Energy balance in fusion and fission

[PAllen]

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]

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]

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

The conversion factor is about 931.5 mev, so about 5.6 mev.

 

[Drakkith]

The conversion factor is about 931.5 mev, so about 5.6 mev.

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

 

[PAllen]

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]

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]

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]

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.

 

[Drakkith]

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 acquire all the energy I expended to get it out of the Earths gravity well in the first place.

 

[PAllen]

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 acquire 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]

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]

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]

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.

 

[Drakkith]

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.

 

[Dan Tibbets]

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]

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]

Dan, I don't really understand your description of the strong force and how it "leaks" out. To my knowledge the effect is identical to the van der waals force except that the latter involves the EM force. A small part of the strong force extends past the nucleon itself and binds it to other nucleons that are close by. I don't see how a lone proton or neutron doesn't have this "leakage". (Also you are saying the strong force betwen nucleons is due to leakage of energy, but it is actually a leakage of force.)

In heavy nuclei that undergo fission, the stored electromagnetic energy should be MORE, correct? As you have many more protons all repelling each other. Unless I am misunderstanding what you mean by EM energy in this case.

Also, the nuclear binding energy per nucleon is an average. The amount will change as you add or remove nucleons. (I'd expect it to be lower at first and increase as you remove more and more nucleons as you have more repulsive EM force initially)

 

[Drakkith]

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?

 

[PAllen]

How is he saying that?

The 'binding energy per atom' of H-F is greater than H2O, therefore the latter is not exothermic.

 

[Drakkith]

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.

 

[PAllen]

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.

 

[Drakkith]

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?

 

[PAllen]

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.

 

[Drakkith]

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?

 

[PAllen]

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.

 

[Drakkith]

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.