Nuclear signatures to be expected from Rossi energy amplifier

by Jacques Dufour
CNAM Laboratoire des sciences nucléaires, 2 rue Conté 75003 Paris France

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Abstract: the nuclear signatures that can be expected when contacting hydrogen with fine nickel powders are derived from thermal results recently obtained (Rossi energy amplifier). The initiation of the reactions (either by proton or neutron capture) is not discussed and considered as true. Proposals are made to check the process either by radiation emission measurements or by elemental analysis (ICP-MS)

1. Intruduction

In a recent paper [1], results are presented on vast amounts of energy (kWh) generated by contacting Hydrogen at pressures of tens of bars and temperatures round 400°C, with nickel powder (with an unspecified additive). No harmful radiations were measured, which is attributed to the presence of a lead shield absorbing γ emission occurring during the run and to the very short period of the instable species formed during the run and decaying after shut down. The efficiency of the process is very high (Eout/Ein up to 400). These levels of energy production strongly points to a nuclear origin. The proposed process [1] would be proton capture by the nickel nuclei. The coulomb barrier problem is suggested to be solved by the strong screening of the electrons. Another solution has been proposed [2]: virtual neutrons formation, reacting with the Nickel nuclei. This solution is also proposed in [3] with a very elaborate justification. In this paper, the capture of a neutron or a proton by a nickel nucleus is accepted as real. The consequences of these captures are analyzed (using very well documented nuclear chemistry data [4], [5]) and proposals are made for precise verification of the process invoked.

2. The neutron or proton capture by Nickel

The reactions paths for these 2 routes finally ends up to the same stable products (59Co, 61Ni, 62Ni, 63Cu and 65Cu) and are summarized in Table 1 and 2 below.

The energy release (see Table 3) occurs mostly by de-excitation through γ emission of the intermediate excited Ni* compound nucleus. The characteristics of this γ emission (depending upon the levels of the excited nucleus), are very well known [4]. This represents (on average) some 8 MeV (balance after deduction of the energy required for the “virtual neutron” formation, i.e 0,782 Mev). The remaining comes from the decay of the ground states of the radioactive intermediate species formed (59Ni, 63Ni, and 65Ni). Data for intermediate radioactive species are from [5].

The energy is released in a way very similar to the neutron capture route, with a lower release from the de-excitation of the intermediate excited Cu* compounds nucleus (some 4 MeV, see Table 3). The remaining half comes from the decay of the ground states of the radioactive intermediate species formed (59Cu, 59Ni, 61Cu and 62Cu). Data for intermediate radioactive species are from [5].

3. Evaluation of the reaction rates

One experiment (Type B) presented in [1], has yielded 3768 kWh for an energy input of 18.54 kWh (between March 5, 2009 and April 26, 2009). This is a net power produced of some 3 kW during some 4.5*10ˆ6 seconds.

From Table 1 and 2, the energy released per Ni atom (averaged by the isotopic composition of the Nickel) has been calculated under following hypothesis:

– the captures (proton or neutron) have the same probabilities whatever the Ni isotope is. This is a first approximation. For the neutron capture route, following cross sections (barn) are measured: 58Ni:4.6, 60Ni:2.9, 61Ni:2.5, 62Ni:15 and 64Ni:2.9). – the subsequent reactions with formed products are not taken into account (too low concentration to have any significant effect).

– decay energy of nucleus with half life time much longer than the experiment duration have been ignored (59Ni for the proton route and 59Ni, 63Ni for the neutron route)

Table 3 below is thus obtained.

As expected, the 2 routes give similar amounts of energy, mainly de-excitation for the neutron route and half de-excitation, half decay for the proton route.

The proton or neutron capture rate can thus be evaluated as:

4. Evaluation of the γ emission rates

The de-excitation of a compound nucleus resulting from neutron capture is very well documented [4]. For nickel, 1 capture gives rise to 2.66 emission of γ photon, with an energy repartition fi given by Table 4:

For the proton capture route, less data are available. To get a first order of magnitude of the γ emission coming from the de-excitation of the primary nucleus formed, the same number of photons per proton capture with the same energy repartition as for the nickel has been taken into account, with of course an average value half the one for nickel (1.79 MeV compared to 3.58). The second half of the energy comes (in the form of γ photons) from the short live β+ emitters: associated γ emission, bremsstrahlung of the positron and annihilation radiation. The average energy of these photons is taken to be in the 0.75 MeV range, thus less penetrating. The energy repartitions ƒ’i (Table 5 ), have been evaluated according to the photon production rate in the proton capture route given below.

Finally, the γ photon production rate for both routes has been evaluated as follows:

5. Effect of lead shielding on expected γ emission

For a poly-energetic beam of photons, the attenuation Ι / Ι0 , resulting from a thickness d of  lead, is:

In [6], the quantity μ/ρ (cmˆ2/g) is given for photon energies from 1eV to 20 MeV. This gives for lead (ρ=11.34 g.cmˆ-3), the absorptions coefficients μi (tables 4 and 5).
Finally, following relations were used to evaluate the attenuation of the beam for increasing values of d (ƒi from table 4 and ƒ’i from table 5):

Figure 1 gives the transmitted γ activity (as log10(I.sˆ-1), as a function of the lead thickness d.

As expected, the lead shielding is more efficient in the proton capture route. Even in that case and for 40 cm of lead, the transmitted activity is still 10ˆ6 sˆ-1. The corresponding value is 3*10ˆ10 sˆ-1 in the neutron capture route.

An important point must be stressed: in the above calculation, the emitting nuclear source is considered to be concentrated in one single point, which is of course not the case. To get a realistic evaluation of the expected flux of photons, it is supposed that the Nickel powder is contained in a cylindrical reactor, diameter 2 cm and length 100 cm (Outer surface 628 cm2). At 1 meter from this tube, shielded by 40 cm of lead, the photons flux is thus ≈1 sˆ-1 cmˆ-2  for proton capture and ≈ 5*10ˆ3  sˆ-1 cmˆ-2  for neutron capture.

6. Final products and residual activity after shut down

The number of stable atoms i formed at the end of the experiment (time T) is:

For radioactive atoms with a disintegration constant λi, the number of atoms formed at T is:

For short life atoms (59,61 and 62Cu-proton capture- 65Ni-neutron capture-), the asymptotic limit is reached well before T and the number of atoms at T is:

For long life atoms (59 and 63Ni-neutron capture and 59Ni-proton capture), the final product at T can be considered to be 59 and 63Ni on the one hand and 59Ni on the other.

Table 6 and 7 summarize the various atoms formed at the experiment shut-down.

Table 6

Table7

7. Residual activity after shut down

For both routes, short live species are formed: 65Ni for neutron capture and 59,61 and 62Cu for proton capture (see Tables 1,2,6 and 7). Their concentrations at shut down Ni(T) are given in Table 6 and 7. Their activity decreases as Ni(t)=Ni(T) eˆλit (t=0 at shutdown). Table 8 gives Ni(T) at shutdown (after duration T of the experiment) and the remaining atoms at t=7200 s (2 hour after shut down) Ni(7200)=Ni(T)eˆ-7200λi and hence the residual activity at that time:

Table 8 and 9 below give the residual activity 2 hours after shutdown. The energy of the main characteristic gammas are given in keV and the branching ratios in % (between brackets).

Table 8

Table 9

It can be seen from Table 8 (neutron capture), that 2 hours after shutdown, the activity of 65Ni is still 1.3×10ˆ13  sˆ-1. For proton capture (Table 9) the corresponding activity of 61Cu is still 4.2×10ˆ14  sˆ-1.

As for the emission during the run, the emitting nuclear source is considered to be concentrated in one single point, which is of course not the case. If, as supposed previously, the Nickel powder is contained in a cylindrical reactor, diameter 2 cm and length 100 cm, the total weight of nickel is some 1260g (apparent density 4, volume 300 cm3). If 3 cm3 of the powder is placed against a germanium detector, the activity would be reduced to some 10ˆ11/10ˆ12 sˆ-1 and characteristic radiations could be measured (annihilation radiation for 61Cu and characteristic gammas (see Table 8) for 65Ni).

8. Transmuted products formed

If the total amount of nickel supposed to be processed is some 1260g, corresponding to 21,7 mole or 1.30×10ˆ25 atoms, a tentative mass balance can be made.

For both routes, the isotopic composition of the Nickel is not significantly altered. For both routes, a sizeable amount of “quasi stable” 59Ni is produced, that represent more than 500 ppm atoms of the starting nickel. This is far beyond the precision of Mass spectrometry and could thus be easily detected. In the neutron capture route, “quasi-stable” 63Ni could also be detected (50 ppm atoms).

As regards the isotopic ratio of copper63Cu/65Cu=2.244, it should increase in the proton capture route (the copper produced has a ratio of 3.92). It should decrease in the neutron capture route (no 63Cu is produced). The copper produced represents some 7 ppm atoms in the neutron route and some 37 ppm atom in the proton route. Starting from Nickel powder containing round 1 ppm Copper should give reliable indications on the process.

9. Conclusion

Strong nuclear signatures are expected from the Rossi energy amplifier and it is hoped that this note can help evidence them.

It is of interest to note that in [3] a mechanism is proposed, that strongly suppresses the gamma emission during the run (it is the same mechanism that creates very low energy neutrons, subsequently captured by the nickel. This does not suppress the emission after shut-down, which should be observed, together with the transmutations described above.

by Jacques Dufour

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REFERENCES:

[1] S. Focardi and A. Rossi “A new energy source from nuclear fusion” Journal of nuclear physics

[2] L. Daddi “Virtual neutrons in orbital capture and neutron synthesis” Journal of nuclear physics

[3] Widom Larsen “Theoretical Standard Model rates of proton to neutron conversions near metallic hydride surfaces” New energy time Widom-Larsen portal

[4] A. Bauer “Protection contre les rayonnements” Commissariat à l’énergie atomique ISBN 2-7272-0102-8

[5] IAEA. ENSDF. Nuclear data service, 2009. http://www-nds.iaea.org/

[6]-NISTX-Ray-Mass-attenuation-Coefficients-Lead http://physics.nist.gov/PhysRefData/XrayMassCoef/ElemTab/z82.html

97 comments to Nuclear signatures to be expected from Rossi energy amplifier

  • Francesco Celani

    Dear Eng. Andrea Rossi,
    Your results look very interesting.
    For a better understanding, it can be important to know the excess power density You get (e.g. in the “old” experiment of 3kW), i.e. how many Watt of excess power You produced (more or less constantly isn’t important) for gram of Nickel.
    I think that sharing such scientific information will not damage Your (very important) Patent.
    Many thanks in advance,
    Francesco Celani

  • Andrea Rossi

    Totally agreed.
    A.R.

  • Enrico Billi

    Could be interesting include the Pico-chemistry process in the simulation software, i am curious to see the pubblication and may be start to develop the methods for this physics
    Enrico Billi

  • Andrea Rossi

    Dear Jacques,
    I agree upon all. I think we can learn from you.
    About your questions:
    1- My company’s strategy is at the moment to manufacture the thermal generators, maintaining confidential the nature of the additives/catalyzers.
    2- Because I would like to see published your paper as soon as possible and I am pushing the reviewer, therefore I think it will be published within the first week of September: it is not a dead line, just a target. Probably you’ll see it published before that term.
    Warmest Regards,
    Andrea

  • Jacques Dufour

    Very good, indeed. We are obviously following 2 parallel and complementary paths: you are aiming at commercializing a process (hence the patent) and I am aiming at introducing and developping pico-chemistry in science (hence the publications). To understand pico-chemistry, I am using much less common metals than Nickel, probably not worth filing a patent (which is not my prime objective). When I gain expertise in this new field, we could work togeher, under a secrecy agreement, to improve your own additive and define the key points of a process based on it. I agree that we should meet, sometimes round the end of this year. Meanwhile let’s keep in touch. 2 questions:
    1/ is your business plan to commercialize the additive (feel free not to answer this question).
    2/ why are you mentionning the 7th of september as a dead line to publish my paper?

    Warm regards

    Jacques

  • Andrea Rossi

    Very Good, My Friend: you are confirming the usefullness of the patent . I thank you and soon we will have to meet. People able to replicate my work has the skills necessary to work together with my company. Yes, there are very big risks if you have found the right additives, and as I always said these kind of experiments must be done only by professionals, due to the toxicity of Ni powders and to the explosivity of H2, also indipendently from the nuclear reactions.
    When you will have detailed results, please send them to the Journal . I will be glad to discuss them and also we will work together, I hope.
    About your second article, the reviewer has promised to give me it back by August, and also already said it is very important. It will be published, I hope, in August and anyway before the 7th of September. Then we will leave it in first position for at least one month. I deem it extremely important.
    Again, thank you.
    Warmest regards,
    Andrea

  • Jacques Dufour

    Dear Andrea,

    I am just starting a series of experiments to shed some light on the results of contacting hydrogen with some selected metals and under selected conditions (pico-chemistry). I operate with gram quantities of metals using a reliable calorimetric system. I have obtained very sizeable amounts of unexplained energy production and I am in the course of confirming and extending all that. But I think important to already send a warning to those who would like to embark in such experiements: the energy release can increase with time in an uncontrolled way (unless you shut down the experiment). So be careful!

    I will have detailed results in a few months, that we can then discuss.

    Warm regards.

    Jacques.

    P.S What is the point of the review of my second article?

  • Andrea Rossi

    Yes, I agree with you, Dr Dufour. Is very interesting. I am waiting for instructions to know how I can help.
    Warmest Regards,
    Andrea

  • Andrea Rossi

    Very good, I wait for your proposal. Just let me know if I can help you and how.
    Warmest regards,
    Andrea

  • Enrico Billi

    Dear Andrea, the application i can developed is just a test code for simulating a nuclear detection system. It is still a research application, it is not enough developed to be a product, i will be more interested to a collaboration with Leonardo Corporation, i can develop the application under your specific request for your special purpose. I found the geiger detector specifications on internet website of the supplier, so i will develop the geometry of the geiger counter, in these days i am abroad for a business trip so if you want to ask me any question you can do to my email i already send you some documents.
    Warmest Regards,
    Enrico Billi

  • Dufour Jacques

    Dear Enrico Billi

    It will be very interresting to compare the results of your simulations with the data usualy considered for the absorption of g rays due to their interaction with matter. If I understand correctly, you are aiming at simulating the interaction of the g rays coming from the experiment with the Geiger detector used (a proportional detector detecting radiations from the gas ionisation created by the incident g Ray). So I think you have to first define what kind of g ray spectrum you are expecting from the experiment. I am very interested by your results.

    All the best.

    Jacques Dufour

  • Andrea Rossi

    Dear Enrico, I know that the diameter of the geiger is 10 mm. Which other info do you need?
    Warm regards,
    Andrea

  • Andrea Rossi

    This is very useful. From my side you have an absolute GO!. Just let me know what can I do for you and how can I help you.
    Is it possible to buy your simulation package? Leonardo Corporation is intesested to your work.
    Warmest regards,
    Andrea

  • Enrico Billi

    With my simulation package i can develop an application in C++. I can define the geometry, the materials, the electromagnetic fields, the elementary interaction between the particles and the materials. The simulation start when i create a first radiation ray, it can be all elementary particle, by now i didn’t have enough time to develop a complete simulation, with all the geometry and the new possible physical phenomena, but with the size of the detectors i can simulate the behaviour of g-rays throught the target and see how many counts will be detected, may be we can discuss in details on email or other ways.
    Warm Regards
    Enrico Billi

  • Andrea Rossi

    Dear Enrico, reading again this comment of yours again I agree. Can you give me a description of the simulation you are thinking about? I can replicate it if you give me a precise description and discuss with you the results. Maybe Jaques Dufour is interested too, I think.
    Warmest regards,
    Andrea

  • Andrea Rossi

    I agree, so, let’s go!
    Yours Andrea

  • Dufour Jacques

    I completely agree with Enrico Billi. We should strictly adhere to the basic principles of quantum mechanics, that is duality wave/corpuscule, which translates into the Hamiltonian of a system and is summarized by the Heisenberg uncertainties.The Hamiltonian can be solved exactly in certain simple systems for instance the hydrogen atom. For more complicated systems, physicist just say they cannot mathematically solve the problem and so use approximations wich can be experimentally validated. This is probably what we have to do for our problem.

    Jacques Dufour

  • Dufour Jacques

    I fully agree with Enrico Billi. We should strictly adhere to the basic principles of qua

  • Andrea Rossi

    Very good,
    I approve. What do you propose to do, exactly?
    I will help you,
    Andrea

  • Enrico Billi

    Looking for a theoretical explanation of these low energy nuclear reaction , i collected a lot of publications from famous journals and also from unknown websites. Most of the are very complicated and start saying some principles of quantum mechanics are wrong. I am a reasonable man, what is known couldn’t be wasted just to proof a specific phenomena. I hope we can find some experimental proof, so we can get some useful information about the reactions inside the Ni. One of my professors usually say “Only Jesus Christ can do wonders”, so if something happen in the system i hope the simulations or some specific experiments can give us the useful informations.
    Warm regards and good work,
    Enrico Billi

  • Andrea Rossi

    I agree perfectly,
    Andrea

  • Dufour Jacques

    Dear Andrea Rossi,

    Thank you for your nice comment.
    I agree with you that some exsplanation/model must be found in the field. The best model will be a model that correctly predicts the main characteristics of what is observed. It needs not, at the beginning, be a complex mathematical construction (as the ones we are now using to explain the slightest details of things that were discovered many decades ago). After all, the Bohr model is now considered as wrong (or at least incompatible with present ideas in quantum mechanics), but it has been of very great help to get to the point we are now in atomistic physico-chemistry!

    All the best with your research.
    Jacques

  • Andrea Rossi

    Dear Dr Dufour,
    Very interesting, as usual, your comment. I always have something to learn when I read what you write.
    By the way:
    I agree with the opportunity to make the “Billi Simulation”.
    About the screening by the electrons as a mean to overcome the Coulombian barrier: what you say is true, but we have to find a theoretical hypothesys which can explain what we get, and as usual, as it happens with the MFP stuff, you get conclusions which crash against rigorous calculations, so that you have to find an explication anyway, hoping further calculations will explain what you imagine.
    I too am not at all convinced about the fact that electrons screening is an explication, although shaky, and I am searching a more solid model. But the more I try to understand, the less I understand, sincerely.
    I am curious to read the comment of Dr Billi about your one.
    Warmest Regards,
    Andrea

  • Dufour Jacques

    Dear Dr Rossi and Billi,
    3 comments on your recent discussions:
    -Compton scattering will reduce the intensity of the g flux from the reactor, because part of the energy of the photons is transferred to the electrons of the screen (contrary to Raleigh diffusion where the photon is only deviated).
    -The idea of simulating a “Widom-Larsen environment” as suggested by Dr Billi seems interesting.
    -As regards the screening by the electrons as a means to overcome the Coulomb barrier, I would suggest that you have a look to this article (in English): R.Balian, J.P Blaizot and P. Bonche “Cold fusion in a dense electron gas” J. Phys. France 50 (1989) 2307-2311.In the case of F-P experiment an electron concentration 6000 times the electron concentration in palladium would be required to achieve the claimed reaction rate. Oviously, in the case of Ni-H (where the coulomb barrirer is much higher) the situation is worse. I know the work of Hucke and alt. and I know the experimentally measured increase of reaction rates when decreasing the deuteron energy. But the enhancement factor (as defined by Huke), should be some 10 power 20 to be efficient at low energy, which is questionnable (compare with the electron density required from first quotation).
    Best regards
    Jacques Dufour

  • Andrea Rossi

    Thank you.
    This is an interesting site. I suggest to the readers of the Journal of Nuclear Physics to take a look at this:
    http://arxiv.org/abs/0803.1071
    Thank you again.
    Andrea Rossi

  • Enrico Billi

    Quantum mechanical ab-initio simulation of the electron screening effect in metal deuteride crystals
    http://arxiv.org/abs/0803.1071

  • Andrea Rossi

    Thick of the target 9 mm, distance of the detector from the target 200 mm, position in front of a square terget, single detector.
    I didn’t receive the link.
    Thank you very much, I really appreciate your comments.
    Warmest regards,
    Andrea

  • Enrico Billi

    Ok, i can set up a simulation but i need to know some details like the thick of the target you want to test, the distance between the target and the detector, the position of the detector or detectors, if you will use more than one. I sent to the Journal also a link of a new publication of a first theoretical study of the migration of conduction electrons of a metal to deuterium atoms when is loaded inside the lattice, very interesting!
    Warm Regards,
    E.B.

  • Andrea Rossi

    Thank you. Really appreciated. We can stay in touch by personal email, if you want.
    Warm regards,
    Andrea

  • Enrico Billi

    If you will use Co-60 the g-rays have about 1.3MeV of energy, with this energy the Compton Scattering will be more important compared with Photo-Electric effect and Couple Creation (e+e-). The g-rays will be scattered but not absorbed by the target.
    If the target layer is not thick (<1cm), most of g-rays will pass throught the layer without interactions, so the spectra and the energy of the gamma will not be affected.
    At first you can try to use different materials, like Ni and Ti, to check how the measure of the counts of g-rays is affected. Then you can give different voltages (with and without H loading) to the layer and check if there are differences of the counts of the g-rays.
    May be we can discuss more in details by email and also i can develop the simulation of the g-rays with my simulation tools so i can give you some usefull data.

  • Andrea Rossi

    Gent. Dr Enrico Billi:
    Ho due notizie per Lei:
    1- We are preparing in our USA laboratory a test about your idea.
    Have you suggestions to give us?
    2- The article regarding Lithium is on the first page of the Journal Of Nuclear Physics. The merit of this publication on our Journal is yours, and we thank you again for having reviewed it for us.
    Warmest Regards,
    Andrea Rossi
    Andrea Rossi

  • Andrea Rossi

    Dear Dr Enrico Billi,
    Your idea is worth a test. What do you think, Dr Dufour?
    Andrea Rossi

  • Enrico Billi

    Dear Jacques Dufour and Andrea Rossi, may be i am flying away with fantasy, but if the theory of Windom and Larsen is correct(http://arxiv.org/abs/cond-mat/0509269v1), so every hard gamma and low momentum neutrons are absorbed or captures into the surface of Ni lattice filled by protons and heavy electrons, in theory it could be used like a radiation shield, especially for gamma radiation. May be an experiment with Co60 as gamma radiation source could be done with one thin layer made in the same way of your reactor Ni-H system. But may be it is just a silly idea, i will looking for your reply.
    Warm regards,
    Enrico Billi

  • Andrea Rossi

    WARNING: WE CONTINUE TO RECEIVE ANONIMOUS COMMENTS. THIS PRACTICE, INDIPENDENTLY FROM THE NATURE OF THE COMMENT, IS DEEMED UNACCEPTABLE FROM THE JOURNAL OF NUCLEAR PHYSICS.
    THEREFORE, FROM NOW ON, ALL THE COMMENTS WHICH WILL NOT BE SIGNED CORRECTLY AND COHERENTLY WITH THE SOURCE OF THE EMAIL WILL BE PUT IN THE TRASH.

  • Andrea Rossi

    Dear Sir:
    1- Our plant is now in the USA, and our factory is not open to the curiosity of the public, for safety and security reasons.
    2- The reason is complex and derives from the balance of the binding energy of the nucleons during the operation of the reactor. The process in this reactor has nothing to do with the nuclear reactions in a Supernova. Anyway, if you go through the articles of Daddi, Dufour, Focardi Rossi published on the Journal Of Nuclear Physics you should find an answer.
    A.R.

  • Hello!

    I read something about this setup, I’m a mechanical engineer. I have two questions:
    1) Is it possible to visit the facility in Bondeno? I don’t live far away, I’d be very curious to see it.
    2) AFAIK, nuclear fusion process is exotermic only up to iron. To get heavier elements than iron (like copper from nichel), energy must be given to the system (like in the supernovae). How do you explain this?

  • Andrea Rossi

    The Enrico Bill’s presentation of an important article of chinese and japanese scientists will be online by the end of next week: we are late 7 days for the check of the formulas.

    Sorry for the late
    Andrea Rossi

  • Andrea Rossi

    The new article of Jacques Dufour, now under peer-reviewing, is a breakthrough. It will be published on the 1st of July and I really suggest a deep study on it. There is to learn, I promise. I am sure the reviewing will pose no problems; if so, I will fight to publish it.
    Andrea Rossi

  • Dufour Jacques

    Dear Dr Rossi,
    The results you obtain could be understood in the frame of what I call pico-chemistry.
    I have sent a paper on the subject last Friday to the Journal of Nuclear Physics.
    Best regards. J. Dufour

  • Andrea Rossi

    Very interesting.
    A.R.

  • Enrico Billi

    Dear Andrea Rossi:
    i will not setup a real reactor, but a virtual reactor in my Linux OS with the simulation software Geant4 (CERN), it is the best for EM radiation simulation and detection efficiency studies, so i can check the behaviour of single gamma rays in a virtual reactor and check if adding new models in the simulator I can reproduce your results. Thank you very much for your advices but by now they are not necessary.
    Warmest regards,
    Enrico Billi

  • Andrea Rossi

    Dear Dr Billi:
    We never found gamma rays out of the reactor, being thermalyzed. Nevertheless, what you are doing is intersting. Be careful manipulating Ni powders ( are very dangerous) and Hydrogen, for obvious reasons.
    Warmest regards,
    Andrea Rossi

  • Enrico Billi

    I can run a Monte Carlo simulation of the gamma radiation should comes out from your reactor. I read the publication of Focardi, Il Nuovo Cimento Vol 111A N.11 November 1998, i can take the features of the CELL B to simulate the possible responce of a gamma detector

  • […] su Rossi-Focardi Prego di andare a vedere al piu' presto questo link: LINK Questa la conclusione di Dufor: […]

  • […] teorica da J.Dufour Prego il sig. Camillo di andare a vedere al piu' presto questo link: LINK Questa la conclusione di Dufor: […]

  • Andrea Rossi

    Yes, I agree.
    Andrea

  • Dufour Jacques

    Very interesting. Your observations mean that the average energy of the gammas you produce is lower than could be expected from proton or neutron capture. There might be an explanation on which I am working. Let’s keep in touch.

    Best regards.

    Jacques Dufour

  • Andrea Rossi

    Very interesting. After 2 hours we still have thermal emission, but still we don’t have radiations out of the reactor. Of course, the lasting of thermal emission means that readiation (gamma) continues to be thermalized. We ( Focardi and me) are convinced that neutrons don’t reach the energy to exit the nuclea, with some exception, which is thermalized in 20′,also because our shielding od boron and lead is much lower that you calculated.
    Warm Regards,
    Andrea Rossi

  • Dufour Jacques

    Thank you for your interest in my analysis. The 2 routes I have studied have a common point: part of the energy released (quasi totality for the neutron route and half the totality for the proton route), comes from the de-excitation of the primary excited nucleus formed, Ni* or Cu* (infortunatly less datas are available for proton capture than for neutron capture hence the assumptions I made). Nethertheless, in the proton capture route, a sizeable thickness of lead is needed to supress the emission, very probably in the range of tens of cm. The proton route seems to be confirmed by your observation of the anihilation radiation. What is the level of activity you observe 2 hours after shut down?
    As regards the neutron route, I suggest that you have a look at the Widom-Larsen approach I quoted in my paper. Ultra low momentum neutrons formed should not be seen outside because immediatly captured by the Nickel. Gamma emission during the experiment should be strongly suppressed, but emission of gammas from short live species should be observed after shut down.
    Finally my feeling is that the energy productions you report are real. A convincing explanation is to be found. I am working in that direction.
    P.S As regards the weight of nickel under experiments, are the assumptions I use, correct.

    Best regards

    Jacques Dufour

  • This insight of Dr Dufour is very interesting. It merits a deep analysis. We never found emission of neutrons outside the reactor, with exception of once, when the reactor has gone out of control because we tried to raise further the pressure. In that case we got a strong emission of neutrons, detected from the bubble neutron revelators. But in average exercise we never had an emission of neutrons. Our shield is made by 20 mm of boron, 20 mm of lead. The water flow has not shielding power towards the neutrons. We detected emission of gamma rays, which have been thermalized in the water flow, and we have strong evidence of matter -antimetter reaction
    (positrons + electrons), with production of energy through the annichilation, but we didn’t get neutrons. We think it is because we don’t reach the energy level necessary to the neutrons to get out of the nucleus. Nevertheless the analysis of Dr Dufour is extremely interesting and we want to go through it.
    Andrea Rossi

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