I Some confusion with the Binding Energy graph of atoms

[JohnnyGui]

TL;DR

Fusing atoms up to Fe-56 releases binding energy but also requires energy to overcome the charges of protons. Where does that energy go after fusion?

My question is about the following graph:

I keep on reading that fusing atoms up until Fe-56 doesn’t cost energy and only releases binding energy. However, I understood that fusing atoms also require energy to overcome the positive charges of the protons. Where does that energy go after fusion? Does it go into the mass of the newly fused atom, escape as heat or is the released binding energy shown in the graph actually the net energy after subtracting the required fusion energy?

I secretly want it to be the latter; that the binding energy shown in the graph for each atom is the net energy after subtracting the fusion energy that is needed to overcome the charges.

If not, it gives me another issue of confusion which I find best to explain after I first know the answer to my initial question.

 

[QuarkyMeson]

The graph represents what is on the axes- mass number against the average binding energy per nucleon. The average binding energy per nucleon represents how much energy you need to supply to unbind that nucleus completely. It's an intrinsic property of the nucleus in the ground state. It doesn't include the energy you need to put in to overcome the Coulomb barrier to fuse the lighter elements together.

So when you fuse the products up to around FE-56 you have more binding energy per nucleon, That difference in total binding energy is released to the outside as kinetic energy and radiation. This is why it's possible for fusion of lighter elements (than FE-56) and fission of heavier elements to release net energy.

Whether or not you get net energy out is the total difference in all the outputs minus the inputs.

Also I think I missed this but it's important- any kinetic energy you put in to get your nuclei over the Coulomb barrier for fusion isn't lost. After fusion, essentially all energy is still accounted for in the final particles. The Coulomb barrier mainly affects how likely the reaction is, not whether it’s exothermic once it happens.

Now that said, a real fusion reactor isn't one fusion event or an isolated system in practice. You need to be able to keep your plasma in a fusion favorable state for long enough to both reclaim and exceed the input energy you've spent in creating and maintaining that state to create useful power - all while dealing with the several loss channels that exist that carry away energy you can't harness to spin a turbine. Anyway. Hope that helps.

 

[JohnnyGui]

@QuarkyMeson Thank you, I still have some confusion though based on the following:

It doesn't include the energy you need to put in to overcome the Coulomb barrier to fuse the lighter elements together.

Okay. If the graph only shows the binding energy that you need to break atoms apart, I'd understand why it doesn't include the input energy to overcome the Coulomb barrier since this input energy is needed for fusing elements instead.

But does that mean that, in the case of calculating the net energy that you get for fusing atoms together, you'd have to look up the difference in binding energy shown in the graph, and then subtract the input energy needed to overcome the Coulomb barrier yourself?

 

[QuarkyMeson]

@QuarkyMeson Thank you, I still have some confusion though based on the following:



Okay. If the graph only shows the binding energy that you need to break atoms apart, I'd understand why it doesn't include the input energy to overcome the Coulomb barrier since this input energy is needed for fusing elements instead.

But does that mean that, in the case of calculating the net energy that you get for fusing atoms together, you'd have to look up the difference in binding energy shown in the graph, and then subtract the input energy needed to overcome the Coulomb barrier yourself?

The energy released is simply the difference in masses between the products and reactants. $Q = (m_1+m_2 - \sum M_{products})c^2$. The energy you put in, lets call it $E_{input}$ is still there, so the total energy you measure now is roughly $Q+E_{input}$ ignoring radiative losses after your fusion event.

The Coulomb barrier only impacts how often reactions occur, or the nuclear fusion cross section. This is the classical picture, you still need to account for quantum effects in practice. Further, like I said before, this is a snap shot of one fusion reaction, when building a reactor you need to account for losses you aren't able to capture to spin a turbine.

 

[JohnnyGui]

The energy released is simply the difference in masses between the products and reactants. Q=(m1+m2−∑Mproducts)c2. The energy you put in, lets call it Einput is still there, so the total energy you measure now is roughly Q+Einput ignoring radiative losses after your fusion event.

So the $E_{input}$ gets released along with $Q$? And that's what the graph also includes?

How can $E_{input}$ be added to the released energy $Q$ while it's energy that costs you, not energy that you receive in addition to $Q$?

 

[QuarkyMeson]

It doesn't get released, it was there all along generally as kinetic energy. No, again, the graph only includes binding energy Q.

$E_{input}$ doesn't go away in this case because energy is conserved. This activation energy is simply helping the reaction to occur, it doesn't get used up.

 

[JohnnyGui]

It doesn't get released, it was there all along generally as kinetic energy. No, again, the graph only includes binding energy Q.

$E_{input}$ doesn't go away in this case because energy is conserved. This activation energy is simply helping the reaction to occur, it doesn't get used up.

Okay, one last question if you don't mind.

When fusing Fe-56 to get As-75, obviously there's a net energy loss as shown. Based on your previous statement:

Whether or not you get net energy out is the total difference in all the outputs minus the inputs.

What type of energy input outweighs the output in this case such that energy is lost from fusing Fe-56 to get As-75?

 

[QuarkyMeson]

Because you're fusing elements after the peak in binding energy your Q will be negative. You need to make up that budget with energy from another source, such as $E_{input}$ which, up to |Q| will then become part of the new nucleus rest mass.

 

[JohnnyGui]

@QuarkyMeson

Sorry but I'm struggling with your two statements which I find to be contradicting.

Whether or not you get net energy out is the total difference in all the outputs minus the inputs

Here it states that the net energy is the output minus the input. This should mean that energy input costs you and therefore you subtract it from the output to get the net energy out.

Einput doesn't go away in this case because energy is conserved. This activation energy is simply helping the reaction to occur, it doesn't get used up.

Here it states that the input energy does not get used up and stays in the particles. It doesn't need to be subtracted from the output $Q$. It is conserved, but how is it conserved? I'd assume the Coulomb barrier uses up that energy input but apparently that's not the case.

Are there two different forms of energy inputs you are talking about in those 2 statements?

 

[QuarkyMeson]

@QuarkyMeson

Sorry but I'm struggling with your two statements which I find to be contradicting.



Here it states that the net energy is the output minus the input. This should mean that energy input costs you and therefore you subtract it from the output to get the net energy out.

What I mean by this is, either you get a positive Q or negative Q and that determines if a given reaction is exothermic or endothermic.

Here it states that the net energy is the output minus the input. This should mean that energy input costs you and therefore you subtract it from the output to get the net energy out.

Here it states that the input energy does not get used up and stays in the particles. It doesn't need to be subtracted from the output . It is conserved, but how is it conserved? I'd assume the Coulomb barrier uses up that energy input but apparently that's not the case.

Energy input to make the reaction more favorable doesn't cost you anything, roughly. The Coulomb barrier is a conservative potential, mechanical energy is then conserved.

 

[JohnnyGui]

Energy input to make the reaction more favorable doesn't cost you anything, roughly. The Coulomb barrier is a conservative potential, mechanical energy is then conserved.

Okay, so the energy input that you subtract from Q is different from the one to overcome the Coulomb barrier.

So in the case of fusing Fe-56 to get As-75, what is the type of energy input that costs you resulting in a negative Q? Surely it's not the input to overcome the Coulomb barrier since that doesn't cost you. So what is the exact cause in the Fe atom that requires a type of energy input that costs you?

 

[QuarkyMeson]

What's costing you energy in that case is your negative Q. The product has a lower binding energy (more mass) than its reactants. The reaction is endothermic, you need to put energy into it to make it happen. That energy is coming from the kinetic energy before the fusion event. Whatever energy you input to make the reaction more favorable can then be used to "make up the difference."

In fact, if you have a very negative Q you could in theory supply enough energy to make the reaction favorable in terms of the Coulomb barrier, but still never achieve fusion because you're still below |Q| you need to make up.

 

[Astronuc]

Okay. If the graph only shows the binding energy that you need to break atoms apart,

The graph shows the binding energy per nucleon, so it is the minimum energy need to remove a nucleon from the nucleus (on average). A gamma ray of sufficient energy may cause the nucleus to eject a neutron.

The Q value for a given reaction will indicate whether a reaction is endothermic or exothermic; the former means the reactants will have less kinetic energy than input, while the latter indicates the reactors have more kinetic energy. The Q value does not address the kinetic energy to overcome the Coulomb barrier nor the effect of atomic/nuclear recoil.

In stellar nucleosynthesis, most of the nuclides beyond Fe/Ni are produced from neutron capture in the so-called slow (S) or rapid (R) processes. Some heavier elements may be formed during core collapse of stars.

References:
https://www.science.org/doi/10.1126/science.aau9540
https://link.springer.com/article/10.1007/s00159-022-00146-x
https://ned.ipac.caltech.edu/level5/Sept16/Rauscher/Rauscher4.html

Our Sun is currently burning, or fusing, hydrogen to helium. This is the process that occurs during most of a star's lifetime. After the hydrogen in the star's core is exhausted, the star can burn helium to form progressively heavier elements, carbon and oxygen and so on, until iron and nickel are formed. Up to this point the process releases energy. The formation of elements heavier than iron and nickel requires the input of energy. Supernova explosions result when the cores of massive stars have exhausted their fuel supplies and burned everything into iron and nickel. The nuclei with mass heavier than nickel are thought to be formed during these explosions.

https://cosmicopia.gsfc.nasa.gov/nucleo.html