More Isotopic Analysis of MFMP Glowstick Fuel/Ash Published by Univ. of Missouri Lab

A new set of isotopic analysis results has been released on ash from the Martin Fleischmann Memorial Project’s Glowstick 2 test, and fuel/ash from the MFMP’s Glowstick 3 test. Below is a link to the spreadsheet published by the Analytical Chemistry Group (ACG) at the University of Missouri Research Reactor (MURR®)

Some interpretation and notes by the testers can be found at the MFMP’s Quantum Heat website here. This is an excerpt:

“The basic message of the results is that, to the ability of our quadrupole ICP-MS to measure, the Li isotope ratios are effectively the same in all four samples that contain major amounts of Li. We do not have a “natural” Li standard with certified isotope ratios, and since Li is a light element, the ratios can vary a bit in nature. Thus I had to assume a natural 7Li/6Li ratio in the standard and compare its isotope ratios with those from the samples, keeping the Li in the standard and sample solutions at an approximately constant concentration. Effectively, it appears that (7Li/6Li) in all samples is a bit higher than the standard.”

“Also, the Ni in all samples appears to be isotopically natural. All ratios measured are within one standard deviation of natural Ni, except for one (61Ni/58Ni) measurement in Vial 46, run 2, which was within two standard deviations of the natural ratio.”

“Please keep in mind that the elemental concentrations were measured without internal standard and may only be considered approximate.”

  • magicsnd1

    GS4 reheat now running with the PA1000 monitoring power. The data is streaming at:

  • builditnow

    Thanks to all those who donated to MFMP to help buy a Tektronix PA1000 power analyzer. $768 was donated and Alan has proceeded with the purchase off ebay for approximately $2200. The purchase will enable the glow stick experiments to be automated, enabling more experiments and longer run times. there is an automation forum at, thanks again.

    • Stephen

      Great news! Good luck with setting up the automation.

    • Bob Greenyer

      In Alan’s own words

      “The GS4 cell pressure has dropped to zero (gauge) after 10 days idle. If it continues dropping, permeation through or reaction with the mullite will be confirmed as the cause. Conversely, stopping at ambient pressure will indicate a leak big enough to admit air.

      The PA1000 shoud be delivered today. After the pressure issue is resolved, I’ll connect it to the intact cell and compare its readings to those of the DAQ system. I’ll initially reheat the cell to 200 °C and hold for a day or so, then bump to 500 °C, or whatever profile is recommended from recent knowledge.”

      Thankyou to everyone that donated!

  • builditnow

    It’s a blind study, the labs don’t know what is in the vials they received to help avoid any bias affecting the results.
    When all the lab studies are back, the information about what was in each vial will be released, referred to here as “the key”. Only at that point can we make sense of the results. We will have to wait patiently for the lab results to come back.

  • Ged

    No idea! The data here is not at all like the Earth Tech data. It is very different, and sees little to no differences in any samples really (except one run of one of the samples). Without the key, we don’t know what is what, but the UM data seems the same levels as only one of the Earth Tech samples (~7.2 ratio for Ni60/NI62). Compare this to the Earth Tech data where two of the three samples were around 6.78.

    Again, we really need the third group’s data so we can get the key released.

    • Bob Greenyer

      More isotopic and also elemental data may come from the lab that Earthtech commissioned.

      We must thank Earthtech for taking on this extra work.

      • Ged

        They really are doing a bang up job, going above and beyond. Looking forward to what they find!

        • Bob Greenyer

          Each analysis costs around $1000 – and it is for everyone’s benefit, a great donation in both time and money as that is before review of data.

  • Ged

    I didn’t think to actually check that number, thank you!

    That is very, very interesting indeed. That takes away the idea that the Parkhomov Ni was different from normal, and adds to the idea that the lower numbers seen by Earth Tech are the two Ash. However… this is 100% pointless speculation without the key. Even worst, this leaves the large disparity between Earth Tech and UM. Earth Tech’s methods seem more professional, and the actual data output is presented in their report, which we don’t have from UM yet. But that doesn’t mean one is right. Heck, UM may have measured the same sample three times on accident for all we know (doubtful).

    We really need an expert in isotope analysis to look at the two and tell us what is going on and which is more reliable. But first, we probably need that key.

  • Ged

    Earth Tech’s results are markably different from UM’s, both for nickel and lithium ratios. The methods seem quite different too, so without a deep technical understanding, I can’t say why we see such a change–also true without knowing what samples are what exactly so as to know what is actually being measured.

    I do notice however that the Earth Tech was measuring concentrations at least as far down as 1 ppb (the NIST standard was at 1 ppb), while UM was measuring between 1 ppm to 100 ppb it looks like. That could mean that the Earth Tech method was far more sensitive, but it really depends on the machines and their dynamic ranges versus resolution. We’ll need some experts in this to tell us more.

  • Mike Henderson

    Please check my math, but I believe the experiment has to generate a LOT of anomalous heat (multiple MW-hrs) before isotope changes will be measurable.

    Each atomic reaction releases a few MeV. Like 3 to 6 MeV or so. Lets use 5.

    Each 1 MeV is about 4 x 10-17 Watt-hours. So 5 MeV = 2 x 10-16 Watt-hours per reaction.

    Let’s imagine an experiment in which 10kW-hrs of anomalous heat were released. In that test, 5 x 10^13 reactions took place. 10,000 W-hr / 2 * 10-16 W-hr per reaction= 5 * 10^13 reactions.

    Let’s suppose each reaction burns one atom of Lithium6.

    5 * 10^13 atoms / 6.023 * 10^23 atoms per mole = .84 * 10^-10 moles of Lithium6 were burned.

    Lithium6 weighs 6 gm / mole.

    The amount of Lithium6 that was consumed is 5 x 10^-10 grams.

    • Ged

      I agree.

      Let’s go one step further and look at how the ratio would be expected to change here, using the 12.18 supposed ratio for Li7/Li6.

      If we had 10 grams of lithium, then we should have ~0.821 grams of Li6, and the rest is Li7. So if we take out 5 x 10^-10 grams of Li6, the ratio would change to 12.180000007.

      • Mike Henderson

        Good idea, but you may be a few decimal places off. If I recall correctly, the reactor load was only about 2 grams and most of that (90%) was Nickel. So let’s assume .2 gm of LiAlH4, but only 6 / 38ths by weight of that is Li. So our reactor contains .032 gm of Li … of that, 1 / 13.18 is Li6 = .024 milligrams.

        Subtract 5 * 10^-10 grams of Li6 from the reactor and the ratio change is still quite small: 12.1800025

        So if the ratio need to shift by, say, 0.1 to be detectable, the reaction needs to generate 10,000 * .1 / .0000025 = 400 MW-hrs. This seems quite a bit higher than anything Parkhomov was able to generate.

        • LuFong

          The Lugano report says their 32 day test generated a net gain of 1.5MWh.

          • Mike Henderson

            I am hoping someone can point out a flaw in my logic. For example, if each reaction releases .5 MeV of energy (not 5) as heat, then the number of atoms involved increases by a factor of 10 and isotope changes are that much closer to being detectable.

            Do you see any incorrect assumptions, bad math, of incorrect logic?

            • Ged

              There are other modes of energy absorption/transfer, rather than just heat release. Endothermic reactions could absorb a lot of the energy release.

              In particular, in dueterium-tritium fusion reactors, lithium is required to sustain the reaction. Li7 reacts endothermically with a single neutron released by D-T fusion, absorbing energy away from fusion reactions while generating more T. However, this does not consume the neutron, and these endothermic Li7 reactions are essential for regenerating neutrons absorbed by other elements in fusion reactions, allowing the larger bulk reaction to continue. I don’t know how much energy is lost (not all of the bulk reaction, obviously, just a “cut off the top”).

              Similar modes could be happening here, particularly since Li7 is so abundant. Without a full understanding of all the reaction modes going on, we can’t know the NET expected energy release, only the GROSS based solely on Li6+H fusion.

              • Sanjeev

                That’s exactly what I was trying to say above before reading your post down here !
                I agree….

            • LuFong

              Your back of the envelope calculations seem OK to me. I did a quick cross check against Lugano and it seems to me that Li6 production still should be below detectable levels but I don’t really know (and am too lazy to research it). I thought you or another person could give more detail. I do note that the uncertainty in the Li6 measurements seem to be well beyond the 10^-8 grams Li6 consumed.

              This leaves your assumptions which again I don’t really know much about.

            • Sanjeev

              Have you considered the possibility that some transmutation pathways involved in LENR might not produce energy or can be possibly endothermic ?
              IMHO, your assumption is that 100% of the energy is coming from mass deficit, which lacks a solid evidence. May be most of it is lost in some interplay of “particles” (an outdated term) ?
              Anyhow, I like what you are pointing to and I also wish someone who is deep into such analysis (with some first hand knowledge of LENR) comes along and clears it up.
              Trying to figure out such energy/mass balances without exactly knowing whats going on inside the E-Cat is a hit or miss job at most.

              • Ged

                Not to mention a bit of that “mass deficit” energy can go into the creation of other subatomic particles, such as neutrinos, muons, pions fast/slow neutrons that escape, and other exotics typically created in fusion reactions. Some of the energy of the fusion becomes the mass of those particles, and then the rest becomes the “mass/energy” of photons, which is the source of the heat (thermalysed photons).
                With the proper detectors, it would be possible to measure the amount and thus mass of the subatomic byproducts, allowing that amount of known energy to be removed from the net results. But yeah, that would be ridiculously way down the road.

          • Mike Henderson

            I had performed similar arithmetic on Lugano values, and 1.5 MW-hr heat seemed to match up with masses and isotopic changes … but only when I included the conversion of Ni58, Ni60, and Ni61 to Ni62. However, I was puzzled because that would have required an awful lot of neutrons … more than I could account for in the fuel.

            The current line of thinking seems to be that Li6 is the fuel, not Ni so my reasoning above leaves Ni out of the equation. Now it still does not add up, but for different reasons.

            • Ged

              The reactor body itself has plenty of neutrons. It just means we don’t know all the reactions, or constituents of the reaction. If we assume a reaction is occurring, and the neutrons aren’t adding up, then further reaction pathways must be analyzed to square it away with the energy. This is how nuclear reactions were figured out in the first place–adding up the energy and tracking the nucleons, no different than your great work here ;). It’s a fun exercise at least!

    • LuFong

      The Lugano report mentioned a net energy gain of 1.5MWh so scaling your calculations by 150 (1.5MWh/10KWh) we get 7.5 x 10^-8 grams Li6 consumed. Is this within the limits of detection?

      • Mats002

        Very good question! Hope someone skilled in the art can answer.

  • anomal change?

    • Ged

      Well, aside from not knowing what samples this group got (a GS3 fuel/ash, and GS2 ash?), it’s hard to say. But they did fine a significant difference in Ni61/Ni58 in one sample.