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Is this the reason why the obviously wrong calculation that fusion ends with iron is hammered into everyone?

by Gauge_Irrahphe 6y ago

XorNot 6y ago
Why is it obviously wrong?
- gct 6y ago
  You can fuse above iron but you spend energy to do it instead of releasing energy I guess?
- adrian_b 6y ago
  I do not know what the other poster had in mind. While fusion does not end at iron 56 (whose nucleus has the lowest mass per nucleon), the quantities of heavier nuclei produced by fusion decrease very steeply with increasing atomic mass, becoming low beyond the isotopes of nickel, very low beyond the isotopes of zinc and negligible beyond the isotopes of germanium. The elements heavier than zinc are produced mostly or completely by other kinds of reactions, e.g. neutron capture.
- deleted 6y ago
  [deleted]
- Gauge_Irrahphe 6y ago
  Because heavier elements have higher binding energies than iron.
  - jjk166 5y ago
    It's the binding energy per nucleon that matters. The nuclei you're producing needs to have a higher binding energy than the nuclei you're fusing do on their own. There is nothing you can fuse with iron, or any heavier nucleus, that will produce a stable element with greater binding energy than its components. Nickel-62 is actually the most stable nucleus, with the most binding energy per nucleon, but stellar nucleosynthesis ends with Nickel-56 which decays into Iron-56, which has the lowest mass per nucleon. All nuclei heavier than Nickel-62 are produced endothermically.
    - Gauge_Irrahphe 5y ago
      You are confusing nucleon (neutron or proton) with nuclide or nucleus. You still get extra energy from fusing a nucleon with nickel-62
      - jjk166 5y ago
        No, I'm not. Nuclei fuse. A proton is just a hydrogen-1 nucleus, and neutron capture isn't fusion.
        Gauge_Irrahphe 5y ago
        OK, Which weighs more? A nickel-62 nucleus and a proton, or a copper-63 nucleus?
        jjk166 5y ago
        You don't get a copper-63 nucleus if you hit a nickel-62 nucleus with a proton, you get a copper-61 nucleus and two neutrons in the reaction: 62Ni(p,2n)61Cu Copper-63 has spin of 3/2 while nickel-62 has a spin of 0, so you can't simply add a proton (whose spin is 1/2), you need at least 3 fermions to participate in the reaction. The Copper-61 and two neutrons of course have a higher mass than the Nickel-62 and the proton.
        Gauge_Irrahphe 5y ago
        That doesn't make any sense, the difference between nickel-62 and copper-63 is one extra proton. It's also bullshit, since the differences in spin get equalized by neutrinos.
        Gauge_Irrahphe 5y ago
        Also note that this closely corresponds to what we see: A star would be expected to burn by fusing hydrogen into heavier nuclei, which shed helium as alpha particles once they get too heavy, eventually ending up with mostly lead and helium. Which either undergoes a fission event which fissions the heavy metals into iron, or stays the way it is, a white dwarf. No need for "exotic" physics. And it also explains the persistent failure of fusion energy, as it tries to build upon nonexistent mechanisms made up to confuse people trying to build nuclear weapons.
wsinks 6y ago
My understanding was always that fusion _in stars_ ends with iron. Maybe that's why people simplify and assume fusion stops entirely?
- adrian_b 6y ago
  For fusion to be possible the reaction must be exothermic, which means that the average mass per nucleon in the resulting nucleus must be smaller than the average mass per nucleon in the reactants. Until the iron 56 isotope, the average mass per nucleon decreases quickly, then it increases slowly with increasing atomic mass. For reactants heavier than iron 56, fusion is not possible, but for some reactants lighter than iron 56 it is possible to fuse into nuclei a little heavier than iron 56, belonging to the few elements placed immediately after iron, where the average mass per nucleon has not increased much yet, so it is still lower than that of many of the lighter elements.
  - Gauge_Irrahphe 5y ago
    >the average mass per nucleon in the resulting nucleus must be smaller than the average mass per nucleon in the reactants. That doesn't equal to the binding energy per nucleon though, as you add a nucleon.
    - azernik 5y ago
      And then immediately on the addition of a hydrogen, the resulting nucleus is suddenly in a higher-energy (lower binding energy) state than some non-hydrogen splitting of it. And since most of the results of hydrogen hitting something are unstable, this happens almost immediately, leaving the total process looking more like fission (a proton hitting a nucleus, and producing two nuclei of smaller mass than the original). Which is, in fact, how uranium fission works, only with neutrons rather than protons.
hcrisp 6y ago
Iron is at the peak of the curve of binding energy [0]. Fusion releases energy by joining atoms up to it, fission releases energy by splitting atoms down to it. [0] https://images.app.goo.gl/oRwsxxpJLRocjgSA6 https://images.app.goo.gl/oRwsxxpJLRocjgSA6
- Cd00d 6y ago
  Yes, the binding energy curve is the correct reply. You can overshoot iron, but you'll decay back to it.
- Gauge_Irrahphe 6y ago
  Only that this is false, or more precisely intentionally misleading. The total binding energy increases all the way up. See carefully the description: It's binding energy per nucleon. It's the amount of energy you get by fusing one more proton with the nuclide, which peaks with iron, but it never goes to negative, ever.
  - XorNot 6y ago
    Any system capable of fusion is going to yield more energy from turning into Iron then turning into anything else. If you fused into something bigger, it would still be more efficient to split that atom and fuse it's nucleons into more Iron. Atoms are mostly empty space - chemical lattices are mostly empty space - so in something like a star you can make super-dense iron and turn anything else kicking around into more iron. EDIT: I feel like the problem here if you're assuming that the size of an atomic nucleus is anything other then negligible compared to it's electron shells. There's no situation under which a non-neutron star will "fit" more of a heavier nucleus in, compared to just densifying iron.
    - Gauge_Irrahphe 5y ago
      >If you fused into something bigger, it would still be more efficient to split that atom and fuse it's nucleons into more Iron. Of course, but that doesn't easily happen, as the nucleons are already bound together into heavier nuclides. In fact I guess that might happen in supernovas: The main sequence burns into lead, which is then reconstituted into the iron-nickel mixture that we know from iron meteorites, which also releases energy due to the higher average binding energy per nucleon, but needs extreme conditions to occur.

Is this the reason why the obviously wrong calculation that fusion ends with iron is hammered into everyone?

XorNot 6y ago

Why is it obviously wrong?

gct 6y ago
You can fuse above iron but you spend energy to do it instead of releasing energy I guess?
adrian_b 6y ago
I do not know what the other poster had in mind. While fusion does not end at iron 56 (whose nucleus has the lowest mass per nucleon), the quantities of heavier nuclei produced by fusion decrease very steeply with increasing atomic mass, becoming low beyond the isotopes of nickel, very low beyond the isotopes of zinc and negligible beyond the isotopes of germanium. The elements heavier than zinc are produced mostly or completely by other kinds of reactions, e.g. neutron capture.
deleted 6y ago
[deleted]
Gauge_Irrahphe 6y ago
Because heavier elements have higher binding energies than iron.
- jjk166 5y ago
  It's the binding energy per nucleon that matters. The nuclei you're producing needs to have a higher binding energy than the nuclei you're fusing do on their own. There is nothing you can fuse with iron, or any heavier nucleus, that will produce a stable element with greater binding energy than its components. Nickel-62 is actually the most stable nucleus, with the most binding energy per nucleon, but stellar nucleosynthesis ends with Nickel-56 which decays into Iron-56, which has the lowest mass per nucleon. All nuclei heavier than Nickel-62 are produced endothermically.
  - Gauge_Irrahphe 5y ago
    You are confusing nucleon (neutron or proton) with nuclide or nucleus. You still get extra energy from fusing a nucleon with nickel-62
    - jjk166 5y ago
      No, I'm not. Nuclei fuse. A proton is just a hydrogen-1 nucleus, and neutron capture isn't fusion.
      - Gauge_Irrahphe 5y ago
        OK, Which weighs more? A nickel-62 nucleus and a proton, or a copper-63 nucleus?
        jjk166 5y ago
        You don't get a copper-63 nucleus if you hit a nickel-62 nucleus with a proton, you get a copper-61 nucleus and two neutrons in the reaction: 62Ni(p,2n)61Cu Copper-63 has spin of 3/2 while nickel-62 has a spin of 0, so you can't simply add a proton (whose spin is 1/2), you need at least 3 fermions to participate in the reaction. The Copper-61 and two neutrons of course have a higher mass than the Nickel-62 and the proton.
        Gauge_Irrahphe 5y ago
        That doesn't make any sense, the difference between nickel-62 and copper-63 is one extra proton. It's also bullshit, since the differences in spin get equalized by neutrinos.
        Gauge_Irrahphe 5y ago
        Also note that this closely corresponds to what we see: A star would be expected to burn by fusing hydrogen into heavier nuclei, which shed helium as alpha particles once they get too heavy, eventually ending up with mostly lead and helium. Which either undergoes a fission event which fissions the heavy metals into iron, or stays the way it is, a white dwarf. No need for "exotic" physics. And it also explains the persistent failure of fusion energy, as it tries to build upon nonexistent mechanisms made up to confuse people trying to build nuclear weapons.

wsinks 6y ago

My understanding was always that fusion _in stars_ ends with iron. Maybe that's why people simplify and assume fusion stops entirely?

adrian_b 6y ago
For fusion to be possible the reaction must be exothermic, which means that the average mass per nucleon in the resulting nucleus must be smaller than the average mass per nucleon in the reactants. Until the iron 56 isotope, the average mass per nucleon decreases quickly, then it increases slowly with increasing atomic mass. For reactants heavier than iron 56, fusion is not possible, but for some reactants lighter than iron 56 it is possible to fuse into nuclei a little heavier than iron 56, belonging to the few elements placed immediately after iron, where the average mass per nucleon has not increased much yet, so it is still lower than that of many of the lighter elements.
- Gauge_Irrahphe 5y ago
  >the average mass per nucleon in the resulting nucleus must be smaller than the average mass per nucleon in the reactants. That doesn't equal to the binding energy per nucleon though, as you add a nucleon.
  - azernik 5y ago
    And then immediately on the addition of a hydrogen, the resulting nucleus is suddenly in a higher-energy (lower binding energy) state than some non-hydrogen splitting of it. And since most of the results of hydrogen hitting something are unstable, this happens almost immediately, leaving the total process looking more like fission (a proton hitting a nucleus, and producing two nuclei of smaller mass than the original). Which is, in fact, how uranium fission works, only with neutrons rather than protons.

hcrisp 6y ago

Iron is at the peak of the curve of binding energy [0]. Fusion releases energy by joining atoms up to it, fission releases energy by splitting atoms down to it. [0] https://images.app.goo.gl/oRwsxxpJLRocjgSA6 https://images.app.goo.gl/oRwsxxpJLRocjgSA6

Cd00d 6y ago
Yes, the binding energy curve is the correct reply. You can overshoot iron, but you'll decay back to it.
Gauge_Irrahphe 6y ago
Only that this is false, or more precisely intentionally misleading. The total binding energy increases all the way up. See carefully the description: It's binding energy per nucleon. It's the amount of energy you get by fusing one more proton with the nuclide, which peaks with iron, but it never goes to negative, ever.
- XorNot 6y ago
  Any system capable of fusion is going to yield more energy from turning into Iron then turning into anything else. If you fused into something bigger, it would still be more efficient to split that atom and fuse it's nucleons into more Iron. Atoms are mostly empty space - chemical lattices are mostly empty space - so in something like a star you can make super-dense iron and turn anything else kicking around into more iron. EDIT: I feel like the problem here if you're assuming that the size of an atomic nucleus is anything other then negligible compared to it's electron shells. There's no situation under which a non-neutron star will "fit" more of a heavier nucleus in, compared to just densifying iron.
  - Gauge_Irrahphe 5y ago
    >If you fused into something bigger, it would still be more efficient to split that atom and fuse it's nucleons into more Iron. Of course, but that doesn't easily happen, as the nucleons are already bound together into heavier nuclides. In fact I guess that might happen in supernovas: The main sequence burns into lead, which is then reconstituted into the iron-nickel mixture that we know from iron meteorites, which also releases energy due to the higher average binding energy per nucleon, but needs extreme conditions to occur.