KitemanSA wrote:D Tibbets wrote: Note : If the protons (hydrogen nuclei) are not being fused into heavier elements like deuterium, helium, etc. that are lighter than iron, but it is being added to nickel- directly or through a conversion to a neutron intermediate, these are generally endothermic reactions.
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Weights in Isotopic Mass (u)
63Cu = 62.9295975
62Ni = 61.9283451
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Delta = 1.0012524
Proton= 1.00782503207
The proton weighs MORE than the delta between 62Ni and 63Cu. The difference is converted into energy. The reaction is EXOthermic. Please remember that. This is the second time folks on this forum have got it wrong. It is just that 62Ni is more stable than 63Cu so the conversion is not likely to stick. TYPICALLY, if a Ni absorbes a proton, it will just spit it back out again. Seems something has to interrupt that "spit it out" process to make the proton stick. (Same with a neutron I believe).
This solution bothered me. Was N1 62 to Cu63 an exception to the binding energy limits? It appears so when looking at the mass/ energy difference comparison. But after some research, I conclude that this is a misunderstanding on your part. As pointed out in the quote below, any nucleus will have a shortfall between it's mass and the mass of its constituents. This difference is the binding energy. This ratio maximizes at iron- thus it's position at the peak of the binding energy curve.
http://hyperphysics.phy-astr.gsu.edu/hb ... ucbin.html
"Nuclei are made up of protons and neutron, but the mass of a nucleus is always less than the sum of the individual masses of the protons and neutrons which constitute it. The difference is a measure of the nuclear binding energy which holds the nucleus together."
Thus when building elements heavier than Fe this binding energy / mass ratio drops. Another way too look at is that while the total mass/ energy in heavier elements is greater, the harvestable energy through nuclear processes is decreassed. Your decreassed mass of Cu63 comparred to Ni62 + proton does not mean that you have released free energy (exthermic), it means that the cupper63 has less total energy (E=MC^2) than the Ni62 +proton. ie: you had to put energy into the system to make the atom.
If you annialate a Cu53 nucleus with an anti Cu63 nucleus you would get more energy that from annialating a Ni62 nucleus. But when you add the proton- anti proton annialation to the Ni62 , the total energy released would be greater (due to the mass difference represented by the decreased binding energy of the Cu63.
It is complex and I can easily become confused, but the simple answer is that no , the Ni62 +P to Cu63 is not exothermic!, it is endothermic (you have to put enenergy into the system to drive the reaction)*. This is a hard fact of physics. To overcome this in the Rossi claimed reaction, you not only have to crerate new physics, but ignor or change this basic understanding of nuclear binding energy.
Some other links describing neucleosynthesis:
http://kencroswell.com/Copper.html
http://iopscience.iop.org/0004-637X/710/2/1557
http://en.wikipedia.org/wiki/Stellar_nucleosynthesis
http://en.wikipedia.org/wiki/Nucleosynthesis
* The annihilation example is not clear in my mind, it may be misleading. Another way of describing it is that you have a bag of protons and neurtons- this energy- mass is unchanging and depends only on their numbers. This energy is irrelevant in terms of the energy needs to build or break down these nucleii. The pertinent energy is the binding energy. Hydrogen has the least binding energy, iron has the most, Metals heavier than iron again have lower binding energies. This translates to any reaction going towards iron can release binding energy, those going away can only consume energy.
ie: the available binding energy in a Cu63 nucleus may be more than the binding energy of a Ni62 nucleus, but it is less than the binding energy of a Ni62 nucleus and a proton added together- less total available energy, so energy has to be added to the system to create the Cu63. This energy is represented by the difference in binding energy, or in your example the mass differences. When you use the mass of the proton it is it's true rest mass, but when you have the proton in a nucleus with other protons and neutrons, it's apparent mass will be less. The difference is the binding energy. I don't know how neutrons fit into this picture. I suspect that they are irrelevant for this narrow discussion. The binding energy is involved with the strong force, while neutrons are only involved with the weak force(?).
It is difficult to conceive of a proton having a binding energy when it is not actually bound to anything. This is essentailly a binding energy of ...f 1 ( I started to say zero, but I think that would represent pure energy). There is a lot of territory between the binding energy of a proton, perhaps represented by the glueons that binds the quarks together, and pure energy. Perhaps using other reactions where there is more than one nuclear component is more revealing (like deuterium or above reactions). A useful viewpoint is to turn the typical nuclear binding graphs (like the one in the first link) upside down. Then the graph would be showing the amount of energy releasable as opposed to the amount of energy retained. And remember you are not comparing the releasable energy between Cu63 and Ni62, but the total of Ni62 plus hydrogen. The heavier elements will release energy compared to iron if they fission or lose mass through some other radioactive process, but you need to remember that it took as much or more energy to create these heavier elements in the first place. Also remember that isotopes can contain excess energy, at least temporarily in the form of excited isomers. But again, these isomers had to be created in the first place through energy input.
Dan Tibbets
To error is human... and I'm very human.