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I have two questions, been trying to get definite and intuitive answers to them for some time so hopefully you can help me:

1) I understand both the strong force and binding energy but what is the relationship between the two?

2) What actually causes energy to be released when nuclei fuse or split? In my high school textbook it says changing the average binging energy between nucleons causes energy to be released... could you please make this clearer?

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Well, there are only two types of energy: the kinetic energy associated with motion and the potential energy associated with an interaction (i.e. a force). For example, if the force is gravity, the potential energy is the gravitational potential energy. In the case of the strong force in the nuclei, the associated energy is called the binding energy.

Let say we have the following nuclear reaction $$_{Z_1}^{A_1}X_1+_{Z_2}^{A_2}X_2\longrightarrow _{Z'_1}^{A'_1}Y_1+_{Z'_2}^{A'_2}Y_2+_{Z'_3}^{A'_3}Y_3+_0^0\gamma$$ then the difference of masses between the Xs and the Ys is converted to kinetic energy of the Ys and to energy of $\gamma$ (this also happens in chemical reactions).

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  • $\begingroup$ While "there are only two types of energy: the kinetic energy associated with motion and the potential energy associated with an interaction" is arguably true I hesitate to state it so boldly in a context where it may be seen by casual readers. The number of "kinds" of energy that exist depends on the choice of scale for probing a system. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jul 14, 2013 at 13:26
  • $\begingroup$ @dmckee I mean this at the fundamental level. Every kind of energy (chemical, nuclear, electric...) is one of these two. $\endgroup$
    – user5402
    Commented Jul 14, 2013 at 13:37
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    $\begingroup$ Yes, I understand, but to make the claim true you have to examine every system at subnuclear scales and you have to take the mass of the fundamental particles to be a kind of potential energy (which is easy enough with, say a $Z^0$, but a little less clear with an electron). Telling that to a layman or a student just starting out will engender confusion because we routinely talk of other kinds of energy. On the whole, I don't like that formulation, but I haven't voted down because it is arguably true. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jul 14, 2013 at 13:41
  • $\begingroup$ @dmckee Well this may be true even for an electron. I would like to read your formulation of my first paragraph. $\endgroup$
    – user5402
    Commented Jul 14, 2013 at 14:01
  • $\begingroup$ I would simply address the issue of the scale on which you probe a system (I am a particle physicist after all). $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jul 14, 2013 at 14:07

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