D Tibbets wrote: KitemanSA wrote:DancingFool wrote: Dude. That's what makes Rossi's work so, you know, revolutionary.
It is also what makes DT so, you know, wrong. Add a proton, binding energy zero, to
ANYTHING (well, any nucleus) and you release energy. DT can't seem to understnad that. Don't know why that is.
Sigh, true, I indeed have not described it well, but the conclusion is not mine. It is that of several Nuclear fusion texts and physics web sites, which I have repeatedly quoted, and linked to.
No, your incorrect conclusion is due to your assiduous mis-reading of such texts.
D Tibbets wrote: Yet another probably confusing analogy. Hydrogen reacts with a lot of other elements to release chemcal energy (like with nickle). Consider hydrogen reacting with nitrogen. One hydrogen added to one nitrogen yields NH, and some exothermic energy/ heat. Add another hydrogen and you get NH2 and more heat. Add another hydrogen and you get NH3 and some more heat. Add another hydrogen and you get NH4, but now heat is absorbed, its is endothermic. Add some other combinations and you get additional heat absorption, the molecule is unstably and can fission , releasing energy in the process- an explosive ( comparable to the heavier elements that can undergo nuclear fission to release energy) .
Totally beside the point. Not analogous at all.
D Tibbets wrote: As I have mentioned repeatedly, You cannot only consider a single combination of nucleons, but the mechanisms by which they combine- which is the derivation of the nuclear packing fraction, or nuclear binding energy per nucleon - This IS THE DETERMINING CONSIDERATION.
Absolutely true... for stability. Not for energy release.
D Tibbets wrote: The total binding energy per nucleus, is not the determinate of the energy release.
Yes, it is. Very simply and is the thing you refuse to understand.
D Tibbets wrote: It is the energy contained within the nucleus , not the energy released by the nucleus as it forms.
This statement is incorrect. The binding energy, per nucleon or total, it the energy released per nucleon or total. Learn that and you will finally get it.
D Tibbets wrote: If you add a proton to Cl35, Fe 56, Ni 58, Ni62, Cu63, Ur 235, etc you are indeed adding energy to the product nucleus.
Dan, you are adding ~14MeV per nucleon and absorbing ~5-7MeV into the nucleus, leaving ~7-9MeV to be released. The amount RELEASED is the binding energy.
D Tibbets wrote: Ignoring stability and probability issues, this is irrelevant to the energy output of the reaction. They are two completely separate issues. The total binding energy of the nucleus can be considered as the stored or potential energy of the nucleus.
No, the binding energy is the amount RELEASED. It is analogous to GRAVITATIONAL binding energy. Gravity binds things together RELEASING a lot of energy in the process.
D Tibbets wrote: The nuclear binding energy per nucleon, can be considered as the surplus energy briefly stored by the formed nucleus which must be shed as heat/ kinetic energy through several possible pathways- gamma, beta, alpha, daughter product KE, etc.
That is exactly what I said above. "Must be shed" = released. It is the energy RELEASED.
D Tibbets wrote: So, why does the hydrogen (proton) not shed energy (exothermic) by itself. It is because I doens't have any energy to shed.
Why doesn't a small asteroid, ~infinately far away from a black hole not shed any gracvitational binding energy by itself. Because BINDING is and act of bringing TOGETHER of two or more items. One single item can't BIND. Duhh!
D Tibbets wrote: The glueons hold the quarks together so well, that it is an entirely different ball game. Even then, some have argued that protons decay to a lower energy state, though they have failed to detect any experimental evidence of this.
An asteroid ~infinitely far from a black hole is at the lowest state too.
D Tibbets wrote: The proton is the lowest energy state when considering quarks. Nickel is the lowest energy state when considering nucleons (protons and neutrons in combination).
Lowest state of WHAT? SPECIFICALLY.
D Tibbets wrote: It is the tightest, most condensed state achievable for a nucleus. As you approach this state, whether from the low side or the high side, there is excess energy liberated- heat. As you move away from this state, you are adding energy to the system (note that I said system, not the nucleus alone), thus heat is absorbed.
Right, but whan you start with a proton, you are ALWAYS moving toward that state. Even if the other particle is UnUnNilium, adding a
proton (or neutron) moves toward a MORE bound state.
D Tibbets wrote: Another analogy. A positive ion gains potential energy and loses kinetic energy as it moves away from a cathode. The energy gained or lost depends on the charge of the ion, not it's mass. A pos hydrogen ion or a 1 + gold ion gains or loses the same amount of kinetic energy despite the fact that the gold ion has a lot more mass/ energy stored in its nucleus. The velocity is different but that is irrelevant. If the velocity was the determinate of the energy exchange the total binding energy would be important, but the velocity is only a portion of what determines the kinetic energy of an object. It and mass essentially cancel out so that the interacting charges is what determines the acceleration. The nuclear binding energy per nucleon is comparable to this charge. It is maximum at Ni62 and represents the greatest delta energy that can be imparted- kinetic energy as it is approached, and potential energy as it is leaving.
Sorry, your analogy doesn't make any sense to me.