dkfenger wrote:Solar fusion reaction, from <
http://library.thinkquest.org/3471/fusion.html>
4H1 + 2e => He4 + (gammas+neutrinos)
delta-E = [(4)(1.007825u) - 4.002603u][931 MeV/u] = 26.7 MeV
Apply the same math to the Hydrogen-Nickel reaction: (masses from Wiki, feel free to provide alternate numbers if you have a better source.)
H1 + Ni62 => Cu63
delta-E = [1.007825u + 61.9283451u - 62.9295975u][931 MeV/u]
delta-E = [0.0065726][931 MeV/u]
delta-E = 6.11 MeV (positive, same as above.)
Stellar evolution - when a star runs out of Hydrogen, it has to fuse heavier stuff. Iron, Nickel and higher do not produce energy when fused to other nuclei of similar weight (ie, what you have left once the Hydrogen, Helium, etc have run out). But as long as there is Hydrogen left, and things to fuse it to, energy production is still possible. It is not the presence of Iron that dooms stars, it's the lack of anything lighter.
Both parts of your response is wrong. You are talking about missing mass. That always goes up as you go to hevier ellements. But to equate that to the energy balance is a fallicy. Otherwise, ehy do heavier elements break down. If you are extracting heat from the reactions, you are going to elements with lower potential energy. You are not creating energy, you can only transfer the energy between the aviable total of kinetic energy and potential energy in the system. If heavier elements (>Ni62) are relaesing KE then they have lost potential energy. Thus they are more stable, and they will not breakdown- decay. There would be no limit on the mass of an element. Uranium would be one of the lighter elements possible. Atomic weights of 23 trillion would be possible and preferred from an energy viewpoint. The missing mass goes up, of course. But this missing mass is primarily in the form of the Strong force, and the electromagnetic force. They have opposite effects on the stability (potential energy) of the nucleus, they grow at different rates and there is a crossover point (there has to be , the slopes of the graphs cross) At this cross over point the attractive Strong force balances against the electromagnetic force. There is almost zero energy that can be converted into KE. the nuclei on either side have more stored energy, thus moving towards Ni62 releases KE, moving away absorbs KE.
Again (sigh) rember that protons and neutrons have no or almost no aviable potential energy by themselfs. It is the incorperation of these nucleons into heavier nuclei that changes the energy picture- through the gluon of photon mediators of the Strong and Electromagnetic effect.
You seem to assume that adding a alpha particle or heavier nuclei is a completely different process than adding protons or neutrons. It isn't . What is important is the potential energy of the resultant product/ nuclei. If the product has a lesser potential energy , then KE is released. If the potential energy is greater (less negative by convention) then it is thermodynamically impossible for KE to be released, it is an endothermic process. The binding energy per nucleon graphs, and tabular data shows this.
Who ever developed this nomenclature did a disservice. The issue that is paramount is the KE of the product. If it is less (more negative) than the sum of the potential energy of the reactants, then KE is released. Think of the binding energy per nucleon as the measure of the potential energy of the nucleus. Here you have to reconize the the negative sign that is used by convention for potential energy. The graph makes more sise if it was inverted. Also the baseline (zero ) should be set at the proton. A Ni62 with a potential energy of ~ minus 8.9 MeV this is a value that reflects the balance between the opposing forces, not the absolute sum of the opposing forceswhich would be the missing mass). Add a proton with 0 potential energy yields a Cu63 with ~ minus 8.8 MeV. The proton with zero potential energy is neutral to the energy balance., but the Cu 63 has a higher potential energy (less negative). That this nucleus hass more energy tied up in the nucleus means that KE has to be absorbed for the reaction to occur, it is endothermic.
Your stellar arguement is a catch 22. You say heavier nuclei fuse together and perhaps but protons do not contribute to the process because they have all been consumed in earlier steps. Of course this means that Cu63 could not be produced in nature through proton absorption. That means it could only exist due to heavy nuclei fusioning together. Neucleosynthesis is a complicated process, but even working very hard with various convolutions you would have a very difficult time getting to all of the elements/ isotopes.
And actually, the hydrogen in the star is only consumed to a small percentage. The core may be depleted, but the higher layers have plenty of hydrogne that becomes aviable in the final stages when the mixture and temperatures change rapidly. Nucleosynthesis progresses through three or four major processes. Proton absorption, an mostly neutron absorption through a slow or rapid process. then there is the alpha particle absorption which is another major pathway. During supernovas the neutron absorption and to a lesser degree proton absorption are more significant because of the mixing and the profuse fusions that are occurring in the superheated plasma of the exploding star. with the elevating temperatures and the zoo of various neutron producing fusions and rapid decays. These reactions are ongoing not because they are leading to lower potential energy states, thus releasing KE, they are occuring (the endothermic ones) because there is plenty of excess KE available, not from exothermic fusion processes, but due to runaway gravitational energy release. It is this gravitational collapse that provides the energy that drives the nucleosynthesis past Nickel/ iron. All elements heavier than Ni62 are batteries. Potential energy (specifically EXCESS electromagnetic energy) has been stored in them- an endothermic process, and this energy can be released through fissions, and decays of various types. But only if the product/s have a net lower potential energy. While the binding energy per nucleon graph is confusing for the way it is setup and defined, this aspect of the Ni62 nuclei being the most stable (thus the least potential energy) nucleus and on the peak of the graph illustrates the direction of energy flow without ambiguity. This is not my conclusion. Look at any graph. If energy direction is indicated, it always points towards Ni62 (or iron in some older graphs) for exothermic reactions. Note this is a generalization- there is some jaggedness to the graph for various reasons, but Ni62 is always the endpoint with the greatest stability. Also note that the slope of the graph (plus or negative) is very flat in this region. This goes with the relatively little energy change as you progress from one element to the next in this region. This fits with what I said earlier about stellar evolution. Read the graph. Everyone accepts that hydrogen to Helium4 is a very energetic reaction and provides a lot of heat for the star. This is reflected by the position of the elements on the chart. The same applies to uranium fission to it's daughter products. The differences in the height of the graph reflects this. Again note that the height of the graph changes very little between Ni and Cu. This reflects the significantly smaller energy yields when converting between them. If there was a KE difference of ~ 8 MeV between Ni62 and Cu63, they would reflect this on the graph. Basic graph interpretation immediately revels when when you are off*.
*Again. The energy difference is the sums of the binding energy per nucleon for the reactants and products. For the questioned reaction it is Ni62 with ~8.9 MeV + proton with 0 MeV yields Cu63 with ~8.8 MeV. The proton with 0 binding energy does not effect the energy balance, so the difference is ~ 0.1 MeV. Binding energy per nucleon is defined as the energy input needed to tear off one nucleon. Of course this definition results in zero for a proton. You cannot rip off a nucleon, so the proton has zero potential energy in this system ( things change if you are talking about Quarks, but that is an unneeded complication. If you do the same comparison with Uranium fusion, the two daughter products sum together as they have non zero binding energies per nucleon, but the ~ 2 neutrons produced are ~ 0 MeV binding energy per nucleon, so they are not contributing to the energy balance comparison. They do have considerable KE imparted to them by the reaction, but the driving energy came from the energy difference between the other reactants. It does not imply that the neutrons contributed to the energy, but are merely a passive carrier of the resultant energy release.
Dan Tibbets