1. May 10, 2015

### davidbenari

My question relates mostly to molecular and nuclear binding energy.

I'm troubled by the statement that "binding energy is the energy required to disassemble a bound system". And that "systems prefer to be in a lower energy state".

My troubles with the first question are these:

The energy required to disassemble a nucleus for example has to be, I think, something completely different from nuclear fission. For example, disassembling a nucleus would mean separating its components "manually and carefully" not bombarding it with neutrons. A neutron bombardment would provoke the release of energy because of an inherent mass defect of the initial and final states. But this seems to contrived, I think. What exactly is "disassembling a nucleus"? How are the initial and final states exactly described? If you tell me the final state is the one where the nucleons have zero KE, then why isn't a fission process also considered a disassembling process, once the nucleons have stopped moving?

Also, when nuclear/molecular systems bind together they release energy in the form of light/heat. Is this energy the binding energy? You would have to input the released energy during binding to separate the nucleus?

My troubles with the second question are these:

I'm used to the fact that systems want to be in a lower energy state because they are immersed in a field of some kind. In nuclear discussion people are talking as if $Mc^2$ where some type of potential energy, and therefore the mass defect accounts for the lower potential state. But I think this makes no sense. What force field are you associating with $Mc^2$?

Why arent electric and nuclear force interactions considered in this release of energy during fusion/fission? Why would the bound state represent the favorable one?

Like for instance, consider this equation:

$Mc^2 + BE = \sum_i m_i c^2$

These are the only energy considerations that are taken into account in a fusion process. But why isn't there in general a $U$ which would represent the total potential energy of the system due to force interactions?

One last question:

Both fission and fusion have less mass in their final states. Can I engineer a process where I go fission>Fussion>fission>fusion>fission.... and end up with no mass at all? Why not?

This is partly the reason why I think disassembling doesn't mean fission. Disassembling shouldn't have less mass in the final state (i think)

Last edited: May 10, 2015
2. May 10, 2015

### Staff: Mentor

The mechanism does not matter, energy conservation makes sure the required energy is the same independent of the path you take.
There is no such thing.

The initial state is the nucleus you consider at rest, the final state has the same number of protons and neutrons, but far apart from each other, all at rest. For this abstract consideration you don't need any additional particles. If you actually want to split a nucleus you need the help of other particles, but that is not the point.

Right.

Systems do not have their own will.
They are the whole reason why binding energy exists. I don't understand that question.
The binding energy is the energy you are looking for.

In general, fusion of small nuclei releases energy, and fission of very large nuclei releases energy. Reversing those processes (splitting small or fusing large nuclei) costs energy. The most tighly bound nuclei (binding energy per nucleon) are around iron and nickel. Both fusing and splitting those needs energy.

3. May 10, 2015

### davidbenari

I don't get what you are denying here. I'm pretty sure there is less mass after a nucleus has disintegrated during fission. Am I wrong in that?

Well take for instance electronic transitions in atoms. There you do consider the coulomb potential in order to calculate, say, the energy of the photon released. Here there is no U term whatsoever. I wonder why?

I don't see how it wouldn't matter. I'm told that during fission there is less mass. But when disassembling the nucleus I'll get more mass to get back to an original state**. (**Maybe thats my misconception).

4. May 10, 2015

### Staff: Mentor

Or more. It depends on the nucleus.

You don't, you take the energy levels of the electrons (which can be calculated using the coulomb potential and quantum mechanics). The binding energy is equivalent to those energy levels.
This is wrong in general.
This is true in general if the disassembly is complete (not counting the dineutron).

5. May 10, 2015

### davidbenari

How can it be wrong in general? My textbook derives the energy released during fission via the disintegration energy Q , with $\Delta m c^2$ by noting that because of kinetic energy of the released nucleons, the bounded mass is greater.

Well.... What I mean is that you calculate the energy of the photon as $E_i-E_f$ where each $E$ is the sum of all its types of energies in that state (intial or final). When calculating binding energy (or the energy released as heat and light), your only considering mass energy. Your not considering potentials of any kind!

6. May 10, 2015

### Bandersnatch

@davidbenari it all depends on what kind of a force field you're dealing with. If it's an attractive force, you end up with particles that gain (kinetic) energy as they get closer and require energy to get apart. If the distance in question is infinity, the energy gain/requirement is called binding energy, as it means energy needed to completely separate the two particles. In other words, the potential energy of the system is being transformed into kinetic energy of its components or vice versa.

If the force is repulsive, as in e.g., two equal charges and the Coulomb force, then you end up with particles that release energy (accelerate) as they get apart. Again, the energy gained during such separation to infinity is the binding energy (with the opposite sign here), and it's the same energy needed to be supplied to overcome the electrostatic repulsion and bring the two particles together.
Here, the binding energy is the energy necessary to bind the particles, rather than the energy released in such binding - all due to the repulsiveattractive nature of the force in question.

With nuclear reactions, the energy, and whether it's gained or lost, comes from the interplay of two fundamental forces: electromagnetic and strong nuclear.
In fusion reactions, you get the energy released due to the nuclear strong force field being attractive. You need high initial energy (hot plasma) to overcome the long-range repulsive EM forces, but once close together, the short-range, attractive strong force field is where the energy release comes from. In fission reactions, the energy release is from the repulsive nature of the EM force between protons being larger than the attractive strong interactions, with some probabilistic quantum effects thrown in to account for the randomness in fission half-life.

The change from gain to loss of energy (i.e. negative to positive binding energy) at around the iron/nickel nuclei, is due to the fact that the strong force, while much stronger than EM force at short ranges, falls down with distance much faster, and around the size of the nuclei in the neighbourhood of iron the accumulated repulsion of all the protons in the nucleus overcomes the attractive strong force holding them together. That is also why it is energetically more favourable for nuclei to have more neutrons than just in a 1:1 proportion to protons - every neutron helps hold the nucleus together via the strong force, while not contributing to the repulsive EM force due to the lack of EM charge.

Similar with gravitational force. It being attractive, binding energy gets released (i.e. KE) as the particles under its influence get together.

All of these binding energies can be translated into a loss or a gain of rest mass of the system via the E=mc^2 equation. It's just that with the nuclear reaction the change is the most significant.

Last edited: May 10, 2015
7. May 10, 2015

### Staff: Mentor

If you fission helium-4 into 2 protons and 2 neutrons, you have to add energy to the system. The resulting fission products have a larger mass than the initial nucleus.

Mass and rest energy are equivalent. Your atoms with different energy levels have a mass difference as well - it is so tiny that you neglect it, but it is there.
You can calculate the binding energy based on potentials for the strong and the electromagnetic force, but this is complicated so it won't be done in your class (or only with a very simplified model).

8. May 10, 2015

### davidbenari

Maybe I'm being repetitive but if nuclear binding energy is the energy required to disassemble a nucleus (say 2 protons 2 neutrons) ––– then if I administer that energy to the bound system (which has less mass than the nucleons separate), I'll get the original masses back? Not necessarily? If not then where did the supplied energy go? Kinetic energy, like in a fission process?

As to this: I don't really see a mechanism for the release of energy here. In a fission reaction , where the nucleons shoot out with kinetic energy you can pose the equation

$Mc^2 = \sum_i \gamma_i m_i c^2$

and conclude that $Mc^2>\sum m_i c^2$. This would mean that energy has been converted to some other form other than mass-energy, DUE to the kinetic energy of the particles that were blasted off.

In the fusion reaction, where my final state is the stable (ideally not moving) bound state, what equation could I pose? Why is there a mass defect?

9. May 10, 2015

### Staff: Mentor

Yes.

10. May 10, 2015

### davidbenari

So why do I get a mass defect (in apparently some) fission reactions? What other interaction is there , or whats going on?

11. May 10, 2015

### Staff: Mentor

You start with two protons and two neutrons, all separated. The sum of their masses is a certain number, call it ΣM1 (= 2Mp + 2Mn).

They fuse together to form a helium nucleus (alpha particle). Its mass is a certain number M2 which is smaller than ΣM1. The difference ΣM1 - M2 = ΔM is the mass defect. During the fusion process energy (ΔM)c2 leaves the system somehow, maybe in the form of gamma-ray photons.

Now you add energy (ΔM)c2 to this bound system. It separates into two protons and two neutrons, all separated. The sum of their masses is the same as you started with, ΣM1.

Last edited: May 10, 2015
12. May 10, 2015

### davidbenari

Ok, here there is no mass defect in the disassembly.

But in a fission process there is a mass defect. (Maybe there are exceptions as mfb told). What different thing occurred in this process for there to be a mass defect and not gaining ∑M1 back?

13. May 10, 2015

### Staff: Mentor

If you disassemble the nucleus to what you had before, you always get back what you had originally.
All nuclear reactions are reversible in principle.