93143 wrote:chrismb wrote:Sorry to point out, 'fraid to say I think you've got that the wrong way around too!
Nope. Binding energy is negative. Therefore it either
increases towards zero or
decreases away from zero. My own personal convention...
D Tibbets wrote:From this I am guesstimating that the average energy needed to force a proton past the Coulomb barrier would in this example be ~ 6.5 MeV- thus it is endothermic by this reasoning.
...so where does the extra mass go? The listed nuclear masses are rest masses.
Any extra energy you add to try to catalyze the fusion reaction just shows back up
on top of the energy of the reaction itself once it happens.
Think about it this way. Theoretically, quantum tunneling could result in a fusion at an extremely low CoM energy, in the eV or less. That's exothermic, right? So if you add 6.5 MeV to make it go faster, what eats that extra 6.5 MeV? Nothing; you get it back at the end. So the reaction is still strongly exothermic.
Remember, the Coulomb barrier is a barrier in the sense that it goes up and then down again, like a wall or a berm, rather than just up like a hill. That's why quantum tunneling is possible. You can't just arbitrarily add the Coulomb barrier on top of the rest energy.
The graphs of quantum tunneling I have seen are not bell shaped curves. The coulomb repulsion graphs are square, the tunneling are curved to a peak then stable. Other nuclear reactions may lead to a decline, not the coulomb barrior aspect.
Considering a nuclear fusion collision as elastic is confusing for me. Most Coulomb collisions are perhaps elastic, but if a collision results in fusion, there is interaction, and the binding energy of the nucleus goes up. a lot or a modest amount depending how heavy the final nucleus is. As the nucleon (proton) approaches the nucleus id is decellerated by the repulsive electrial force. This could be considered a loss of energy from the protons or nuclei standpoint. I can see it as being elastic from a COM standpoint. But, this completely negates the kinetic energy needed to overcome the coulomb repulsion. Instead of using a 200KeV proton aimed at a stationary B11 to yield a 8.7 MeV fusion reaction, you would essentially have a 8.9 MeV fusion reaction since the the kinetic energy of the proton would be conserved. The energy of the reaction is represented by the KE pf the products, and any radiation that may be emmited.
This results in total energy always increasing with fusion , irregardless whether the input KE of the proton is 1000KeV or 20,000 KeV. This reintroduces the argument that if this was the situation, then stars would continue to burn past fe/Ni. The universe would be a very different place. It also, I think, precludes the possibility of heavy nuclei fission releasing energy. The two opposite processes in the same nucleus cannot both produce energy without violating conservation of energy(?). In this case you could have a perpetual free energy machine, merely by oscillating back and forth between two isotopes.
If I have to abandon my KE of the input nucleon as reaching and exceeding the energy of the nuclear binding energy for the reaction meaning that the net reaction is endothermic, then my previous arguments must be reinstated, unless you can otherwise explain to me how the universe could exist, or at least how both fission and fusion can be exothermic with the same reactants and products. If you illustrate a reaction with arrows going both ways. The reactants must have more energy than the products, visa versa, or the energy balance is zero. You cannot have more energy on both sides of the equation. This is essentially what your interpretation of the Nuclear Binding Energy graph is saying if you ignore the turning point (where the slope of the graph reverses at Ni62).
Dan Tibbets
To error is human... and I'm very human.
Obviously you have a very open mind concerning Rossi.