I have read that there is an increase in the amount of the binding energy from the reactant nuclei and the product nucleus during nuclear fusion and this causes the decrease the internal energy (potential and the positional energy ) which in turn causes a decrease in the rest mass of the nucleus. This "lost"mass converts to energy. Here,Shouldn't the repulsive force of the protons balance the strong nuclear force in the reactant and product nuclei so that the nucleons don't experience any increase in the binding energy? Since proton repulsion increases as the binding energy is increasing....
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$\begingroup$ If my theory is wrong , kindly correct it and mention the source of energy in nuclear fusion. $\endgroup$– Scientific CoDec 30, 2021 at 19:07
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2$\begingroup$ It's a combo of electromagetic, weak, and strong forces. Depends on the exact configuaration. Maximum binding occurs at iron. As posed you are asking for an explanation of huge amounts of nuclear physics. $\endgroup$– DanDec 30, 2021 at 20:42
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1$\begingroup$ What "increase in the binding energy"? As you said earlier, energy is emitted when the reactants fuse. Also, why do you expect the electrostatic repulsion to be balanced by the nuclear attraction? If those forces were perfectly balanced, the nucleons wouldn't be bound together. $\endgroup$– PM 2RingDec 30, 2021 at 20:55
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1$\begingroup$ Does the same logic make you think chemical reactions cannot be exo- or endothermic? $\endgroup$– J.G.Dec 30, 2021 at 23:13
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1$\begingroup$ You seem to be equating force with energy. They aren't the same thing. $\endgroup$– ProfRobDec 31, 2021 at 9:56
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3 Answers
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The protons' mutual electric repulsion does indeed try very hard to prevent the two of them from ever touching, but at extremely small separations the (short-range and attractive) strong force kicks in, overwhelms the electrostatic repulsion, and binds the protons together. To make them stick together for good, you do have to add a couple of neutrons to the recipe to serve as a sort of glue- which also bind via the strong force, but have no repulsive electrical charge to overcome in close encounters.
Note also that the strong force is what holds quarks together inside the radius of the proton or neutron. The residual strong force which "bleeds outside" that radius is the force which actually binds the nucleons together.
answered Dec 31, 2021 at 1:39
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$\begingroup$ Increase in the amount of the binding energy from the reactant nuclei and the product nucleus during nuclear fusion and this causes the decrease the internal energy (potential and the positional energy ) which in turn causes a decrease in the rest mass of the nucleus. This "lost"mass converts to energy. Is this the case of the energy released during nuclear fusion? $\endgroup$ Dec 31, 2021 at 4:52
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See the correct answer by niels nielsen.
Just to add some details : as its name implies, the so-called "strong force" is potentially much stronger than the electrostatic repulsion, by a factor of (very roughly) one hundred. Most of it holds quarks together within the protons and neutrons.
Just what "bleeds out" is still large enough to overcome electrostatic repulsion and bind protons and neutrons together in nuclei. This "bleeding out strong force" is not that strong anymore. Indeed for light nuclei, (deuterium, helium, carbon, nitrogen, oxygen, ... up to sulfur) even this "bleeding out strong force" is still larger enough than electrostatic repulsion that the numbers of protons and neutrons are equal, which is what the strong force favours. But this is not so overwhelming larger than electrostatic repulsion anymore. Indeed for heavier nuclei, electrostatic repulsion (which grows like the square of the number of protons) fights for fewer protons than neutrons, and does start to compete with the strong force (that only grows like the total number, not its square) and indeed you see the the ratio of protons to neutrons decreases as the mass of the nucleus grows : already 18 versus 20 for argon, 26 versus 30 for iron, 82 versus 126 for lead, 92 versus 146 for Uranium 238 (I am only mentioning even numbers of both protons and neutrons because with odd numbers there are, well, odd behaviours...)
Eventually, with increasing number of protons, electrostatic repulsion wins. Indeed, lead is the heaviest stable nucleus, all heavier ones are radioactive (well, bismuth is so weakly radioactive that one can consider it as stable too, for all practical purposes). And the lifetime of heavier and heavier nuclei is shortened by electrostatic repulsion. At some point, there will be no bound nucleus at all, making the creation of new artificial heavier elements a harder and harder task which will eventually fail to produce new ones at all.
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$\begingroup$ Increase in the amount of the binding energy from the reactant nuclei and the product nucleus during nuclear fusion and this causes the decrease the internal energy (potential and the positional energy ) which in turn causes a decrease in the rest mass of the nucleus. This "lost"mass converts to energy. Is this the case of the energy released during nuclear fusion? $\endgroup$ Dec 31, 2021 at 4:52
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Indeed the stability of a nucleus is dictated by some sort of balance between Coulomb repulsion and the residual strong nuclear force.
The binding energy is the (negative) energy at the minimum of a potential energy curve - where the gradient, which equates to some kind of net force, is zero. But the depth of that minimum and the equilibrium radius is determined by the shape of the potential energy curve, which isn't universal and depends on the number and type of nucleons. Since the fused nucleus is different to the nuclei that fused to produce it, then the potential energy curve is different too.
Why does the shape of the potential curve change? Because the Coulomb repulsion and strong nuclear force components have very different dependencies on inter-nucleon separation: the Coulomb force acts at long range, only between protons and is always repulsive; the strong nuclear force is short range so it essentially only acts between adjacent nucleons (protons or neutrons) and is attractive beyond about $10^{-15}$ m but strongly repulsive inside that (in broad-brush terms).
