D Tibbets wrote:KitemanSA wrote: I wrote: Yes, and except where they assume stellar evolutionary process (which are not necessarily the same as solid state processes) they all agree with my statements.
You provided as the first link in your prior post a page called "FurryElephant". I propose we discuss that page section by section. To begin:
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So this has been a pretty boring (read uncontentious) beginning to the section by section. Any concerns? Shall we go on?
Dan, shall we go on?
I don't understand your point . The link you refer to
http://www.furryelephant.com/content/ra ... -defect/Is indeed named that. I don't see why that has any significance. The text explains things fairly well.
Yes, I know it is named that, and in the prior post I quoted I had started to go thru that page section by section. Did you read my prior post? Do you want to ontinue discussing that page section by section?
Then he wrote: As fas as stellar evolution, I don't know how much solid state (within nickel crystal latticeses) varies as far as probability of coulomb shielding, but it would not change nuclear reactions once the reactants get within range of the strong nuclear force, unless you are proposing new physics.
No new physics, just physics in a different regime.
As you point out below, while a star is using "fusion" the H runs out in the core where the Fe generally is, so such reactions are unlikely to start with. As I stated in my prior post, in stars, the only ways to shed the energy produced by the improbable p-Ni reaction is either to release gamma radiation (low prob) or to spit out (eject) the proton again and "pretend it never happened" (high prob, they call this part of the scattering cross-section). In a lattice, two things are different. First, unlike a star, there are about as many H as Ni. Second, the inner electron shells are filled and so there is another path to shed the energy, a path called "internal conversion" (IIRC). Basically, there is a significant probablilty that an electron will be inside the nucleus at the time of the reaction and will carry off a good deal of the energy. Thus, there will not be enough energy remaining to eject the proton. As such, the scattering cross section goes down and the reaction cross section goes up. My recollection is that the probability of IC varies with isotope. It may be that 56Ni doesn't use IC so doesn't "react" while 62Ni has a high probability of IC so the reacion takes place frequenctly. When it does, it excites the nucleus which must shed the energy by ejecting the proton, emitting gamma, or some other mechanism. It is that "other mechanism" that may be significant in lattice reations. But in any case, the p:Ni reaction is exothermic.
Then he wrote: Stars are actually believed to proceed with nucleosynthesis through multiple pathways. As the star heats up much of the nucleosynthesis of the light elements proceeds as proton absorption, neutron absorption and alpha absorption. Once the iron/ nickel limit is reached in massive stars, the core is mostly iron and there is little hydrogen available within the core. But, when the star explodes the iron/ nickel mixes with huge amounts of hydrogen in the outer layers of the star. Now there is plenty (actually a large excess) of protons available, At the temperatures and pressures within this expanding fireball a whole zoo of nuclear reactions occur. Very many neutrons are also produced. , but don't forget that the cost of providing these neutrons must be factored into the final balance. Many of the light element fusion reactions that produce neutrons are endothermic. The heavier elements are built through alpha absorption, neutron capture, and yes proton capture. Iron plus iron nuclei fusion is a vanishingly small occurrence. The neucleosynthesis primarily proceeds by small nuclei absorption - protons, neutrons, and alphas. As has been pointed out in many provided links, these are endothermic reactions past Ni62 .
Please show me ONE link that says that p:Ni is endothermic. Just one.
Then he wrote: Have you watched the video?
Not yet. Don't have access at work and have limited time at home.
Then he wrote:How can you justify exothermic heavy element fission, if the energy output from adding nucleons to progressively heavier nuclei is always exothermic. You could cycle between say nickel and uranium, gaining energy from both pathways.
Yup, but the point to remember, it is not medium+medium>large and then large>medium+medium. It is medium+p+n+p+n...>large and then large>medium+medium. It is the +p+n+p+n part that releases the energy (storing some excess as it goes until the L>M+M when some of that excess is released.
Then he wrote:You now have an inexhaustible free energy system.
Only so long as you have an inexhaustible p&n supply.
Then he wrote:There has to be a tipping point where the energy balance reverses, otherwise nothing makes sense.
The "tipping point is ~Ni, but H is NOWHERE NEAR that tipping point. H (e.i., p) is a potent source of potential energy.
Then he wrote: I don't know why you insist on the idea, that just because a heavier nucleus has more mass/ energy (binding energy) that it must always release more energy as it forms.
I didn't insist on it, Einstein did thru that little wequation E=mc². You should be arguing with him.
Dan, I really don't understand you. You have done the numbers yourself (wrongly but when corrected you acknowledge the fact. Your own corrected numbers showed that Ca+p has about the same binding energy as Ni+p. Believe your own (corrected) numbers, dude!
Then he wrote:These are two related but separate issues. Again, it is the average energy of the nucleons contained within the nucleus that determines the energy balance,
No, it is the average that determines the stability, the "urge" to change, the stability. It is the SUM that determines the energy released.
Then he wrote: not the total binding energy.
IBID
Then he wrote:
As often stated if you pulled all of the nucleons apart from the parent nucleus, the total binding energy applies. But you are pulling apart (or adding) one or a few nucleons to the outside of the nucleus where the strong force is weaker.
That define HOW MUCH of the protons potential is release, not THAT it is released.
Then he wrote:Some of the internal nucleons are very tightly bound, others are barely clinging to the surface of the nucleus. This relationship is governed by the very short range of the strong force in competition with the much longer range electromagnetic force. It turns out that the average energy per nucleon predicts things, and this is represented by the experimental finding that Ni62 is the peak/ turnaround point of the energy flow. It is the most stable, tightly bound possibility.
It is the most tightly bound PER NUCLEON. But if you can get a proton to stick, it will release about 8.7MeV when combined with Ni. Do the math. It is quite simple. You've done it before (with a glitch but you know what you did wrong there). Do it again. You will find that p+anything releases energy (except MAYBE for p+4He). Try it, I beg you. LEARN BY EXPERIENCE! Do the math.
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