Energy balance in fusion and fission

  • #1
I have argued with others on the Talk Polywell forum about the Nuclear Binding Energy per Nucleon vs the Total Nuclear Binding Energy per Nucleus graphs (which is the NBE/Nucleon * the atomic mass number). Despite referenced statements that, while the total binding energy increases continuously with larger nuclei, it is the binding energy / nucleon (which peaks at 62Ni) that determines the energy balance.
I have argued that the first graph is the appropiate tool for predicting the exothermic or endothermic nature of a reaction, and have given multiple references like this one-

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

,along with arguments about stellar evolution, and the need for some mechanism for both fusion and fission being exothermic under the appropriate conditions. Using the Total Binding per Nucleus graph as the predictor, I believe this is impossible. You need the minimal potential energy state/ most stable nucleus represented by 62Ni to allow this. Despite this I am ignored and / or belittled. Specifically, the discussion started when I claimed that the reaction of 62Ni + proton ---> 63Cu is an endothermic reaction and thus the Rossi Cold fusion claims were impossible based only on this basic reality.

Presuming I am correct, I doubt further arguments will help (though if any one knows of a layman's level presentation that spells this out clearly, it might help). The empirical mass formula and the opposing natures of the strong force and electromagnetic forces have not helped.

What may help is if someone with good physics credentials can briefly state their opinion on the issue, and specifically on the above reaction. It would be more difficult to ignore the conclusion if stated by an authority.

Thanks, Dan Tibbets
 

Answers and Replies

  • #2
PAllen
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It does appear to me that this reaction would be slightly exothermic. Practical realization is a separate issue.

For your example, the mass of copper 63 is:

62.9295975

the mass of Nickel 62 is:

61.9283451

and the mass of a proton is:

1.007276466812

Thus the reaction releases a small amount of energy: the energy equivalent of .006 amu.

The obstacle to this happening in a cold environment is not that it is endothermic, but that the proton can't get close enough unless it is moving fast - high temperature. Deuteron-Deuteron fusion is certainly exothermic, but that doesn't mean it happens via cold fusion.
 
  • #3
Shoot, I just lost a long reply that explained things completely.
Oh well, I will try again with a more brief response.

The total binding energy is derived from the mass deficit. There is not consideration of the nature of the energy that makes up this missing mass. There are two primary considerations. The attractive strong force binding energy, and the repulsive electromagnetic force binding energy. They both make up the missing mass. As the bound nuclei increases in contained nucleons, the strong attractive force saturates ( approaches 0). It always increases but in smaller increments. At the same time the electromagnetic repulsion also grows, but it does not saturate as fast because it has a greater range than the strong force. Because of this there is a point where the opposing forces balance out. At this point the maximum differential in attractive force has been reached. Any further growth will increase the repulsion more than the attraction. This point is where the nucleus is the most compact, densest, most stable, and has the lowest potential energy. This point is 62Ni.
Any nuclei larger or smaller than this are less stable, and have more potential energy. Remember that minimal potential energy implies that the maximum of kinetic energy has been extracted. Going towards this point yields energy, while going away requires energy input. This is why some graphs of the nuclear binding energy per nucleon (like in the provided link) are labeled with arrows that indicate the direction of exothermic energy flow.

If you use the total binding energy per nucleus (which is the total binding energy or missing mass (or better called the total bound energy) you are ignoring the fact that this bound energy is actually made up of two opposing forces, and the net effect is the balance between them. The only time the total binding energy would apply is when you tear all of the nucleons off- all or nothing. When you are considering pulling off (or adding) single or a few nucleons, the balance is determined by the binding energy of those nucleons. The total binding energy assumes the binding energy per nucleon is a constant, and thus the energy balance is only due to the number of nucleons. This is not the case. It is like traveling from the sea shore to a destination at a higher elevation, with a straight even sloped road in between. But, as illustrated by the binding energy per nucleon graph, there is a mountain in the way. You spend more energy as you climb the mountain, and get it back as you descend- opposing fusion and fission energy yields (or costs if you reverse directions) that is determined by the height and location of the mountain peak, not by the start and ending elevations alone.
Yes, you can still gain considerable energy, with an end product of a heavier than 62Ni nucleus in comparison to a collection of unbound nuclei, like hydrogen, but some of the steps are yielding energy, while others are costing energy. This is obvious in the binding energy per nucleon graph.

How could you explain exothermic fission, if you assume fusion is always exothermic as implied by using the ever increasing mass deficit/ total binding energy. Why are nuclei limited in size? If you always get energy out by combining nucleons (lower the potential energy of the product) why are there not stable nuclei with 5,000 or 78,000,000 nucleons (neutron stars is a different issue)?

The universe as we know it, star evolution and our existence would not be possible without a reasonable point where this lowest potential, most stable nucleus condition exists as a tipping point. This has bee measured as occurring at 62Ni.

PS: This is shorter than the original version :)

Dan Tibbets
 
  • #4
PAllen, concerning probability issues, they are circumvented by Rossi's claim of a super secret catalyst that eliminates the coulomb barrier.

The question of cold fusion despite the conventional coulomb barrier , whether real or not, at least when discussing deuterium - deuterium fusion by atypical pathways, there is at least a reasonable possibility of exothermic reactions. But, my contention is that Rosi's claim of exothermic fusion starting from 62Ni is impossible from an energy balance point of view. If he was talking about some other isotope of nickel, there might be some wiggle room, but not with 62Ni.

Dan Tibbets
 
  • #5
PAllen
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PAllen, concerning probability issues, they are circumvented by Rossi's claim of a super secret catalyst that eliminates the coulomb barrier.

The question of cold fusion despite the conventional coulomb barrier , whether real or not, at least when discussing deuterium - deuterium fusion by atypical pathways, there is at least a reasonable possibility of exothermic reactions. But, my contention is that Rosi's claim of exothermic fusion starting from 62Ni is impossible from an energy balance point of view. If he was talking about some other isotope of nickel, there might be some wiggle room, but not with 62Ni.

Dan Tibbets
Sorry, but you are not correct. In some cases, with minimal (or even no) energy release on absorbing a particle, a nucleus would fission (releasing more energy), and this would be treated as fission rather than fusion. However, in this case Cu 63 is stable. If you should a beam of protons of the right energy at a Ni 62 target, you will get Cu 63 with release of energy.

Fission is exothermic when the the mass of the products is less than the mass of the starting nucleus. In this case, there is nothing for Cu 63 to fission into that is exothermic. For something higher up in atomic weight, there are many possible splits that are exothermic; but not for Cu 63.
 
  • #6
PAllen
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Note that nucleosynthesis stops not really at iron-56, but at nickel-56, which is what is produced in the chain of helium 4 captures from silicon. The next element in the chain would Zinc 60, but this reaction is endothermic. If the nickel-56 had time to decay into iron-56, then some additional fusion products would be possible. However, collapse + supernova happens way before the Nickel 56 can decay. The nickel-56 blown off then decays into iron-56. Heavier elements are formed in the supernova by neutron capture.

If all of this didn't happen so fast, you could get e.g. Nickel 60 by exothermic alpha capture from iron 56.
 
  • #7
PAllen said:
"Sorry, but you are not correct. In some cases, with minimal (or even no) energy release on absorbing a particle, a nucleus would fission (releasing more energy), and this would be treated as fission rather than fusion. However, in this case Cu 63 is stable. If you should a beam of protons of the right energy at a Ni 62 target, you will get Cu 63 with release of energy."

Fission is exothermic when the the mass of the products is less than the mass of the starting nucleus. In this case, there is nothing for Cu 63 to fission into that is exothermic. For something higher up in atomic weight, there are many possible splits that are exothermic; but not for Cu 63."

There is a flaw in your reasoning. When you say 63Cu is stable, do you mean it is more stable than 62Ni, despite multiple statements to the contrary. Whether 63Cu, is relatively stable with a made up half life of 100 billion years, does that mean it is more stable than 62Ni with a half life of 500 billion years? These are made up numbers to illustrate my perspective.
A 63Cu may be relatively stable. The key word here is 'relatively'. and the linking of stability to the lowest possible potential energy state.
Do you accept that 62Ni is reported as the most stable, most densely packed and lowest energy nucleus possible? Again, this means this nucleus has the lowest potential energy possible. Just because 63Cu has a very long half life, does not mean that the 63Cu has a lower potential energy than 62Ni. The Gibbs free energy (nuclear physics equivalent?) can apply to the likelihood of a decay. The nucleus can be very stable, despite it not not yet having decayed into the absolute lowest potential energy state that is possible, does not imply that this nucleus is the lowest potential energy state.

Consider this. A mole of 63Cu does have more total binding energy than a mole of 62Ni (ignoring the opposing natures of the makeup of this binding energy). But 63 grams of 62Ni has more total binding energy than 63 grams of 63Cu. This is a back door approach to saying that the binding energy per nucleon for a specific isotope is the important measure, despite the confusion of a greater mass deficit in the heavier isotope. This is consistant (I think) with the equivalent units. The binding energy per nucleon graph is exactly the same as the binding energy per mole. On a molar 62 grams of 62Ni/ 62 g/ mole has a higher binding energy than 63 grams of 63Cu/ 63g/mole. The grams cancel out, so 62Ni has more binding energy/ mole than one mole of 63Cu.
. The proton does not have any binding energy, so the 62Ni has the all of the binding energy per unit of weight. Do this comparison with other isotopes on both sides of the 62Ni peak. On a weight basis 62Ni has the most total binding energy per unit of mass. This is another perspective, but I think it is equivalent to the binding energy per nucleon graph where 62Ni is the peak. This again illustrates that the 62Ni is the most densely packed form of nuclei that can exist before entering the degenerate matter realm.


This quote sums it up. It is repeated in various sources. Note that 56Fe is used. There has been some evolution of the thinking of what is the most stable, densest isotope. It depends on some definitions. 56Fe is more abundant than 56Ni, as and endpoint for steller exothermic fusion reactions, but this because of the pathways involved with stellar nucleosynthesis and possibly other issues.

http://en.wikipedia.org/wiki/Nuclear_binding_energy

"Once iron is reached—a nucleus with 26 protons—this process no longer gains energy. In even heavier nuclei, we find energy is lost, not gained by adding protons. Overcoming the electric repulsion (which affects all protons in the nucleus) requires more energy than what is released by the nuclear attraction (effective mainly between close neighbors). Energy could actually by gained, however, by breaking apart nuclei heavier than iron."


If you wish to see a number of references about the nuclear binding energy per nucleon graphs, look here. Click on the graphs, then close the graph image to link to the articles themselfs.

http://www.google.com/search?q=nucl...OJaaNsQLt6f2UCA&ved=0CEcQsAQ&biw=1016&bih=591


This link is the best simple explaination I have found about this issue:

http://www.furryelephant.com/content/radioactivity/binding-energy-mass-defect/

Even this has not been convincing, thus my excursion to a different form to hopefully obtain a different perspective.

Unfortunately, thus far the entrenched positions have not yet been breached.

PS: As I fumblingly tried to describe.above. Your comparison is between 62 grams of 62Ni, and 63 grams of 63Cu. The differences in the binding energy per nucleon of these neighbors in this region is not enough to overcome this mass advantage, but when you make molar comparisons, which is an alternate label for the binding energy per nucleon -Y axis. it becomes more apparent.

Dan Tibbets
 
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  • #8
PAllen
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There is a flaw in your reasoning. When you say 63Cu is stable, do you mean it is more stable than 62Ni, despite multiple statements to the contrary. Whether 63, is relatively stable (likw a half life of 100 billion years, does that mean it is more stable than 62Ni with a half life of 500 billion years? These are made up numbers to illustrate my perspective.
A 63Cu may be relatively stable. The key word here is 'relatively'.
Do you accept that 62Ni is reported as the most stable, most densely packed and lowest energy nucleus possible. Again, this means this nucleus has the lowest potential energy possible. Just because 63Cu has a very long half life, does not mean that the 63Cu has a lower potential energy than 62Ni. That the Gibbs free energy( nuclear physics equivalent) is not easily overcome has little to do with the energy balance.

Consider this. A mole of 63Cu does have more total binding energy than a mole of 62Ni (ignoring the opposing natures of the makeup of this binding energy). But 63 grams of 62Ni has more total binding energy than 63 grams of 63Cu.. The proton in the reaction has 0 binding energy, so the 62Ni has the all of the binding energy per unit of weight. It is not shared with the free proton. Do this comparison with other isotopes on both sides of the 62Ni peak. On a weight basis 62NI has the most total binding energy per unit of mass. This is another perspective, but I think it is equivilant to the binding energy per nucleon graph where 62Ni is the peak.but again illistrates that the 62Ni is the most densely packed form of nuclei that can exist before entering the degenerate mater relm.

This quote sums it up. It is repeated in various sources. Note that 56Fe is used. There has been some evolution of the thinking of what is the most stable, densest isotope. It depends on some definitions. 56Fe is more abundant than 56Ni, as and endpoint for steller exothermic fusion reactions, but this because of the pathways involved with stellar nucleosynthesis and possibly other issues.

http://en.wikipedia.org/wiki/Nuclear_binding_energy

"Once iron is reached—a nucleus with 26 protons—this process no longer gains energy. In even heavier nuclei, we find energy is lost, not gained by adding protons. Overcoming the electric repulsion (which affects all protons in the nucleus) requires more energy than what is released by the nuclear attraction (effective mainly between close neighbors). Energy could actually by gained, however, by breaking apart nuclei heavier than iron."


If you wish to see a number of references about the nuclear binding energy per nucleon graphs, look here. Click on the graphs, then close the graph image to link to the articles themselfs.

http://www.google.com/search?q=nucl...OJaaNsQLt6f2UCA&ved=0CEcQsAQ&biw=1016&bih=591


This link is the best simple explaination I have found about this issue:

http://www.furryelephant.com/content/radioactivity/binding-energy-mass-defect/

Evn this has not been convincing, thus my excursion to a different form to hopefully obtain a different perspective.

Unfortunately, thus far the entrenched positions have not yet been breached.

Dan Tibbets
Copper 63 is stable, period. There is nothing it can decay to with release of energy.

I know all about binding energy per nucleon, mass per nucleon, etc. All of that is irrelevant to the energy balance of a particular reaction in the abstract (e.g. where you can posit a beam of protons of the right energy hitting a target). Binding energy per nucleon says a lot about what processes will occur with any measurable rate in a plasma or stellar interior - but this is separate from whether a particular reaction is exothermic or endothermic.

Your aim was to argue that proton + nickel 62 -> copper 63 is requires net input of energy. This is simply false. Copper 63 is a lower energy state than nickel 62 *plus a proton*. If a proton has enough energy to reach the Nickel nucleus, the reaction will proceed with release of energy (gamma plus KE Copper 63 will be greater than KE total for proton and Nickel 62).
 
  • #9
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You might get more responses if you clarify, very briefly, your question (if you have a question) or the point you wish to discuss. Not many people are going to wade through all the prose, and I confess you lost me several posts ago.
 
  • #10
PAllen
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One way to see that the statement that "Ni-62 + proton -> Cu-63 is exothermic" is not inconsistent with the statement that you would essentially never see this as a fusion reaction (while you certainly could if you shot a proton beam at a nickel target) is the following observation:

In any environment with lots of Ni-62 and protons at high temperature and density, the reaction chain of 4 protons to He-4 is energetically favored by a factor of 4, and does not have a huge coulomb barrier. Thus in any fusion favoring environment, the Nickel reaction would be insignificant, probably undetectable.
 
  • #11
PAllen , the claim that Cu63 has higher binding energy per nucleus is valid, but that is not the question. The question is if this is a proper measure for energy flow. My contention. is that it is not. The relevant measure is the binding energy per nucleon.
In my previous post I included two links. The latter states that the binding energy/ nucleon is the same as the binding energy per mole (it is proportional) just multiply by avagadros number. Unless you contest this contention, it is obvious that on a molar basis the binding energy per mole is greater for 62Ni.

As an exercise , using your analysis, compare any heavier nucleus you like to 62 Ni , There will always be excess total binding energy compared to 62Ni which you interpreted as exothermic. Try 235Ur. It has a lot more total binding energy per nucleus. Does building up to it release energy overall and/ or in the intermediat steps? If there is intermediats steps where things reverse, where is it ( hint- it is 62Ni. If you get exothermic energy while building all the way to Uranium, how can any of these reactions reversed (fission) release energy?

The total binding energy is the sum of the absolute values of the opposing energies, but the stability/ potential energy of the nucleus is the difference between these two opposing forces.

PS: Are you claiming the half life for 63Cu decay is infinity? The only nucleus that might fit this is hydrogen (a Proton). Even that was challenged by some theories, but the experiments have failed to show evidence of this. In any case the statement most stable in this context refers to the BOUND nucleus with the least potential energy.

Dan Tibbets
 
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  • #12
JeffKoch,
The basic question is if the binding energy per nucleon, which peaks at 62Ni (or 56Fe in some considerations) the appropriate way to predict the direction of energy flow (endothermic or exothermic), and the quantity of the energy absorbed or emitted when nucleons are added or removed from near or far neighbors on the chart.

A list of representations of this is provided by this link:

http://www.google.com/search?q=nucl...OJaaNsQLt6f2UCA&ved=0CEcQsAQ&biw=1016&bih=591


This link provides the data in tabular form:

http://www.nndc.bnl.gov/amdc/masstables/Ame2003/mass.mas03 [Broken]

Dan Tibbets
 
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  • #13
PAllen
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PAllen , the claim that Cu63 has higher binding energy per nucleus is valid, but that is not the question. The question is if this is a proper measure for energy flow. My contention. is that it is not. The relevant measure is the binding energy per nucleon.
You keep ignoring the fact that to ask about whether the reaction is exothermic you need to compare the complete starting and ending states. You need to compare whether Ni-62 and a proton is a higher or lower energy state that Cu-63. It is trivially true that Cu-63 is energetically favored over Ni-62 plus a proton.
In my previous post I included two links. The latter states that the binding energy/ nucleon is the same as the binding energy per mole (it is proportional) just multiply by avagadros number. Unless you contest this contention, it is obvious that on a molar basis the binding energy per mole is greater for 62Ni.
This links were irrelevant. Your argument above is irrelevant.
As an exercise , using your analysis, compare any heavier nucleus you like to 62 Ni , There will always be excess total binding energy which you interpreted as exothermic. Try 235Ur and it's primary fission daughter products. The Ur will have higher total binding energy (BE/nucleon * A) so how did the fission of Uranium release energy?
This is simply false, as unfortunately many of your statements are.

Consider one typical U-235 fission reaction:

n + U-235 -> Kr-92 + Ba-141 + 3n

The mass on the left is 236.052595, the mass on the right is 235.86656. Thus the reaction proceeds exothermically. Same method answers all such cases.
The total binding energy is the sum of the absolute values of the opposing energies, but the stability/ potential energy of the nucleus is the difference between these two opposing forces.

PS: Are you claiming the half life for 63Cu decay is infinity? The only nucleus that might fit this is hydrogen (a Proton). Even that was challenged by some theories, but the experiments have failed to show evidence of this. In any case the statement most stable in this context refers to the BOUND nucleus with the least potential energy.

Dan Tibbets
To decay there must be a decay path. There is none for Cu-63 that can occur without input of energy [not just energy barrier - there is no combination of products with lower total energy than Cu-63]. One does not normally bring in hypothetical proton decay into discussions of half life of nuclei.
 
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  • #14
PAllen
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To decay there must be a decay path. There is none for Cu-63 that can occur without input of energy [not just energy barrier - there is no combination of products with lower total energy than Cu-63]. One does not normally bring in hypothetical proton decay into discussions of half life of nuclei.
Let me add a caveat to this. An isolated Cu-63 atom is strictly stable. However, a reasonable mass of Cu-63 does have a lower energy configuration, thus over essentially infinite time, it could re-arrange itself, if only by quantum fluctuations. If you will, the reaction would be:

62 Cu-63 -> 63 Ni-62 + 62 e+ + 62 neutrinos

This would release energy. However, this has no bearing on anything else I've said.
 
  • #15
Drakkith
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Isn't the key here that it would take MORE energy to cause the Hydrogen to fuse with the Nickel than the reaction itself actually releases? Trying to shove a proton towards 28 other protons is going to take a good amount of energy. I think it is obvious that all reactions where a proton fuses with a nucleus releases energy if you only look at the different masses, but the process of getting it to fuse is where all that energy goes.

So while a Proton + Nickel fusion would release energy, it would not release more energy than it takes to cause the fusion in the first place.

Or perhaps I am mistaken?
 
  • #16
PAllen
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Isn't the key here that it would take MORE energy to cause the Hydrogen to fuse with the Nickel than the reaction itself actually releases? Trying to shove a proton towards 28 other protons is going to take a good amount of energy. I think it is obvious that all reactions where a proton fuses with a nucleus releases energy if you only look at the different masses, but the process of getting it to fuse is where all that energy goes.

So while a Proton + Nickel fusion would release energy, it would not release more energy than it takes to cause the fusion in the first place.

Or perhaps I am mistaken?
All fusion involves overcoming coulomb repulsion. That's why high temperature / density are required. It's a matter of degree - for Nickel you have enormously higher barrier than for light elements. However, if you have a proton with enough energy to get close to the nickel nucleus, it can react to produce copper, and the reaction products will return all of the proton's kinetic energy plus more. That's all it means to say the reaction is exothermic.

Now, to ask if it is a plausible source of energy, you need many other conditions, which will defeat this reaction for practical use. First, the temperature needed would be absurd, even compared to other fusion reactions. Second, as I mentioned in an earlier post, this reaction would compete with other reactions that are so favored energetically and with lower barrier, that, for practical purposes this reaction wouldn't occur (but not because it is endothermic - that is simply false).

Note that if you continue hitting a target with protons off appropriate energy, you will reach a point where after absorbing a proton, fission will occur. But this is way beyond copper.
 
  • #17
Drakkith
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However, if you have a proton with enough energy to get close to the nickel nucleus, it can react to produce copper, and the reaction products will return all of the proton's kinetic energy plus more. That's all it means to say the reaction is exothermic.
Are you sure that the energy produced would actually be more than the energy required to cause the fusion? I'm asking because a proton in a nucleus is always less massive than a proton not in a nucleus, so the whole missing mass thing doesn't really apply. For example, adding 2 protons to U-238 turns it into P-240. But the mass difference between the 2 is only 2.0030309 u, whereas 2 protons are 2.014552933 u.
 
  • #18
PAllen
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Are you sure that the energy produced would actually be more than the energy required to cause the fusion? I'm asking because a proton in a nucleus is always less massive than a proton not in a nucleus, so the whole missing mass thing doesn't really apply. For example, adding 2 protons to U-238 turns it into P-240. But the mass difference between the 2 is only 2.0030309 u, whereas 2 protons are 2.014552933 u.
I never said that the energy released by the absorption itself would be greater than the proton KE needed for approach - only that if the proton had enough KE to get close, the reaction would occur with release of energy. Spontaneous fission or other rapid decay may then occur (for your example, not for Cu-63).

However, I was mistaken about one key point. The momentum of the result nucleus will be initially equal to the momentum of the incoming photon (in the nucleus rest frame), then a gamma ray will be released carrying the energy from the mass deficit away. However, it is not necessarily true that all of the kinetic energy of the proton will be returned as nucleus KE plus gamma ray. This cannot be determined just from mass and binding energy comparisons (or binding energy per nucleon). To determine if the reaction is endothermic in this systemic sense, you need to know the energy to overcome the coulomb barrier, and compare this to the released energy (from absorption). I don't know a figure for this offhand. For one thing, it depends on the distance needed for nuclear attraction to kick in (from this, it is straightforward to calculate) but I don't know what this distance is.

Note that if the crank fusion process the OP is trying to refute (good!) has a magical way to overcome the coulomb barrier (which they claim), then the above issue is irrelevant (and the reaction would be guaranteed to be a net energy gain).
 
  • #19
Drakkith
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I never said that the energy released by the absorption itself would be greater than the proton KE needed for approach - only that if the proton had enough KE to get close, the reaction would occur with release of energy.
Isn't that the whole point of this thread? Determining whether or not the reaction would release net energy or not? That would have to include the KE of the proton.
 
  • #20
PAllen
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Isn't that the whole point of this thread? Determining whether or not the reaction would release net energy or not? That would have to include the KE of the proton.
Yes, of course. The thread got side tracked by two things:

1) OP focus on binding energy per nucleon, which is not directly relevant to the energy balance of a particular reaction.

2) My focusing on whether the reaction would be energetically favored to occur once the proton was close to the nucleus, e.g. whether a proton beam would produce this reaction. I forgot to consider how inelastic this process was, and that it was possible for very little of the initial proton KE to be returned.

So, yes, the real question comes back to the coulomb barrier and related issues.

1) The coulomb barrier should make the crank process impossible on its face.

2) If you had a plasma of Ni-62 and protons of sufficient temperature for protons to be absorbed, then:

a) independent of energy balance, the reaction would be very rare because proton->...>He-4 fusion would enormously favored, energetically and by cross section (largely due to enormously lower coulomb barrier).

b) If a few such reactions occurred, then they might actually cool
the plasma (this is what I mistakenly thought wouldn't happen). This is due completely to the difference between the energy to overcome the coulomb barrier and the energy released by absorption.

3) If the crank mechanism could magically bypass the coulomb barrier (which they claim) to make the reaction occur at low temperatures, then, it also follows that the reaction would necessarily be exothermic. The absorption still releases the same energy; if you haven't expended energy to overcome the coulomb barrier, you get all this energy for free. Of course this is nonsense.
 
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  • #21
PAllen, First the Binding energy per nucleon graph/ tables show the binding energy of the bound nucleons within that nucleus. It has nothing to do with any free protons that may be around. It takes so much energy to tear off one nucleon. It takes less energy to tear off one nucleon from 63Cu than it does 62Ni. The torn off nucleon has zero binding energy, it is now irrelevant. Take this comparison. Take a sea of free protons and neutrons, magically combine 63 of them together to make a 63Cu and take 62 of them to make a 62Ni. Which has the highest binding energy? Now take the nuclei apart, again which represents the larger energy difference. The free nucleons are irrelevant because they all have zero binding energy. A billion free protons have 0 binding energy. A billion and one free protons also has 0 binding energy. It is only the bound nuclei that provides a magnitude of the binding energy- it is afterall defined as the potential energy stored in the nucleus (actually negative potential energy) that has to be overcome in order to remove one nucleon...

considering the Coulomb barrior as a measure of the kinetic energy of the merging particles that are merging seems straight forward. But , what about quantum tunneling. This can significantly lower the effectiver barrior- the KE of the approaching particles can be lower. But once merged, the binding energy is only a result of the potential energy of the nucleus- it does not change. The residual KE is conserved in the velocity of the merged particles, but is not part of the binding energy. If you acceppt muon catalyed fusion, then the KE of the parent particles may actually be quite low. The Coulomb barrier only represents the resistance to the particles getting close enough that the short range Strong force can become dominate. The electromagnetic repulsion between bound protons, competes against the Strong force within the bound nucleus, but the barrier is now irrelevant. What is relevant is the competition between the now bound repulsive electromagnetic force and the short range Strong force. As electromagnetic repulsion energy builds up, you are increasing the potential energy stored in the bound nucleus. You are loading the spring. The strong force is an attractive force- sometimes considered a negative potential energy. As a consequence, the negative potential energy becomes less (more negative) and KE is thus released, until such time as the strong force approaches saturation. Meanwhile the electromagnetic repulsion continues to increase (adding positive potential energy)- storing energy. As is stated repeatedly in various references. The electromagnetic positive potential energy added with the addition of a nucleon becomes incrementally larger at 62Ni. From this point on while building larger nuclei, you are adding more electromagnetic positive potential energy than negative Strong force potential energy. The overall potential energy of the nucleus no becoming more positive. This implies that KE has to be input for the process to proceed.

PAllen, you apparently saw my post before I edited it. The distance between the fission nuclei is too great to represent the issue. If you wish to suffer through my explanation, I will elaborate. The shorter range comparison of adding a neutron to the Ur illustrates the point. Using your method the 236 Ur produced from 235Ur + neutron would be a fusion that released energy
Run a series of comparisons. 63Cu to 64Zn, or Co, then proceed to heavier elements. You will find that using your method of using BE/Nucleon *A will always result in greater binding energy in the heavier neighbor, unless you umpd far enough in A that the BE.nucleon matches the value of the A ( such as in you Ur fission example. So long as you stay with close neighbors though, you would have to conclude that nucleosynthesis , even up to Uranium would release energy. Can you accept that? Calculate your 235 Ur fission reaction, and use all of the significant figures you can get out of the table linked to. Does your approach give the proper quantitative values compared to published yields? Use the BE/ Nucleon values without multiplying by A. , or if you prefer multiplying by A, then dividing by A (which cancels out A, but accounts for molar bookkeeping). does this give a more acurate measure of the energy yields?

Dan Tibbets
 
  • #22
Drakkith
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Dan, I think pallen is saying that if you ignore the coloumb barrier, then the fusion always yields energy, which is true. But as post #20 says, he was not considering that barrier in his statements. The only way for net energy gain would be to ignore the columb barrier in this case I think.
 
  • #23
No, at least in my understanding. The KE of the incoming particles has nothing to do with the binding energy of the product. It does effectthe probability of the reaction. The Ke of the proton approaching a nucleus at rest, is not absorbed into the nucleus, except as heat/ vibration. This may actually be enough to tear the nucleus apart once past a few MeV - Oppenheimer- reaction. Usually this kinetic energy is conserved by the kinetic energy of the product nucleus (it picks up speed).

Let me try this approach. The nuclear binding energy per nucleon is that energy that must be applied to tear off one nucleon from that specific nucleus. To tear off a second nuclei you have to look up the new binding energy for that nucleus. You can use the total binding energy that is used to completly tear apart the nucleus. by assuming that these changing binding energies per nucleon can be reduced to an average energy per nucleon. Then multiplying by the atomic wt will give the energy change. This works when all of the nucleons are removed, as the average applies approriatly. But this does not predict energy flow when each step is different. This is a restatement of the mountain analogy.

Consider - tearing a single nucleon off of 63Cu requires 8.75 MeV of energy.
Tearing off one nucleon off of 62Ni requires 8.79 MeV. These numbers do not actually represent the energy difference between Ni and Cu. This energy difference is the difference between these two energy states 8.79 MeV - 8.75 MeV = 0.04 MeV.
Now you have to figure which direction the energy flow is between these two nuclei.
You can use the Binding Energy per nucleon to determine this if you keep track of what the graph is representing. It represents the energy needed to dissasemble the nucleus, thus the potential energy of the nucleus. It is easier to consider this as a negative potential energy but not required. On the chart which nucleus has the highest binding energy (negative potential energy) in relation to removing one nucleon and one nucleon only? 62 Ni (or 56Fe in some older charts) is the answer.
Why is 62 Ni at this peak- simply because the strong force attractive energy has reached the maximum energy differential over the repulsive electromagnetic repulsion. This differential is less in 63Cu. Note that the strong force binding energy has still increased, but this differential between these competing forces has decreased. If this was not the case there is no way that fission could release energy, if it was only based on the total strong force binding energy, or even the sum of the strong force plus electromagnetic energy tied up in the bound nucleus, unless you recognize and account for the opposing nature of these forces.

Dan Tibbets
 
Last edited:
  • #24
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Dan,

I think you are hunting around to try to work out how the strong nuclear force balances the electromagnetic force. As a nucleon enters we have exothermic work done by the strong force, and endothermic work done by the electromagnetic force. For what I have read, both are into the multi-10's of MeV, but are finely balanced to just a few MeV at this point in the nucleon-energy curve.

However, all we need to look at is the mass change. This fully summarises that energy balance. The mass difference IS the difference in work done by one to the other.

To illustrate the point, and hopefully close your 'concern' on this, I am posting here the cross-section for the 62Ni(p,g)63Cu reaction. (This is a bit like asking if God exists, and I give you His phone number. We know the cross-section of this reaction. It is something we can really talk about, which means it must exist!!)

62ni_pg.jpg



To further emphasise the point, I will plot that graph again with the 62Ni(p,n)62Cu reaction added. You can see that once the proton gets over 5 MeV energy or so, it starts knocking off neutrons.

62ni_pgn.jpg



I think the whole matter you are raising is this energy balance; exothermic work done by the strong force, and endothermic work done by the electromagnetic force. It is always like this - these do not 'reverse' at any particular point in the periodic/isotopic table. Whether there is net work one way or the other is dependent on what is reacting, and reactions with the lightest isotopes of all, neutrons, protons, and deuterons, remain 'net exothermic' for a few 'Z' above 62Ni. That's just the way it is. The graphs above shows this is a reality. If the reality doesn't match your theory, then it's either an issue with reality, or with your theory!
 
  • #25
PAllen
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I will clarify one more point. I originally claimed that all the kinetic energy (KE) of the proton would be returned in the reaction, i.e. the KE of gamma and copper nucleus would be greater than than the initial KE of proton, by the amount of the mass defect. Drakkith challenged me on this. Without any need to account for details of coulomb field, or quantum tunneling or the like, it is easy to establish that my initial claim is exactly true.

Total energy must be conserved in any reaction. Total energy for particles 'some distance from each other' (beyond where coulomb potential energy is significant) is simply energy equivalent of mass plus kinetic energy (a photon is pure kinetic energy since it has no rest mass). In the expressions below, mass(x) is defined as rest mass of x. At the moment the proton reaches the nucleus (by any combination of classical coulomb penetration or quantum tunneling), you have an excited state of Cu-63. After emitting a gamma photon, you will have ground state Cu-63 of the rest mass given in my post #2, plus the photon, each carrying energy and momentum. Let KE(x) be kinetic energy of x. By conservation of energy we have:

mass(Ni-62) c^2 + KE(Ni-62) + mass(proton) c^2 + KE(proton) =

mass(Cu-63) c^2 + KE(Cu-63) + E(photon)

which leads to:

KE(Cu-63) + E(photon) = [mass(Ni-62) + mass(proton) - mass(Cu-63)] c^2 + KE(Ni) + KE(proton

QED.
 

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