Mass Defect to calculate the stability of a nucleus

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4everphysics
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This is just a simple conceptual question.

When we try to calculate a nuclear binding energy of some nucleus,
we get the mass defect(del M) and find the binding energy by using
(del M)c^2 right?

Well, what I do not understand is this.
For instance, let's take a formation of a flourine-19.
Flourine formation(out of neutrons and protons) would be exothermic because del M is negative, but why would energy be released to the surrounding if also the same amount of energy is being used to bind the nucleus together?

Meaning, if some (x)J amount of energy is being used as a binding energy of flourine, the mass would convert to that energy and will be thus used to bind the flourine nucleus (and if that energy is being hold onto by the nucleus as a binding energy, it should not be released to the surrounding. no?)

I hope I have phrased my question right.
Thank you.
 
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Binding energy is energy required to break something apart, not to put something together. I know it sounds backwards, but it is consistent with classical picture of holding things together. If there is a force pulling two objects together, you must do work to pull them apart. Conversely, you can make them do work by letting them pull together closer. So the binding energy is released when the nucleus is formed, not absorbed.
 
So if the nucleus forms a "rope" when it is formed by converting the mass to the "rope",
why is seemingly "additional" energy released to the environment as well? when it is used up by forming the "rope"?
 
That's just not how it works. Picture two powerful magnets sticking to each other. The sound they make as they smash into each other is basically the release of the binding energy. There is nothing there that requires additional energy.
 
Magnets that snap together have opposite charge. Maybe the binding energy is the energy required to hold together like charged nucleons...
 
Look at it this way. The mass defect in a hydrogen atom is -13.6 eV, relative to a free proton and electron.