Rossi: Helium Found in E-Cat Reaction [Update

There is a very interesting detail provided by Andrea Rossi on the Journal of Nuclear Physics today, in a response he gave to a question from Pekka Janhunen regarding the detection of helium.

Pekka Janhunen
April 8th, 2015 at 7:54 AM
Dear Andrea,
There exists an endothermic reaction Li7(alpha,n)B10 which has cross section 0.03 barn at 5 MeV and 0.25 barn at 7 MeV. I also read in wikipedia (article Neutron_source) that there exist americium-lithium neutron generators based on this reaction. If there are energetic alphas (energy about 17/2=8.5 MeV) in the reactor as per your paper’s suggestion, then there should also be some free neutrons generated from lithium-7 by this reaction. But Bianchini saw no neutrons. This looks like a dilemma.
With mysterious regards, /pekka

Rossi responded:

Andrea Rossi
April 8th, 2015 at 8:40 AM
Pekka Janhunen:
We found that calculating properly the Mossbauer effect the understandable dilemma you propose ( that obviously was clear to us from the beginning) could be resolved. I underline “could be”.
We are still studying on this. I think the keys are 7Li3+p -> 8Be4 -> 2 Alpha -> Mossbauer Effect-> Heat + (2 Alpha + 2e)-> 2He
Maybe we are wrong, but this is the reconciliation we propose. And this is the core of the Rossi Effect, we suppose. Now we are working on the calculation of the He produced, that we have already found.
This will be the object of a next Cook-Rossi paper. The nuclear model of Prof. Norman Cook suits perfectly my effect.
Warm Regards,
A.R.
(emphasis added)

I believe this is the first time that Rossi has mentioned the detection of helium, which is a definitive marker of nuclear fusion reactions are taking place. It looks like there will be at least one more paper to come from Norman Cook and Rossi, which will deal with the subject.

Dear Frank:
The Lady E-Cat 1MW is stable and the COP is very high, because we are mostly in ssm mode. I am writing from inside the computers container.
I have learnt much from Prof. Cook; I have read his book 12 times and every time I got something. His book and the book of Greiner Maruhn are my Linus blanket and I have biult up using them my personal Nuclear Physics synoptic system, to have fast access to the information I need when I have some intuition during our experiments, also here in the 1 MW plant. I always have them in my briefcase. We have joined our ideas to reconcile the Lugano results; we decided to continue our theoretical collaboration.

  • Bob Matulis

    How does this reconcile with the Lugano report? My understanding is that He production was never mentioned in the Lugano report. I would be surprise if an attempt was not made to detect He during the test.

    • Andreas Moraitis

      The Lugano team analyzed the powder, not the gaseous reaction products. Any helium would have escaped at the latest when they opened the reactor.

      • Bob Matulis

        Agreed it would have escaped. My question is why would they test it in a way that would let the He escape? Neutrons, He and radiation are among the first things people check for in fusion reactions.

        • Julian Becker

          To me it sound more and more similar to the reaction seen by the guys from Airbus. They also detect neutrons and radiation. Bad news for the safety of the Ecat for home use if there is radiation and bad for the fact that it kind of has some sort of exhaust fume. Helium…

          • bachcole

            The radiation might be a problem, but the amount of helium would be absurdly small, certainly not enough to raise one’s voice. Helium, as you probably know, is chemically inert. And the idea of being asphyxiated is completely blown away by the fact that if we got that much helium, asphyxiation would be the least of our problems. Your section of the city would be black glass.

          • NT

            Long before the glass effect everyone’s voice would sound like Donal Duck…

          • Axil Axil

            Using a hydride to produce hydrogen provides a chemical based fail safe to make sure the temperature of the reactor is above 500C before the reaction begins. this makes sure no radiation is produced.

        • Ivan Idso

          Correct me if I am wrong, Laguna was a sealed chamber that ran for several weeks. If it were producing much He at all then it would build up pressure in the reaction chamber, unless it were also being consumed?

          • Axil Axil

            In a number of experiments where the pressure gauge was available, the gas pressure decreased to a significant fraction of its maximum reading.

          • Pekka Janhunen

            With 5 GJ of energy produced and with 17 MeV per reaction, the experiment would have generated 1.8e21 helium atoms. Assuming reactor volume 20 cm3, that would corresponds to about 18 bar helium pressure at 1200 C temperature. After cooling down the helium pressure would be about 3 bars above atmospheric.

  • artefact

    Rossi was asked if he found Helium many times. His answer was always that he could not give details of the reaction. I’m glad that this seems to change a bit now.

  • Ged

    Well now. Let’s see what Pekka thinks of this.

    Rossi is right that to show this mechanisn more conclusively, we must accurately measure the amount of He evolved compared to that calculated for excess heat quanta.

    Thank you for this post, Frank. He evolution is the best indicator of an LENR reaction, and that Rossi has seen it is interesting and good to know. Constrains theory space.

    • Pekka Janhunen

      It still seems to me that if their theory predicts (as I understood that it does) 8.5 MeV alphas, it would also predict some neutrons. In AR’s answer, I don’t understand how the Mössbauer effect is related to thermalisation of alphas.

      • Ged

        Do you have a sense of what shielding is needed to block neutrons of this energy, and what would be needed to detect them? Perhaps detection failure has been due to rarity versus detector cross sectional area. I have no idea, so, very curious to know what is needed to definitively say, or if the past neutron detection work is already sufficient to rule them out.

        • Pekka Janhunen

          Neutrons are difficult to shield against. Basically one first has to moderate them by letting them collide elastically multiple times with low-Z material, then absorb them in layer of B10 or Li6 or other neutron-absorbing isotope (because only moderated neutrons are absorbed efficiently). Neutron shields are always thicker than the Lugano HotCat was, because their elastic mean free path is typically centimetres. I don’t know the detector area, but the unit of measurement was counts per minute. I’m pretty sure that no neutrons in the detector means essentially no neutrons produced anywhere across the main energetic pathway. There could be some side reactions such as piezonuclear fusion that could produce some tiny amounts of neutrons that might disappear in the background, but that’s another matter.

          • Ged

            Really interesting. Thank you for the insights, Pekka!

          • Pekka Janhunen

            Another way to think about is that if they make neutron sources based on alpha active isotopes and Be9 or Li7 to convert some of the alphas to neutrons, …, I mean, such device is so lower power that it’s not even a tiny bit warm, yet it’s a neutron source, surely produces at least so many neutrons that a detector has no trouble seeing them. The HotCat is a kilowatt class source, much larger than any lab-scale alpha source.

            I must warn, however, that I’m not any kind of expert in radiation and have no direct experience about it.

  • SG

    Although words can be interpreted in different ways, I’m pretty sure Rossi is referring to the He being found, not the calculation.

  • ecatworld

    To dispel any doubt in what Rossi meant:

    Steven N. Karels

    April 8th, 2015 at 2:11 PM

    Dear Andrea Rossi,

    You posted “Now we are working on the calculation of the He produced, that we have already found.” So you have experimentally verified the production of helium in the eCat operation?

    Andrea Rossi

    April 8th, 2015 at 4:01 PM

    Steven N. Karels:
    Yes.
    Warm Regards,
    A.R.

  • ecatworld

    Rossi has confirmed — see comment above.

    • Edac

      This confirmation is excellent news.

  • Andy Kumar

    Helium is convincing proof of Rossi effect. To properly honor the dawn of new Rossi Era, we should adopt a new secular calendar starting the years with the publication of Rossi’s recent paper. The years will be counted as BR and AR (Before and After Rossi).

  • Alan DeAngelis

    Li(7) + p > 2He(4) 17.3 MeV !!!
    https://www.youtube.com/watch?v=drhq4frGo60

  • Hank Mills

    I have a favor to ask the E-Cat World community.

    Please read the following article and answer the question: where does the magnetic field of a moving charged particle come from?

    http://pesn.com/2014/10/24/9602555_Late-Night-Speculative-Raving_From_Hank-Mills/

    • Alan DeAngelis

      See alpha voltaic cell at 0:40.
      https://www.youtube.com/watch?v=Dy0kHQASsX8

    • Hank,
      A magnetic field, or any field, or any force, doesn’t consume energy until it performs work. So no energy is needed for a loaded particle to produce the magnetic field in itself. What you should look into is what happens when the field actually performs a work. I guess that would slow down the particle, but I’m not sure.

      • Chris, Italy

        I’m quite sure it’s just gotta slow it down… unless there is a force on the charged particle that, in turn, performs work on it.

        😀
        I sure hope that particle ain’t about to explode; I hope it’s only carrying freight with it. Or maybe it ain’t an even bet for gamblers, like for spin up vs. down in a Stern-Gerlach magnet…..
        😀

        How do you say charged and loaded in Svenska? English is an ever odder language than it.

    • Chris, Italy

      Before I find the time to read through all that and write more, I’ll just say that the magnetic and electric fields are, in a sense, the same thing.

      Th Maxwell equations are Lorentz covariant, even though the most familiar forms they get written in doesn’t make it so abvious. There is no essential difference between the charged body moving past you and you moving past it (the opposite way, of course, just like the trees and fields that whizzzzzzz past the train you’re sittng on). Now it’s not so obvious why this causes the same field to have both its “electric” and “magnetic” aspects; it takes an in-depth understanding of special relativity and how electromagnetism is formulated.

      Let’s put it more simply: It “comes from” basically the same “where” as the magnetic field surronding an electric wire with electric current flowing along it (if this is of any help).

  • Eyedoc

    Maybe no gamma rays EXTERNALLY detected b/c of Mossbauer

  • There has long been a generic rule amongst those working around neutrons. Depending on whether the neutrons are fast or slow defines the danger zone. The safest place to be with fast neutrons is in between your lab partner and the source, the dangerous place to be with slow neutrons is behind your lab partner. We are of course 70% water.

    As an aside for those working in tritium labs it’s always good to see the table with the 24 pack of Coors Light just outside the evacuation exit with the instructions “In case of tritium exposure, drink.” Of course there were a lot of tirtium evac. drills late on Fridays. http://www.atom-ecology.russgeorge.net

  • Omega Z

    Maybe Rossi is finally acknowledging the helium factor at this time because he is getting so close to putting a product on the market. i.e.- He is far enough along that it no longer matters if competitors know..

  • runningman97

    This is important information and something I suspected may be found. To me it supports Widom-Larsen theory or something similar.

  • Pekka Janhunen

    Yes, one litre of helium corresponds to running the Lugano reactor for one year, or 15 tons of TNT.

  • Donk970

    Now do the opposite, pull the car off the magnet. You return the original potential energy of the system.

    • LCD

      Not really the question. Where does the energy of the gravitational field come from? By that analogy you would be losing “gravity” every time something got attracted to say a planet but it’s quite the opposite.

      Hold a metal toy car in the air with a magnet. Now the magnetic field is doing work against the gravitational field, constantly.

      The answer for the magnet question is that it eventually demagnetizes. But even that answer is not really satisfying with gravity

  • Obvious

    Si-C-O-Cl-Fe-Mg-Mn = blood and tears, and bits of grindstone.

  • Alan DeAngelis

    Pardon me Hank. I thought that question was leading up to the question of how to generate electricity from alphas. Here’s the answer to your question.
    https://www.youtube.com/watch?v=D8TyClG8d5k

  • LCD

    Helium is going to be so hard to detect unless it is a very strong signal. Just ask Mckubre.

  • TPaign

    In regards to 7LI3 in an endothermic reaction, with 2He as a product, here’s some history
    http://en.wikipedia.org/wiki/Castle_Bravo#Cause_of_high_yield