Does the nuclear binding energy increase and decrease simultaneously during nuclear fission or nuclear fusion?
Hello, in theoretical nuclear physics the binding energy (per nucleon) becomes lower after nuclear fission or nuclear fusion of atomic nuclei because the difference in the total energy of the two original nuclei (in fusion) or the original nucleus (in fission) and their products becomes ever larger (this only applies if more energy is released than expended).
This is a graph of binding energy in theoretical nuclear physics. (The binding energy is negative here, of course, because it shows the difference in the total energy of the individual nucleons and the atomic nuclei.)
In practical nuclear physics, however, binding energy is the energy required to split an atomic nucleus into its individual nucleons. However, the binding energy increases after fission or fusion of atomic nuclei.
Here is a graph showing the binding energy in practical nuclear physics (binding energy increases with fission or fusion)
Let's take another example: If we fuse two atomic nuclei during nuclear fusion, for example, we could say that the binding energy is released during the fusion (the binding energy becomes lower, as in theoretical nuclear physics). However, this increases the stability of the atomic nuclei because the individual nucleons have a more favorable energy ratio, which in turn means that the binding energy increases. (More stability = more energy required to break the nucleus down into individual nucleons (practical nuclear physics))
This means that the binding energy gets higher and lower at the same time, right?
If that were the case, I would be very confused. Can someone enlighten me?
Thanks in advance!
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Light cores become more stable and lighter by fusion, i.e. emit energy. The emitted energy is the binding energy, it is now quasi negative and would have to be fed back to obtain the original state before the fusion.
Severe cores are unstable and disintegrate with energy output to stable iron. They therefore emit energy by splitting and become lighter. Here, too, the binding energy is the now lacking energy which would have to be fed back for the original state (ganese core).
Adding to better understanding: The iron core is most stable and has energetically the lowest and most favorable level for it. It cannot provide energy through fusion or split. Energy would have to be inserted both for further cleavage and for further fusion, namely the binding energy which was previously released in the reversed processes on the way to the iron.
Hello leo93247,
The graphs are practically mirror-image relative to one another (mirrored on a horizontal axis, more precisely on the zero line) and are intended to test the same in principle.
Imagine, for example, two graphics, one of which is the position of a basement under the soil level and the other, as High from there to reach the ground level. Two sides of the same coin.
Finally understood! Thanks for always writing me such detailed answers. :
One more question: However, the binding energy is not the energy which holds the atomic nucleus together, but the energy which is required to disassemble the atomic nucleon. So good. But what is called the energy that keeps the nuclear core together? Is this the strong nuclear power, and if so is it part of the overall energy of an atomic nucleus?
However, or the strong nuclear power is what holds the atomic nucleus together. As a result of this attracting force, the nucleons lie practically in a potential pot whose energy level is below the level of the environment, i.e. the binding energy is the potential energy of the nucleons in the atomic nucleons. And it’s negative.
This is the other side of the same coin.
Imagine you’re parterre, and here we set the zero level of gravity potential.
You want to get a box from the basement that’s 3 m deeper. In other words, the height is −3 m in the basement.
Together with the box, you have the total mass m = 100 kg, and with the gravitational field strength g = 9.81 m⁄s2 there is a weight of m∙g = 981 N.
In the cellar you have the potential energy Ep = m∙g∙h = −2943 J.
So you have to do the work of +2943 J to get up from the basement.
Thank you.
I don’t understand. We have discussed this several times, most recently here
The binding energy is NOT lower, but HÖHER. The binding energy is the energy required to change the core, i.e. to split or fuse. The amount of change in binding energy is the mass defect of the nuclear reaction, which then becomes free as energy.
You don’t understand what I want. Here the English Wikipedia article about this:https://en.wikipedia.org/wiki/Nuclear_binding_energy. Read the beginning, hopefully you understand what I mean with the question and don’t always write the same answer that doesn’t help me.
That’s exactly what I wrote:
It’s just…
https://www.schoolphysics.co.uk/age16-19/Nuclear%20physics/Nuclear%20structure/text/Binding_energy_per_nucleon/index.html
well explained. The reason why one sometimes assumes the “negative” mode of view is that it is well expressed that nature is striving to an energetic minimum that is achieved precisely in the sole of the cow. In the overall balance, this makes no difference, it is merely a sign in the invoice.
Simply enter energy negative graph and go to images
There are even sides where this is apparently explained with the negative binding energy. I’m looking at this again.
The interesting thing is that the corresponding area is also very controversial among the authors of the article, as you can read in the discussion page. In the article “negative binding energy” is no longer present. The equivalent is access since the amounts of the respective binding energies are the same. Depending on whether the binding energy is subtracted or added in the equation, the same comes out. So far, I couldn’t find the graphics with negative binding energy, where did you get it?
In nuclear physics, the nuclear binding energy is considered a negative number. In this context it represents the energy of the nucleus relative to the energy of the react nucleons when they are infinitely far apart. Both the experimental and theoretical views are equivalent, with slightly different emphasis on what the binding energy means. Look at this again. That is the whole point of my question.
Now please read the other part with the theoretical nuclear physics.