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How can 30% of nickel in Rossi’s reactor be transmuted into copper?

by Dott. Giuliano Bettini
Retired. Earlier: Selenia SpA, Rome and IDS SpA, Pisa
Also Adjunct Professor at the University of Pisa
Adjunct Professor at Naval Academy, Leghorn (Italian Navy)

Abstract
In the present article I would like to answer a question posed by L. Kowalsky in a recent paper: how can 30% of nickel in Rossi’s reactor be transmuted into copper? “Everything should be made as simple as possible, but not simpler”, says a guy. I apologizes if I am too simplistic here.

Introduction
The interest on Andrea Rossi’s Nickel-Hydrogen Cold Fusion technology is accelerating [1]. However, Rossi says that about 30% of nickel was turned into copper, after 6 months of uninterrupted operation. Kowalski [2]. says that “this seems to be impossible because the produced copper isotopes rapidly decay into Ni”. But how it works?

How it works
Following Focardi Rossi [3]. a Ni58 nucleus produces a Copper nucleus according to the reaction

Ni58 + p → Cu59

Copper nucleus Cu59 decays with positron (e+) and neutrino (ν) emission in Ni59 nucleus according to

Cu59 → Ni59 + ν + e+

Then (e+) annichilates with (e-) in two gamma-rays

e- + e+ → γ + γ

Starting [3] from Ni58 which is the more abundant isotope, we can obtain as described in the two above processes Copper formation and its successive decay in Nickel, producing Ni59, Ni60, Ni61 and Ni62. Because Cu63, which can be formed starting by Ni62, is stable and does not decay in Ni63, the chain stops at Ni62 (i.e. Cu63). Each process means some MeV.

Of course how can a proton p gets captured by the Ni58 nucleus? (and subsequent Ni59, Ni60, Ni61 and Ni62). Following Stremmenos [4]. a neutron-like particle, an electron proton pair, a mini-atom, a proton masked as a neutron, gets captured by the Ni58.

If the masked proton becomes a neutron the result is Ni59.
In order to have Cu59 (increase of atomic number from 28 to 29) the electron (of the masked proton) gets ejected from the nucleus. The masked proton becomes a proton.

The same process holds for all the subsequent transformations, until Cu63.
It remains to be understood the issue of the gamma radiation in the MeV range.

Numbers
I am an electronic engineer, so I need easy numbers in order to understand.
However “Everything should be made as simple as possible, but not simpler”, says a guy. Maybe I am too simple here.
Let’s calculate.
 
MeV for each Ni transformation
I read that starting from Ni58 we can obtain Copper formation and its successive decay in Nickel, producing Ni59, Ni60, and Ni62. The chain stops at Cu63 stable.
For simplicity I assume all the Nickel in the reactor in the form Ni58.
For simplicity I suppose for each Ni58 the whole sequence of events from Ni58 to Cu63 and as a rough estimate I calculate the mass defect between (Ni58 plus 5 nucleons) and the final state Cu63.
Ni58 mass is calculated to be 57.95380± 15 amu
The actual mass of a copper-Cu63 nucleus is 62.91367 amu
Mass of Ni58 plus 5 nucleons is  57.95380+5=62.95380 amu
Mass defect is 62.95380-62.91367=0.04013 amu
1 amu = 931 MeV is used as a standard conversion
0.04013×931 MeV=37.36 MeV
So each transformation of Ni58 into Cu63 releases 37.36MeV of nuclear energy.
 
 
Nickel consumption
According to many blogs in the Internet “One hundred grams of nickel powder can power a 10 kW unit for a minimum of six months”.
How much of Ni58 should be transformed, in six months of continuous operation, in order to generate 10 kW?
I follow a procedure outlined in [2].
10 kW is thermal or electrical (?) power. The nuclear power must be larger. Assume a nuclear power twice:
20 kW = 20,000 J/s = 1.25 x 10**17 MeV/s.
Each transformation of Ni58 into Cu63 releases 37.36MeV of nuclear energy.
The number of Ni58 transformations should thus be equal to (1.25 x 10**17)/37.36 = 3.346 x 10**15 per second.
Multiplying by the number of seconds in six months (1.55 x 10**7) the total number of transformed Ni58 nuclei is 5.186 x 10**22.
This means 5 grams.
The order of magnitude is not exactly the same but seems to be plausible. This means also 5 grams of Nickel in Rossi’s reactor transmuted into (stable) Copper after six months of continuous operation at the rate of 10 kW.
 
Conclusions
Rossi says that about 30% of nickel was turned into copper, after 6 months of uninterrupted operation. At first glance this seems to agree with calculations based on simple assumptions.
 
References

 

841 comments to How can 30% of nickel in Rossi’s reactor be transmuted into copper?

  • björn

    Professor Rossi.
    Congratulations on your commercial success.
    There is some confusion regarding the waste copper produced from spent Ni.
    Professor Focardi says the Cu isotopes produced in the reaction are not the natural ratio, while Kullander and Essen claim a sample of spent fuel privided by you contained Cu 63 and 65 in the natural ratio?
    This is confusing for me, please explain.
    Yours sincerely, Björn Eriksson, Dingle.

  • Bertil Nilsson

    Dear mr.Rossi

    As I have understood it water passes between the stainless steel vessel and the copper tube. Is the lead shielding outside the copper tube?

    Thx for you excellent attitude to all questions here.

  • Andrea Rossi

    Dear Renaissance Man:
    I cannot answer to your questions for confidential reasons, and what I could say you can find in my former answers in this post. Sorry, nothing more than that I can say.
    Warm regards,
    A.R.

  • Renaissance Man

    Dear Mr.Rossi

    What are your thoughts and theories about the isotopic ratio of the copper in the spent fuel? Why do you think it is so close to the isotopic ratio found in natural copper?
    Warm regards
    Renaissance Man

  • Andrea Rossi

    Dear Mr Stephen Shorland:
    No, it is not possible.
    Warm Regards,
    A.R.

  • Stephen Shorland

    Hello Dr.Rossi,

    Can your apparatus be used to transmute elements into Gold and/or Silver and if so,at what rate?Personally,I think the only cure for the next phase of the World financial crash that is coming will be a ‘Gold Standard’ or a floating Gold reference price along with other precious metals.Actually,your invention might be a large part of the cure but we need sound money too that can’t be manipulated in an irresponsible way: ie it can’t just be conjured onto a balance sheet by the stroke of computer keys.Are your apparatus and a precious metal money happy bedfellows!?

  • In my recent post the formula of the reaction should read Ni58 + Ni58 + 3H = Cu63 + Fe56. Of course subsequent calculations are made with Fe56.

  • A most useful tool for investigating synthesis-decay chains is available here.

  • Andrea Rossi

    Dear Mr William:
    1- I wanted to say that to treat the powders is confidential because difficult
    2- the Ni processing system increases 10% the cost of Ni
    3- Cu cannot enter the reactor: Cu is what the water tube is made of, inside the water tube there is the reactor, which is tightly sealed.
    Warm regards,
    A.R.

  • Andrea Rossi

    Gent. Sig. Fabio Sanzani:
    Lei ha perfettamente ragione, il fattore energetico e’ indipendente dagli utilizzi. Ovviamente, e’ ben noto che il ciclo di Carnot ha un’efficienza del 35% circa, ma questo vale per qualsiasi fonte energetica. Anche se si fa energia elettrica con il petrolio occorre produrre circa il triplo di energia termica rispetto all’energia elettrica prodotta. Il fatto e’ che l’energia termica prodotta con questa tecnologia ha un costo dieci volte inferiore nella peggiore delle ipotesi. Alla fine, il rapporto di convenienza non cambia, basta fare elementari conti economici per rendersene conto.
    Quindi Lei ha ragione.
    Cordiali saluti,
    A.R.

  • fabiosanzani

    Egregio Ing. Rossi,
    innanzitutto mi complimento per lo straordinario lavoro che sta facendo e mi scuso se scrivo in italiano.
    In merito ai risultati del bilancio energetico finale della reazione, semplicisticamente . watt energia elettrica in input e kcal (watt) energia termica in output, ho notato in alcuni commenti (+o- economici) in rete su un eventuale fattore di riduzione del quantitativo di energia dell’output (per la trasformazione dell’energia termica in elettrica), parificando in tal modo il tipo di energia input-output ma abbattendo in qualche misura il bilancio finale energetico finale.
    Credo che tali considerazioni sia del tutto opinabili sia perché l’utilizzo finale potrebbe essere esclusivamente in energia termica ed anche perché il rendimento effettivo del processo di trasformazione tra le forme di energia è in costante affinamento.
    Mi scusi per la divagazione, più economica che fisica.
    Cordiali saluti
    Fabio Sanzani

  • William

    Mr. Andrea Rossi,

    What do you mean by the last part of this statement?

    “The treatment of the powders is part of the invention and is confidential so far: the difficult part of it stays in the low operational cost.”

    Is it difficult to keep the processing cost low?

    Is the processing of the nickel powder an expensive process?

    How much does the processing add to the cost of the nickel?

    One final question: Can you state for the record that no part of the copper tubing enters the reactor vessel?

    Thank you again for being willing to answer so many questions. In the minds of many people I know this adds tremendously to your credibility. Your openness is FAR different than some other researchers who did not openly discuss their technology.

    Sincerely,
    William

  • Andrea Rossi

    Dear Mr Ivan Moho:
    Months,
    Warm Regards,
    A.R.

  • Ivan Moho

    > In due time we will make a joint communication

    Are you able to give a very rough estimate (or rather, order of magnitude) for the time until the announcement date?
    Is it a matter of days, weeks, months or more (i.e: a year or more) ?

    Thanks,
    IM

  • Andrea Rossi

    Dear Dr Joseph Fine:
    The stainless steel I use is AISI 316 L, which is an alloy of Ni,Cr,Fe: no Cu, so I do not think the wall of the readtor has been “sputtered”. The reason of the slight delta of the isotopes of Ni depends on the treatment we make to the Ni before the reactions, so that at the end of the work substantially 62 and 64 Ni have reacted and the final composition returns close to the normal, not exactly of course.
    The treatment of the powders is part of the invention and is confidential so far: the difficult part of it stays in the low operational cost.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Craig:
    The Customer of us is a kind of Customer you need to have a written authorization of to talk about him. In due time we will make a joint communication. We will manufacture together a network of plants to sell the energy. I am very happy for this, because I am extremely indebted with the USA, where I got my rebirth, and we will make here new jobs and a useful work, so I will give back part of the help I got, as it is my duty. I always give back what I get, turned into energy. It’s my job.
    Warm Regards,
    A.R.

  • Dr. Rossi, dr. Gillis, dr. Fine,
    I am disappointed if Iron is just a contamination from the reactor walls, because I had just developed an idea in order to justify the production of the same amount of Iron and Copper.
    Here follows the idea. It is just al-kīmiyā’, (χημεία), chemistry.

    Perhaps I’m wrong here. Please check.
    In my paper I supposed, ultimately, a sort of free circulation (free exchange) of masked protons, in which these masked protons are:
    (A) “freely” produced by (p + e-) Hydrogen;
    (B) “relatively free” to transform into neutrons.
    In these hypotheses suppose the “pseudo-chemical reaction” (for simplicity I assume all the Nickel in the reactor in the form Ni58):

    Ni58 + Ni58 + 3H = Cu63 + Fe54

    In this reaction, taking into account point (A), (B), there is a balance of protons p, neutrons n and electrons e. The reaction, once reached I suppose a threshold in temperature and pressure, is exothermic.

    Now calculate the mass defect between (Ni58 + Ni58 + 3 nucleons) and the final state (Cu63+Fe56). Because of the hurry, I’m not at all sure of the calculations. Please check.
    2 Ni = 115.8706858 amu
    Cu+Fe = 118.864535 amu
    2 Ni +3n = 118.8706858 amu
    Mass defect is 0.0061508 amu ie 0.0030754 amu for each transformed Ni.

    In my paper the calculated Mass defect was 62.95380-62.91367=0.04013 amu for each transformed Ni, ie 13 times larger.

    Conclusions.
    With respect to my paper we have:
    -13 times more Nickel consumption, 5 x 13 = 65 grams (ie “65 grams of nickel powder can power a 10 kW unit for a minimum of six months”);
    -65 grams of Nickel transformed into Copper;
    -the same amount of Iron, 65 grams of Iron.

  • [...] little snippets of information about his plans and activities. An interesting piece of information was posted on April [...]

  • Dr Rossi,

    Are you able to tell us any more regarding the nature of the contract you mentioned was signed in the USA yesterday? Can you perhaps indicate whether this was a contract from a customer to manufacture supply e-cats or if this was with a supplier for materials? Any information would be appreciated as there is growing excitement around the commercial rollout of your technology.

    Many Thanks,

    Craig
    (Editor)- Free Energy Truth

  • Joseph Fine

    Dr. Rossi,

    In the earlier version of the E-Cat, Copper isotopes found in used fuel had a ratio of [Cu-63/Cu-65] = 1.6 (vs 2.24 in natural copper). According to Profs. Essen and Kullander, the Mini E-Cat (50 cc) version did not have such an isotopic variation in the Copper ‘ash’. Is this due to design modifications, or just that the mini E-cat has not run long enough?

    Below is the chemical composition of the Stainless Steel alloy used. Could one of the ingredients in the reactor wall be part of the reaction process?

    http://www.stainless-steel-tube.org/316-Chemical-Composition.htm

    J.F.

  • Andrea Rossi

    Prof. Brian Josephson ( a Nobel Prize) continues to honour us with his attention.
    Warm regards,
    Andrea Rossi

  • Andrea Rossi

    Dear Dr Gillis HRG:
    As usually, you make throughly insight: you are right, Cr has not been found, but it is difficult to know if this depends on the fact that Cr is less in the alloy of the AISI 316 L, maybe better bound, not to be taken off. I am sure Fe comes not from reactions, so I think Cr is just better bound (also being less in percentage in the Ni-Cr-Fe alloy).
    Warm Regards,
    A.R.

  • HRG

    Dr. Rossi:
    Please pardon my nit-picking; but if your reactor walls are made of stainless steel then there should have been chromium in the spent fuel at levels comparable to that if iron (if the iron is just a contaminant from the walls). Chromium is present on the surface of stainless steel. I did not see any mention of chromium in the analysis of the spent fuel from the latest demo. Was any chromium found? There was a lot of iron present. If little or no chromium is found then possibly the iron was formed in the reaction. Or possibly it was added to the initial fuel charge for some purpose?

  • Brian Josephson

    To clear up the confusion which seems to have arisen in some people’s minds from my brief account of the expt. that I have been conducting in connection with the Rossi reactor: in more detail, this involves the following steps:

    1) write to the editor of Nature making him aware of the existence of the Rossi reactor, and suggesting it may be important enough that the Journal should publish something about it

    2) keep watching Nature to see if anything appears in it

    3) at the same time, keep looking out of the window to see if any flying pigs can be seen going past

    4) note whether 2 or 3 happens first

    Not a particularly dangerous expt., but I thank all those who have contacted me regarding the dangers should a slightly different expt. be done instead.

    Brian Josephson

  • Andrea Rossi

    Dear Mr Bal Herrera:
    I agree with you on this: theories that challenge traditional physical rules and that are not able to give birth to useful apparatuses are just a loss of time. This is why I privileged the experimntal construction of an apparatus, after the initial theory, to allow a new theory to be funded on solid experimental results. I share totally your philosophy of decentralized power plants. That’s our strategy. Maybe in future you will help us: you are getting education in a very solid school.
    Warm Regards,
    A.R.

  • Congratulations for your invention to you Dr. Rossi, Dr. Focardi and the rest of your team. I am pursuing a master in Nuclear Fusion Science, obviously in “hot fusion”. Nevertheless I believe in the success of alternative technologies and the challenging to our traditional laws of physics. I remain skeptical to many theories that obscure the understanding of nature. Your attemps are a light in this darkness. I study in Belgium and in some months I will be studying in Sweden, in the same institute where the Prof. Hanno Essen is working! I hope that your technology will beat the thermonuclear fusion projects like ITER, which are sold like “inherently safe and cheap” but that have the intention to centralize the creation and commercialization of electricity worldwide, which in my opinion is not good because large amounts of people will be dependant on, let’s say ONE huge plant to meet their needs. So, my best regards and if you need free help with your projects you can open a “volunteering” section in your webpage, I will be the first to go and work for your noble ideals. Bal.

  • Andrea Rossi

    Dear Ivan Moho,
    Thank you for the suggestion, I will surely pass it to our informatic.
    Warm Regards,
    A.R.

  • Ivan Moho

    Dear Mr. Rossi,

    I’m sorry if this is off-topic here, but I think it’s time for your blog to have one or more fixed sections devoted to general enquiries about your Energy Catalyzer, suggestions or any other off-topic LENR-related matter. It’s starting to become very hard and time consuming to follow all discussions and new messages (which more than often contain useful information!) scattered in every thread made so far (23 of them as I’m writing) as there are no means for ordinary users to receive notifications when new messages get posted.

    Please consider discussing this with your web administrators.
    In alternative, the blog could be extended with better functionalities. As it uses WordPress (as far as I can see), it should be easily done (furthermore, at no cost!) by your web admins.

    Again, sorry for this off-topic message.

  • Andrea Rossi

    Dear Mr Bruce Fast:
    Wrong: yesterday we signed a very important contract in the USA .
    Warm Regards,
    A.R.

  • Bruce Fast

    Dr. Rossi,

    I recently read that Dr.Hanno Essén, and Dr.Sven Kullander were invited to investigate your energy catalyser. Their report is most compelling. I am thoroughly convinced that your device is for real. It is very exciting. Thank you for letting such distinguished academics report on your technology. The world will soon be listening, but so far the North American world remains oblivious.

  • Andrea Rossi

    Dear Dr Gillis HRG:
    The walls of the reactor are made of stainless steel, copper free. Yes, I have understood why scaling up we have more difficulties to have a flat curve of Delta T. Also the theory is consolidating. I am learning a lot in this period, I learnt a lot from the Professors of The Universities of Bologna, Stockolm and Uppsala ( in alphabetic order, of course) and from the People of DOE and DOD in the USA. From them there is really to learn. They say 10-20 words and from those words I get a universe of informations. In these last 2 months we made substantial evolution, after every test I redesigned and remade the reactors. Today I am in the USA factory of Leonardo Corporation where I signed a contract of tremendous importance. As soon as I will be allowed to announce it, believe me, it will be extremely important.
    Warm regards,
    A.R.

  • HRG

    Dr. Rossi:
    Is it possible that those reactor walls may also have contained some copper, as a contaminant, which then got into the spent fuel? It may be useful to rule this out.
    Congratulations on your most recent independent verification. Hopefully it will help to get your patents granted.
    Any idea why scaling up the device leads to reaction spikes?
    Perhaps soon we will have a world with lots of cheap energy- – and rusting windmills.

  • Andrea Rossi

    Dear Prof. Joseph Fine:
    I think iron is just a contamination from the reactor walls.
    Anyway, usual good insight from you, thanks,
    A.R.

  • Joseph Fine

    After operating for some time, Iron was found in the Mini E-Cat (or E-kitten). Was Iron present at startup or was it formed during operation – as Copper was?

    If Iron was not present at the start of operation, where did it come from?

    I think some Iron was present at startup. One alternative is that Nickel can capture a “masked” proton, then eject two protons. That would be fission, of a sort. But that is not very likely. Iron probably is coming from a lighter element.

    J.F.

  • Enrico Billi

    This simplified example can give just a first estimation of the energy released, in the original hypothesis the 30% of 58Ni became 63Cu with 5 serial capture reactions of these “screened proton”. The cross section of such reaction should be big enough to allow 5 different reaction with the same nucleus. But it’s interesting if we think 30% is the same percentage of the other isotopes of Ni is about: 60Ni(26%), 61Ni(1%) and 62Ni(4%) (These are natural percentage). If we use these as starting point to obtain 63Cu, the different isotopes need the following number of serial capture reactions: 3 miniatom for 60Ni, 2 miniatom for 62Ni and only 1 capture reaction for 63Ni. Here the number of screened proton capture reactions are 6 not only 5, but they can occur partially in parallel between the different isotopes, and not necessary in series for all the nucleus of 58Ni.
    Best regards.

  • I mean: 37.4 MeV per reaction (not 34.7, sorry).

  • I’ve found interesting to compare estimates of my paper with those in the travel report by Hanno Essén and Sven Kullander, 3 April 2011, “Experimental test of a mini-Rossi device at the Leonardocorp”, Bologna 29 March 2011.
    Participants: Giuseppe Levi, David Bianchini, Carlo Leonardi, Hanno Essén, Sven Kullander, Andrea Rossi, Sergio Focardi.

    In the report they said:
    “6 grams of nickel (assuming that we use one proton for each nickel atom) are about sufficient to produce 24 MWh through nuclear processes assuming that 8 MeV per reaction can be liberated as free energy”.

    Let’s calculate and compare.
    Hanno Essén and Sven Kullander report.
    1 Wh = 3,600 J
    24 MWh = 8.64 10**10 J = 5.4 10**23 MeV
    5.4 10**23 MeV/ 8 Mev = 6.75 10**22 Nickel athoms
    About 6 grams

    In my paper I considered 20 kW for 6 months = 86 MWh, and 34,7 MeV per reaction
    86 MWh/24 MWh = 3.6 times more energy to be produced
    34.7 MeV/8 MeV = 4.33 more energy for each reaction
    Nickel grams 6 (3.6/4.33) = 5 grams

    OK.
    Obvius, but interesting.

  • Joseph Fine

    To all,

    I just saw the following article in the Swedish publication “New Technology” or ‘Ny Teknik’. (English version, Swedish version and Italian translation)

    The link to the article is here:

    http://www.nyteknik.se/nyheter/energi_miljo/energi/article3144827.ece

    There is a new Energy “Kitten” or mini E-Cat which was recently tested.
    The link to the pdf file is here.

    http://www.nyteknik.se/incoming/article3144960.ece/BINARY/Download+the+report+by+Kullander+and+Ess%C3%A9n+%28pdf%29.

    This reopens the question of the power density of an E-Cat (or E-Kitten).

    J. F.

  • Andrea Rossi

    Dear Joseph Fine:
    In October, when we will start the 1 MW plant ( and you will be invited to attend) I will release the theory. For now I just publish all the comments related with maximum respect, but without reviewing, for obvious reasons. Maybe from these comments can be born competition: I wouldn’t mind.
    Your insights are always very interesting.
    Warm Regards,
    A.R.

  • Joseph Fine

    Dr. Rossi

    I apologize for a premature comment I made about the possible decay modes of Ni-64, Cu-64 and Zn-64 in a recent message.

    There probably is no ping-pong between these nuclides since Ni-64 and Zn-64 are stable according to isotopic data.

    ( I had said that Zn-64 can decay to Cu-64 and Cu-64 can decay back to either Zn-64 or to Ni-64. )

    However, Cu-65 still has to come from somewhere.

    Thanks to an unintended clue from Prof Kowalski’s paper (which mentioned Ni-61) I thought of a mechanism that can produce Cu-65 – by the decay of Zn-65.

    That is, I think Nickel-61 could capture an alpha and become Zn-65. (Which is unstable and decays to Cu-65).

    So where does Ni-61 come from? Possibly from Fe-57, of course, via another alpha.

    And where does the Fe-57 come from? Either via the decay of Cobalt 57 or another alpha-capture by Chromium 53.

    And where do all the alphas come from? That is the real question, is it not?

    I suppose you could use Helium as well as Hydrogen – but this is undesirable as Helium is not an abundant resource.

    Another possibility is to use a few elements that begin with B. That is, Beryllium or Boron. (Maybe even Bismuth?)

    A new idea popped into my head this morning. Perhaps, Zn-64 can become Ga-65
    (Gallium) and decay via electron capture to Zn-65 and then decay to Cu-65.

    This would be an alternate path, but where does all the Zn-64 come from?

    Joseph Fine

    http://education.jlab.org/itselemental/iso031.html

  • Andrea Rossi

    Dear Mr Bjorn:
    It is important that Prof. Brian Josephson keeps me informed of the particulars he is going through. I will try to help. For me will be easier to help after the patent will be granted, because now I have the hands bound. Anyway, I will do all I can to help. Just write to my personal email address, which prof. Josephson knows. Please remember to handle the Ni powders with proper gloves, goggles, masks, clothes. I know you are very high level pro, but be prudent. Forbid smoking in the area where you are working and be sure the room is well ventilated.
    Warm Regards,
    A.R.

  • björn

    Dear professor Rossi.

    Have Brian Josephson told you that he will try to replicate the process?
    Do you have any advice for him?

    “By the way, I’m now doing an experimental investigation — which will happen first, pigs flying, or Nature publishing something relating to the Rossi reactor?”
    http://www.physicsforums.com/showthread.php?t=484427&page=4

  • Andrea Rossi

    Joseph Fine is always intriguing. Anyway he has preannounced a correction that we will publish as soon as we will receive it.
    A.R.

  • Dear Realista,
    thank you, you are right, of course the correct formula is
    1 MeV = 1,61 10**13 J
    I’ve simply made calculations using this formula in Kowalsky [2]
    P1 = 16 kW = 16,000 J/s = 10**18 MeV/s
    and scaling it from 16 kW to 20 kW.
    With this, the amount of copper reduces to 5 grams.
    How much this agrees with the true measures made by Rossi?

  • Jarek

    Unfortunately there is a problem with explanation of combining electron+proton into neutron to avoid Coulomb repulsion – energy of neutron is larger by 0.78MeV.
    Quantum mechanics allows to ‘borrow’ such energy, but uncertainty principle says that only for times like 4*10^-22s. Velocity of proton in room temperature is about 1000m/s, so this way we can only explain last attometers – it’s definitely not enough …

    To understand LENR, we indeed need electron screening charge of proton, but it should be rather between proton and nucleus.
    It occurs that taking into consideration electron’s magnetic moment, classical approximation of electron’s behavior is literally bouncing from nucleus – or between nucleus and proton as required in this case.
    These classical trajectories were considered for a few decades by Michal Gryzinski ( http://en.wikipedia.org/wiki/Free-fall_atomic_model ), he became cold fusion enthusiast after P&F experiment and had explaining article in Nature a month after .
    Here is his lecture with animations of such bouncing electron trajectories: http://www.cyf.gov.pl/gryzinski/teor5ang.html
    With best wishes,
    J.D.

  • Joseph Fine

    Dr. Rossi, Prof. Kowalski, Dr. Bettini:

    A mechanism that can produce Cu-65 is the decay of Zinc-65 (to Cu-65).

    Although not rapid, it is very reliable. Zn-65 is unstable and decays to Cu-65 by electron capture. [1]

    We can discuss whether Electron capture (or K-shell capture) is the same as a Positron emission. However, if there are no positron emissions, there is no electron-positron annihilations.

    As there is no Zinc to start out with, as far as I know, Zinc-65 may be synthesized. That is another topic that needs to be discussed further.

    This gets into the entire subject of whether the “Phantom of the Catalyzer” is a “masked proton”, a low-momentum neutron or something else. [2]

    Dr. Bettini commented that Ni-58 can accept a sequence of 5 “masked protons” to form Ni-63 which decays to Cu-63.

    That’s all well and good, if it occurs, but the real question is where is the Cu-65 coming from?

    “Think Zinc!” And if there is no Zinc to start with, what is the Phantom of the Catalyzer?

    See:

    http://en.wikipedia.org/wiki/Table_of_nuclides_(complete)

    OR

    http://en.wikipedia.org/wiki/Table_of_nuclides_%28complete%29

    An Amazing Coincidence!

    Is it my fault that the left and right parentheses symbols are %28 and %29 ? (as in Elements 28 and 29)?

    Joseph Fine

    References:

    [1] Jefferson Lab Isotopes: Elements 28, 29, 30

    a) http://education.jlab.org/itselemental/iso028.html
    b) http://education.jlab.org/itselemental/iso029.html
    c) http://education.jlab.org/itselemental/iso030.html

    [2] Hydrogen/Nickel cold fusion probable mechanism

    http://www.journal-of-nuclear-physics.com/?p=338#comments

  • Realista

    Dr Bettini:
    I think there was a bug:
    20 kW = 1,25 x 10 ** 17 MeV / s.
    not: 1,25 x 10^18 Mev/s
    it changes everything.

  • Andrea Rossi

    Dear Dr. Bettini:
    You told us that you are not professor, I apologize for our mistake. We will correct asap.
    Warm Regards,
    A.R.

  • I’ve received an e mail from dr. R. Little.
    Here the e mail and the answer.
    Dear dr. Little

    thank you very much for your e mail.
    You see, I’ve invented this concept of “masked proton” only in order to
    explain in a simple way the proton capture by Nickel.
    Unfortunately I am only a Radar Engineer (as I said to ing. Rossi “I AM NOT
    PROFESSOR”) but it seems to me very interesting the “reverse beta process”
    you are speaking about. It seems to me that they are looking for this, to
    explain the proton capture in LENR.
    Why you dont’t post a paper in Journal of Nuclear Physics?

    Warm regards
    Giuliano Bettini

    —– Original Message —–
    From: “Reginald Little”
    To:
    Cc: “Reginald Little”
    Sent: Monday, April 04, 2011 4:10 PM
    Subject: Prior Model

    Dear Prof ,
    I write in regards to your recent publication:
    How can 30% of nickel in Rossi’s reactor be transmuted into copper.

    I would like to note my prior model for forming the masked proton for proton
    capture in

    Magnetocatalytic adiabatic spin torque orbital transformations for novel
    chemical and catalytic reaction dynamics: The Little Effect

    And also in:
    On the Enhanced Reverse Beta Processes
    in Graphene-Iron Composite Nanostructures
    at High Temperatures in Strong Magnetic Field

    Thanks,
    Reginald B. Little

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