If binding energy (from mass defect) is essentially the strong nuclear force maintaining the nucleus, why is it released?

I am aware that during nuclear fusion for light elements and nuclear fission for heavy elements, the resultant elements have less mass than the original reacting elements (ie the mass defect) because of the differences in their binding energies per nucleon. This is what allows them to release energy.

Now, from my understanding from reading online, this mass defect comes into existence because of the strong nuclear force — some of the mass of the nucleons is used up to form the strong nuclear force to keep the nucleus intact.

So, if the mass defect is essentially equal to the binding energy of the nucleus, and this energy is equal to the strong nuclear force holding the nucleus together, why is this energy released as usable energy after the fission/fusion? Isn't this the energy keeping the nucleus intact? How can it be released and still leave the nucleus under the influence of the strong nuclear force?

• If gravitational potential is essentially gravity holding the planet together, why is it released when we drop a book onto the floor? Apr 11, 2019 at 21:45
• @JonCuster when you rose the book, you had to use energy to do that work against gravity. That energy you gave then manifested into potential energy in the book, which then turns to kinetic energy when you dropped it. So you gave something (work) and got kinetic energy out of it in the end. In the case of binding energy, the nucleons are giving their mass and getting the strong nuclear force - the binding energy. So then, why is there usable energy in the end? Shouldn't it just be a conversion of mass to strong nuclear force with no left over energy to cause heat?
– Andy
Apr 11, 2019 at 21:52
• The nuclei that can fission split into pieces further down the binding curve - nature made them in a way that they can fall to the ground, and that energy needs to be released. Similarly, fusing (some) nuclei allows them to form a nucleus more tightly bound, so again, you release the potential energy. Apr 11, 2019 at 21:57

You seem to assume that the energy required to keep the nuclei „intact“ is proportional to the number of nucleons, which is (roughly) the same before and after the fission/fusion.

In general, physical models of the nucleus are not easy to understand. But the so called „liquid drop model“ provides some intuition of the matter. https://en.m.wikipedia.org/wiki/Semi-empirical_mass_formula.

It that model, one factor is that the surface area of the nuclei before and after the fusion/fission is not equal. But this influence the total energy because the nucleons at the surface „feel“ a different binding as those inside the nucleus, comparable to the molecules in a drop of water. Like in water, this produces surface tension, also called surface energy.

In this way, including many more effects, the water drop model reproduces the masses of atomic nuclei quite nicely.

What is holding the nucleus together is fact that the binding energy is negative. The more negative the binding energy is, the more stable the nucleus is.

In fusion or fission reactions, the final nuclei are more stable than the original ones, and thus have more negative binding energy. The balance, which is positive (negative minus more negative) is released.

The details about exactly how negative this energy is, Harmut Braun gave you some explanations, which are basically correct. But it is in fact much more complicated than that. These details are important to compute exactly how much energy is released, but not in order to answer your question.

The main idea is, if you start from weakly bound nuclei (deuterium and tritium for fusion vs uranium or plutonium for fusion) and end up with more strongly bound ones (helium vs fission products), energy is released.

Only 1% of the mass of the original nucleus is released as the fission energy, mostly in form of the kinetic energy of the fission elements that fly apart at 3% of the speed of light.

The remaining daughter nuclei have 99% of the mass of the original nucleus, but the daughter nuclei are still intact meaning they still have strong interaction inside them that keeps them in one piece.