# Energy balance in fusion and fission

Drakkith
Staff Emeritus
Can someone calculate the minimum energy needed for one proton to overcome the coulomb barrier of 62 Ni? I tried but I don't have the knowledge required.

cmb
Can someone calculate the minimum energy needed for one proton to overcome the coulomb barrier of 62 Ni?
I think you'll find the 'classic' Coulomb barrier energy for this reaction is ~5.65MeV.

Once you are above it, you tend to knock neutrons off rather than help fusion.

The energy of the incoming projectile is, of course, added to the gross output of the reaction emissions, bearing in mind that we are talking of the centre-of-mass energies.

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Drakkith
Staff Emeritus
I think you'll find the 'classic' Coulomb barrier energy for this reaction is ~5.65MeV.

Once you are above it, you tend to knock neutrons off rather than help fusion.

The energy of the incoming projectile is, of course, added to the gross output of the reaction emissions, bearing in mind that we are talking of the centre-of-mass energies.
How is the energy of the projectile added to the reaction? Wouldn't that mean that ALL elements could produce energy via fusion?

PAllen
2019 Award
How is the energy of the projectile added to the reaction? Wouldn't that mean that ALL elements could produce energy via fusion?
In some sense, almost all of the more stable isotopes of elements have available exothermic 'fusion' reactions (meaning they can absorb a proton with release of a gamma ray, with result KE greater than input KE). However, the cross section for such reactions is very low, the result may decay very fast (releasing more energy).

To produce energy in a potentially practical way, you consider not only cross section of the given reaction, but that of competing reactions (as well as the temperature required and the yield).

On a related note, U-235 hit by a neutron may exothermically produce the long lived isotope U-236 instead of splitting. Fortunately, the cross section for this is low. If it were higher, it could make U-235 impractical for fission, swamping it with low yield (but still exothermic) production of U-236, without self sustaining production of more neutrons.

Drakkith
Staff Emeritus
Do we know the amount of energy released when a proton is fused with Nickel 62 to form Copper 63?

PAllen
2019 Award
Do we know the amount of energy released when a proton is fused with Nickel 62 to form Copper 63?
Yes, I gave it in post #2.

Drakkith
Staff Emeritus
Yes, I gave it in post #2.
Ah ok, 0.006 amu. I'm not sure how to convert that to electron volts though.

PAllen
2019 Award
Ah ok, 0.006 amu. I'm not sure how to convert that to electron volts though.

Drakkith
Staff Emeritus
So the best reaction rates need almost as much energy applied to the reactants as you would get out of the reaction, correct?

PAllen
2019 Award
So the best reaction rates need almost as much energy applied to the reactants as you would get out of the reaction, correct?
In this case, yes. Don't know if it is coincidence.

Drakkith
Staff Emeritus
Sounds like it nearly perfectly obeys the binding energy curve. Input about 5.6 MeV and get the same amount as output, resulting in no net release in energy. Unless I am misunderstanding something of course. What I would be interested in seeing is calculating the same thing for a heavier element.

cmb
I suspect the co-incidence is connected with 62Ni sitting at the top of the binding energy/nucleon curve. As a new nucelon [proton] has to 'bring its own energy' along, as it were, then one might expect it to be so. In other words, this shows that exothermic energy arises because of the change of the hydrogen's binding energy, rather than the nucleons of the 62Ni (because 62Ni is already at the dead bottom of the nuclear energy barrel).

...That explanation is a bit of a guess at the co-incidental energies....

cmb
Sounds like it nearly perfectly obeys the binding energy curve. Input about 5.6 MeV and get the same amount as output, resulting in no net release in energy. Unless I am misunderstanding something of course.
Yes, you are misundertanding it. If you put in a 5.6MeV proton, you are more likely for it to fly straight into the nucleus and knock off a neutron.

If it did happen to fuse to 63Cu (I don't know if it could, with that level of excitation), then it'd have an excitation state of 11.2MeV.

PAllen
2019 Award
Sounds like it nearly perfectly obeys the binding energy curve. Input about 5.6 MeV and get the same amount as output, resulting in no net release in energy. Unless I am misunderstanding something of course. What I would be interested in seeing is calculating the same thing for a heavier element.
No, as demonstrated in post #25, if inputs have KE of 5.6 Mev, outputs (almost all in gamma ray in center of momentum frame) will have 11.2 Mev.

Drakkith
Staff Emeritus
I don't see it as a coincidence, I see it as a direct result of the binding energy/mass per nucleon of the nucleus.
Edit: Hrmm, ignore this while I wrap my head around the last few posts.

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Drakkith
Staff Emeritus
If the ouput in energy ALSO includes the input energy as well, then I don't understand the issue here. A proton that fuses with ANY nucleus always has less mass, and so would release energy no matter what.

Also, how is this energy conserved here? How could you get it back if you have to overcome the coulomb barrier? It sounds like you are saying that if I throw a ball at a million Km/h from the surface of the earth into the sun that it would somehow aquire all the energy I expended to get it out of the Earths gravity well in the first place.

PAllen
2019 Award
If the ouput in energy ALSO includes the input energy as well, then I don't understand the issue here. A proton that fuses with ANY nucleus always has less mass, and so would release energy no matter what.
This isn't true. It appears to be when you look only at masses the more stable isotopes of each element. If you look at less and less stable, more proton rich isotopes, you would definitely see a point where adding the proton would not result in mass decrease. At that point, the proton could not be bound at all.
Also, how is this energy conserved here? How could you get it back if you have to overcome the coulomb barrier? It sounds like you are saying that if I throw a ball at a million Km/h from the surface of the earth into the sun that it would somehow aquire all the energy I expended to get it out of the Earths gravity well in the first place.
Look at it this way: Suppose there is no binding energy for the additional proton, but it has enough energy to graze the nucleus and then bounce away. As the proton gets closer, it gets slower (let's be classical for simplicity). Its KE is being converted to potential energy (positive - anti-binding). At moment of grazing (supposing no momentum at this point e.g. center of momentum frame), all KE of the proton has been converted to potential energy or excitation of the composite system. The mass of the composite system would be greater than the nucleus+proton rest masses by KE/c^2. Now the proton bounces away - it accelerates as it leaves (coulomb repulsion), carrying away the excitation energy. Assuming a no-binding energy elastic collision, the proton leaves with all the energy it started with. If you add binding energy to this picture, that means that the ground state is less massive that nucleus plus proton rest masses, so the excitation is greater, so the energy carried away as reaction KE must be greater than initial proton KE. Re-read #25, which derived all this from conservation of energy.

Now, for your earth/sun example, the baseball starts with negative potential energy (gravity well; baseball is bound) and positive KE. As it escapes earth, it has lower KE and near zero PE, and higher mass than it had before you threw it. More simply, at the moment your threw it, its total energy was: mc^2+KE-PE , where m is rest mass 'at infinity', outside of any gravity well. When it escapes earth, it has exactly the same total energy, except the KE' = KE-PE, where KE' is KE on escape from earth (and we suppose PE=0 at this point). When it reaches the sun, it has the same total energy, only now the KE is much larger, and also the magnitude of negative PE is much larger. Energy is perfectly conserved throughout, changing form along the way.

Drakkith
Staff Emeritus
So why would the KE of the proton need to be given off. Wouldn't it be stored as potential energy that could be converted into kinetic energy if you rip the proton from the nucleus?

Drakkith
Staff Emeritus
This isn't true. It appears to be when you look only at masses the more stable isotopes of each element. If you look at less and less stable, more proton rich isotopes, you would definitely see a point where adding the proton would not result in mass decrease. At that point, the proton could not be bound at all.
Ok. I'm not sure how that applies to my statement though. IF it fuses with the nucleus it will have a lower mass right? As an example see my first post on page 2 dealing with adding a proton to Uranium. If adding a proton would NOT result in less mass then your saying it wouldn't bind in the first place right?

PAllen
2019 Award
So why would the KE of the proton need to be given off. Wouldn't it be stored as potential energy that could be converted into kinetic energy if you rip the proton from the nucleus?
Because we know the ground state mass of the nucleus. Any mass/energy above that is excitation, and will (soon) be emitted, it some form. If the proton itself got re-emitted (for example), you would be starting with an excitation of 11.2 Mev. Of this, 5.6 Mev would be needed to overcome the binding force, reaching the state of an unbound nearby proton, then 5.6 Mev is delivered to it as KE as it flys off. If, instead, the proton remains bound, 11.2 Mev is available to emit as a gamma ray to reach the ground state.

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Drakkith
Staff Emeritus
Because we know the ground state mass of the nucleus. Any mass/energy above that is excitation, and will (soon) be emitted, it some form. If the proton itself got re-emitted (for example), you would be starting with an excitation of 11.2 Mev. Of this, 5.2 Mev would be needed to overcome the binding force, reaching the state of an unbound nearby proton, then 5.6 Mev is delivered to it as KE as it flys off. If, instead, the proton remains bound, 11.2 Mev is available to emit as a gamma ray to reach the ground state.
Alright. Let me go dive into potential energy and such and mull this all over. Thanks for the explanations, its starting to make a *little* sense.

cmb
Don't forget to look up the Centre of Mass issue also. Energy-wise it makes less and less difference as the 'projectile' and 'target' masses are increasingly different. That is, a comparatively very heavy particle could be treated as 'stationary' before and after a collision, as an approximation, but it is not quite true. For 'light-on-light' isotope fusion, it makes all the difference.

CMB, we may need to agree to disagree. Your representation of the crossection and the KE of the approaching proton needs to overcome the Coulomb repulsion are both reasonable but irrelivant to the energy balance between 62Ni and Cu63.

I at one time thought the Coulomb repulsion and the speed needed to overcome it was relavent, but I abandoned that approach. Even with Rossi's unfounded claim of a catalyst that allows fusion a a few hundred degrees C, the reaction is not exothermic.

Assuming elastic coulomb collisions, the energy needed to overcome the electromagnetic repulsion may be considerable, but the Crossection only describes the probabilities/ rate of a fusion event at different energies. Whether the particles fuse or bounce off of each other the kinetic energy of the pair is unchanged (if binding energy is ignored, and other nuclear reactions like knocking off a neutron).
What is significant is the binding energy per nucleon. Again the mass deficite is made up of mostly two forces the attractive strong nuclear force (which can be described as a negative potential energy) and the repulsive electromagnetic force( which in a bound nucleus can be described as a positive potential energy)

The sign of the slope for these two forces do not reverse. But the relative magnitude of both forces as a portion of the total energy does change. At a cross over point (62Ni) the rate of growth of the repulsive electromagnetic force overtakes the strong force attractive energy growth rate. Using the negative potential energy of the strong force, this continues to grow and stores more and more negative potential energy as the nucleus grows up to 62Ni. Past 62NI the strong force is still being accumulated but in decreasing quantities per nucleon added. The repulsive electromagnetic force is continuing to grow, now at a faster rate and this results in a now progressive decrease in the negative potential energy of the nucleus.
Simple physics- Any bound nucleus lighter or heavier than 62Ni has more potential energy (remember the potential energy is lowest for 62Ni it is a negative value). As you progress with nucleosynthesis towards 62Ni you are extracting potential energy- which means you are releasing kinetic energy (exothermic). As you proceed with nucleosynthesis past 62Ni you are always increasing potential energy- which means you are consuming KE (endothermic). This is why other terms like the most condensed nucleus, the greatest packing fraction, the most stable nucleus are used as synonyms for the 62Ni nucleus. Some have difficulty with negative energy terms, but I've seen in subsequent posts here that this does not seem to be a problem. Recall thet the Nuclear binding energy per nucleon is defined as the energy needed to extract one nucleon from a specific nucleus. It is not a measure of the total energy or missing mass within that nucleus.

A more reasonable depiction of the pertinent relationship might to represent the potential energy of the nucleus (in terms of that energy absorbed or released in fusion/ fission reactions) 62Ni has been found to have the smallest potential energy. Place this at the origin (0,0 point) of the graph. Compare this to any other nuclei. They will always be higher on the chart- have more potential energy. You cannot have an exothermic reaction when the end product has more potential energy than starting reactant.

I have heard repeatedly that the binding energy of the proton makes all of the difference, but that is wrong. The proton (or neutron for that matter) is not bound. It is undefined in this system, in calculus terms, I believe it is a limit where the equation no longer applies. This is confusing, where does the binding energy come from then? Well, I have heard the strong force is a leakage of the quark binding Gluon energy. The bound nucleus borrows this attractive energy and releases it as heat/ KE but only when there is a bound state. In a proton there is no leakage of this force, thus no missing energy. Keep in mind this energy release is based on the 'glueon' energy leaked. But this is not nessisarily released from the bound nucleus. Some has to be retained to overcome the electromagnetic repulsion, otherwise the nucleus would immediatly fly apart. In fission of heavy elements, as the nucleus breaks apart the stored electromagnetic energy is not as great, and thus more of the strong force stored energy can be released to compensate, thus energy is released in heavy element fission. It can be described in several ways. In this example the electromagnetic force is considered only as that in proportional opposition to the strong force. This interaction is driven by the nuclear radius, the strength and range of these forces.

Also, considering intermediates like an excited isomer does not change the energy balance between the initial and end products.

Dan Tibbets

PAllen
2019 Award
CMB, we may need to agree to disagree.
...
Dan Tibbets
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.

Drakkith
Staff Emeritus