By prof. Christos Stremmenos
After several years of apparent inaction, the theme of cold fusion has been recently revitalized thanks to, among others, the work and the scientific publications of Focardi and Rossi, which has been conducted in silence, amidst ironical disinterest, without any funding or support. In fact, recently, practical and reliable results have been achieved based on a very promising apparatus invented by Andrea Rossi. Therefore I want to examine the possibility of further development of this technology, which I deem really important for our planet.
Introduction
I will start with patent no./2009/125444, registered by Dr. Ing. Andrea Rossi. This invention and its performance have been tested and verified in collaboration with Prof. Sergio Focardi, as reported in their paper, published in February 2010 in the Journal of Nuclear Physics [1]. In that scientific paper they have reported on the performance of an apparatus, which has produced for two years substantial amounts of energy in a reliable and repeatable mode and they have also offered a theoretical analysis for the interpretation of the underlying physical mechanism.
In the history of Science, it is not the first time that a practical and reliable apparatus is working before its theoretical foundation has been completely understood! The photoelectric effect is the classic example in which the application has anticipated its full theoretical interpretation, developed by Einstein. Afterwards Einstein, Plank, Heisenberg, De Broglie, Schrödinger and others formulated the principles of Quantum Mechanics. For the interactive Nickel/Hydrogen system it would be now opportune to compile, in a way easily understood by the non expert the relevant principles and concepts for the qualitative understanding of the phenomenon. Starting with the behavior of electrically charged particles in vacuum, it is known that particles with opposite electric charge attract themselves and “fuse” producing an electrically neutral particle, even though this does not always happen, as for instance in the case of a hydrogen atom, where a proton and a electron although attract each other they do not “fuse”, for reasons that will be explained later. On the contrary, particles charged with electric charge of the same sign always repel each other, and their repulsion tends to infinity when their distance tends to zero, which implies that in this case fusion is not possible (classical physics).
On the contrary, according to Quantum mechanics, for a system with a great number of particles of the same electric charge (polarity) it is possible that a few of them will fuse, as for instance, according to Focardi-Rossi, in the case of Nickel nuclei in crystal structure and hydrogen nuclei (protons) diffused within it, Although of the same polarity, a very small percentage of these nuclei manage to come so close to each other, at a distance of 10-14 m, where strong nuclear forces emerge and take over the Coulomb forces and thus form the nucleus of a new element, either stable or unstable.
This mechanism, which is possible only in the atomic microcosm, is predictable by a quantum-mechanics model of a particle put in a closed box. According to classical physics no one would expect to find a particle out of the box, but in quantum mechanics the probability of a particle to be found out of the box is not zero! This is the so called “tunneling effect”, which for systems with a very large number of particles, predicts that a small percentage of them lie outside the box, having penetrated the “impenetrable” walls and any other present barrier through the “tunnel”! In our case, the barrier is nothing else but the electrostatic repulsion, to which the couples of hydrogen and nickel nuclei (of the same polarity) are subjected and is called Coulomb barrier.
Diffusion mechanism of hydrogen in nickel: Nickel as a catalyst first decomposes the biatomic molecules of hydrogen to hydrogen atoms in contact with the nickel surface. Then these hydrogen atoms deposit their electrons to the conductivity band of the metal (Fermi band) and due to their greatly reduced volume, compared to that of their atom, the hydrogen nuclei readily diffuse into the crystalline structure of the nickel, including its defects. At this point, in order to understand the phenomenon it is necessary to briefly describe the structure both of the nickel atom and the nickel crystal lattice.
It is well known that the nickel atom is not so simple as the hydrogen atom, as its nucleus consists of dozens of protons and neutrons, thus it is much heavier and exerts a proportionally higher electrostatic repulsion than the nucleus of hydrogen, which consists of only one proton. In this case, the electrons, numerically equal to the protons, are ordered in various energy levels and cannot be easily removed from the atom to which they belong. Exception to this rule is the case of electrons of the chemical bonds, which along with the electrons of the hydrogen atoms form the metal conductivity band (electronic cloud), which moves quasi freely throughout the metal mass.
As in all transition metals, the nickel atoms in the solid state, and more specifically their nuclei, are located at the vertices and at the centre of the six faces of the cubic cell of the metal, leaving a free internal octahedral space within the cell, which, on account of the quasi negligible volume of the nuclei, is practically filled with electrons of the nickel atoms, as well as with conductivity electrons.
It would be really interesting to know the electrons’ specific density (number of electrons per unit volume) and its spatial distribution inside this octahedral space of the crystal lattice as a function of temperature.
Dynamics of the lattice vibration states
Another important aspect to take into consideration in this system is the dynamics of the lattice vibration states, in other words, the periodic three dimensional normal oscillations of the crystal lattice (phonons) of the nickel, which hosts hydrogen nuclei or nuclei of hydrogen isotopes (deuterium or tritium) that have entered into the above mentioned free space of the crystal cell.
It could be argued that the electrons’ specific density and its spatial distribution in the internal space of the crystal structure should be coherent with the natural frequencies of the lattice oscillations. This means that the periodicity of the electronic cloud within the octahedral space of the elementary crystal cell of Nickel generates an oscillating strengthening of shielding of the diffused nuclei of hydrogen or deuterium which also populate this space.
I believe that these considerations can form the basis for a qualitative analysis of this “NEW SOURCE OF ENERGY” and the phenomenology related to cold fusion, including energy production in much smaller quantities and various reaction products.
Shielding of protons by electrons
In the Focardi-Rossi paper the shielding of protons provided by electrons is suspected to be one of the main reasons of the effect, helping the capture of protons by the Ni nucleus, therefore generating energy by fusion of protons in Nickel and a series of exothermic nuclear reactions, leaving as by-product isotopes different from the original Ni (transmutations). Such shielding is one of the elements contributing to the energetic efficiency of the system. From this derives the opportunity, I think, to focus upon this shielding, both to increase its efficiency and to verify the hypothesis contained in the paper of Focardi-Rossi. Of course, what we are talking of here is a theoretical verification, because the practical verification is made by monitoring the performance of the apparatus invented and patented by Andrea Rossi, presently under rigorous verification by many independent university researchers.
In my opinion, the characteristics of the shielding of the proton from the electrons should be defined, as well as the “radiometric” behavior of the system.
In other words, the following two questions should be answered:
- Which is the supposed mechanism that overcomes the powerful electrostatic repulse (Coulomb barrier) between the “shielded proton” and the Nickel nucleus?
- For what reason there is almost no radiation of any kind (experimental observation), while according to the Focardi and Rossi’s hypothesis there should have been some γ radiation (511 KeV) produced by the predicted annihilation of the β+ and β- particles that are being created during the Fusion?
I believe that some thoughts based on general and elementary structures, data and principles of universal scientific acceptance, might shed some light to this exciting phenomenon. More specific, I refer to Bohr’s hydrogen atom, the speed of nuclear reactions (10-20 sec) and the Uncertainty Principle of Heisenberg.
I will take Bohr’s hydrogen atom as a starting point (figure 1a), which stays at its fundamental state forever in the absence of external perturbations, due to De Broglie’s wave, accompanying the sole electron.
As stated before, in contact with the metal, these atoms lose their fundamental state, as their electrons are being transmitted to the conductivity band. These electrons, together with the “naked nuclei” of hydrogen (protons), form a freely moving cloud of charges (plasma at a degenerate state) inside the crystalline lattice. That cloud is being defused through the surface to the polycrystallic mass of the metal, covering empty spaces of the non-canonical structure of the crystalline lattice, as well as the tetrahedral and octahedral spaces between the molecules. As a consequence, the crystalline structure is covered by “delocalized plasma” (degenerate state), which is consisted by protons, electrons produced by the “absorbed atoms” of hydrogen, as well as by the electrons of the chemical valence of Nickel of the lattice, at different energy states (Fermi’s band). (Fig. 2)
Fig.1b
In this system, if one considers the probability of the creation inside the crystalline lattice of temporary (not at the fundamental state) “pseudo-atoms” of hydrogen with neutral charge, for example at a time of the order of 10ˆ-17 sec, then that possibility is not completely ill-founded. (Fig 1b)
Fig.2
According to the Uncertainty Principle of Heisenberg, the temporary atoms of hydrogen will cover during that small time interval Δt, a wide range of energies ΔΕ, which means also a wide range of atomic diameters of temporary atoms, satisfying the De Broglie’s condition. A percentage of them (at fist a very small one) might have diameters smaller than 10ˆ-14 m, which is the maximum active radius of nuclear reactions. In that case, the chargeless temporary atoms, or mini-atoms, of hydrogen together with high energy but short lived electrons, are being statistically trapped by the Nickel nuclei at a time of 10ˆ-20 sec. In other words, the high speed of nuclear reactions permits the fusion of short lived but neutral mini-atoms of hydrogen with the Nickel nuclei of the crystalline lattice, as during that short time interval the Coulomb barrier (of the specific hydrogen mini-atom) does not exist.
Afterwards, it follows a procedure similar to the one described by Focardi and Rossi, but instead of considering the capture of a shielded proton by the Ni58 nucleus, we adopt the hypothesis of trapping a neutral temporary atom, or a mini atom, of hydrogen (with a diameter less than 10ˆ-14 m) which transforms the Ni58 nucleus into Cu59 (copper/59, short lived isotope*).
It follows the predicted “β decay” of the nuclei of the short lived isotope of copper, accompanied by the emission of β+ (positrons) and β- (perhaps the electrons of the mini atoms trapped inside that nucleus during the fusion). These particles are being annihilated with an emission of γ radiation (two photons of γ of energy 511 KeV each, for every couple of β+ and β-).
In other words, whoever has experimented with this system should have suffered the not-so-harmless influence of those radiations, but that never happened. The radioactivity measured at the experiments is almost zero and easily shielded.
In any case, a rigorous, in my opinion, theoretical approach for the interpretation of that phenomenon with quantum mechanical terms, would give clear quantitative answers to the above stated models. With my Colleges of theoretical chemistry, we are already planning to face the problem using the time-depended quantum mechanical perturbation theory, bearing in mind the following:
- The total wave function (of the nucleus and the electrons) of temporarily, non-stable states.
- The total time-depended Hamiltonian, for temporarily states.
- Searching for the resonance conditions at that system.
Such an approach had a successful outcome at a similar problem of theoretical chemistry and we hope that it will be valid in this case as well.
Let’s go back to the intuitive, with ideal models, approach, in order to give a qualitative explanation for the (almost) absent radiations of the system, by using:
- First of all the Boltzmann’s distribution (especially at the asymptotic area of high energies).
- The photoelectric effect
- The Compton effect
- The Mössbauer effect
We have already mentioned that from the temporary mini atoms of hydrogen, the ones with diameter less than 10ˆ-14 m, have a larger probability of fusion. But, in order for them to be created, high energy bond electrons should exist at the “delocalized plasma” of the crystalline lattice.
1. Boltzmann’s statistics:
There are reasons to believe that the H/Ni system, at first at temperatures of about 400-500oC, contains a very small percentage of electrons in the “delocalized plasma” with enough energy to create (together with the diffused protons), according to the wave-particle duality principle, the first temporary mini atoms of hydrogen, that will trigger the fusion with the nickel nuclei and the production of high energy γ photons (511 KeV).
2. Photoelectric Effect:
It is not possible, the HUGE amount of energy (in kW/h), that the Rossi/Focardi reactor produces, as measured by unrelated scientists in repeated demonstrations (at one of them by the writer and his colleagues, Fig 3), to be created due to the thermalization of the insignificant number of γ photons at the beginning of the reaction.
Fig.3
I believe that, as stated above, these photons are the trigger of fusion at a multiplicative series, based on the photoelectric effect inside the crystalline structure.
The two γ photons can export symmetrically (180°) two electrons from the nearest Nickel atoms. The stimulation, due to the high energy of γ, concerns electrons of internal bands of two different atoms of the lattice and has as a prerequisite the absorption of all the energy of the photon. A small part of that energy is being consumed for the export of the electron from the atom and the rest is being transformed into kinetic energy of the electron (thermal energy).
The result of that procedure is to enrich the “delocalized plasma” with high energy electrons that will contribute multiplicatively (by a factor of two) at the progress of the cold fusion nuclear reactions of hydrogen and nickel and at the same time transform the hazardous γ radiation into useful thermal energy.
3. The Compton Scattering:
It gives the additional possibility of multiplication, this time due to secondary photons γ, in a wide range of frequencies, as a function of the angular deviation from the direction of the initial photon of 511 keV. That has as a result the increase of the export of electrons, due to the photoelectric phenomenon at the crystalline mass, in many energy/kinetic levels, which gives an additional possibility of converting the γ radiation into useful thermal energy.
4. The Mössbauer effect:
It gives another possible way of absorbing the γ radiation and transforming it into thermal energy. It is based on the principle of conservation of momentum at the regression of the new Cu59 nucleus/ from the emission of a γ photon. Relative calculations (Dufour) showed that this mechanism has an insignificant (1%) contribution.
It follows that, according to given data, the Photoelectric phenomenon and the Compton Effect, could explain the absence of radiations in the Focardi-Rossi system, which, from the amount of producing energy versus the consumption of Ni and H2, as well as from the experimental observation of element transformations, lead undoubtedly to the acceptance of hydrogen cold fusion.
ACKNOWLEDGEMENTS: The author wishes to acknowledge Aris Chatzichristos for the contribution in formulating this paper in English
References:
(1)www. journal-of-nuclear-physics.com /Focardi Rossi/ (A new energy source from nuclear fusion)
* I believe that the phasmatometric tracing of copper is the most definitive sign of nuclear fusion: From the relative bibliography (HANDBOOK OF CHEMISTRY AND PHYSICS, 66TH edition), it follows that the stable non radioactive isotopes of nickel are the following five:
58, 60, 61, 62 and 64. These, when fused with a hydrogen nucleus, are being transmuted relatively to Cu-59, Cu-61, Cu-62, Cu-63 and Cu-65. From these isotopes of copper only the last two (Cu-63 and Cu-65) are not radioactive, i.e. they are stable. The other three Cu-59, Cu-61, Cu-62, are being transmuted again to Nickel, with an average life expectancy of some hours and the most unstable Cu-59 in 18 seconds.
By prof. Christos Stremmenos
Dear Mr Charlie Zimmerman:
1- I cannot give this information
2- No, I did not say that. There has been a misunderstanding. Is correct what you thought.
3- Ni 59 doesn’t exist. It is a typo. We buy regular Ni powder, then we make a treatment of it wich changes the isotopical composition. In that paper I referred to the powder as we buy it, not to the composition of the powder after the treatment we make. In any case, the composition of Ni, as we buy it, is well known: 58 (67,88%), 60 (26,23%), 61 (1,19%), 62 (3,66%), 64 (1,08%).
After that, we change it.
I do not think you misunderstood, I think some typo is in the translation.
Warm Regards,
A.R.
Dear Mr Daniel Garcia:
1- we are working on this. Sterling engine is not fit for us: it needs over 700 Celsius degrees.
2- Please contact us in November for commercial issues.
Warm Regards,
A.R.
Dear Mr. Rossi,
I am so excited about your invention that I often go back and watch older information. I was watching the Jan 15th video, with English subtitles and was confused by some comments in it compared to more recent information you provided:
1) You said that gamma radiation may be hidden by the extremely complex internal geometry of the device. Did you mean physical geometry or the geometry of the reaction, ie, Nickel lattice?
2) You said that a rare isotope of copper is produced. Can you elaborate on this more? I thought only NI62 and NI64 are reacting to produce copper 63 and 65.
3) You said that you ran a reactor for 6 months with NI59. You didn’t mean pure NI59 did you? Plus, since you have said that only NI62 and NI64 react, then how did this reactor with 59 work?
Obviously some things are confusing me. Potentially, it was just bad translation in the subtitles or my own misunderstanding of comments you have posted here.
Best Regards,
charlie zimmerman
Dear Mr Herald Patterson:
Thank you for your questions, here are the answers:
1- Yes: like Flash Gordon! Seriuosly: what happens inside the reactor is influenced only by what is inside; outside there is only cooling and thermalization
2- Gamma have been regularly measured by us
3- Analysys of powders are the evidence of the transmutation
4- Wrong
5-Beta decay has nothing to do with my process, Widom Larsen theory has nothing to do with my process
6- I am looking for October too, my friend.
Warm Regards,
A.R.
Dear Mr. Rossi,
Thank you for refuting those two metropolitan legends. Unfortunately, there are other legends and unfounded rumors circulating around the internet as well. I will list a few of them here in case you would like to comment on them, and put an end to some ridiculous speculation that is taking place on the internet.
1) Other than the catalysts, hydrogen pressure, the special processing of the nickel powder, and the heat added to the system by the resistors there is some “other” factor that is critical to making the system work. For example, a source of radio frequency radiation to stimulate the processes inside the reactor vessel.
2) No gamma radiation is actually produced inside of the reactor vessel. They claim you will not let independent scientists measure the gamma radiation inside the reactor *not* because the signatures detected could reveal the patent pending catalysts, but because no gamma radiation would be found.
3) No nickel is actually transmuted into copper. They try to connect this to the lack of gamma radiation, to support their idea that some extraordinary but totally *non-fusion* process is taking place.
4) That you no longer think any form of fusion is taking place. They claim because you use the term Low Energy Nuclear Reactions, you non longer think a fusion reaction is taking place between the nickel and hydrogen.
5) Others claim there is no radiation being produced, except from beta-decay. Some push this idea to support a pet theory they religiously proclaim all over the net called, “Widom Larsen” theory.
I wish people would just take you at your word, instead of trying to twist the truth to support their own pet theories and ideas.
If you wish to comment on any of the above, I will do my best to spread your answers on the net to counter act the rumor-mongering taking place.
Thank you for all your work and willingness to interact with us.
I’m looking forward to October!
Herald
Mr Rossi, above all, congratulations.
Have you try to produce electricity with the E-cat and a Stirling generator? How are you thinking to transform the heat in electricity for domestic use?
Do you need a distributor in Madrid? 😉
Good Luck with your invention!!
Dear Mr Tomasz Rojewski:
In all the test has been preliminary checked the absence of radioactive emissions. Of course there cannot be Ku. We do not use radioactive materials and do not leave radioactive waste.
Warm Regards,
A.R.
Dear Andrea Rossi
Did tests in sweeden exclude a hiden nuclear reaction like curium?
ART:
Talking of ART, there are two metropolitan legends which are walking around:
1- We do not know the theory behind the operation of our apparatus: false, I know the theory , and will release it after the international patent will be granted. We could not arrive to produce our E-Cats, with their constant operation, without knowing the theory. One year ago I was not sure, now I’m pretty confident.
2- There will be a new public test somewhere (Greece, or Italy, or USA, or Sweden, etc): again, no more public tests will be made, the sole tests we make are the tests of the modules of the 1 MW plant which will go in operation in October in Greece, and obviuosly such tests are made with closed doors.
Warm Regards,
Andrea Rossi
Dear Mr Giuliano:
The quantity of Ni we use is irrilevant respect the production. Should all the energy of the world be made with this system, only the 1% of the world production of Ni would be consumed. Remember that 1 gram of matter makes 23 x 10^6 kWh.
Warm Regards,
A.R.
Egregio Ing. Rossi, complimenti per la sua invenzione!
Lei non crede che, per merito della futura applicazione industriale, ci potrà essere una corsa al Nickel?
Se Lei asserisce di aver testato anche altri metalli ma di essersi soffermato sul Nickel per questione di facile approvvigionamento e costi, non pensa che i prezzi di questo metallo potrebbero decollare?
Distinti saluti
Giuliano
ROSSI AND FOCARDI’S ENERGY
CATALYZER
Perhaps this time we have a chance
to turn a new leaf: perhaps
we can get busy turning
nickel into copper
in the presence of hydrogen.
But what else, in addition to heat,
comes out – neutrons,
gamma-rays, beta particles? Something
gets pumped up, probably something
probably decays to something else
along the way – oh, now the words
are flying! Open it up, turn loose
the physicists of hell
to scratch
their heads and measure
this and that.
Dear Brian Ahern,
You stated: “What is not widely known is that when nickel is processed below 15nm it causes the H atoms to go into chaotic motion.”
Can you explain more about Hydrogen’s chaotic motion including why the smaller particles would create that effect, what that would mean regarding its activity within the nickel lattice, and especially how you think such motion might affect the reaction process?
Dear Mr Andrew Wright:
We do not use deuterium.
Warm, Regards,
A.R.
Dear Mr. Andrea Rossi
I am wondering if the hydrogen gas you use contains some deuterium which is being preferentially reacted, or is any deuterium filtered out within the nickel lattice because it has a larger size of nucleus than hydrogen?
Is there any energy advantage in your reactor if you used pure deuterium for say larger power station type reactor applications? I ask as deuterium has always been put forward as a fuel for fusion in the hot fusion area.
I think you have found a way to use ordinary hydrogen gas so perhaps deuterium is an irrelevance in your reactor.
Apologies if the questions aren’t valid – I’m not a scientist. I’m looking forward to reading about your success with the larger unit you are building.
Dear Raul Heining
I believe that the hydrogen molecules need to undergo dissociation: the Ni favors this process.
But the miniatom not involves ionization.
Regards
Lino
Dear Dr. Rossi,
congratulations from Austria for the historic discovery ! I firmly believe this device will change our world for the better.
Starting October 2011 I would like to get involved with the distribution of this device in my home country, so please would you send me the contact information of your sales-representative? My goal is to become a local dealer for the E-Cat device here in Austria.
Thanks a lot and I wish you great success!
Gregor
Dear Mr. Andrea Rossi
I don’t understand how it work.However i really hope that your machine will come out on market soon.
Best Regards,
Dol
Dear Mr daniel De Francia:
Yes, indeed.
Warm regards,
A.R.
Dear Mr. Andrea Rossi,
So, the water does not get into contact of the nickel powder nor the catalyst, right?. That’s truly different from all other low energy nuclear devices up to now, is that it?
Best,
Daniel.
I understood the idea of shielding with protons with electrons closer than Bohr radius. What I do not understand is how the nickel as a catalist dissociates the H2 molecules and ionizes the H atoms (13.6 eV required).
Regards.
raul
Mr. Gilbert Schmidt;
In 1957 I worked for a company called Inland Testing Labs.We built a 60000 Curie Cobalt60 facility to test for gamma radiation effects on many materials.The walls of the facility were 12 feet thick made of dense concrete.In the center of the facility which was a circular room 15 feet in diameter was a circular 4 foot diameter water well 20 feet deep containing an elevator that went from the bottom of the well up to the floor of the facility. On top of the elevator was placed a three foot circular container three foot tall holding the Cobalt60 material.To enter the room safely the container had to be dropped to the bottom of the well.A special lead glass window was used to observe inside the facility from the outside,when the radiation source was elevated.As I recaall the gamma counter began measuring when the source was about 15 feet below the floor of the facility.The tests were done only with the source fully elevated and the level of radiation exposure was controlled by time,dropping the source after a specific period of exposure.This was the most consistant method for comparing radiation doses to effects on exposed materials.We irradiated food for the Army,beer for a brewery,airplane parts for a nuclear powered airplane under development for the airforce,cigarettes for a manufacturer,live pigs for a human exposure program,semiconductor devices and many more materials as requested.The only gamma radiation dose of interest was the fully exposed source output.
Dear Mr Daniel De Francia MTd2:
Absolutely not: catalyzers are inside the reactor, water is outside.
Warm Regards,
A.R.
Dear Andrea Rossi,
I was wondering about this: does any particles of the catalyzer or the nickel powder gets carried by the flowing water and pollutes the environment?
Best,
Daniel.
Dear Prof. Christos Stremmenos
I agree to your hypothesis:
Dynamics of the lattice vibration states
Another important aspect to take into consideration in this system is the dynamics of the lattice vibration states, in other words, …
The vibration into the lattice is considered in my paper How zitterbewegung contributes for cold fusion in Pamela Mosier-Boss experiment.
See the link:
http://peswiki.com/index.php/Article:_How_zitterbewegung_contributes_for_cold_fusion_in_Pamela_Mosier-Boss_experiment
In 2009 I made a suggestion to Pamela Mosier-Boss, in order to increase the amplittude of the resonance into the lattice and to accelerate the cold fusion reactions, by using deuteron gas into a buble crossed by an electric discharge, in her experiment.
The photons emitted by the deuteron gas would increase the vibration of the deuterons captured by the palladium lattice (the vibration would growth by resonance).
Look at the figure:
http://peswiki.com/index.php/Image:BAAAfig5a-coldFUSION-pamelaMOSIERboss.JPG
Unfortunatelly Pamela replied to me:
“Dear Wladimir,
Like many, we have very few funds and resources. But we will consider your suggestions and see what we can do as time and money permits.
Regards,
Pam ”
Dear Readers
Here is the link to my research paper, which I mentioned earlier.
http://www.rudramweb.com/documents/Excess_neutron_%20shell_Model_of_Nuclie.pdf
Thanking you with warm regards
Bhagirath Joshi
Dear Mr. Gilbert Schmidt and Dear Mr. Lino Daddi
Please refer to the link bellow for absorption coefficient of water at various energies.
http://physics.nist.gov/PhysRefData/XrayMassCoef/ComTab/water.html
I hope this helps.
Dear Mr Gilbert Schmidt:
I want not to comment the article to which you refer, which has not been written by me, and the Author will answer you. Just, to avoid ambiguities, I want to point out that when my reactors are not in operation there is not gamma emission. Besides, when it is in operation gamma rays are turned into heat and the gamma radiation measured ouside the reactor respects the limits imposed by the law (0.2 microSievert/h). We do not use radioactive material and we do not produce radioactive materials, as evidenced in years now of repeated tests with our reactors.
Just to avoid misunderstandings.
Warm Regards,
A.R.
” When not in use, the gamma ray “source” is stored in a pool of water which absorbs the radiation harmlessly and completely. ”
Can someone tell us how many inches of water is needed to be safe ??
To Bhagirath Johshi,
Do you want to confirm that the excitation energy of the molecule of water is about 511 keV (instead of 511 ev?).
Thanks
Lino
Luke:
In the previous excerpt the important line is the following:
‘When not in use, the gamma ray “source” is stored in a pool of water which absorbs the radiation harmlessly and completely. ‘
Sorry for including the whole paragraph.
Thanks
Dear Luke Mortesen
Please see this link.
http://uw-food-irradiation.engr.wisc.edu/Process.html
Here is the excerpt.
“Gamma rays with specific energies normally come from the spontaneous disintegration of radionuclides. Naturally occurring and man-made radionuclides, also called radioactive isotopes or radioisotopes, are unstable, and emit radiation as they spontaneously disintegrate, or decay, to a stable state. The radionuclide used almost always for the irradiation of food by gamma rays is cobalt-60. It is produced by neutron bombardment in a nuclear reactor of the metal cobalt-59, then doubly encapsulated in stainless steel “pencils” to prevent any leakage during its use in a radiation plant. Cobalt-60 has a half-life of 5.3 years. This technology has been used routinely for more than thirty years to sterilize medical, dental and household products, and it is also used for radiation treatment of cancer. Radioactive substances emit gamma rays all the time. When not in use, the gamma ray “source” is stored in a pool of water which absorbs the radiation harmlessly and completely. To irradiate food or some other product, the source is pulled out of the water into a chamber with massive concrete walls that keep any rays from escaping. Medical products or foods to be irradiated are brought into the chamber, and are exposed to the rays for a defined period of time. After it is used, the source is returned to the water tank.”
However still you can not stay too close to the system.
I hope this helps.
Professor Stemmenos,
Clarification:
Ni 58 to CU 59 to Ni 59
Are you saying in your paper that there is something special about the AR reactor in absorbing gamma rays? No doubt some contributes to the reaction as you say.
What percentage of the gammas would you expect to be absorbed on the exit from the reactor on a Cu 59 to Ni 59 decay?
Unless something special is happening, only a small portion of the 511 KeV would be absorbed by the crystalline lattice on their way out from the reactor. And like you said, not the most healthy place to be.
Thanks,
Luke
I am reading the May 25 Article and than I looked at my analysis.
If you look at the Isotope and abundance of Ni , It is evident that the most readily available isotope of 58Ni with .6808 abundance has two excess neutron and is unstable with half life E+ 18 a. When you refer to the shell structure I proposed, the instability is coming from the S0 shell, which is not spin balanced, since in my model any excess neutron first occupies the center and than other shells. Since Ni is even Z =28 , the stability of the nuclei comes from the paired p-n outer shell, since it is spin balanced, thus allowing more than three isotopes and the other weak forces can work to create more isotopes or elements through beta capture, emission, proton capture etc, since the neutron shell is open.
Once the H+ permeates the ni, it tries to pair up with other neutron or turns in to additional neutron and adds to the stability of Ni.
This is a more simplified answer to the actual mechanism involved, but it proves the reaction.
The experimental results on Ni and Sn and cross section calculation deviation from the standard model , are easily explained with the help of my model.
I am attaching the article here on cross section studies done. The most interesting point to note here is mentioned in the conclusion, which talks about neutron shell.
Please refer to this article.
http://www.nucl.phys.tohoku.ac.jp/fusion03/proc/liang.pdf
Thanking You
The water molecule has the excitation energy around 511 Kev. Here is the experimental data and explanation link.
http://www-als.lbl.gov/index.php/contact/167-isotope-and-temperature-effects-in-liquid-water-probed-by-soft-x-rays.html
The photons are instantly absorbed and the result is heat. In fact water is used in many application to curtain the gamma radiation.
Andrea,
I fully understand.
Many thanks,
Per
I have submitted the paper “Excess neutron Shell Model of Nuclei” when it is published, it will throw the light on the exact mechanism. However, in the absence of that, I will explain that the Cu has only two stable isotopes. 63 CU and 65Cu. since Cu has Z = 29, the stability comes from the Excess neutron shell. The P-N shell is highly unstable. Therefore, as soon as 59Cu is formed, it decays in to 59 Ni through positron ( beta positive) emission and Proton turns in to neutron, which moves further down and occupies the available position on the Excess neutron shell. This will be fully explained as soon as my Paper is published in this Journal.
Thanks
Concerning the presence of Cu in Ni samples after conducting the experiment – have you eliminated the possibility of Cu diffusing from a hot wire ohmic heater, if it is inserted inside the reaction chamber. Many heating elements contain copper, and if so, that might explain the natural Cu isotopes ratio found, as I understand, in the Swedish Lab.
Best wishes,
–Baruch
Dear Mr Per:
I am very sorry, but you put very important questions which are very confidential. It is not a pleasure for me to answer “I can’t answer”, but it is for me mandatory.
Warm Regards,
A.R.
Dear mr Paterson,
The above graph is based on data mesured, at the date shown on it, during a test of Defkalion team on e-cat
Regards
Christos Stremmenos
..nel caso venga tutto confermato perchè non chiamate l’apparato e/o il sistema di produzione:
E-Italy..sarebbe un omaggio all’Italia…a presto..
Rossi, Henk
Thanks for your responses.
If enrichment is involved, is there any particular reason you change the isotopic content of the initial Nickel powder so it results in a used powder containing copper with natural isotopic distribution?
From what I understood from the article in NyTeknik, Uppsala also examined the initial powder showing natural isotopic distribution, is this misunderstood?
Thanks and good luck with your work,
Per
Since there is no radioactivity detected before and after the operation, the most simple candidate reaction would be:
Ni-62 or Ni-64 + H -> Cu-63 or Cu-65
Which could be in line with a comment of A. Rossi that only a small percentage of the Ni reacts to copper, leaving the rest unaffected. Abundance of these Ni isotopes is 3.6% and 0.9% respectively.
Secondly, I have seen several somewhat contradictory statements about radio emission during operation:
1) No radiation whatsoever detected
2) Burst of gamma ray’s detected only at start and shutdown
3) Gamma radiation throughout operation detected, but levels are low and easily screened
I think number three is easiest to comprehend; in addition one should note that levels can be kept very low because much energy is released per reacting atom / few atoms are reacting per unit of time. On various descriptions of the E-Cat it is revealed that lead shielding is placed around the reactor; this both explains why no radiation was detected: it was screened, and suggests that indeed radiation is emitted: why else would one apply lead screens?
Regards,
Bastiaan.
AR,
Is Professor Stemmenos available to answer questions and continue dialogue on this Forum? His participation would be welcome.
As I understand it, he is free to speculate on the theoretical aspects, whereas you have important reasons to keep your hypothesis quiet until October.
Best,
-Luke Mortensen
As Per mentioned the Uppsala analysis seems to indicatate that no nuclear reactions took place.
But the actual charge in the reactor is not be the pure nickel powder. It is probably enriched and treated with a catalyzer.
This is the real trick of the trade, and for good reasons Mr. Rossi is very carefull to not disclose this secret.
So, since we do not know the actual composition of the initial charge no conclusions can be drawn from the Uppsala analysis.
Dear Mr PER:
To answer to this question I ‘d have to enter in confidential particulars regarding the charge and the operation.
Your observation is correct, in absence of more explication.
Warm regards,
A.R.
This problem wqas solved in 1983 by MIT professor Keith Johnson. The screening of nuclei was completely accounted for and enormous vibrational modes of H were defined and subsequently verified experimentally.
What is not widely known is that when nickel is processed below 15nm it causes the H atoms to go into chaotic motion.
I think I read the graph wrong. I see it was not a 13 hour test now. I apologize for my error.
Prof. Stemmenos,
I enjoyed reading the paper. However, I have a very serious question.
Does figure number three represent a system in self sustain mode? If so, when did you witness this test? Where did you get the data? Where did the test take place? If you have any additional information you can share please do so.
A self sustaining test of Rossi’s system over 13.5 hours is very important! It could help many of us who support Rossi’s technology to counter the claims of many naysaying skeptics.
Thank you.
Herald
Sincerely,
Any comments to the preliminary results of the element analysis and isotopic analysis at the Ångström Laboratory in Uppsala, which showed natural isotopic composition of the nickel and copper. The initial nickel-powder also had natural isotopic composition which some take as evidence that no nuclear reactions occured? What is your view?
“Both measurements show that the pure nickel powder contains mainly nickel, and the used powder is different in that several elements are present, mainly 10 percent copper and 11 percent iron. The isotopic analysis through ICP-MS doesn’t show any deviation from the natural isotopic composition of nickel and copper.”
kind regards,
Per