Merged Cold Fusion Claims

[JoelKatz]

Brilliant! But evil. Do you have any more info on that?

I have tried to find additional information about it, but I can't remember much more than that and haven't been able to find information from search engines and the like. The only other thing I remember is that I think -- unless I'm confusing this with another incident -- that he did a demonstration in Houston around 1986 or so. I've been trying to figure out what keyword to search for to try to find more information.

 

[Mr.D]

Therefore this fusion is energetically favorable, radiating about 100 keV per nucleon or 6.2 MeV total.

What's the penetration depth of gamma and neutron (and whatever else) radiation in lead at those energies?

no radiation escapes due to lead shielding.

... making some reasonable assumptions on the dimensions of said shielding? (ie O(<1m)), what kind of limits can be put on their claims?

 

[ben m]

What's the penetration depth of gamma and neutron (and whatever else) radiation in lead at those energies?

Neutron emission is unlikely IMO, but I'll try to find out The actual gamma-ray spectrum would be (usually) a cascade of many of 63Cu's characteristic energies, rather than a single > 6 MeV monster gamma ray. This available energies can be found at http://www.nndc.bnl.gov/chart/chartNuc.jsp --- search for 63Cu, then click on "list of levels".

Anyway, high-energy gamma rays like this will fall off exponentially as they go through matter; in at 6MeV, in lead, you block about half the rays after 2cm, 3/4 after 4cm, 7/8 after 6cm, etc.

 

[Hindmost]

Binding energy per nucleon of the large nuclei is not the conserved quantity in this problem. Total binding energy (all nucleons) is conserved. The binding energy per nucleon of the unbound state (p+62N) is 545259.118/63 = 8654.906 MeV, while the binding energy per nucleon of the bound state (63Cu) is 551381.562/63 = 8752.088 MeV. (Or multiply both sides by 63 to get the answer I gave before.)

Therefore this fusion is energetically favorable, radiating about 100 keV per nucleon or 6.2 MeV total.

Any proton gamma reaction is going to have a positive Q value in this range on the nuclide chart. Including the threshold energy and the probability of the reaction, this isn't going to be favorable, however. If this was a truly energy producing reaction, old stars and asteriods would have a lot of copper 63.

glenn

Nitpick: you values are in KeV for BE/n

 

[Jrrarglblarg]

I don't understand all that you guys are talking about, but thanks. It seems like an important part of analyzing this claim.

 

[Hindmost]

I don't understand all that you guys are talking about, but thanks. It seems like an important part of analyzing this claim.

if you have a specific question, I may be able to answer it.

glenn

 

[ben m]

Any proton gamma reaction is going to have a positive Q value in this range on the nuclide chart. Including the threshold energy and the probability of the reaction, this isn't going to be favorable, however. If this was a truly energy producing reaction, old stars and asteriods would have a lot of copper 63.

Maybe I'm missing your meaning. The reaction has a positive Q and is exothermic. That does not mean that it has a high cross section at a given temperature.

Neither the exothermic-ness, nor the cross section, by themselves, tell you how much you expect this reaction to occur in any particular environment--you'd have to run the whole network of reactions. In relevant astrophysical contexts (i.e. supernovae) that necessarily includes lots of photodisintegration, and I have exactly zero intuition for what how the 63Cu/62Ni ratio depend on the Q-value and the cross section. (Keep in mind 62Ni(n,gamma) also generates 63Cu via a beta decay, and important step in s-process nucleosynthesis.)

And none of this matters for Rossi's poor deluded investors, since at lab temperatures the 62Ni(p,gamma) cross section is zero point zero times zero to the zeroeth zeroness. But the Q value *is* positive.

 

[Hindmost]

Maybe I'm missing your meaning. The reaction has a positive Q and is exothermic. That does not mean that it has a high cross section at a given temperature.

Neither the exothermic-ness, nor the cross section, by themselves, tell you how much you expect this reaction to occur in any particular environment--you'd have to run the whole network of reactions. In relevant astrophysical contexts (i.e. supernovae) that necessarily includes lots of photodisintegration, and I have exactly zero intuition for what how the 63Cu/62Ni ratio depend on the Q-value and the cross section. (Keep in mind 62Ni(n,gamma) also generates 63Cu via a beta decay, and important step in s-process nucleosynthesis.)

And none of this matters for Rossi's poor deluded investors, since at lab temperatures the 62Ni(p,gamma) cross section is zero point zero times zero to the zeroeth zeroness. But the Q value *is* positive.

I suppose I am looking at it from an overall reaction/probability basis--not exactly pure physics. An individual P, gamma reaction would be positive assuming the threshold energy is low enough. The threshold energy to get the proton to "fuse" with the nickel has got to be very high and it will obviously cause other reactions that I would believe would be more probable. Any reaction that kicks out a nucleon would always have a negative Q value. So, I look at the reaction as being net negative--and completely ridiculous to even consider obviously.

glenn

 

[Horatius]

--and completely ridiculous to even consider obviously.

glenn

Well, yeah, obviously.

 

[Jrrarglblarg]

if you have a specific question, I may be able to answer it.

glenn

I'm a carpenter, I play with electronics at an amateur level and I took high school chemistry in 1985. I wouldn't know where to begin; but I know that particle physics is important for a "fusion" claim.

I'm like a dog watching tennis. Keep swatting the ball, it's fun to watch. If I get hold of it it'll just get all sloppy and nobody will wanna play with it anymore.

 

[Hindmost]

Well, yeah, obviously.

to the most casual observer

 

[Dancing David]

I'm a carpenter, I play with electronics at an amateur level and I took high school chemistry in 1985. I wouldn't know where to begin; but I know that particle physics is important for a "fusion" claim.

I'm like a dog watching tennis. Keep swatting the ball, it's fun to watch. If I get hold of it it'll just get all sloppy and nobody will wanna play with it anymore.

The JREF welcomes dogs, the sloppery balls often get the most traction and generate the best questions!

 

[Jrrarglblarg]

Woof! Woof!

I won't run onto the tennis court, but since you've tossed me the ball I guess I'll chew it for a minute and bring it back to you.

So if the proton from Hydrogen joins the protons from nickel to make copper, this is going to create an isotope of copper, since the Hydrogen doesn't bring along it's own neutron. Yes?

I did a smidgen of research and find that there are quite a few isotopes of nickel; the nickel being used in this device may be a mix of various. Same goes for copper. The claim for this device also states the copper is stable.

So I guess the slobber on this tennis ball is:
Could any isotope of nickel have a half life long enough to get into his mix, and would the addition just a proton create an isotope of copper with a half life long enough to be considered "stable" by this researcher?

How much heat would be produced by one atom doing this? Can that be calculated?

 

[ben m]

Could any isotope of nickel have a half life long enough to get into his mix, and would the addition just a proton create an isotope of copper with a half life long enough to be considered "stable" by this researcher?

Yes, there are only two stable isotopes of copper, 63 and 65. (All other isotopes are short-lived, half lives less than a day). Both of these would be daughters of Ni-p fusion from 62Ni and 64Ni, both also stable.

If you fused with any other nickel isotope, you'd get a beta-unstable copper isotope which would decay; some go to heavier nickel isotopes, some go to zinc.

How much heat would be produced by one atom doing this? Can that be calculated?

About 6 MeV (10^-12 joules) per event.

 

[Jrrarglblarg]

http://peswiki.com/index.php/Directory:Andrea_A._Rossi_Cold_Fusion_Generator

Eng. Andrea A. Rossi and Professor Sergio Focardi of the University of Bologna, have announced to the world that they have a cold fusion device capable of producing more than 10 kilowatts of heat power, while only consuming a fraction of that. On January 14, 2011, they gave the Worlds' first public demonstration of a nickel-hydrogen fusion reactor capable of producing a few kilowatts of thermal energy. At its peak, it is capable of generating 15,000 watts with just 400 watts input required.

They don't use the term "cold fusion" do describe the process, be refer to it as an amplifier or catalyzer process.

Focardi states:

"Experimentally, we obtained copper; and we believe that its appearance is due to the fusion of atomic nuclei of nickel and hydrogen, the ingredients that feed our reactor. Since hydrogen and nickel 'weigh' with less, copper must have released a lot of energy, since 'nothing is created or destroyed.' Indeed, the 'Missing Mass' has been transformed into energy, which we have measured: it is in the order of a few kilowatts, two hundred times the energy that was the beginning of the reaction." [1]

10000 Watts at 10E-12 joules per event = 833333333333333 events = 1.3838149009188530942101184545555e-9 grams of copper produced?

 

[Hindmost]

Of course they left out the part how this type of fusion would take a few billion degrees.... They just said, well, we got copper, so it must have been fusion. They should have also detected some very energetic gamma rays as well--and they are very easy to detect.

glenn

 

[Jrrarglblarg]

Yeah, but they're doing it with Hydrogen! At 80 bars! Surely that means fusion, right?
[/sarcasm]

 

[ben m]

10000 Watts at 10E-12 joules per event = 833333333333333 events = 1.3838149009188530942101184545555e-9 grams of copper produced?

That's an amount produced per second. Run it for four months (10^7 s) and you have something that's trivial to detect.

 

[mike3]

Apparently, Andrea A. Rossi (I think his web site is at journal-of-nuclear-physics.com) is now claiming to have a device that takes 400W in and produces 15KW out. It's roughly the size of a large suitcase and claimed to be able to run for six months, powered on about 1 gram of nickel.

peswiki.com/index.php/Directory:Andrea_A._Rossi_Cold_Fusion_Generator

I'm guessing it's an investment scam, as these things usually are. But he's claimed an unusually tight timetable, which usually makes it hard to make much money on this kind of thing.

Does anyone have any links to skeptical articles on this particular claim or Andrea Rossi generally?

Nickel is at bottom of NBE curve. Bottomed out = no energy. BS. QED.

EDIT: Now I see, the claim is to fuse nickel with hydrogen, hmm... But I don't think this'll work because you've got to get over that Coulomb barrier somehow.

 

[Evilgiraffe]

I think there is a perfectly plausible explanation that doesn't invoke fusion whatsoever. Heating in a reducing atmosphere is the perfect way to remove the oxide layer that is present on all metals. Indeed, the reduction of NiO with hydrogen at 800^oC is exothermic to the tune of 13kJmol^-1. Not massive, but a respectable energy release. This explains their "missing mass", oxygen is being removed by the hydrogen as water, and the energy output.

If they find better results using nano-nickel, instead of macroscopic granular nickel, this is also explained by reduction of an oxide layer. The smaller the particles, the more surface per unit mass hence more oxide.

I don't know where the copper comes from. Most likely it was present as a contaminant in the original nickel samples.

ETA. Typical assay of nickel includes 50ppm copper, easily detectable by any number of techniques. Linky