Excess neutron shell model of Nuclei

by Bhagirath Shantilal Joshi
Msc Solid state Physics, Gujartat University, Gujarat, India
MS Computer Engineering, Univ of Lowell, Lowell, Mass, USA

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Abstract

Review of the periodic table and existing research on isotopes of the various elements of the periodic table is conducted and the attempt is made here to visualize the process of element formation.

The role of Neutron in element formation is investigated and found to be vital for existence of elements . In the new light, the new model of nucleus is proposed which explains the stability of the  nuclei and reason for multiple stable isotopes of elements.

Review of the Periodic Table

The inspection of the periodic table of elements reveal an interesting fact that for all elements other than Hydrogen, for element to be stable, number of Neutrons are always greater or equal, to the number of Protons in the nucleus.  The periodic table is analyzed with respect to the abundance in nature for the elements.

The fusion of Hydrogen to produce Helium can be envisioned as follows:

1H + 1H + 1H + 1H →  2H + 2N + 2e+ (1)

where 2e+ later annihilating to produce additional gamma radiation.

Since tremendous force is needed to keep four protons together during the formation of helium and emit 2 positrons, it is unlikely.  The mass of neutron is more than the mass of proton and this fact  alone negates the likeliness of the above scenario.

In another scenario, the formation of helium may be envisioned as four neutrons coming close together to form a stable nuclei of Helium as follows:
4N → 2N + 2P + 2e (2)
Here the mass of Neutron is higher than the Proton.  The Neutron releases electron to form proton. Also justifies the reason for the  Proton to be lighter than the Neutron.
Thus it is assumed that
  1. Neutrons are the building blocks for elements in nature.
  2. In the elements , other than hydrogen, neutron and proton form a pair (np) and keeps distinct identity.
  3. Excess neutrons stay at the center of the nuclei but keeps their distinct identity.
Using the above assumption the periodic table is analyzed by finding excess neutrons for all  stable isotopes of elements as follows:
Excess neutrons = Atomic mass – 2P where P is the number of Protons for the element
The relative abundance of Isotopes is obtained from the research papers and wikipedia and is included in the Table 1, 2 ,3. The relative abundance is indicative of the preferred state for the element (nuclei) in nature.
The following facts are found as a result of analysis:
For Odd atomic number nuclei:
  • Abundance of isotope is close to 100% when excess neutrons are odd count.
  • If the excess neutron count is even, the isotope is radio active.
  • Majority elements have only one Isotope.
  • Only Potassium has three isotopes, the one with 2 excess neutron is radio active with 0.001 relative abundance.
For Even atomic number nuclei:
  • Elements He, C, O, Ne, Mg, Si, S are close to one in abundance with no excess neutron and maximum three isotopes.
  • Element Be has one excess neutron and one isotope, 100% abundance.
  • All other higher elements require four or more excess neutrons for the element to be abundant in nature and has up to 10 stable isotopes.
  • Element Ca is exception with 0 excess neutron and 97% abundance. However it is unstable with >E+21 a Half life.
Atomic number 43 (Tc) and 61 (Pm) has no stable isotopes in nature.
Thus from the above analysis, it looks like that all nuclei, stable or otherwise,  prefers to keep at least one neutron at the center of the nuclei from its excess count. The remaining excess neutrons stay very close to the center keeping its own identity. The other neutrons pair up with each proton and stay close together to the proton (like heavy hydrogen) but maintains its separate identity.

The nature prefers, for more complex nucleus, more neutrons than protons and creates a delicate balance to form a stable nuclei.

This balance of forces is so critical that in case of element F (the stable isotope is with  9 proton and 10 neutrons) The isotope 18F with 9 Protons and 9 Neutrons, with in 20 minutes decays and forms 18O which is stable with 8 protons and 10 Neutrons and gives up a e+ positron to convert proton to neutron and e+ and e- reaction produces Gama radiation.

Equation   18F → 18O + e+

Thus the existence of Neutron is vital to the existence of the universe itself, because without neutron, elements may not have been possible and hence the intelligence as we know today.

Structure of Neutron

The fact that Gravitational binding forces (Fg) of masses in the nucleus needs to be more than the destructive electromagnetic forces (Fem) created by the electrically positive environment of the nucleus,  (Fem < Fg) excess neutrons are required in the nuclei. However, even at the center of the nucleus neutrons keep their distinct identity, rather than lumping together to form one heavy neutron. Therefore, neutron can not be just a fuzzy mass but a well defined structure able to hold induced charges with precise demarcation of boundary with an insulating layer, more like a free standing capacitor. Which suggests that at least three particles with two sets of characteristics are required to create a neutron. For our model here, that has to be two particles capable of holding charges and one particle which provides insulation between these particles, similar to that of dielectric layer in the capacitor. Thus in this postulated model the neutron looks and behaves neutral for all practical purposes. However, within nuclei keeps separate and distinct identity and does not combine with other neutrons to form a one central heavy body.

The proposal here defers from the model independent analysis predicting the structure having negative charge on the surface, Positive charge in the next lower level and than neutral  mass. My model suggests that neutron is a free standing charged capacitor. This helps in keeping the distance between excess neutron in the nuclei. Otherwise there is nothing stopping these neutrons from lumping together to form a mini neutron star with in the nuclei. This model of Neutron is justified from the known fact that Neutron is heavier than Proton.

For elements other than Hydrogen, the nuclei becomes a chaotic environment. For Example, in case of

4He, two protons require their space and wants to be separate from each other as far as possible, and two neutrons are caught between the two protons as shown in figure 1.  From the reference frame of protons it is envisioned that Neutrons spin on its axis at very high rotational speed. One clockwise and other anticlockwise.

Model of nuclei

A new model of nuclei is proposed as shown in the figure 3. A spherical shell of excess neutrons with one neutron at the center of the shell surrounded by paired proton neutron (pn) shell.

For the stable nuclei system the electromagnetic forces needs to be balanced and gravitational forces maximized.  From the example of 4He above, it is envisioned that  elements are built, in this model, by Heavy Hydrogen nuclei (pn) as a building block and just enough excess neutrons to provide needed gravitational force for stability.

Energy levels (orbitals) in pn outer shell follows the similar shell structure of electrons, with K,L,M,N,O being primary shell and s, p, d, f, g sub shells with similar total charge particle capacity. However for excess neutron shell it differs, where a single neutron stays at the center of the nuclei when in excess. The energy level for that central neutron in this model is called “Foundation neutron” (Fn).

The Table 4 shows the placement of neutrons in each shell.

Nuclei Stability Analysis using above model

For Odd atomic number nuclei:

In the ‘p-n shell’ the outermost pair has no symmetry, however the excess neutron shell is symmetrical for all odd atomic number elements giving the stability and abundance. Referring to table 2 and table 4 , isotopes, with one excess neutron, has relative abundance in nature of 1 or close to 1,  The excess neutron takes the place of Fn.  The isotopes with 2 excess neutrons are all radio active which can be attributed to the asymmetry of neutron in unfilled K shell, which can hold up to 2 neutrons. When the K shell is completely filled as in the case of 41K, the nuclei is stable. Table 2 outlines isotopes and abundance for majority of elements, all follow the same principle and model fits perfectly with the exception of 14N with atomic number 7. Iodine has 21 excess neutrons, as we see in table 4, the outer most excess neutron shell is symmetrical, making it stable. All other elements follows.

For Even atomic number nuclei:

The stability is obtained from the pn shell. However, excess neutron shell  becomes asymmetric and depending on which sub shell, there is a room for additional excess neutrons and hence exhibits many stable isotopes with relative abundance. Theoretical probability calculation may prove this fact.

Instability and radio activity in heavy elements:

The neutron proton (np) pair in nuclei has capability to grow indefinitely. However as the heavier elements are built the inner excess neutron shell is large enough to interact with lower np shells, thus giving instability to the nuclei. e.g. for 238U there are 148 total neutrons and 54 excess neutrons.  A nuclei with 111 excess neutron will completely fill up the O shell. All elements after 209Bi are radio active, this shows that interaction of neutrons form excess neutron shell(inner shell) with pn shell is destructive when the N shell is half filled. Thus there is a upper limit to the size of the  inner excess neutron shell and this may be the reason for non existence of super heavy stable elements in nature.

Conclusion

This proposed model of nuclei, explains the stability and relative abundance of the nuclei. The nuclei(except for H and He) in this model is a shell with in a shell, where an innermost shell structure is formed with excess neutrons with one Fundamental foundation neutron (Fn) and the rest arranged in shell structure. Protons, form pair with neutron and form shell enclosing this excess neutron shell as shown in figure 3. The stability of the nuclei is a direct result of symmetrical placement of neutrons in the outer most shell of excess neutron shell. The same shell is responsible for many stable isotopes in even atomic number nuclei. The additional isotopes are possible if the outer shell of excess neutron shell holds odd number of neutrons. The nature of neutron is postulated here, the proof of which depends on future experiments.

References

Numerous Scientists’ and scholars’ exhausting work in development of periodic table, investigation of Isotopes is utilized and due credits are given. The list is too large to print here, but due credits are mentally given to all scientists.

by Bhagirath Shantilal Joshi

796 comments to Excess neutron shell model of Nuclei

  • Wladimir Guglinski

    Dear Bhagirath Joshi

    what are the two charged particles of your neutron model ?

    Has the third particle a mass ? Or is it massless ?

  • alex viroli

    I’m an Italian electronic engineer and I really hope you succeed and maybe set your company in Italy

  • Enzo de Angelis

    Caro Ing. Rossi,
    non riesco a fare la mia prenotazione di un E-Cat Home sul sito e-Cat.com
    Può chiarirmi cortesemente come fare visto che ho più volte provato a inviare il format senza successo e mi segnala sempre che non accetta caratteri non alfabetici mentre in realtà sono tutti alfabetici?
    Grazie,
    cordiali saluti

  • I posted a story at PESN about how the alternative media has been ignoring the E-Cat, even as mainstream media has been giving it some coverage.

    Alternative Media fails ‘free energy’ promotion when it should take an activist, leadership role – Given the potential of ‘free energy’ to: 1) provide affordable back-up power in the case of a short- or long-term grid failure, 2) provide clean, cheap, distributed energy, 3) create zillions of jobs to reinvigorate the world’s economy, 4) instill hope in depressed times; the alternative media should be playing more of an instigating, leadership roll; not ignoring the topic. (PESN; November 25, 2011)

  • Wladimir Guglinski

    Dear Bhagirath Joshi

    The nuclear model of current Nuclear Physics has not a structure. The nucleons move chaotically within the nuclei.

    In my new nuclear model there is a central 2He4, which produces gravity “strings” (flux of gravitons), and they capture the nucleons. The nucleons captured by the strings form hexagonal floors, having the 2He4 in the center.
    So, in this point my nuclear model is similar to that proposed by you: the nucleons (deuterium) are distributed along a “shell”. So, the nucleus is empty, having only a central 2He4, and it has a shell of deuterons and excess neutrons.

    In your model the shell is also formed by deuterons (P-n shell), but there are excess neutrons in the center of the nucleus.

    Do you think is there a chance to make an experiment, so that to discover what really exists in the center of the nuclei ?

    Do you think the physicists have never tried such sort of experiment because they do not believe that it is possible to exist “something” in the center of nuclei?

  • Wladimir Guglinski

    … and Schrödinger wins the duel… again… ?

    According to Heisenberg, a theory had to be developed free of metaphysical concepts. When Schrödinger discovered the zitterbewegung in the Dirac’s equation of the electron, he interpreted it as a physical phenomenon, according to which the electron would be moving with helical trajectory.

    But a helical trajectory is a metaphysical concept, and from Heisenberg’s philosophical viewpoint it could not be accepted by the scientific criterion.

    So, while Schrödinger believed that Physics must be developed by keeping some fundamental physical concepts underlying the phenomena, unlike Heisenberg considered it disagree to the scientific criterion, from which any metaphysical concept would have to be eliminated in the development of a theory, in order that Theoretical Physics would have to be developed from mathematical concepts only.

    In July-2010 a new experiment had showed that Schrödinger wins the duel: the experiment shows that photons have helical trajectory.

    This new experiment was published in Phys. Rev. Letters in July 2010, under the title “Unveiling a Truncated Optical Lattice Associated with a Triangular Aperture Using Light’s Orbital Angular Momentum”

    So, this new experiment with the light shows that, concerning the zitterbewegung concept, Quantum Field Theory is wrong. The successor of Quantum Mechanics is not correct, since QFT does not consider the zitterbewegung as a physical helical trajectory of elementary particles.
    More details here:
    http://www.zpenergy.com/modules.php?name=News&file=article&sid=3209

    Other duel Schrödinger X Heisenberg is concerning the interpretation of the quantum states. Schrödinger and Einstein believed that a pure state is a physical property of system, much like position and momentum in classical mechanics. Unlike, Heisenber, Bohr, and others, were sure that a pure state has only a statistical significance, akin to a probability distribution in statistical mechanics.

    Now a new theoretical work, published in 14 November 2011, is suggesting that Schrödinger and Einstein were right.
    In a paper entitled The quantum state cannot be interpreted statistically, the authors show show that, “given only very mild assumptions, the statistical interpretation of the quantum state is inconsistent with the predictions of quantum theory.”
    http://xxx.lanl.gov/abs/1111.3328

    Probably Schrödinger believed that quantum state is a physical property of the system because he suspected that there is a connection between the quantum state and the helical trajectory of the particles.

    So, it seems that Schrödinger was right… again !!!

    And new experimental finding, new discoveries, and new theoretical researches are suggesting that Quantum Mechanics, after all, is wrong, or at least incomplete.

    Sure that, beyond the spread of Rossi’s eCat worldwide, big surprises are waiting for us in 2012.

  • Nick Pourmi

    Dear Andrea Rossi,

    The demand for your products will be huge. Don’t bother with sign-up lists.

    In my opinion, the best strategy to maximize the selling price for the future 10kW E-Cat home units would be to provide customers a 30-Day Money Back Guarantee. If they are dissatisfied with your product or service for any reason, they should receive a full refund. Yet, in this way, making them paying dearly, you may ponder the avalanche of requests to cope with while supplying capacity is still inherently too weak, as in any inception.

    At this point, can you envisage such a possibility when commercial sales of E-Cat home units will be available?

    Esteem and Respects,
    Nick Pourmi

  • Bhagirath Joshi

    Dear Wladimir Guglinski

    Your Question:

    November 25th, 2011 at 4:58 AM
    Dear Bhagirath Joshi,
    there are only two stable helium isotopes: 2He3 and 2He4.
    So, we have 3 questions:

    1- 2He3 has no any excess neutron. Unlike, it has one proton excess.
    So, it could not be stable, according to your theory.
    How does to explain it ?

    2- 2He4 has no excess neutron, and it is the most stable of helim isotopes. Why ?

    3- According to your theory the one excess neutron of 2He5 would have to occupy the center of the nucleus, and so it would have to be more stable than 2He4.
    Why is not 2He5 stable ?

    Answer:

    The even number light nuclei up to calcium are very vulnerable to gravitational forces and uneven mass distribution. He in particular , in my model envisioned as two np pairs for the most stable and 1p and 1np pair for 3He and rotating around like a dumb bell with center of mass perfectly at the center of the dumbbell for 4He and 1/3 way from np pair for 3He. The relative abundance of 3He is .00001 in nature.

    Just from the rudimentary calculations of Fg for 4He and 5He , the initial Delta Fg is 4 times larger. The introduction of neutron at the center pulls in np pairs by making impossible orbit. and hence unstable.

    From the quantum mechanical calculation, the orbit shrinks to smaller than the neutron and np pair wavelength.

    And this is where, I also think that the planks constant is a variant, as you proposed in your theory of neutron.

    Thanks
    Bhagirath Joshi

  • Andrea Rossi

    Dear Mark Saker:
    We are still working on all these issues.
    Warm Regards,
    A.R.

  • Mark Saker

    Dear Andrea Rossi,

    Have you had the gaskets changed and operated the plant at 1MW yet since the purchase and if so how often and how well is it operating?

    Has the 1MW plant been retrofitted to use the new liquid to attain 450 celsius or has this happened only on a single e-cat?

    If the 1MW is operating at the higher temperature, have you attached the output to a turbine yet to create electricity. How much electricity did it generate?

    Thankyou

  • Wladimir Guglinski

    Mr. insight wrote in November 25th, 2011 at 8:00 AM:

    Dear Bhagirath Joshi,
    Why does your model accuracy imply that those experiments will not find EDM? “

    Dear insight,
    I suppose it is because in his model B. Joshi considers the neutron as composed by two charged particles (and a third NO CHARGED particle which provides insulation between these particles).

    A finite EDM can only exist if the centers of the negative and positive charge distribution inside the particle do not coincide.

    The neutron model of Particle Physics is composed by three quarks (d,u,d). So, it’s hard to believe that the centers of three charged particles may coincide. Mainly because the charge of quark up is +2/3, and the charge of quark down is -1/3.
    Therefore, from the quark model (d,u,d) we would have to expect a no null neutron EDM.

    But two charged particles may have the centers of their distribution coincident, and so a neutron composed by two particles may have null EDM.
    It is the case, for instance, of the neutron model formed by proton+electron, proposed in Quantum Ring Theory.
    As the proton has charge +1, and the electron has charge -1, so the center of their charge distribution can be coincident.

  • Andrea Rossi

    Dear bernie Koppenhofer:
    The word of mouth will be our main marketing carrier: millions of customers against hundreds of plumbers trying to stop the Niagara Falls with bags of sand.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Giovanni Nobili:
    Thank you very much for your suggestion, I forwarded it to our informatic.
    I agree totally with your considerations regarding the wannabe “competition”: I will have competition when I will find products offered in the market; before that all we have is wannabe competition based on attempts of copy. Nevertheless, the tiny results that somebody has acheived starting from the publication of our patent are the evidence of the fact that we gave enough information in our patent to reproduce my effect, even if with an efficiency much smaller, due to the fact that the wannabe competition has not the know how of the catalizers. This is the reason why the wannabe competitors are extremely aggressive in their attempts to steal technology forcing us to yield more and more information through more and more tests: every test I make will never be enough for them: should I run 100 meters in 8 seconds on the water in the middle of a stormy ocean, they would say, you bet: ” Rossi is a hoax! At 62 he isn’t even able to swim !”
    They make chatters, we make plants.
    Warm Regards,
    A.R.

  • insight

    Dear Bhagirath Joshi,
    You say ” There is insulating layer of empty space between proton and electron in Hydrogen atom”. So why cannot the insulating layer you say is inside the neutron be of empty space too?
    You say “The fact that Gravitational binding forces (Fg) of masses in the nucleus needs to be more than the destructive electromagnetic forces (Fem) created by the electrically positive environment of the nucleus, (Fem < Fg) excess neutrons are required in the nuclei". What would happen if Fg were not more than Fem?
    You say "There are experiments underway to find the neutron EDM. If my model of neutron is accurate, it will not be possible for them to find EDM the way the experiment is setup". Why does your model accuracy imply that those experiments will not find EDM?
    Regards

  • Wladimir Guglinski

    Dear Bhagirath Joshi,
    there are only two stable helium isotopes: 2He3 and 2He4.
    So, we have 3 questions:

    1- 2He3 has no any excess neutron. Unlike, it has one proton excess.
    So, it could not be stable, according to your theory.
    How does to explain it ?

    2- 2He4 has no excess neutron, and it is the most stable of helim isotopes. Why ?

    3- According to your theory the one excess neutron of 2He5 would have to occupy the center of the nucleus, and so it would have to be more stable than 2He4.
    Why is not 2He5 stable ?

  • Giovanni Nobili

    Buongiorno Dott.Rossi,

    con la presente volevo solo consigliarLe di creare (se possibile) una discussione in rilievo (in modo tale che sia sempre a vista) nella pagina principale del journal of nuclear phisics per parlare dell’E-cat, in modo da lasciare liberi gli altri argomenti. Tutto questo per maggior leggibilità. Attualmente si incrociano discussioni riguardanti ad esempio la massa dell’elettrone con altre che riguardano esclusivamente l’e-cat. Per il resto Le auguro ogni successo economico dalla Sua splendida idea e iniziativa, in barba ai competitors tardivi (teorema Artù-Merlino: se Merlino con la magia dice ad Artù che esiste una soluzione ad un problema – seppur complicato o np-completo, Artù è immensamente facilitato a trovarla – troppo comodo!).
    Cordiali saluti,

    Giovanni Nobili

  • Bernie Koppenhofer

    Mr. Rossi: THANK YOU! It will be an honor to have the first E-Cat in Fort Collin, Colorado. In this age of the internet, “word of mouth” marketing is powerful. I promise to use “word of mouth” to accurately describe my experience using your historic and critically important new invention.

  • Bhagirath Joshi

    Dear Insight:

    Your Question:
    November 24th, 2011 at 9:41 AM
    Dear Bhagirath Joshi,
    – Why do you need to depict the neutron as a capacitor with an insulator between the positive and negative charge(s)? Do you fear that the charged particles you say are in it collapse one onto the other? So, could you also be saying that there should be an insulator layer between the electron and the proton in the hydrogen atom?
    – You writes in your paper: “The fact that Gravitational binding forces (Fg) of masses in the nucleus needs to be more than the destructive electromagnetic forces (Fem) created by the electrically positive environment of the nucleus, (Fem < Fg) excess neutrons are required in the nuclei." so it is clear that you are talking about gravitational forces as enough to keep the protons together. So Raul Heining's objection is well-founded.

    Answer:

    There are experiments underway to find the neutron EDM. If my model of neutron is accurate, it will not be possible for them to find EDM the way the experiment is setup. If they are able than my model of neutron would be wrong. However, the mechanism I described still remains.

    Secondly, There is insulating layer of empty space between proton and electron in Hydrogen atom.

    Thirdly, what I am saying is that gravitational forces are playing part along with other forces, and my model considers that, to show that there is a upper limit to isotope and element building.

    Thank you for nice question.
    Bhagirath Joshi

  • Andrea Rossi

    Dear Todd Burkett:
    Happy Thanksgiving to you and all our US Readers!
    I agree with you,
    Warm Regards,
    A.R.

  • Bhagirath Joshi

    Dear Wladimir Guglinski

    Your Question:
    November 24th, 2011 at 7:23 PM
    Dear Bhagirath Joshi,
    Some nuclei do not fit to what you state:
    “The fact that Gravitational binding forces (Fg) of masses in the nucleus needs to be more than the destructive electromagnetic forces (Fem) created by the electrically positive environment of the nucleus, (Fem < Fg) excess neutrons are required in the nuclei.”

    For instance, oxigen has 3 stable isotopes: 8O16, 8O17, 8O18.

    8O16, with fewer neutrons (only 8), is the most stable. It is a magic number.

    8O19 is not stable: it decays in 27 seconds.
    According to your theory, 8O19 would have to be more stable than 8O16, since it has 11 neutrons.

    Answer:

    Precisely that is the point I am making in the paper. That the gravitational force which is responsible for stability also makes the nuclei unstable and hence we can not build stable isotopes by adding more and more neutrons.

    There are two distinct separate sources of stability for the nuclei, One for even number and other for the odd number nuclei.

    The stability of the even number nuclei comes from the np shell in my model , which are always spin balanced and gives better stability to the nuclei, allowing for more than one stable isotope. Since all the excess neutrons reside at the center of the nuclei, as more and more neutrons are added, at one point the gravitational force starts interfering with the np shell and makes the nuclei unstable.

    In case of odd number nuclei, the np shell is not spin balanced, and becomes unstable quickly, and thus it has no more than three stable isotopes and in many cases only one. The stability for these odd number nuclei comes from the excess neutron shell.

    Various tables are provided for review in the down load, for every element. I also have included table for excess neutron energy levels.

    In case of Hg (80 protons) seven isotopes are possible, with 202Hg being the most stable. In case of Gold (79 Protons) there is only one stable isotope with 39 excess neutrons. pls note that np shell of the Gold is not spin balanced. But neutron shell is spin balanced and critically balanced, one less neutron or one more neutron makes it unstable.

    Thank you very much for the nice observation.

    Sincerely,
    Bhagirath Joshi

  • Todd Burkett

    Happy Thanksgiving !
    I am really excited about the introduction of this technology to the world!
    The next few years are going to be very exciting thanks to your diligent efforts.

  • Wladimir Guglinski

    Dear Bhagirath Joshi,
    Some nuclei do not fit to what you state:
    The fact that Gravitational binding forces (Fg) of masses in the nucleus needs to be more than the destructive electromagnetic forces (Fem) created by the electrically positive environment of the nucleus, (Fem < Fg) excess neutrons are required in the nuclei.

    For instance, oxigen has 3 stable isotopes: 8O16, 8O17, 8O18.

    8O16, with fewer neutrons (only 8), is the most stable. It is a magic number.

    8O19 is not stable: it decays in 27 seconds.
    According to your theory, 8O19 would have to be more stable than 8O16, since it has 11 neutrons.

  • Nick Pourmi

    Dear Andrea Rossi,

    Now I know Defkalion uses the KaCO3 catalyst!!!
    And I admire you more and more…

    Nick Pourmi

  • Mark

    Hey Andrea, I had the same thoughts as Bernie:-) It’s not a new thought because that strategy is used to “invalidate” psi experiments and deny the reality of parapsychological phenomena. Any chance on a second give away … 😉

    Best wishes.

  • Nick Pourmi

    Dear Andrea Rossi,

    Thank you for your sense of humour and observation. Of course I meant K2CO3!
    It was just a joke taken ad litteram from the internet gossip (sic).

    The first question was serious and I appreciate your answer perfectly legitimate.

    Warm Regards,
    Nick Pourmi

  • Andrea Rossi

    Dear Nick Pourmi:
    1- until the patents will not be granted, no confidential information will be released
    2- he,he,he..
    Warm Regards,
    A.R.

  • Nick Pourmi

    Dear Andrea Rossi,

    1. Will you give away your E-Cat secrets to MIT?
    2. (Rhetorical question) Is your secret catalyst KaCO3? – my guess, keep reading…

    “There are no secrets better kept than the secrets everybody guesses.” (George Bernard Shaw)

    Admiringly,
    Nick Pourmi

  • Andrea Rossi

    Dear Bernie Koppenhofer:
    I will send you as a personal gift a 10 kW E-Cat, as soon as we will put them in the market, to give a prize to your intelligent attitude upon this issue. Your comments regarding the tests is undispudedly the most intelligent .
    Warm Regards,
    A.R.

  • insight

    Dear Bhagirath Joshi,
    – Why do you need to depict the neutron as a capacitor with an insulator between the positive and negative charge(s)? Do you fear that the charged particles you say are in it collapse one onto the other? So, could you also be saying that there should be an insulator layer between the electron and the proton in the hydrogen atom?
    – You writes in your paper: “The fact that Gravitational binding forces (Fg) of masses in the nucleus needs to be more than the destructive electromagnetic forces (Fem) created by the electrically positive environment of the nucleus, (Fem < Fg) excess neutrons are required in the nuclei." so it is clear that you are talking about gravitational forces as enough to keep the protons together. So Raul Heining's objection is well-founded.

  • Bhagirath Joshi

    Dear raul heining

    Raul: Dear Joshi, I still cannot understand why you take in consideration gravitic force in the nucleous.

    Answer: As you can see from the research papers I mentioned earlier, two key issue emerge. 1) Neutron Binding energy and 2) Resonance at Low energies of nuclei. As you know , even minute force applied in resonance to a strongest system will induce catastrophic results. The Binding energies in my Model for neutrons ( a charge less particle) does not have luxury of electromagnetic forces. At a distance of the order of -17 to -18 meters for the particles keeping distinct identity, the gravitational binding force is around 31 eV. The experiments research paper from Los Alomos lab puts the Binding Energy of neutron at 7.3 eV. The np shell has the strongest binding force. However you can visualize that inner neutron core can be destabilized easily. What my model suggests that even though quantum mechanics is at work, one can not discard the Gravitational forces. It is like considering fourth order vibration and resonance analysis in construction of a Skyscraper.

    In other words, If some one puts together the hamiltonian of the system as described in my model with correct boundaries, the solution will definitely prove the point.

    Regards

  • Bernie Koppenhofer

    Hank Mills: Re: Another test

    The super skeptics simply have to get over it and realize LENR is a reality.

    After the test you propose and the minor inconsistencies that the super skeptics will create to make that test “inconclusive”, what will be the illogical reasons for next test and the next test and the next test. More testing is a huge waste of Leonardo Corporation’s time, energy and money.

    Rossi is taking the only logical path in his circumstances, build E-cats. Let his customers decide the uses and economical advantages.

  • Andrea Rossi

    Dear George (Jim) Hebbard PE:
    Yes, we have a water treatment. Good question.
    Warm Regards,
    A.R.

  • George (Jim) Hebbard PE

    I figure that at 7000 watts for 6 months, normal tap water will deposit something over 20 pounds (9 kilograms) of solid material in the E-Cat heat exchanger. Normal developing world, often river or reused water, will be much worse.

    Does the E-Cat system include a deionizer system for the feed water?

  • raul heining

    Dear Joshi I still cannot understand why you take in consideration gravitic force in the nucleous.
    In this situation, this force should not be considerated because is much lower intensity than, for instance, electromagnetic force and strong force, even only in face of electromagnetic force.
    Regards
    raul

  • Andrea Rossi

    Dear Tiziano Cecconi:
    If yu like, you can adhere to the waiting list writing to
    info@leonardocorp1996.com
    When we will launch the E-Cat in the market, you will be advised and will receive the offer: at that point you will be free to order or to cancel with no charge at all.
    Warm Regards,
    A.R.

  • Tiziano Cecconi

    Buon giorno sig. Rossi,
    scrivo da Firenze, lavoro nel settore dell’edilizia e delle energie rinnovabili, e probabilmente pochi si augurano quanto me che la sua invenzione possa concludersi nel migliore dei modi e con la più ampia diffusione possibile. Ho sentito di una prevendita dell’E-cat, mi interessa e vorrei saperne di più. Può darmi notizie?

  • claudio m.

    Dear Bhagirath Joshi,
    I read your paper and I found it very interesting. Maybe you already know what I’m saying but in Italy there is a physicist which seems following almost the same theory you do. He is Roberto Monti and this is his website:
    http://www.scienzaperlamore.it/contStd.asp?lang=it&idPag=427
    Maybe you can have a link with him … who knows.
    Whith my best regards
    cm

  • Bhagirath Joshi

    Dear insight:
    You completely misunderstood the point. The hamiltonian as currently describing the nuclei completely ignores weak forces. What my model does is that it introduces weak forces in the equation with boundaries. No one is disputing the forces produced in nuclear reactions as proposed by Einstein, but than there are other issues which needs to be understood e.g. 1) why there is no stable isotope of hydrogen with 100 neutrons ? 2) What makes the isotopes of elements unstable and radio active? 3) Why removal of a single neutron from the system induces positron emission? 4) What makes every element after Bismuth radio active and unstable? 5) Why all the nuclei of a radio active element does not disintegrate at the same time? 6) What is the theory behind half life of radio active element? 7) How does the free protons and neutrons behave in a solid?
    These are the questions and we are trying to find the answers. Your input in this endeavor will be greatly appreciated.
    Sincerely, Bhagirath Joshi

  • Andrea Rossi

    Dear Renato:
    I do not know.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Mahler:
    I already answered. We have not time for further public tests. We test only the units for the Customers.
    Warm Regards,
    A.R.

  • Mahler

    Gentile ing. Rossi,
    le segnalo la lettera aperta che le ha appena scritto Brian Josephson tramite Focus:

    http://www.focus.it/scienza/e-cat-and-cold-fusion-open-letter-to-andrea-rossi_C12.aspx

    Cordiali saluti.

  • Bhagirath Joshi

    Dear Raul:
    Strong Nuclear force is not actually forgotten, However we should not forget that the standard model has no explanation for non existence of stable isotope of element Tc and Pm, 43 protons and 61 Protons system. If you arrange these elements in my model you will immediately know why?

    Regards

  • Renato

    Dear Dott. Rossi,

    Congratulations for the work you did and do.

    Then some questions.

    The power consumption of the e-cat is rated at 1/6 of the thermal energy produced.

    Do you expect that the amount of consumed electric energy will or could be reduced in the future?

    Cordiali Saluti,
    R.S.

  • Bhagirath Joshi

    Dear Raul:

    Here is another paper published in 2004, which states and I quote

    Accelerated beams of 132Sn (98% pure) were used to measure sub-barrier fusion cross sections of 132Sn+64Ni induced reactions down to
    energies for which the cross section was as low as 3mb. The measured excitation function shows large enhancement in cross section that
    could not be accounted for by coupling of inelastic and transfer channels. A simple computational model for calculating sub-barrier fusion
    cross sections between neutron-rich heavy nuclei reproduces the very large enhancement observed for this system.”

    and

    ” Fusion between heavy ions at sub-barrier energies is the subject
    of intense study since it was discovered that the magnitude
    of these fusion cross sections far exceeds the expected
    values based on quantum penetration of the one-dimensional
    Coulomb barrier [1–3]. Present views hold that this enhancement
    is due to the complexity of the colliding nuclei. ”

    Sub-barrier fusion of neutron-rich nuclei: 132Sn+64Ni
    D. Shapiraa, J.F. Lianga, C.J. Grossa, J.R. Beenea, J.D. Biermanb, A. Galindo-Uribarria, J. Gomez del Campoa,
    P.A. Hausladena, Y. Larochellec, W. Lovelandd, P.E. Muellera, D. Petersond,
    D.C. Radforda, D.W. Stracenera, and R.L. Varnera
    aPhysics Division, Oak Ridge National Laboratory,
    Oak Ridge, Tennessee 37831, USA
    bPhysics Department AD-51, Gonzaga University,
    Spokane, Washington 99258-0051, USA
    cDepartment of Physics and Astronomy, University of Tennessee,
    Knoxville, Tennessee 37966, USA
    dDepartment of Chemistry, Oregon State University,
    Corvallis, Oregon 97331, USA
    Recibido el 20 de enero de 2004; aceptado el 23 de mayo de 2004

    Regards

  • Bhagirath Joshi

    Dear Raul Heining

    Your observation is absolutely correct. Here is the excerpt form a internal report published and now made public
    “APPLIED NUCLEAR DATA RESEARCH AND DEVELOPMENT QUARTERLY PROGRESS REPORT”
    April 1 – June 30, 1981
    Compiled by P. G. Young, Los Alamos, LA-9060-PR Progress Report
    UC-34C

    ” The neutron transmission coefficients used in Eq. (1) were calculated from
    the deformed optical model parameters described in the previous section. To
    calculate gamma-ray transmission coefficients , we applied the Brink-Axel giant
    dipole resonance (GDR) mode16 normalized to the ratio of the experimental values“’
    for the average gamma-ray width ( = 0.084 eV) and S-wave resonance
    spacing ( = 7.3 eV) at the neutron binding energy. ”

    The key point here is the resonance at 7.3 eV of s wave resonance.

    The internal neutron shell of my model puts upper limit to number of isotopes of an element and to the heaviest stable nucleus in nature. Also the ability of nuclei to capture ultra cold ( very small kinetic energy ) shows that weak forces are playing major part.

  • insight

    Dear Raul Heining,
    in fact the author seems to be convinced that it was the gravitational force that made A and H bombs so destructive weapons.

  • Andrea Rossi

    Dear Kim Patterson:
    Great things are happening in the USA in these times concerning our work.
    Warm Regards,
    A.R.

  • Kim Patterson

    Into the Eye of the Tiger.

    Good Luck Andrea with the Massachusetts State Senate meetings with Massachusetts government leaders and MIT officials.

    Respect
    Kim

  • raul heining

    Strange, strong nuclear force is here forgotten as it is weak force.
    Regards
    raul

  • Hank Mills

    Cremating Thermal Inertia With A Custom E-Cat Test

    Andrea Rossi has stated that any qualified customer who wishes to purchase an E-Cat (Energy Catalyzer) plant, will have the opportunity to perform a test they can customize and control. What testing methods will the next E-Cat customers utilize? Perhaps a test that forever puts to rest the absurd hypothesis of the thermal flywheel effect!

    By Hank Mills

    The test of the one megawatt E-Cat (Energy Catalyzer) test on October 28th was a success, and pushed cold fusion technology into the commercial market place. One important aspect of the test, was that the customer’s representative was allowed to conduct the test in whatever manner he so desired. During the even, he chose how to measure the input/output, what equipment to utilize, what mode to run the plant in (with input or in self-sustain mode), and many other aspects of the test. Very encouragingly, Andrea Rossi has stated that all qualified customers will be allowed to have the same opportunity to perform their own customized test, before finalizing their purchase.

    In my opinion, by allowing customers to perform such pre-purchase testing, he has yet again demonstrated his willingness to prove beyond any doubt, that he is offering real, legitimate, and working cold fusion products. A scammer, hoaxer, or con-man — which many of his jealous competitors falsely claim him to be — would not make such an offer to potential customers. My hope is that his future customers will take the opportunity to perform such tests, and share the results they obtain.

    What I think is particularly exciting of the prospect of such testing (in addition to Rossi obtaining additional satisfied customers) is the opportunity to squash some of the false arguments the pseudo-skeptics have been spreading.

    I see a common theme emerging in the forum posts, websites, and communications of the hardcore cynics, and Rossi’s competitors. This theme is the *false* idea that a “thermal flywheel” or “thermal inertia” effect is responsible for the energy produced during self sustain mode. To be blunt, their hypothesis is dead wrong, for a multitude of reasons.

    First, the E-Cat is composed of materials that have a low specific heat, or the ability to store thermal energy. Lead, which composes most of the weight of the E-Cat, has an extremely low specific heat. There is simply not enough material in an E-Cat to store the amount of heat produced during self-sustain mode.

    Secondly, during the warm up phase, cooling water is being circulated through the E-Cat. The water is constantly removing heat from the module. The cynics tend to ignore this fact, because they don’t want to admit to the fact that energy cannot be used twice. Even if the E-Cat could store a large amount of heat (which it cannot), it cannot magically be both storing the heat during the warming phase, AND releasing it during self sustain mode. Energy cannot be used TWICE — this would break the laws of thermodynamics.

    Thirdly, cooling water is being pumped through the E-Cat during self-sustained mode, which is constantly taking heat out of the system. So in addition to the massive surface area of the module which contributes to heat loss, there is another active cooling process taking place. If excess energy was not being produced by cold fusion reactions, the output temperature would drop *immediately* upon power to the resistors being cut off. Instead, the temperature of the output water (liquid or steam) stays constant, or sometimes rises. This completely eliminates the possibility of the thermal inertia (thermal flywheel) effect.

    Fourth, at the test of the one megawatt plant the system self-sustained for five and a half hours. During this time the output temperature stayed almost constant, and occasionally increased for brief periods. There is zero possibility that thermal inertia could explain this. None. Zilch. Nada. Those pseudo-skeptics that think thermal inertia can be responsible for five and a hour hours of self sustained heat production, are like those who did not believe the Wright Brothers achieved flight, despite seeing their invention flying in the sky.

    For these reasons and others, the arguments of the naysaying skeptics and snakes (jealous competitors) are totally invalid, and a shocking example of flawed thinking. At worst, their statements are a blatant example of information warfare against Rossi’s company, Leonardo Corporation. They are telling lies and attempting to diminish the significance of the extended runs in self sustain mode, because they cannot handle the fact their technology is not capable of the same performance.

    The really sad part, is that some individuals who are sincerely supportive of the E-Cat technology (who are not competitors) have listened to them, and the nonsensical misinformation of the skeptics is spreading.

    A Proposed Test

    I have an idea for a new type of test that could be performed, that could instantly put to rest the ridiculous notion of thermal inertia. Simply put, I think it would throw a huge monkey wrench into the plans of the hardcore cynics and Rossi’s competitors, that keep pushing the irrational idea that the E-Cat is only releasing “stored” energy during self sustain mode.

    The following is the idea for the test. Please note this is not a test of an entire one megawatt plant — because only two reactors or modules are utilized. However, I think the following test would be a good one for any customer to perform, in addition to testing the entire one megawatt plant they intend on purchasing.

    (Please note that the following is simply a short and rough description of the proposed test. A full test protocol would be more detailed, and should be prepared by a professional engineer.)

    — text box —

    1) Put two E-Cat modules of the same design side by side. Any design of module could be used, as long as they were each of the same model of E-Cat.

    2) Fill the first module (A) with hydrogen under the highest SAFE pressure possible, high enough so that the self-sustain mode would be particularly robust.

    3) Fill the second module (B) with ordinary atmosphere. The reactor could would contain the “charge” (nickel, catalyst, etc), but no hydrogen. This would be the “control” unit, that would produce no excess energy.

    4) Setup both systems so that each would receive the exact same electrical input, and the same flow rate of water. Use appropriate methods to measure the electrical input, and the flow of water throughout the test.

    5) Place thermocouples at the inlet of each reactor, inside of each reactor, and in the water (air or steam) exiting each reactor (the output).

    6) Start logging the temperatures via an automatic data collection system, and turn on the electric resistors in both module (A) and (B).

    7) Once it is absolutely certain (no doubt) that self sustain mode has “took” in module (A) which contains hydrogen, turn off the resistors in both module (A) and (B). Then turn on the radio frequency generator in both modules.

    8) Keep recording data. During this time period module (A) in self sustain mode should output a continual flow of liquid water or steam, that maintains a fairly constant temperature, or actually rises in temperature. However, the output from the control module (B) — with no hydrogen — should quickly drop in temperature.

    9) Keep the test going for as long as practically possible, to show that long after the control module has stopped emitting heat, the module with hydrogen is continually producing a steady output.

    — text box —

    I sincerely think that if the above test was performed, it would provide rock solid evidence (completely undeniable) that the thermal flywheel effect hypothesis, is total nonsense. In my opinion, it would…

    – Take away a heavily used tool of misinformation from the cynics, and Rossi’s competitors who are spreading lies on the internet. These “snakes” (as Rossi calls them) are pushing this false hypothesis with all their might, and are using it as a weapon in the “war” against the E-Cat. To yank it out of their hands would be a major victory.

    – Expose those who have been spreading the misinformation as total idiots (at best) or at worst, flat-out, blatant liars. It would make people think twice, before listening to anything else the “snakes” may ever have to say.

    – Provide even more evidence (on top of what Rossi has have already provided) that the excess heat produced by an E-Cat system is absolutely massive, and absolutely must be produced by exotic nuclear reactions.

    A test utilizing a “control” module, is the most simple and clear way to demonstrate the power of the E-Cat. Such a test compares a system that should produce zero excess energy (due to containing no hydrogen), to a system that produces large amounts of excess energy due to cold fusion processes. The difference between the performance of the two units would be as opposite as night and day. Also, the results of such a test would be clear cut, and super simple to understand. No one would be able to dispute the results, because long after the control unit had stopped producing output in the form of heat, the actual module with hydrogen would continue producing heat — constantly and steadily.

    There are many good tests that I hope potential customers perform, but I hope one of them will perform a test with a control module (no hydrogen) vs. an actual module (with hydrogen). Although it is already obvious that the thermal inertia hypothesis is total nonsense, such a test would put a final nail in it’s coffin. Actually, I would compare it more to a cremation, because the ashes of the absurd “thermal flywheel” effect would be spread in the wind.

    As Rossi has stated on his blog, there is a war taking place. Since the successful test of the one megawatt plant (which self sustained for five and a half hours) the hardcore cynics and pseudo-skeptics have became more desperate in their attempts to discredit the E-Cat technology. Thus, they are ramping up their tactics, and becoming even more nasty. I would like to see the above test performed — either by a customer or anyone else Rossi would allow to perform it — because it would be a simple way of fighting back against their lies and propaganda in a peaceful manner.

    The test described in this article is not needed to prove the validity of the E-Cat technology, because that has already been done in my opinion. However, it would be tremendously useful towards fighting back against the pseudo-skeptics, and the jealous, envious competitors that have been spreading a false hypothesis. Andrea Rossi is absolutely correct in that the best way to fight against the competitors is for him to sell working plants to customers. But, I think a test with a control E-Cat, would be another valuable weapon in the very real war that is taking place.

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