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# Nuclear Physics; Binding energy vs energy released? watch

1. I'm slightly confused here. I know that once nucleons fuse to make the nucleus or vice versa there is a change in energy as shown in the mass-energy equation.

What I don't understand is which part of this is the usable energy?
Is it that anything over the binding energy is usable energy? e.g. if the binding energy was 33MeV and 35MeV was released during the fission, then 2MeV would be the excess usable energy?

And also, when the term "energy released" is used in terms of fission or fusion, is it referring to the binding energy, the excess usable energy or both?

Sorry one more question, in fusion what is it that does work to fuse nucleons? it's heat right?

2. (Original post by ludd-sama)
I'm slightly confused here. I know that once nucleons fuse to make the nucleus or vice versa there is a change in energy as shown in the mass-energy equation.
What I don't understand is which part of this is the usable energy?
Is it that anything over the binding energy is usable energy? e.g. if the binding energy was 33MeV and 35MeV was released during the fission, then 2MeV would be the excess usable energy?
And also, when the term "energy released" is used in terms of fission or fusion, is it referring to the binding energy, the excess usable energy or both?
Sorry one more question, in fission/fusion what is it that does work to split apart or fuse nucleons? it's heat right?

Binding energy is the minimum energy needed to separate nucleons. You have to put this energy in to break apart the nucleus. This energy would be converted into mass, accounting for why the separate nucleons have a greater mass than the original nucleus.

In nuclear fusion, if you were fusing together every single nucleon to make the nucleus, it would release this amount of energy (equal to the total binding energy). Like in chemistry, if the forward reaction is exothermic, the reverse reaction is endothermic and of the same magnitude.

The more stable a nucleus, the higher its binding energy per nucleon. So on that graph of binding energy per nucleon against the number of nucleons, the closer an element is to Fe-56, the more stable it is.

It's easier to understand that if you go from a more stable to a less stable state, you'd need to put in energy to destabilise it. The opposite is also true, if you want to go from a less stable (lower BE per nucleon) to a more stable (higher BE per nucleon) state, energy would be released.

During fission or fusion, you're making a nucleus with a higher BE per nucleon so energy is released. The difference in binding energies is the energy that is released. Assuming 100% efficiency, all of this energy released would be kinetic energy that heats water to make steam that turns a turbine to generate electricity.

In induced fission, a high temperature and free neutrons break nuclei apart. In fusion, a high temperature helps to overcome the electrostatic repulsion between protons in the nuclei, and a high pressure gets the nuclei close enough that the strong force comes into play.

-Hope this helps
3. (Original post by Laurasaur)
It's easier to understand that if you go from a more stable to a less stable state, you'd need to put in energy to destabilise it. The opposite is also true, if you want to go from a less stable (lower BE per nucleon) to a more stable (higher BE per nucleon) state, energy would be released.

In induced fission, a high temperature and free neutrons break nuclei apart. In fusion, a high temperature helps to overcome the electrostatic repulsion between protons in the nuclei, and a high pressure gets the nuclei close enough that the strong force comes into play.
-Hope this helps
Thanks alot! everything was really helpful, just got a couple questions, referring to things I bolded in the quote.

Firstly, could you explain the idea of needing to put more energy in to destabilise. Does this relate to putting in more energy as mass, so the binding energy per nucleon is less?

and then secondly, I understand the high temperature in fusion, but what does the high temperature in fission achieve?
4. (Original post by ludd-sama)
Thanks alot! everything was really helpful, just got a couple questions, referring to things I bolded in the quote.

Firstly, could you explain the idea of needing to put more energy in to destabilise. Does this relate to putting in more energy as mass, so the binding energy per nucleon is less?

and then secondly, I understand the high temperature in fusion, but what does the high temperature in fission achieve?
I base the destabilise idea on other concepts in physics. Like if you went down the stairs, you'd go from a less stable condition to a more stable one, releasing energy. If you wanted to get up the stairs again, you'd be doing work against gravity so would have to gain energy as GPE to get to a less stable condition. For nucleons, if you wanted to go from a low BE per nucleon to a higher one, you'd be getting more stable and (like going down the stairs) would release energy.

How this relates to the nucleons is a different matter, and is beyond my level of understanding! If nucleons gained energy and therefore mass, I guess it would just mess up the structure and cause it to break apart into nucleons of a greater mass. When they recombine to form a more stable structure, with a decent ratio of protons to neutrons, the nucleons release some of the mass as energy in order to bind again. That energy that gets released can be used by us. Because the ratio of protons to neutrons is different, for whatever reason, the binding energy per nucleon is different (and would be higher if there's a net release of energy).

In fission, the initial neutron would need sufficient kinetic energy to break apart the nucleus. When I say high temperatures, I don't actually know what the temperature required is, it's just an educated guess that it needs to be high for the neutrons to induce fission .
5. (Original post by Laurasaur)
I base the destabilise idea on other concepts in physics. Like if you went down the stairs, you'd go from a less stable condition to a more stable one, releasing energy. If you wanted to get up the stairs again, you'd be doing work against gravity so would have to gain energy as GPE to get to a less stable condition. For nucleons, if you wanted to go from a low BE per nucleon to a higher one, you'd be getting more stable and (like going down the stairs) would release energy.

How this relates to the nucleons is a different matter, and is beyond my level of understanding! If nucleons gained energy and therefore mass, I guess it would just mess up the structure and cause it to break apart into nucleons of a greater mass. When they recombine to form a more stable structure, with a decent ratio of protons to neutrons, the nucleons release some of the mass as energy in order to bind again. That energy that gets released can be used by us. Because the ratio of protons to neutrons is different, for whatever reason, the binding energy per nucleon is different (and would be higher if there's a net release of energy).

In fission, the initial neutron would need sufficient kinetic energy to break apart the nucleus. When I say high temperatures, I don't actually know what the temperature required is, it's just an educated guess that it needs to be high for the neutrons to induce fission .
that stairs analogy helped me make sense of it, thanks for the helpful info, you definitely cleared it all up for me :]

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