D Tibbets wrote: Axil wrote:Here is a post from the nextbigfuture that shows a positive energy production from NI58 to CU63
So each transformation of Ni58 into Cu63 releases 37.36MeV of nuclear energy.
When you go through the calculation, you get positive energy output.
Jonathan Starr wrote: I ran the mass calculations with H = 1.007825032 and all the beta particle included but to no avail there math is correct I got 37.1778MeV the discrepancies being the difference in our mass tables. So despite what they tell you about the curve of binding energy, fusions for elements heavier then Iron can make energy (it surprised me anyway).
Again, the mass defect represents just that, the mass equivalent of a nucleus that is not measured on a scale. That is because it is in the form of energy. The total binding energy in fact.
Right, which is the total energy RELEASED by the nucleus during the reaction.
D Tibbets wrote:Why people believe that this energy represents some released energy is beyond me.
Becasue that is its definition. Why you can't understand that is beyond me.
D Tibbets wrote:It is a part of the nucleus. The nucleus cannot exist without this incorporated energy- it is the binding energy - a part of the nucleus.
No, the binding energy is the energy RELEASED. The remainder up to the ~14MeV is the part that stays with the nucleus as part of it's mass.
D Tibbets wrote:A lead rock sitting on the ground doesn't release any more energy than a lithium rock sitting on the ground. The lead rock has a lot more mass in the form of neutrons and protons and electrons, and binding energy, but that means nothing in this static situation. If you apply kinetic energy (an equal shove on both) it will not have any more energy than the light rock. If you divide up the mass in the lead into that represented by neutrons and protons, and mass deficit or binding energy, the total does not change. The binding energy may have increased, but this is a part of the nucleus, not the energy released in the formation. The total binding energy is simply the missing mass or the change in the balance between the defined masses of the particles in isolation and their mass within a nucleus. The mass- energy total does not change, only the ratio.
Dan, I begin to believe you are hopeless.
Go back to the empirical binding energy equation you linked to before. Notice that the FIRST term is a positive term of about 14MeV and the rest (except the final) are negative. Those NEGATIVE terms are what you are calling (inaccurately) the binding energy. Those terms relate to the amount of mass that DOESN'T go deficit upon a reaction. If no energy was absorbed by stretching bonds and going into surface tension, then each and every nucleon added would release 14MeV. But they don't. They release somewhat less, but still positive. Each and every time you add a nucleon.
D Tibbets wrote: The binding energy per nucleon is a different matter.
The binding energy per nucleon is by definition the amount of matter lost in the sum total of all reactions that make up the nucleus divided by the number of electrons. It is by Einstein's equation, the amount of energy lost (released) by the sum total of reactions divided by the number of nucleons.
DAN LISTEN, DON'T TRY TO MAKE IT ANY HARDER!!! It is just that simple. E=Mc². Mass deficit = energy released. This is called "binding energy". All those other terms (surface tension, couloum repulsion, symmetry) REDUCE the binding energy.
D Tibbets wrote:This experimentally derived table/ graph is less confusing (for me) if you consider it as the packing fraction, or nuclear density. Ni62 is the most tightly packed due to the interactions between the strong but short range Strong nuclear force, and the relatively weak but long range Electromagnetic repulsive force.
Seems it just makes it more confusing to you since you just refuse to get it.
D Tibbets wrote: Nuclear physic sometimes uses a liquid model (or gas) to describe things. As you compact/ increase the density of a fluid, the temperature increases (energy is released), as the fliud/ gas expands heat is absorbed. This is the same for nickel62. It is the most dense collection of nucleons possible for this consideration (neutron stars is different physics).
Dan, if you start condsing something, it may be the mopst compact form, but you continue to release energy from the least compact for if you condense it onto the unit. Same with nuclei. 62Ni may be the most dense, but it plus a proton are MUCH less dense that 63Cu. And it is the
PROTON (or neutron) that releases the energy, NOT the nickel.