEMFdjd19wOUk/edit?usp=sharing Hope these help.

Still waiting for my hosting site Another finding, buried in the sea of data involves the “Orphan Parents”. If you look at the chain of decays that leads to 3He in http://periodictable.com/Isotopes/002.3/index.full.dm.html you will see that there are only 5 “source” Isotopes that lead to 3He. Similarly for 4He http://periodictable.com/Isotopes/002.4/index.full.dm.html which has 10 source Isotopes 6Li has only 2 http://periodictable.com/Isotopes/003.6/index.full.dm.html and 7Li has 5 http://periodictable.com/Isotopes/003.7/index.full.dm.html and so on; 9Be has 3; 10B has 3; 11B has 4; 12C has 6; 13C has 9 You will notice from Paragraph 22 of Isoanal1.pdf that the density of these nuclei decrease as the number increases. It has been known for some time that there are pressure waves reverberating in the Sun. The consequence is that there must be a “convergence zone” in the depths of the star where these interfere and produce pressure peaks. The implication of the data is that it is not the pressure that causes fusion, but the “trough” which allows the source Isotopes to precipitate. The resultant decay produces another pressure pulse perpetuating the cycle. The deeper the trough, the higher the isotope number.

OK I think I have found a suitable site. When I have uploaded the files would you like to check it out before I post the URL in the forum?

There is also large energy-mass of neutrino. And kinetic energy of emitted electron/positron. You can see it in Cloud Chamber and compare with other radioactive fission of other element to see difference. As I said in Paragraph 32 of Isoanal1.pdf, the data is not refined enough to identify the neutrino. As for the kinetic energy, this should show as a discrepancy between the mass equivalent of the emitted frequency and the total decay energy. However I was unable to find any comprehensive list of emission frequencies to establish this. You mixed beta decay+ with electron capture.. http://en.wikipedia....lectron_capture Beta decay+ is positron emission http://en.wikipedia....sitron_emission When electron is absorbed by nucleus (in electron capture), one of protons change to neutron. So atomic number Z is decreased, and mass number A remain the same. There is no need to import yet another electron, as far as I am aware. No I did not mix them up. The data sheets clearly identify electron absorbtion from beta+ decays. For the beya plus an extra electron mass is required to balance the equation Moderators: Is there any way that the data files can be put on a trusted file sharing site?

1. Type “quark mass” into google. You will find http://hyperphysics.phy-astr.gsu.edu/hbase/particles/quark.html where you will find a table with these values and remarks: Up quark mass 1.7 – 3.3 Mev Down quark mass 4.1 – 5.8 Mev The numbers in the table are very different from numbers previously quoted and are based on the July 2010 summary in Journal of Physics G, Review of Particle Physics, Particle Data Group. A summary can be found on the LBL site. These masses represent a strong departure from earlier approaches which treated the masses for the U and D as about 1/3 the mass of a proton, since in the quark model the proton has three quarks 2. Take any beta decay and calculate the change in nuclear mass. For a beta minus decay the nucleus loses two electrons, one of which becomes a shell electron, and the other is emitted For a beta plus decay, the nucleus absorbs a shell electron and imports an external electron.

Para. 31 of Isoanal1.pdf Up Quark = 5 Mev = 8.91x10^-30Kg = 0.0053677AMU Down Quark = 6 Mev = 1.07x10^-29Kg = 0.0064413AMU Difference = 1 Mev = 1.78x10^-30Kg = 0.0010735AMU

There are 41 instances of 2beta-minus and 46 instances of 2beta-plus. Most of these are stated to have half lives in excess of 10^6 years and the majority of 2 beta-plus have negative decay energies. So yes I agree, from a chemical point of view these are stable. The point I wish to make is that these have the effect of blocking the beta decay sequence in the “nuclide number” set and in all cases there is an “orphan” isotope that sits one step away from the node of the sequence (where the decay mode changes from minus to plus). I’m surprised nobody has asked me about the significance of Para 31. CODATA gives the difference in mass between an up and down quark as 0.00107AMU Two electron masses is 0.00110AMU I think the textbooks may need some amendments

First then, the more general issues: “I.. am a bit mislead by your choice of words. Nuclide" Yes. I hit a language problem. In my interpretation for instance of "nuclide 40” I mean any combination of protons and neutrons that add up to 40. “an "Isotope" uses to designate variants of an element with a distinct number of neutrons.” I decided to edit the reply: That is the conventional wisdom. What this analysis shows is that the "set" of isotopes is determined by the "A" number ( the number of nucleons) rather than the "Z" number (the number of protons) “I also wish more detailed captions at the curves, for instance units, explanations of what the axes represent, where numbers start if it's not zero...” That is the practical consequence of pasting the graphs into a Word document. This is where you really need to have copies of the spreadsheets, where this sort of detail can be readily seen “With 40 nucleons, you have Ar and Ca, both stable.” 40Ar is the only fully stable isotope in this group. 40Ca is identified as having a double beta plus decay mode. It has a very long half-life, so from a chemistry point of view it is regarded as stable. There are quite a number of these double beta types, and all have very long half lives, and are regarded as chemically stable. They share two notable charachteristics: where they occur there is always an “orphan” single beta decay isotope that feeds the fully stable isotope, and secondly as shown in para 11 they do not fit in to the stability curves “: Is it the density of the lone atom, the element in its standard form, the nucleus? I suppose the nucleus” Yes it is the density of the nucleus “but why a Rydberg constant then?” Good question. And more importantly it is 2p/Ro this is one of the “discoveries” and I can only speculate: The Rydberg constant defines the spectral properties. There is another paper that I am working on which is too premature to print, but it would appear that there is an absolute permissible increment in spin angular momentum which defines the possibility of an Isotope. What can be inferred is that the increments ensure that every isotope has a distinct spectral signature. We can also infer from the absence of gaps in the graph is that these are all the isotopes that can be. “What is the stable nuclide with 8 nucleons? Or did I misinterpret the curve on page 8 of Isoanal1.pdf?” There isn’t one! Its not just 8 also 5 and a few more “By the way, if someone can explain me why two alphas need an additional neutron to bind, but three and four don't, thanks in advance!” I’m not sure what you mean here “I wouldn't like to appear too critic - your work is impressive, and is the kind of broad compilation needed to extract intelligible patterns.” “Which crystal form did you choose for each element? Most have several ones, not mentioning the compounds of these elements. While we can influence the crystal form, this has no effect on the decay energy nor period. So would the correlation, if significant, indicate that the nucleons organise with caracteristic numbers (not periods neither) that, at least over some interval, resemble the electronic shells?” I used the default description from the database at Periodictable.com. Also from reading various descriptions in Wiki there were several instances where a decay is described as being in two stages (although the data only shows a single figure for decay energy) “You checked for periods in the curves. Older theories wanted to see so-called magic numbers; these worked a bit better than periods, though not satisfactorily, as for instance 56Co is unstable but 56Fe stable. Possibly an interesting extension would be to subtract the contribution of electronstatic repulsion from the mean nucleon mass. This would: - Show more clearly if there are patterns in the nucleon attraction versus their number Not much into magic numbers ! “- Tell if the protons spread uniformly in the nucleus, or group a little bit at the surface” “one difficulty being that big nuclei are not all spherical.” There is possibly a solution to the geometry of the nucleus, but it’s a bit premature to discuss these now.

didnt spot the pun! and yes it took some time to extract the data Can you please correct the spelling in the topic name?

OK, There are 3,150 isotopes. If these are arranged by the Nuclide number, rather than the Atomic (proton) number, distinct patterns appear. There are 293 “nuclide numbers of which only 196 have fully stable isotopes and Each nuclide has only one instance of a “fully stable” isotope. I can't seem to include a table, or attach a partia excel spreadsheet One of the findings is that the stable isotope sits in a “well” of sequential Beta decays. The lower “Z” numbers in Beta- decays, and the Higher “Z” numbers in Beta+ decays. Each nuclide chain is fed by an “orphan” isotope, those with Neutron decays always feed Beta- decay chains; Proton decays always feed beta+ decay chains;

The "Summary"is 34 pages long Will the site accept this? Graham

I have done an analysis of the nuclear properties of Isotpes which you may find interesting. The summary is attached. The 8 data sheets supporting this total about 47 MB! If you contact me at gh_jyhughes@hotmail.com, I can send you copies Graham Tut! now attached Isoanal1.pdf Isoanal1.pdf Isoanal1.pdf

Actually it must be cyclic. If you look at Isotopes, there are a finite number which are "stable" Therefore if Entropy takes its course, and all the stars are burnt out, and all the possible isotope decays take place, then all we are left with is the stable particles. The only force left is gravity. These Isotopes, by definition, are irreducible, so what we are left with is an implosion, which collapses the particles to (presumably) the Planck density. Where (and I speculate) we get a "big bang" reversal, which generates a new universe. Chew on that Occam

It’s gone quiet. There are two things missing from the Standard Model and QCD: Space-time, and Gravity. The key to this question is Density. Consider this table of numbers: It is in .csv format and you may find it useful to put these into a spreadsheet and draw a line graph of the four columns.. -1,-4,-30,-88 -2,-7,-27,-82 -3,-10,-24,-76 -4,-13,-21,-70 -5,-16,-18,-64 -6,-19,-15,-58 -7,-22,-12,-52 -8,-25,-9,-46 -9,-28,-6,-40 -10,-31,-3,-34 -11,-34,0,-28 -12,-37,3,-22 -13,-40,6,-16 -14,-43,9,-10 -15,-46,12,-4 -16,-49,15,2 -17,-52,18,8 -18,-55,21,14 -19,-58,24,20 -20,-61,27,26 -21,-64,30,32 -22,-67,33,38 -23,-70,36,44 -24,-73,39,50 -25,-76,42,56 -26,-79,45,62 -27,-82,48,68 -28,-85,51,74 -29,-88,54,80 -30,-91,57,86 -31,-94,60,92 -32,-97,63,98 -33,-100,66,104 -34,-103,69,110 -35,-106,72,116 The figures are exponentials. The first column is “wavelength” -2 is centimetre waves, -6 is optical wavelengths, up to the Planck limit~-35. Column 2 makes the assumption wave length = wave width = diameter and gives the notional spherical volume of the boson. The Numerical value of the constants is in SI units, and to obtain the conversion values these must be considered against a time base of one second. Therefore the crucial calculation is how many “Boson volumes” will fit into a volume 1 light second in diameter. This gives “space-time – Volume per second” Using the first constant h we can calculate the “Energy Density” and thus the coulomb force exerted by the volume. This is column 3. Now in Column 4 we can use the second constant h/c^2 to calculate the “Mass density” and thus the gravitational force. From the data you can see that for each order of magnitude decrease in wavelength the Coulomb force increases by 3 orders of magnitude, but the Gravitational force increases by 6 orders of magnitude. Most significant is that these curves cross. At a wavelength of 6.8967 x10-21 metres the forces are equal, and at shorter wavelengths, or greater “density” the gravitational force is dominant. The consequence is that at a lesser density the Boson momentum is linear, and is observed as a Photon. At higher densities the momentum is a closed angular momentum, and the Boson is observed as a “gluon” or more properly is the Graviton. The full explanation can be found in the thread; http://www.scienceforums.net/forum/showthread.php?t=42325&highlight=elephant