Energy balance in fusion and fission

  • #51
Drakkith
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Complete, total, hogwash. Your 'argument' amounts to saying: combustion of hydrogen and oxygen is not exothermic because combustion of hydrogen and fluorine is favored more.
How is he saying that?
 
  • #52
PAllen
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How is he saying that?
The 'binding energy per atom' of H-F is greater than H2O, therefore the latter is not exothermic.
 
  • #53
Drakkith
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The 'binding energy per atom' of H-F is greater than H2O, therefore the latter is not exothermic.
I don't really see that in his post. It looks like his is saying that if the binding energy per atom of a chemical reaction product is less than it's fuels, then that reaction should consume energy, not release it.
 
  • #54
PAllen
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I don't really see that in his post. It looks like his is saying that if the binding energy per atom of a chemical reaction product is less than it's fuels, then that reaction should consume energy, not release it.
No, he is ignoring the inputs as a whole, and only looking at Ni-62 vs Cu-63. This is analogous to saying production of water can't be exothermic because production of H-F has higher binding energy per atom.

If you look at inputs as a whole, you see What I've been saying since post #2: binding energy per nucleon of Ni-62 + proton vs binding energy per nucleon of Cu-63, the latter is clearly greater.
 
  • #55
Drakkith
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No, he is ignoring the inputs as a whole, and only looking at Ni-62 vs Cu-63. This is analogous to saying production of water can't be exothermic because production of H-F has higher binding energy per atom.

If you look at inputs as a whole, you see What I've been saying since post #2: binding energy per nucleon of Ni-62 + proton vs binding energy per nucleon of Cu-63, the latter is clearly greater.
Ok, I see what you are getting at now. The comparison to the chemical reactions had me confused a bit. Yeah, you have to look at the inputs to the reaction. If the binding energy is MORE for the end products than the inputs, then the reaction will release net energy.

Edit: I feel like I'm missing something here. Why exactly is the last step in nucleosynthesis considered to be iron if energy can still be released by fusing other things with nickel/iron? Is it that at a certain point ONLY nickel/iron is left in the core and the fusing of these will not result in energy release?
 
  • #56
PAllen
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Ok, I see what you are getting at now. The comparison to the chemical reactions had me confused a bit. Yeah, you have to look at the inputs to the reaction. If the binding energy is MORE for the end products than the inputs, then the reaction will release net energy.

Edit: I feel like I'm missing something here. Why exactly is the last step in nucleosynthesis considered to be iron if energy can still be released by fusing other things with nickel/iron? Is it that at a certain point ONLY nickel/iron is left in the core and the fusing of these will not result in energy release?
In a star, you have a huge amount of hydrogen and helium to start. After all light elements have been fused to nickel/iron, converting any significant part to a higher atomic number element will consume energy. This is not inconsistent with saying: suppose you have a plasma, with about 4 to 5 mev per nucleus, consisting of 1 hydrogen per hundreds of Ni-62 (so the hydrogens can't find each other and fuse to dueterium with higher cross section). Will the hydrogens fuse with Ni-62 producing Cu-63 and releasing energy: yes.
 
  • #57
Drakkith
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In a star, you have a huge amount of hydrogen and helium to start. After all light elements have been fused to nickel/iron, converting any significant part to a higher atomic number element will consume energy. This is not inconsistent with saying: suppose you have a plasma, with about 4 to 5 mev per nucleus, consisting of 1 hydrogen per hundreds of Ni-62 (so the hydrogens can't find each other and fuse to dueterium with higher cross section). Will the hydrogens fuse with Ni-62 producing Cu-63 and releasing energy: yes.
Ok, so once the star reaches the point where the vast majority of it's core is composed of nickel and iron, it simply has nothing else to fuse with that nickel and iron other than nickel and iron themselves. And none of these possible reactions would release energy because the end product has less binding energy (and hence more potential energy) than the fuels. That sound about right?
 
  • #58
PAllen
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Ok, so once the star reaches the point where the vast majority of it's core is composed of nickel and iron, it simply has nothing else to fuse with that nickel and iron other than nickel and iron themselves. And none of these possible reactions would release energy because the end product has less binding energy (and hence more potential energy) than the fuels. That sound about right?
Generally yes. Another way to state it, is that for the hot dense plasma as a whole, there is no lower energy state than essentially all Nicke/Iron.
 
  • #59
Drakkith
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Generally yes. Another way to state it, is that for the hot dense plasma as a whole, there is no lower energy state than essentially all Nicke/Iron.
Excellent, it makes perfect sense to me now.
 

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