occam Posted January 6, 2014 Share Posted January 6, 2014 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 1 Link to comment Share on other sites More sharing options...
imatfaal Posted January 6, 2014 Share Posted January 6, 2014 ! Moderator Note Graham / Occam Per our rules we like to have a summary / abstract of any questions or debating points on the forum - ie members should not have to click away from the site to unknown (and for all we know potentially infectious) places and files for the basic argument. ...Links, pictures and videos in posts should be relevant to the discussion, and members should be able to participate in the discussion without clicking any links or watching any videos... Thus could you post your ideas and what it is that you find interesting in your data. Thanks. Link to comment Share on other sites More sharing options...
occam Posted January 6, 2014 Author Share Posted January 6, 2014 (edited) The "Summary"is 34 pages long Will the site accept this? Graham Edited January 6, 2014 by occam Link to comment Share on other sites More sharing options...
imatfaal Posted January 6, 2014 Share Posted January 6, 2014 Graham No - that would not qualify under any reasonable definition of a scientific abstract or summary. Why do you have 47 meg of data? what if anything new does it show? can it confirm predictions or challenge established theories? Or have you found a new way to present this data, new calculations etc. And bearing in mind how you spelt "properties" are you sure of its initial validity and treatment thereafter - JK So can you find a happy medium between bare bones and 30-something pages? Link to comment Share on other sites More sharing options...
occam Posted January 7, 2014 Author Share Posted January 7, 2014 (edited) 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; Edited January 7, 2014 by occam Link to comment Share on other sites More sharing options...
imatfaal Posted January 7, 2014 Share Posted January 7, 2014 Graham - thanks for that. I hope someone with a little more expertise than I can take a look - maybe you might need to talk about a few more ideas and explaining where this is new. It seems a herculean task done with care and precision - nice one. and btw - for a task that some (really not me - I am way along the spectrum) might claim is a tad obsessive is a document titled "I so Anal.pdf" really appropriate I won't even go near StabAnal.xls Link to comment Share on other sites More sharing options...
occam Posted January 7, 2014 Author Share Posted January 7, 2014 Graham - thanks for that. I hope someone with a little more expertise than I can take a look - maybe you might need to talk about a few more ideas and explaining where this is new. It seems a herculean task done with care and precision - nice one. and btw - for a task that some (really not me - I am way along the spectrum) might claim is a tad obsessive is a document titled "I so Anal.pdf" really appropriate I won't even go near StabAnal.xls didnt spot the pun! and yes it took some time to extract the data Can you please correct the spelling in the topic name? Link to comment Share on other sites More sharing options...
Enthalpy Posted January 8, 2014 Share Posted January 8, 2014 With 40 nucleons, you have Ar and Ca, both stable. I find many curves interesting in your files, but am a bit mislead by your choice of words. "Nuclide" usually designates one couple of proton and neutron numbers, apparently it's their sum in your files. As well, an "Isotope" uses to designate variants of an element with a distinct number of neutrons. 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... Just as an exemple, page 5 of Isoanal1.pdf: Is it the density of the lone atom, the element in its standard form, the nucleus? I suppose the nucleus, but why a Rydberg constant then? What is the stable nuclide with 8 nucleons? Or did I misinterpret the curve on page 8 of Isoanal1.pdf? 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 wouldn't like to appear too critic - your work is impressive, and is the kind of broad compilation needed to extract intelligible patterns. ---------- (Later post) The correlation with the crystal form is astounding. Did you check its statistical significance? 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? ---------- 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 - 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. 1 Link to comment Share on other sites More sharing options...
occam Posted January 9, 2014 Author Share Posted January 9, 2014 (edited) 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. Edited January 9, 2014 by occam Link to comment Share on other sites More sharing options...
Enthalpy Posted January 10, 2014 Share Posted January 10, 2014 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. At least from en.Wiki: "The most abundant isotope, 40Ca, is theoretically unstable on energetic grounds, but its decay has not been observed." So unless you have better sources than Wiki, I read it as "stable until observed to decay" rather than "it must decay, just wait a little". Each nuclide has only one instance of a “fully stable” isotope. I usual wording: only one element is stable for a given atomic mass. For which I see in my usual http://www.webelements.com/ (choose an element there, click on Isotopes then) and checking at wiki's "isotopes of...": 64Ni and 64Zn (64Zn observationally stable, "believed to decay by double beta-plus" never observed) 58Fe and 58Ni (58Ni observationally stable, "believed to decay by double beta-plus" never observed) 84Kr and 84Sr (84Sr observationally stable, "believed to decay by double beta-plus" never observed) 86Kr and 86Sr (86Kr observationally stable, "believed to decay by double beta-minus" never observed) Until one or several of these decays are observed, I'd feel prudent to consider that the nuclides are stable. Link to comment Share on other sites More sharing options...
occam Posted January 11, 2014 Author Share Posted January 11, 2014 (edited) 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 Edited January 11, 2014 by occam Link to comment Share on other sites More sharing options...
Enthalpy Posted January 11, 2014 Share Posted January 11, 2014 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 What do you call Para 31? Does Codata give a mass difference between up and down quark, despite both are unknown (provided "mass" means something for them)? Or what do you call said "mass difference"? Link to comment Share on other sites More sharing options...
occam Posted January 12, 2014 Author Share Posted January 12, 2014 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 Link to comment Share on other sites More sharing options...
Enthalpy Posted January 13, 2014 Share Posted January 13, 2014 OK, claiming a known mass for quarks, and attributing it to codata was the bit in excess for me. There is no science in this operation. I give up. Fare well. Link to comment Share on other sites More sharing options...
occam Posted January 17, 2014 Author Share Posted January 17, 2014 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. Link to comment Share on other sites More sharing options...
Sensei Posted January 18, 2014 Share Posted January 18, 2014 (edited) 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 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. For a beta plus decay, the nucleus absorbs a shell electron and imports an external electron. You mixed beta decay+ with electron capture.. http://en.wikipedia.org/wiki/Electron_capture Beta decay+ is positron emission http://en.wikipedia.org/wiki/Positron_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. Edited January 18, 2014 by Sensei Link to comment Share on other sites More sharing options...
occam Posted January 22, 2014 Author Share Posted January 22, 2014 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_captureBeta decay+ is positron emissionhttp://en.wikipedia....sitron_emissionWhen 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? Link to comment Share on other sites More sharing options...
Sensei Posted January 22, 2014 Share Posted January 22, 2014 (edited) You put a lot of effort in preparing your pdf, and we appreciate it.But you should first learn how to calculate stuff.Otherwise it's quite wasted time (but wasted on good thing).I am pretty confident that you "forgot" to subtract energy of electrons from isotope energy-mass that you took from database of isotopes prior doing calcs.You should calculate just energy of nucleus, not nucleus with electron cloud! Let's analyze case that's showed in this page:http://en.wikipedia.org/wiki/Positron_emission11C → 11B + e+ + νe + 0.96 MeVCarbon-11 has mass11.01143361 u,you have to multiply it by 931.49406121 MeVand receive energy 10257.0850131232single electron has energy 0.510999 MeVCarbon has them 6so E-Eelectron*6 is10254.0190191232 MeV (approximate energy of nucleus of C-11) Boron-11 has mass11.00930544 u *931.49406121 = 10255.1026354069 MeV -0.510999*5=10252.5476404069 MeV Subtract them: 10254.0190191232-10252.5476404069 = 1.4713787162 MeV energy of positron that is escaping nucleus is 0.510999 MeV so subtract it as well will give0.9603797162 MeV And it's decay energy of this isotope. Pretty matching value from article. It's combined kinetic energy of positron plus energy of neutrino. If I wouldn't subtract electrons cloud energy, I would receive 1.47 MeV instead of 0.96 MeV, giving impression that nucleus is absorbing electron from cloud.. For a beta plus decay, the nucleus absorbs a shell electron and imports an external electron. No. In beta decay+, which is positron emission, nucleus is not absorbing shell electron, nor it doesn't need to import electron from environment.. Rather reverse. Carbon-11 is emitting positron, and losing electron from electron cloud because it's no longer hold by electrostatic forces (less protons in nucleus after decay). Positron will eventually annihilate with some electron ionizing other element and producing gamma photons which can ionize even more elements. Edited January 22, 2014 by Sensei Link to comment Share on other sites More sharing options...
swansont Posted January 22, 2014 Share Posted January 22, 2014 Moderators: Is there any way that the data files can be put on a trusted file sharing site? Not to be snarky, but the answer is yes: upload the data files to a file sharing site that you trust. SFN would not be involved in that — it's up to you. Link to comment Share on other sites More sharing options...
occam Posted January 23, 2014 Author Share Posted January 23, 2014 Not to be snarky, but the answer is yes: upload the data files to a file sharing site that you trust. SFN would not be involved in that — it's up to you. 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? Link to comment Share on other sites More sharing options...
swansont Posted January 23, 2014 Share Posted January 23, 2014 I don't think that's necessary. Link to comment Share on other sites More sharing options...
occam Posted January 30, 2014 Author Share Posted January 30, 2014 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. Link to comment Share on other sites More sharing options...
Sensei Posted January 30, 2014 Share Posted January 30, 2014 (edited) Similarly for 4He http://periodictable.com/Isotopes/002.4/index.full.dm.html which has 10 source Isotopes Not quite. Any alpha decaying isotope is source of Helium-4. Alpha particle is Helium-4 nucleus. http://en.wikipedia.org/wiki/Alpha_particle Edited January 30, 2014 by Sensei Link to comment Share on other sites More sharing options...
occam Posted February 3, 2014 Author Share Posted February 3, 2014 Not quite. Any alpha decaying isotope is source of Helium-4. Alpha particle is Helium-4 nucleus. http://en.wikipedia.org/wiki/Alpha_particle 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. Link to comment Share on other sites More sharing options...
occam Posted February 25, 2014 Author Share Posted February 25, 2014 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! Link to comment Share on other sites More sharing options...
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