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

788 comments to Excess neutron shell model of Nuclei

  • Bhagirath Joshi

    Dear Wladimir Guglinski
    December 2nd, 2011 at 7:01 PM

    Your Comment:
    The same we can say about 4Be10, 6C14, 10Ne22, 12Mg26, 14Si30, 16S34, 18Ar38, 20Ca42, 22Ti46, 24Cr50, 26Fe54, 28Ni58.

    All they have null electric quadrupole moment, Q(b)=0.

    My Explanation:
    FYI
    http://www.uni-due.de/physik/wende/englisch/nuclear-moments.pdf

    In the above table Q(b) is not zero for these isotopes. “Excess Neutron shell” is asymmetric for these isotopes And it agrees with my model.

    Sincerely,
    Bhagirath Joshi

  • Wladimir Guglinski

    Bhagirath Joshi wrote in December 2nd, 2011 at 6:53 AM:

    “My Explanation:

    FYI http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/elequad.html

    The Electric quadrupole moment is indicative of the charge distribution. In “excess neutron shell model” we have two shells to deal with, 1) Excess neutron Shell and 2) neutron-proton (np) shell.”

    Dear Bhagirath Joshi,
    actually the electric quadrupole moment Q(b) is caused by two different causes.
    Let’s see them:

    1- FIRST CAUSE:
    The asymmetric charge distribution. For instance, the nucleus 3Li6 has 3 protons (positive charges). They cannot form a symmetric distribution of charge. That’s why 3Li6 has Q(b)= -0,00083
    5B10 is other example, with Q(b)=+0,085

    Here you can apply the definition of elec. quad. mom. mentioned by you:
    FYI http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/elequad.html

    2- SECOND CAUSE:
    The unbalance of masses, due to the addition of neutrons.
    The neutron has no charge. And so it cannot change the charge distribution of charge.
    But an odd neutron can cause unbalance of masses. As the nucleus gyrates, it shakes as happens when an unbalanced wheel of a car has shimmy. And so, the charge distribution of charge oscillates, and this is detected in experiments as electric quadrupole moment.

    Look for instance the proton 1H1. It has Q(b)=0.
    If you add one neutron to the proton, the 1H2 has Q(b)=+0,0028, because the neutron is more massive than the proton.

    1H3 has two neutrons, and it has Q(b)=0, because the two neutrons eliminate the unbalance of masses existing in the 1H2.

    2He3 has two protons and one neutron, and therefore its mass is balanced, and so it has Q(b)=0

    2He4 has two neutrons and two protons, and so it has Q(b)=0

    4Be8 has Q(b)=0

    4Be9 has one odd neutron, and so it has Q(b)=+0,053

    6C9 has Q(b)=0. It has 3 no matched protons. Two of them take symmetric positions with regard to the line about which the nucleus gyrates. The third proton takes position coincident with the line center of rotation. So, there is balance of masses.
    It seems to be impossible to explain why 6C9 has Q(b)=0 from the current Nuclear Physics, since it has a charge formed by 3 protons.

    6C10 is balanced, and so it has Q(b)=0

    6C11 has Q(b)=0,032

    6C12 has Q(b)= 0

    6C13 has Q(b)= 0. It has one odd neutron, and it takes place coincident with the line center about which the nucleus gyrates. So such odd neutron does not cause shimmy in the nucleus rotation.
    Again, it seems to be impossible to explain why 6C13 has Q(b)= 0 from current Nuclear Physics, because there is not in that theory a cause responsible for keeping the neutron coincident with the line of rotation.

    6C14 has Q(b)=0, because there is balance of masses.

    6C15 has Q(b)=0, because two neutrons take symetric positions with regard to the line of rotation, and the third takes a place coincident with that line.

    8O15 has Q(b)=0. It has one no matched proton, which takes place coincident with the line of rotation of the nucleus.

    Stable 8O16 has Q(b)=0
    Excited 8O16 has Q(b)=0 , and we will see the reason ahead.

    Stable 8O18 has Q(b)=0

    Excited 8O18 has TWO DIFFERENT values of Q(b):
    Something very interesting occurs with the excited 8O18.
    If you look in the nuclear table, you realize that it exhibits TWO DIFFERENT values of Q(b):
    -0,010 or +0,020
    -0,07 or -0,05
    -0,11 or -0,08
    -0,05 or 0,02
    This is because the excited 8O18 actually has 3 different structures, because the 3 neutrons can occupy 3 different positions in the hexagonal floor formed by 6 deuteons in the 8O18:
    FIRST STRUCTURE- The two excess neutrons have symmetric positions, and so its Q(b)=0, and not detected in experiments.
    SECOND STRUCTURE- The two neutrons occupy positions no symmetric in the hexagon, but aparted one from the other.
    THIRD STRUCTURE- The two neutrons occupy positions no symmetric in the hexagon, but they are close to each other.

    Now pay attention why the excited 8O16 has Q(b)=0. It’s because it has not excess neutrons, in order to occupy diffent RELATIVE places in the hexagonal floor that forms the structure of 8O16.
    So, even when excited, the 8O16 cannot form an unbalanced structure, because it has not odd neutrons.

  • eernie1

    Derar Guglinski,
    It is a pleasure to discuss scientific topics in this post.Although the commercial and technical aspects of Rossi’s device are interesting it is more enjoyable to read your’s and Johsi’s comments and blogs without having to comb through many other blogs.I present to you more emperical observation of mine for your amusement.(1)All tests where particles are injected into the nucleous seem to result in large energy and secondary particle outputs.Particle accelerators(Beta and proton)inject their particles with much energy and produce multiple photons and a wide variety of particles.Neutron generators and isotopes must have their produced energies modified to effectively enter a specific nucleous(as in the production of the atom bomb)to create fission.From these observations it is easy to conclude that Rossi’s device does not involve direct particle penetration since large quantities of secondary particles are not observed and only a minimal shielding is required.There is enough emperical evidence to show at least for me that Rossi’s device produces excess energy that can only come from the nucleous(other observers have made the same conclusion)but also for me,it shows that the mechanism must be instigated from external forces.
    (2)Since we have an abundance of isotopes that emit particles, with less energy than required to overcome the coulombic barrier from the internal nuclear configuration,the strong forces that keep the protons and neutrons inside the nucleous must be neutralized in some manner to allow these nucleons to exit.Perhaps since these forces are effective over only a small distance,an expansion of the nucleous diameter may allow the nucleons to separate enough to lessen their effective containing force.

  • Wladimir Guglinski

    Bhagirath Joshi wrote in December 2nd, 2011 at 6:53 AM:

    In “excess neutron shell model” we have two shells to deal with, 1) Excess neutron Shell and 2) neutron-proton (np) shell.

    a) In case of 8O18 np shell is symmetric. and “excess neutron shell” is asymmetric . for np shell L , p has four np pairs.

    Dear Bhagirath Joshi ,
    thanks for the explanations, but I did not understand yet.

    In your paper you say:
    “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 ”

    I suppose such condition must be ALWAYS satisfied, otherwise you cannot explain the occurrence of isotopes, as it is your intension.

    Stable 8018 has electric quadrupole moment Q(b)=0
    It has two excess neutrons.

    So, I suppose that one neutron occupies the center, and the other one occupies the n-shell.
    So, its structure is asymetric.

    Then I cannot see how stable 8O18 may have Q(b)=0, unless if the two excess neutrons occupy the same excess n-shell (but in such case the fundamental premise is not satisfied: one excess neutron must be in the center).

    The same we can say about 4Be10, 6C14, 10Ne22, 12Mg26, 14Si30, 16S34, 18Ar38, 20Ca42, 22Ti46, 24Cr50, 26Fe54, 28Ni58.

    All they have null electric quadrupole moment, Q(b)=0.

  • Bhagirath Joshi

    Dear Wladimir Guglinski
    December 2nd, 2011 at 3:44 AM

    Your Comment:
    Electric quadrupole moment of excited 8O18 and stable 3Li7:

    Stable 3LI6 has 3 protons.
    Its electric charge distribution is excentric, that’s why its electric quadrupole moment is not null: Q(b) = -0,00083

    There is a big growth in the excentricity when one excess neutron is additioned:
    3Li7 has Q(b) = -0,0400

    According to the excess neutron shell model, the excess neutron in 3Li7 would have to occupy the center of the nucleus.
    So, when one neutron is additioned to 3Li6, according to the theory there would have to be a decreasing in the electric quadrupole moment.
    Therefore the value of 3Li7 Q(b) would have to be less than -0,00083.

    My Explanation:
    You need to include the gravitational effect and decrease in the orbit radius in your calculation. Addition of neutron will definitely increase the EQM

    Thanking You
    Bhagirath Joshi

  • Bhagirath Joshi

    Dear Vladimir Guglinski
    December 1st, 2011 at 4:59 PM

    Your Comment:
    Bhagirath Joshi, let analyse the excess neutron shell model:

    1- The assymetry in the excited 8O18 is caused by the excess neutron shell.
    But:
    – as the 3Li7 has 7 nucleons
    – and 8O18 has 18 nucleons
    – then one neutron cannot cause in the 8O18 the same assymetry caused by one neutron in the 3Li7.
    – Nevertheless, they have the same value of the electric quadrupole moment.

    My Explanation:

    FYI http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/elequad.html

    The Electric quadrupole moment is indicative of the charge distribution. In “excess neutron shell model” we have two shells to deal with, 1) Excess neutron Shell and 2) neutron-proton (np) shell.

    a) In case of 8O18 np shell is symmetric. and “excess neutron shell” is asymmetric . for np shell L , p has four np pairs. The charge distribution is symmetric . Neutron becomes asymmetric under excitation and EQM is observed.

    b) in case of 3Li7 np shell is asymmetric.

    c) WG: 20Ca41 would have to have null electric quadrupole moment.

    BJOSHI: Let us consider the symmetry of Ca. in my model np K and L shell are completely filled. M shell s, p satisfied but d has two np pairs and shell is asymmetric. Thus exhibiting EQM

    d) WG: The same we can say about the 8O17:
    – Its one excess neutron takes the center (Fn – fundamental position).
    – Therefore its structure is symmetric
    – And it must have Q(b)=0
    – But stable 8O17 has Q(b)= 0,02578

    B Joshi: In nature a trace amount of 8O17 is found , so it is not the most preferred state, However some explanation can be found in my model. One explanation is that the neutron has very small binding energy.

    As I said, need further investigation and development.

    Thanking You
    Bhagirath Joshi

  • Wladimir Guglinski

    eernie1 wrote in December 1st, 2011 at 6:51 PM:

    I am trying to fit my thoughts about the Rossi effect into the part of the model which has been proven through extensive observations which are repeated by many tests.One of these observations is the requirement of tremendous energy levels before the coulombic barrier can be breached.Current hot fusion efforts are indicative of this requirements

    Dear Eernie1,
    as I already said, in the nuclear model of my QRT the nucleus is involved by two concentric fields. The inner field has gravitational nature. The external field is electromagnetic.

    There is a “hole” in the electromagnetic field. That’s why, in special conditions (stimulated by resonance and repulsive gravity) some particles are able to enter into the nucleus.

    The alpha particle 2He4 leaves out the 92U328 with energy 4MeV, while the Coulombic barrier has 8MeV.
    The alpha particle is able to cross the 8MeV barrier with energy 4MeV because the 2He4 leaves out the nucleus crossing the “hole” of the external field which involves the 92U nucleus.
    So, cold fusion occurs in the reverse phenomenon occurred in alpha decay of radioactive nuclei: crossing the Coulombic barrier by passing by the hole in the external field.

    In hot fusion there is need to pierce the Coulombic barrier. This requires very big energy.
    In cold fusion there is no need to pierce the barrier, because some conditions are satisfied so that to allow the particle to enter into the nucleus through the “hole” existing in the external field of nuclei.

    In the core of the Sun only hot fusion occurs, because into the star there are no suitable conditions to occur resonance.
    But I suppose cold fusion occurs in the Sun’s heliosphere, where suitable conditions to occur cold fusion may exist.
    Such cold fusion could explain why the heliosphere (1.000.000°C) is hotter than the surface (6.000°C) of the Sun

  • Wladimir Guglinski

    Electric quadrupole moment of excited 8O18 and stable 3Li7:

    Stable 3LI6 has 3 protons.
    Its electric charge distribution is excentric, that’s why its electric quadrupole moment is not null: Q(b) = -0,00083

    There is a big growth in the excentricity when one excess neutron is additioned:
    3Li7 has Q(b) = -0,0400

    According to the excess neutron shell model, the excess neutron in 3Li7 would have to occupy the center of the nucleus.
    So, when one neutron is additioned to 3Li6, according to the theory there would have to be a decreasing in the electric quadrupole moment.
    Therefore the value of 3Li7 Q(b) would have to be less than -0,00083.

    When we compare 3Li7 and 8O18 the paradox is clear:
    – According to the theory, the excess neutron of 3Li7 would have to occupy the center of the nucleus, and its Q(b) would have to be less than -0,00083.
    – However, 3Li7 has Q(b)= -0,04
    – Stable 8O18 has two excess neutrons. One of them must occupy the shell neutron, and the other one must occupy the center. So, stable 8O18 would have to have Q(b) NOT null. But it has Q(b)= 0.
    – Excited 8O18 has two excess neutrons, one in the shell neutron, and the other in the center. As the number of protons is pair, there is a symmetry of charge due to the protons, and so excited 8O18 could not have the same Q(b) exhibited by stable 3Li7 (which has 3 protons). Therefore excited 8O18 actually would have to exhibit a Q(b) less than -0,00083 (less then that exhibited by 3Li6).

    In Quantum Ring Theory the electric quadrupole moment of excited 8O18 is calculated in the page 140 of the book.
    The values ( x10^-30 m2 ) are the following:

    Experiments: Q(b) = -6,05 or -3,1
    Theoretical calculation: Q(b)= -6,43 or -2,30

  • eernie1

    Dear Gluginski,
    Please accept my omission of your first name as a complement.Everybody should now know who Guglinski is.
    I am a pragmatist but also open minded with regards to the present nuclear model.I have no doubts that the model is not complete.I am sure it will be revised as further investigations by people like you continue.I am trying to fit my thoughts about the Rossi effect into the part of the model which has been proven through extensive observations which are repeated by many tests.One of these observations is the requirement of tremendous energy levels before the coulombic barrier can be breached.Current hot fusion efforts are indicative of this requirements.My theories are directed towards eliminating the requirement for particle penetration into the nucleous because of this fact.It seems to me to make more sense if the external fields which have no problem entering the nucleous can trigger a reaction in the internal particle configurations of specific isotopes.The multitude of radioisotopes which naturally decay with emission of energy indicate to me that these events are quite common in nature and may only need a slight push to be activated.Radioactive isotopes are used in the generation of energy for satellites and perform consistantly for many years.
    To answer your question,I dont think the present nuclear model is completely right but certain aspects can be used to explain how Rossi’s device operates.I prefer to call it (FIND)field induced nuclear decay.

  • Wladimir Guglinski

    Bhagirath Joshi wrote in December 1st, 2011 at 2:56 PM

    Your Question:
    Is it possible to explain such anomaly of the 8O18 from your nuclear model ?

    Answer:
    Yes

    The two excess neutrons take up different positions, one at the center of the Nuclei (Fn – fundamental position) and second on K shell of excess neutron shell. Thus it is assymentric.

    Bhagirath Joshi, let analyse the excess neutron shell model:

    1- The assymetry in the excited 8O18 is caused by the excess neutron shell.
    But:
    – as the 3Li7 has 7 nucleons
    – and 8O18 has 18 nucleons
    – then one neutron cannot cause in the 8O18 the same assymetry caused by one neutron in the 3Li7.
    – Nevertheless, they have the same value of the electric quadrupole moment.

    2- The stable 8O18 has null electric quadrupole moment. It has two excess neutrons.
    In your paper you say:
    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

    So, in the stable 8O18 the two excess neutrons take up different positions, one at the center of the Nuclei and second on K shell of excess neutron shell.
    Therefore the structure of the stable 8O18 is assymetric according to your theory, and so it must have NOT null electric quadrupole moment.
    But the experiments show that stable 8O18 has Q(b)=0.

    It seems the isotope 8O18 does not follow the rules of excess neutron shell model.

    The same happens with some other nuclei:

    3- 20Ca41 has one excess neutron, and so it occupies the center of the Nuclei (Fn – fundamental position).
    Then, according to the excess neutron shell model, 20Ca41 has a symmetric structure (since the one excess neutron takes the center of the nucleus).
    A neutron in the center of the nucleus cannot yield an assymetry in the charge distribution of the nucleus
    Therefore:
    1 – 20Ca41 would have to have null electric quadrupole moment.
    2- However, the stable 20Ca41 has Q(b) = -0,080

    4- The same we can say about the 8O17:
    – Its one excess neutron takes the center (Fn – fundamental position).
    – Therefore its structure is symmetric
    – And it must have Q(b)=0
    – But stable 8O17 has Q(b)= 0,02578

  • Bhagirath Joshi

    Dear eernie1
    December 1st, 2011 at 10:02 AM

    Your Comment:
    Thank you for your prompt answer.I am familiar with the different lifetimes of the external and internal neutrons and that is the point of my argument.What keeps the neutron stable in the nucleous?How difficult would it be under certain conditions present in a specific nickel isotope to disrupt the equilibrium with an external electric field and allow the internal neutrons to decay?This would result in the extra proton and transmutation to copper.

    My explanation:
    The link I sent you is to show that when neutron disintegrates it produces a proton and electron and something else.

    In solid nuclei of the element behave differently. This is the great area of my interest.

    Thanking You
    Bhagirath Joshi

  • Bhagirath Joshi

    dear Eric Ashworth
    December 1st, 2011 at 11:31 AM

    Your Comment :
    Dear Bhagirath Joshi, Having read your paper ‘Excess neutron shell model of nuclei’ I believe I can see where you are coming from.

    my explanation:

    The universe is held together by one force, Gravitational at macro level. My attempt here is to show that even at nuclei level this fact holds. Plasma physicist first tried using hydrogen to fuse in to Helium, but it required so much more energy to do that, they resorted to the isotope of hydrogen with neutron in it(Deuterium) . My question was if the nuclear forces are so powerful than why every element in the universe exhibits so many neutrons in it. Why just protons are not held together with that strong nuclear force? Why nature requires neutrons too. ANd that lead me to propose this model. THe model as proposed also shows that along with strong nuclear forces, gravitational forces are important. And since in my model excess neutrons stay at the center , they do provide sizable gravitational pull on the proton-neutron pair circling the core. (Imagine a solar system with a central body 54 times massive as the planets circling it. Planets will not be able to escape the gravitational pull. Now say in such a system the central body captures slow moving body with mass of one planet. I am sure this will create havoc on the planets circling) the example above is for understanding how gravitational force plays part in my model.

    My model when applied to the “solid state nuclear physics” may give more insight into half life of elements and also cold nuclear fusion.

    In solid state physics one chiefly deals with valance electrons, which collectively behave as cloud and can be controlled. As a result we have all these computers and many devices we enjoy today.

    Same is true for a nucleus in solid. The nucleus is fixed at one location. So one would assume. However, excess neutron have half as much binding energy of valance electron. This creates interesting scenario where these neutrons may also behave like valance electron cloud with neighboring nuclei.

    The hydrogen atom has one electron. Once it is infused in the solid the electron is free to roam around at will and the hydrogen behaves as a free proton and free to interfere with other nuclei and even steal neutron from the outer shell of the other nuclei if excited.

    Imagine the possibility here.

    When Einstein was developing his famous theory, he discarded the solution , which made prediction of receiving message before it is sent, since that is physically impossible.

    Therefore, My model is a one step forward in showing the nature of the universe.

    Thanking You
    Bhagirath Joshi

  • Bhagirath Joshi

    Dear Wladimir Guglinski
    December 1st, 2011 at 7:49 AM

    Your Question:
    Is it possible to explain such anomaly of the 8O18 from your nuclear model ?

    Answer:
    Yes

    The two excess neutrons take up different positions, one at the center of the Nuclei (Fn – fundamental position) and second on K shell of excess neutron shell. Thus it is assymentric.

    Thanks

    Bhagirath Joshi

  • Dear Bhagirath Joshi, Having read your paper ‘Excess neutron shell model of nuclei’ I believe I can see where you are coming from. The problem with physics like any technical subject is language. Especially if you are none academic like myself and trying to relate to the same subject. You make several interesting assumptions (1) Excess neutrons = Atomic Mass. Atomic mass relates to positive gravity. What I refer to in my pattern of neutral energy is the inner cone/core. This I could refer to as the neutron energy, same thing. I refer an energy unit to dimensions of force and flow. This being the relationship between latent and kinetic energy. The neutron is a measurable force of positive gravity. Within the inner core of a mass, such as the Earth or the structure within which the Earth is formed and exists also has an inner core of neutron force. This inner force has a flow and therefore could be termed a ‘kinetic neutron flow’ with a force (lines of force comprised of Aether of a diametrical dimension with a positive gravity value). What has to be taken into consideration is that there is a structuring force of four helical trajectories pertaining to a pattern (previously given). Wladimir’s helical trajectory theory is correct and you are too with regards your neutron theory and your statement “gravity and the existence of the universe”. However and I shall repeat, without a pattern/theory with regards creation and destruction of the physical world nothing will make sense and contradictions will forever be. Mathematics starts with add, subtract, divide and multiply not with calculas. Same goes for atomic physics.

    The neutron force/positive gravity of the inner core of the Earth is due to the two units of power energy of the helical trajectories of the formative structure within which the Earth sits and will exist until it reaches its destination of destruction i.e. the positive apex of the pyramid within the positive zone of the helical trajectories. The neutron force is the inner positive gravity responsible for holding the Earth together but what is the force holding together?. Lets go from the inside out, starting with the inner core. The inner core is composed of highly active lines of neutron force of a vertical dimension that equates as the positive gravity of the mass (it has no dimension being central to the mass i.e. the central line but has a focul point). This is positive neutron force energy. Upon the very active inner core rests a layer of active positive energy of a positive gaseous dimension, labeled gaseous having less kinetic energy than the core and upon this active gaseous dimension rests a less active yet still positive dimension of a liquid dimension or magma. Now upon this liquid dimension lies a neutron dimension of none activity with no flow i.e. a latent neutron force and upon the neutron force of none activity i.e. igneous rock, rests a layer of active negative energy of a dimension of liquidity and upon this liquid dimension rests a more active layer of negative energy of a gaseous dimension. Above this outer electron zone lies the static boundary layer. This is a none material substance that resembles exterior radial dimensions of a kinetic neutron force above which is supported two other dimensions relative to a liquid and gas dimension. What the Earth is comprised of, like all units of energy, is three neutron forces that represent a positive neutron, a neutral neutron and a negative neutron or you could say Positive kinetic, neutral latent and negative kinetic. Positive kinetic vertical/diametrical dimension. Neutral and negative of a horizontal/radial dimension. Between positive kinetic and neutral latent mobile oscillatory curvature forces are present and between neutral latent and negative kinetic mobile oscillatory forces are present. The two oscillatory mobile curvature forces above negative kinetic neutron are mobile oscillatory forces of an extraneous origin that form a link between dimensions of differing cubes within a systemic system. Oscillating mobile curvature forces are caused by helical trajectories of Aether rays that are emitted from the apex of the overriding pryamid within the cube and that permeate and energize the cube. This interaction of energy described is ‘the static and mobile mechanics of energy interaction’ being the science of latent and kinetic energy interaction with regards fission and fusion. This science has become a technology by being embodied and thereby able to demonstrate ‘Unifying field oscillation technology’ an important part of science with regards the macro system of existence. Hope you find this interesting as an introductory overview to the subject in review. Regards Eric Ashworth

  • eernie1

    Dear Joshi,
    Thank you for your prompt answer.I am familiar with the different lifetimes of the external and internal neutrons and that is the point of my argument.What keeps the neutron stable in the nucleous?How difficult would it be under certain conditions present in a specific nickel isotope to disrupt the equilibrium with an external electric field and allow the internal neutrons to decay?This would result in the extra proton and transmutation to copper.

  • Wladimir Guglinski

    Dear Bhagirath Joshi

    The isotope 8O18 has null electric quadrupole moment: Q(b)=0

    But when the 8O18 is excited, it exhibits a no null EQM: Q(b= -0,036 (9) , in barns:
    http://www.uni-due.de/physik/wende/englisch/nuclear-moments.pdf

    The stable isotope 3Li7 has Q(b) = -0,0400 (6)

    So, the excited 8O18 and the stable 3Li7 exhibit practically the same electric quadrupople moment.

    Obviously the 8O18 has an anomaly, because in spite of it is excited, however the isotope has a pair number of excess neutrons, and they have to take places symmetrically with regard to the center of the nucleus (according to current Nuclear Physics), and so the 8O18 keeps its symmetry.

    Actually, from current Nuclear Physics we had to expect Q(b)=0 even for the excited 8O18, because as the stable 8O18 has Q(b)=0, then even when it becomes excited the two neutrons keep their symmetrical distribution regarding to the center of the nucleus (because according to Nuclear Physics there is not any “reason” able to explain why the symmetry could be broken).

    3Li7 has one odd neutron.
    So, its electric quadrupole moment cannot be equal to the 8O18 (according to the model of current Nuclear Physics).

    So, the current Nuclear Physics cannot explain why 3Li7 and the excited 8O18 have the same Q(b).

    Is it possible to explain such anomaly of the 8O18 from your nuclear model ?

  • Wladimir Guglinski

    eernie1 wrote in November 30th, 2011 at 2:17 PM :

    “Dear Guglinski and Joshi,
    I am proud to call myself a pragmatist.”

    Dear Eernie1,
    as you are a pragmatist, then please answer the following question:

    Suppose that the current nuclear model of Nuclear Physics is wrong.
    Do you think is it possible to explain cold fusion by taking such model in consideration?

  • Dear all,

    Here are our latest news on the E-cat:
    Here’s a link to a letter that Prof Christos Stremmenos sent to Ny Teknik with regard to statements from Defkalion GT.

    Letter from Christos Stremmenos

    Kind Regards,
    Mats Lewan

  • Bhagirath Joshi

    Dear eernie1
    November 30th, 2011 at 2:17 PM

    Your Question:
    My question to both of you is have you taken into consideration the effect of external fields in the proposed configuration and interaction of the particles within the nucleous?

    Answer:
    No. That is the subject of “Solid state Nuclear Physics” where collective nature of nuclei is investigated.

    Your Question:
    Is the neutron different when inside or outside the nucleous?

    Answer:

    In free space neutron survives around 10 minutes and disintegrates.

    Here is the link to the detail of neutron disintegration.

    http://www.terra.es/personal/gsardin/news21.htm

    Thanking you
    Bhagirath Joshi

  • eernie1

    Dear Guglinski and Joshi,
    I am proud to call myself a pragmatist.I have always started my thoughts with emperical observations.At least,any theoretical output of mine is reached on some level of a factual basis.Using this technique there are a number of conclusions that I can use to start my investigation into Rossi’s methology for achieving his large COP output from nickel/hydrogen interactions.(1)Particle Physicists have experimentally proven what energy requirements are necessary for any particle to enter the nucleous of an atom.These tests show that the proton or any equivalant charged particle requires an enormous amount of energy to overcome the coulombic barrier with the exception of a small probability of tunneling.The possibility of neutron formation(which can penetrate easily)is highly improbable and is only observed in the nickel reaction in a small limited amount when claimed.Gamma output,which can only come from the nucleous is also limited to a relatively small amount.My conclusion is that,if we accept the generation of excess energy from the nucleous(very probable from test observations)the energy output must be triggered from external electric fields created by injecting either H- or H+ ions into the electron configurations of the nickel atom.These fields then distort the equilibrium energy fields of the stable atomic nucleous which result in nuclear decay mechanisms which produce energy or particles which exit the nucleous and produce the observed excess energy.For example, Kaon decay which produces muons(heavy beta particles)positrons and gamma photons.(2)We know that particals do not need to penetrate the nucleus to have their fields interact with the fields within the nucleous.Observations of many tests tell us that there is for every proton in a stable atom an electron in its surrounding area.Even when we strip all the electrons from the atom(complete ionization)the nucleous will regain the lost negative charges without undergoing transmutation.This occurs in every flourescent light we use every day.Strong evidence of field interactions between external and internal fields of the atoms without particle penetration.I conclude from these observations that the unconstrained atom will attempt to regain its ground state condition when unperturbed by other fields.However if the atomic fields are constrained by other conditions such as surrounding lattice electrons,the injection of other fields(H- H+)into its electronic configuration could cause major changes in the electric fields of the nucleous resulting in enhanced decays of energy levels and rearangement of internal particles.One result could be the decay of a neutron into a proton transmutating a nickel atom into a copper atom.
    My question to both of you is have you taken into consideration the effect of external fields in the proposed configuration and interaction of the particles within the nucleous?Is the neutron different when inside or outside the nucleous?

  • Wladimir Guglinski

    Bhagirath Joshi wrote in November 28th, 2011 at 5:47 PM :

    “Answer:

    My model has np shell. The lowest np shell K, S0 orbit has two np pair. Which is 2He4. Uranium 238 has 54 excess neutrons and are arranged in a interior excess neutron shell. The 7 neutrons in the outer most “excess neutron shell” are in constant collision with lowest np shell. The kinetic energy is about 4.8 to 5 Mev of alpha particle emitted. Since it comes out at a tangent, it did not originate exactly at the center of nuclei , but from an orbit around the center.”

    Dear Bhagirath Joshi
    I think such explanation is no satisfactory.
    Let me explain why:

    Consider a xyz system of coordinate axis within the 92U328 nucleus. And consider that the z-axis is coincident with the axis about which the nucleus rotates.
    Therefore we are considering that the nucleus gyrates about the z-axis.

    It is obvious that, in order to emit a 2He4 particle moving outside the nucleus along a trajectory which starts in the center of the nucleus, such 2He4 particle must be emitted along the z-axis direction.
    Otherwise, if the 2He4 is emitted along any other direction, it will be dragged by the rotation of the nucleus when the 2He4 crosses the Coulombic barrier. And so the 2He4 will leave out the nucleus by a tangential trajectory.

    In the current nuclear model of nuclear physics there is no reason why the 2He4 particle must be emitted ALWAYS along the z-axis direction.
    After all, we cannot find any “REASON” why the 2He4 must be emitted along a preferential direction (according to the current model of nuclear Physics).
    That’s why the prevailing nuclear physics is not able to explain “WHY” the 2He4 moves out the 92U238 along the direction starting from the center of the nucleus.

    In your model, I also cannot see “WHY” the 2He4 is ALWAYS emitted through a preferential direction: along the z-axis.

    In your model the 2He4 would have to emitted along any direction. For instance, consider that the 2He4 is emitted along the y-axis direction. Well, then when the particle crosses the Coulombic barrier, its trajectory will become tangential, no matter if “it did not originate exactly at the center of nuclei , but from an orbit around the center“.
    Because if the 2He4 is NOT emitted along the z-axis direction, it will leave out the nucleus by a tangential direction. ALWAYS!!!

    In current nuclear model, and also in your model, the most of particle would have to be emitted in any direction, different of the z-axis direction. Only a few quantity of 2He4 (less than 5%) would be emitted along the z-axis direction.
    Therefore, more than 95% of 2He4 particles would have to be emitted in tangential direction, according to your model, and also according to the prevailing model of nuclear physics.

    Why is in your model the 2He4 emitted ALWAYS along the z-axis direction ?

  • Andrea Rossi

    Dear Luca Salvarani:
    You are right: in fact, I confirm that all the improvements will be applied to the E-Cats already bought if in the meantime such improvements will have been actually consolidated.
    Warm Regards,
    A.R.

  • Luca Salvarani

    Dear Andrea Rossi

    1- I’m very happy for the latest developments… expecially the opportunity to see the 1 MW plant just sold to a normal customer in operation…. it will be a great promotion for you!
    2- After the great improvements in electric production, thanks to cooperation with Ingegner Fioravanti, can we expect an electric 10W e-cat for home in LESS than 2 years?
    3- Don’t you think the expectable and likely improvements (mainly in performances and costs) for a virgin technology as e-cat could stop some clients from buying your product? Perhaps you need a sale formula that let clients benefit from these improvements. It’s just my humble idea.
    4- Can I ask you how many 1 mw plants have you already sold? I’m eager to know all news.. Sorry for my not perfect english and good luck!

  • Francesco Benetti

    @MK thanks to advise about the petition, I signed up and voted it.
    Kind regards.
    Francesco

  • Andrea Rossi

    Dear Stefano:
    Yes
    Warm Regards,
    A.R.

  • stefano

    Dear Mr Rossi,
    am I in Your 10k list?
    Best Regards!

  • Andrea Rossi

    Dear Mark Saker:
    1- yes
    2- we can reduce the reaction
    3- when we will also be able to supply electricity, it will be a co-generation. The heat will always be useful, to produce either heat or cool. If no use of the heat will be possible, a passive heat exchanger will dissipate the heat, as it happens in normal air conditioners. Bu it will be rare, I suppose.
    Warm Regards,
    A.R.

  • Mark Saker

    Dear Andrew Rossi,

    Will the 1st Generation home e-cat provide only heating, and not electricity? If this is true, is the start-up time of the e-cat quick enough so that it could be used with a timing switch as a normal boiler would, or can you reduce the reaction low enough so it is constantly working, but the reaction can be increased rapidly when required?

    If the 1st generation e-cat will provide electricity as well, then it would want to be on 24/7. What will you do with the waste heat/water when it is not required?

    Many Thanks

    Mark

  • Here’s a compilation of the latest news:

    A Week of E-Cat News Flurry – It’s amazing how much E-Cat cold fusion news is accumulated in just one week, with no less than twelve news sites dedicated to covering it, thus creating these 31+ stories, including news on customers, products, competitor progress, pending scientific validations, and history.

  • Andrea Rossi

    Dear Charlie Zimmerman:
    I confirm that no radiations above the background in relevant measure have been found in the controlled explosive tests. I cannot enter in particulars, because I cannot give information regarding what happens in the reactors.
    Warm Regards,
    A.R.

  • Bhagirath Joshi

    dear Wladimir Guglinski
    November 27th, 2011 at 8:47 AM

    Your Question:

    In your model, as there is not a central 2He4, there are two things to explain:

    1- Why 2He4 are emitted ? Instead of one 2He4, why not two deuterons 1H2? After all, I cannot see why a 2He4 particle would be created in your model, so that to be emitted in the alpha decay.

    2- Why the 2He4 leaves out the nucleus from its center? As there is not a central 2He4 in your model, the alpha particle could be emitted in any direction. And therefore many alpha particles would be emitted along a tangential direction.

    Answer:

    My model has np shell. The lowest np shell K, S0 orbit has two np pair. Which is 2He4. Uranium 238 has 54 excess neutrons and are arranged in a interior excess neutron shell. The 7 neutrons in the outer most “excess neutron shell” are in constant collision with lowest np shell. The kinetic energy is about 4.8 to 5 Mev of alpha particle emitted. Since it comes out at a tangent, it did not originate exactly at the center of nuclei , but from an orbit around the center.

    In my model, np shell has paired neutron proton. Therefore they can always form a localized cluster of 2He4 under some destructive influence.

    It would be of great interest to investigate, why individual proton or proton neutron pair is not emitted?

    In case of elements disintegrating with beta+ (positron) emission, It could have been quite easy to just emit proton. But instead only the positive charge is detached from proton and remaining is kept as a neutron. The explanation for this comes from my model of neutron, where charges are loosely connected to the neutron as well as to proton. These charges can be easily knocked down, to achieve stability of the system.

    The true understanding of behaviour of elements can come only from development of “Solid state nuclear physics”.

    Thanks for the patience, I was a bit busy.

    Bhagirath Joshi

  • Dear Sterling Allen, Your article is absolutely correct but without the understanding of the science which not only applies to cold fusion but anything to do with fission/fusion because to understand fission and fusion properly i.e. creation to destruction, this being the entire energy cycle reveals a massive secret whereupon what can be considered miracles today become reality tomorrow. Problems with regards cellular energy are also wrapped up in a fission/fusion interacting energy cycle. Nothing in life could exist without the interaction of these two major forces. This is the real problem. The working of the E-Cat helps reveal this major mystery. I believe we have to look upon this problem as a game of chess. Perhaps black against white but probably to be more accurate grey against a lighter shade and this is why I stress the need to understand the entire energy cycle. Its the understanding that will contribute towards the proof positive. Hope this is not too confusing. Regards Eric Ashworth

  • Charlie Zimmerman

    Dear Mr. Rossi,

    I was interested in your comments regarding intentionally causing explosions of the device during safety testing. I had previously understood that short half lived radioactive isotopes of Copper and Nickel were rapidly decaying within the device and that this radioactivity was shielded. But, during an explosive event, the radioactive isotopes would be exposed to the environment without shielding before they would have a chance to decay.

    1) Are there short lived radioactive isotopes as in your patent and paper published here?
    2) Do those radioactive isotopes escape during an explosion?
    3) Are you taking proper precautions yourself against such dangers?

    A concerned fan,
    Charlie Zimmerman

  • I’ve posted an editorial that mentions E-Cat as an example.

    Climategate 2.0 — Politics of Science Trumping Left and Right – The same elite scientific establishment that prostituted themselves to promote “global warming” for political reasons, disregarding science (and who support the 9/11 official story); is the same corrupted scientific establishment that has said “cold fusion is impossible.” Don’t buy their lies. They are liars, as proven. Free energy is possible. The universe is full of it.

  • Andrea Rossi

    Dear Jarioslaw Bem:
    While you swap one reactor the others can work.
    Warm Regards,
    A.R.

  • Jaroslaw Bem

    Dear Andrea Rossi

    You told, that one core swap takes 30 minutes. Is that means, in 1 MW reactor swap process takes 30 x 52 modules = 1560 minutes = 26 hrs ?
    Recharge time of fuel, including time for cooling reactor, and again heating process, takes approximately 2 days.
    Will each the swap process makes reactor 2 days out of order ?

    Best regards,
    Jaroslaw Bem (Poland)

  • Wladimir Guglinski

    Eric Ashworth wrote in November 27th, 2011 at 2:14 PM :

    My guess as to why there is no stable isotope of hydrogen is because our present environment is too positive to support 100 neutrons with regards this element.

    Dears Eric Ashworth and Bhagirath Joshi:

    In my Quantum Ring Theory, the reason why there are no hydrogen isotopes with 100 neutrons is the same reason why there are no dineutrons 2n0 in Nature.
    As there is no repulsion between two neutrons, dineutrons 2n0, trineutrons 3n0, fourneutrons 4n0… hundered neutrons 100n0 would have to exist.
    The Heisenberg isospin is not an acceptable explanation, because it is only a mathematical postulate.
    As two neutrons interact through the strong force (and there is no repulsion between two neutrons), only a FORCE of repulsion could be able to separate them.
    A mathematical postulate as the isospin is not able to create a FORCE of repulstion, so that to separate two neutrons tied by the strong force of attraction.

    In my QRT, all the elementary particles have two concentric fields.
    The neutron has two concentric fields. The external field is null (because it is the overlap of the external fields of the proton and of the electron).
    The internal field is not completely null, because there is not a perfect overlap of the proton and electron internal fields, since the electron is not situated in the center of the proton.

    In long distances, there is no repulsion between two neutrons, because their interaction occurs through their external fields.
    But in small distances, as within the nuclei and in the structure of a dineutron, there is repulsion between the internal fields of two neutrons.

    That’s why dineutrons are not stable. And that’s why the repulsions within the nuclei increase with the growth of the quantity of neutrons within them.

  • Dear Bhagirath Joshi, Re: Comments to insight. Good questions asked and there has to be answers if progress in nuclear physics is to made. I have some suggestions. Radioactivity, I believe, is due to an element transmuting due to displacement by its immediate environment caused by the element entering into ever more positive zones within its macro system of existence between its creation stage to its eventual destruction stage. In the overall scheme distance travelled is a factor. Consequently, time becomes part of the equation with regards duration, this being an interpretation of the distance, if you know the velocity. An isotope is an element in a stage of becoming with regards a period of growth due to climatic change. My guess as to why there is no stable isotope of hydrogen is because our present environment is too positive to support 100 neutrons with regards this element. I am assuming that a hydrogen isotope with 100 neutrons would contain a gravity mass that would by interaction in an environment with a specific degree of positive gravity be unable to exist in such a form due to a conflict of forces between the macro system of support and the micro system of the existing isotope. Alternatively, if it were possible to take such an isotope of hydrogen out of this environment into one with less positive gravity then most probably the isotope would become stable. I believe it is the macro environment that controls the micro world. This is why the pattern of the macro system has to be understood in order to understand the micro world of sub atomic physics with regards transmutation. The E-Cat appears to be working well but outside of the laws of known physics today and this is why it is so important to explore all probable explanations. When the law is found the doubting Thomas’s will disappear with regards this new found technology although I know that Andrea does not care a hoot. As he says he is too busy manufacturing. Regards Eric Ashworth.

  • Wladimir Guglinski

    Dear Bhagirath Joshi
    in the alpha decay of uranium 92U238, the experiments have shown that the alpha particle 2He4 is emitted from the center of the nucleus.
    That is, the alpha particles move outside the nucleus along a trajectory which starts up in the center of the nucleus.

    There is now way to explain it by considering the current nuclear model of Nuclear Physics, because the alpha particle could be emitted in any direction. And as the nucleus has a spin, when the alpha particle leaves the nucleus it would be dragged by the spin, and it would leave out the nucleus by a tangential trajectory.
    Also, there is not a satisfactory explanation for the reason why 2He4 particles are emitted. The theorists claim that it is because it is energetically most favorable for the 92U nucleus. But this is not true, because instead of one 2He4 particle, the 92U could emit two deuterons 1H2 (it is easy for the deuteron to cross the Coulombic barrier, because its charge is +1, while the charge of 2He4 is +2).

    In my nuclear model, the central 2He4 has an oscillatory motion along the z-axis within the nuclei. So, when the 2He4 leaves out the 92U nucleus, its trajectory beggins in the center of the nucleus (this is the reason why the experiments always detect the alpha particle moving along a trajectory which starts in the center of the nucleus).
    The alpha particle moves along the z-axis direction when it leaves out the nucleus.

    In your model, as there is not a central 2He4, there are two things to explain:

    1- Why 2He4 are emitted ? Instead of one 2He4, why not two deuterons 1H2? After all, I cannot see why a 2He4 particle would be created in your model, so that to be emitted in the alpha decay.

    2- Why the 2He4 leaves out the nucleus from its center? As there is not a central 2He4 in your model, the alpha particle could be emitted in any direction. And therefore many alpha particles would be emitted along a tangential direction.

  • Wladimir & Bhagirath Joshi, Perhaps this is something to be considered, if the neutron is of major importance why not find out the pattern with regards its position within the framework of a unit of neutral energy. When I refer to neutral energy I am referring to a balanced unit of energy containing a proton zone, an electron zone and a neutral zone ( an atom of substance). If you can establish the geometry then maybe some of the major anomalies in nuclear physics could be overcome.

    How I would like to start is by stating that a theory does exist with regards helical trajectories and that this theory has been embodied by understanding four dimensions of potential energy and three basic shapes in geometry. Pyramid, sphere and cube and relating these shapes to six dimensions comprised of four potentials and two powers. Four potentials with regards the initial structuring process but three potentials when the process is complete due to positive gravity of the structure becoming an attribute of the structure that has been formed. This is not complicated but it will require pen and paper.

    Take a cube and divide it in half from flat to flat. Then put within the top half of the cube one pyramid. Base of pyramid same dimension same dimension as the flat of the cube, apex of pyramid central position of the top flat. Point of compass at central position of base of pyramid, circumference of sphere from apex of pyramid. Base of pyramid separates the positive zone above from the negative zone beneath. Negative zone of absolute fission dimension equals the outer points of the cube. Therefore, make a circle at the outer points from the middle at the base of the pyramid and then take the distance between inner circle and outer circle and make this the measurment of the radius of the the inner cone at the central position at the base of the pyramid and it should resemble a hemisphere within the pyramid at the base. Apex of cone at the apex of pyramid. This cone is the inner core and contains the most positive gravity by being empty within a unit of neutral energy and is responsible for fusing the positive mass together. Therefore, divide the positive half of the cube that contains the pyramid into three sections. The base section is the height of the hemisphere with two sections above of equal height. Put the point of the compass on the outer edge of the hemisphere at base of pyramid and draw a semi circle from one third up to one third down and draw another semi circle from two thirds up to two thirds down and repeat on other side of hemisphere. Now the cube contains six sections within the inner circle and one section above the apex of the pyramid and another section beneath the opposite flat of the cube (eight sections in all). The top section of the pyramid is where absolute fusion produces fission. This loss of energy is from the diametrical dimension that is a product of positive gravity produced by the negative cone. A diametrical dimension is always shorter than a circumferential and therefore, the negative cone at its apex, the one beneath the pyramid not the one within, being in the reverse direction can be considered a ‘black hole’ due to this discrepancy of distance/time factor with regards an energy cycle. Consequently, to stand within such a cube and look towards the apex of its pyramid you would only see a ‘white hole’ behind which in a systemic system would be the ‘black hole’ of a more positive neutral cube.

    This pattern consists of four helical trajectories. This being the minimum to form a neutral mass within, such as a planet, within such a pattern. The lower zone beneath the pyramid should be labeled consciousness formative zone and above the base of the pyramid should be labeled intelligent formative zone. The cosciousness zone is condensing down from massive structure and is comprised of four dimensions of fizzed energy that responds to the ‘black hole at the apex of the negative cone in the lower half of the cube. This position represents an instigating point of origin with regards the positive gravity of the mass that is being created beneath the pyramid. As the structure/mass is drawn within because of the positive gravity it begins to become more solid/defined and develops an electro magnetic field system (EMF) that is able, early on, to support a prehistoric form of life in the consciousness zone of the cube. Only when the planet now enters the positive zone of the cube does it roll over, negative south of the planet is attracted to positive north of the four helical trajectories because of the now independent field system of the mass caused by the fourth aether that could be said to have sunk due the positive gravity of the mass causing it to fuse down and fizz out to form a very defined EMF system of a three potential unit of energy that now represents intelligence with regards a structure and its EMF system. The sudden expansion of the fourth aether by positive fusion force could in theory account for an ice age to occur within a planets development stage. These four Aethers are four vibrations when related to atomic mass but when related to the helical trajectories of electro magnetic forces that are responsible for forming mass they are referred to as rays of a sopecific intensity due to their point of origin with regards to the disintigration of a mass. It appears that fission force energy can only fizz out to a certain distance after which it collapses back due to the increase in positive gravity created by this fizzing activity and.

    It has been postulated with regards to this particular pattern that a systemic system of four solar systems being four neutral cubes form the diametrical dimension of one cosmic system and that four cosmic systems form the diametrical dimension of one greater structure that could be labeled Y and that four Y structures could form the diametrical dimension of an Absolute, with regards containing without, in retrospect to within, an absolute linear force devoid of curvature forces. A solar system contains four neutral planets of varying potentiality between one planet of absolute negativity and one planet of absolute positivity. Therefore, it appears that the absolute energy being external linear forces of a structure are directly connected to the absolute energy of atomic curvature forces within by various potentials of curvature force energy. The linear force is the static boundary layer and is responsible for absolute isolation on the Y structure. Anyuthing within Y is a potential of the absolute and therefore can be considered a molecular force with regards maintaining an identity of a mass. Linear force energy can be considered an exterior radial to a structure. To return to the pattern, six sections within the cube. Top section of pyramid associated to solidity, beneath liquidity, bottom section gaseousness. Exterior to the pyramid above Aether and above the Aether is the ‘black hole ‘ of the more positive neutral cube. The Aether dimension in the sections beneath the pyramid is beneath the gaseousness and this is the ‘black hole’ of the neutral cube. This I believe is sufficient information at the moment and represents an introductory overview of the geometry to interactive neutral cube energy with regards fission and its association to negative gravity and fusion with positive gravity. I have tried to keep it as brief as possible. Regards Eric Ashworth.

  • Bhagirath Joshi

    Dear Wladimir Guglinski

    November 26th, 2011 at 6:26 AM

    Your Question:

    what are the two charged particles of your neutron model ?

    Answer:

    I have not named them yet. Some work is still going on.

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

    Answer:

    Yes they have mass.

  • Andrea Rossi

    Dear Enzo De Angelis:
    The address is correct, maybe you got some busy moment.
    Anyway I got your pre- order now: is accepted already.
    Warm Regards,
    A.R.

  • Enzo de Angelis

    Caro Ing. Rossi,
    Sono spiacente di disturbarti di nuovo ma ho provato a inviare il pre-ordine all’indirizzo infoleonardo1996.com ricevendo il seguente messaggio: Il messaggio non ha raggiunto alcuni o tutti i destinatari.

    Oggetto: Pre ordine E-Cat home
    Inviato: 26/11/2011 18:27

    Impossibile raggiungere i seguenti destinatari:

    ‘info@leonardocorp1996.com’ 26/11/2011 18:27
    550 RCPT address has non-existant domain

    L’indirizzo è corretto o è solo un problema contingente?

  • Italo A. Albanese

    @ Italo & Andrea Rossi:
    For what I know, lead absorbs gamma rays also when is liquid! You just have to made the e-cat vessel waterproof, in order not to leak molten lead.

    Best regards,
    Italo A.

  • Bhagirath Joshi

    Dear Wladimir Guglinski
    November 25th, 2011 at 7:01 PM

    Your Question:

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

    Answer:

    Since the cross section area of “excess neutron” is so small , we do not have a direct tool to measure that other than neutron scattering. Many experiments were done using gold foil and fast neutron source and the slides exists. If we can get hold of those old slides and go hunting might be able to find something that says that there is something at the center. The Excess Neutron shell for Gold is made up of 39 neutrons and hence large enough to leave its imprint. Some mathematical calculations and use of computers in comparing and sorting old slides will speed up the process of hunting. I think we can do it. We just have to ask the permission from Los Alamos or Other lab and get to it.

    Your Question:
    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?

    Answer:
    There were so many other interests of study that they might have kept the issue on the back burner. The model I came up with and conclusion drawn are from their Hard work in isotopes and neutron studies. I think after the failure to find higg’s Boson, they might change the direction.

    The planck’s constant as it exists to day the wavelength of neutron and proton is same as the size of nuclei. If that is accepted , no element will ever exists due to the constant continuous collisions . But the fact is that elements exist and it fails to describe the size of individual particles in the nuclei. This is another argument against keeping planck’s constant constant.

    One another interesting fact emerges that particles once out of the nuclei expand and increase in size in free space.

    My back ground is in Solid state Physics and nuclear and plasma engineering. An atom in free space has its valance electrons tied to the atom. However in solid it is impossible to tag the valence electron to a particular atom. Inside the solid it is just a big electron cloud, which one can push around in a controlled and predictive fashion. That is why it is high time to think and create a branch of Solid Sate nuclear physics. My Model suggests that the excess neutrons have very weak binding energy, even half as much as valance electron binding energy. Hence in solid they behave completely differently. I suspect that is the reason for Half life of radio active element and do not disintegrate all at the same time.

    Since electrons in solid behave collectively as a cloud, if hydrogen is infused, it will behave as free proton. Imagine, how many more discoveries are possible with this new branch of “Solid state Nuclear Physics”

    Your suggestion to start looking for “what is at the center of nuclei” is wonderful. Let us do that.

    Thanking you
    Bhagirath Joshi

  • Andrea Rossi

    Dear Italo:
    We have resolved this problem in a confidential way. Of course this was an issue to be addressed, you are right.
    Warm Regards,
    A.R.

  • Italo

    Dear Ing. Rossi, having reached a 450°C temperature of output steam, I suppose you have reaplaced the lead with another material having an higher melting temperature. is it true?
    Kind regards
    Italo

  • Andrea Rossi

    Dear Enzo De Angelis:
    To enter in the waiting list to buy an E-Cat you have just to send an email to
    info@leonardocorp1996.com
    You will receive an acceptation answer within 24 hours.
    You will have to pay nothing and you will have the possibility to cancel your pre- order when we will inform you that your E-Cat will be ready for delivery. DO NOT SEND MONEY !
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Alex Viroli:
    Thank you.
    Warm Regards,
    A.R.

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