Is this theory correct? I got this theory from a nuclear fusion expert in stack exchange and he told me to confirm it here... The theory:

When two nuclei come closer enough (due to external energy provided) for the nuclear strong force to act, then the kinetic energy of the nuclei is greatly increased and since this is not due to external forces Consider two nucleons close enough so the strong nuclear force attracts them to each other. The nucleons each gain kinetic energy (KE) as the force accelerates them toward each other, and they eventually are BOUND together as two nucleons. Viewing each nucleon separately, it is the work from the force that increases the KE of a nucleon. Viewing the TWO NUCLEONS TOGETHER AS A SYSTEM, the force is internal, and there is no work or heat added to the system so the increase in KE must be a decrease in internal energy. The strong nuclear force caused the decrease in the internal energy. For fusion there is external energy required to push the charged nuclei sufficiently close where the strong nuclear force then comes into play; then the strong nuclear attractive force causes more energy release than the energy to push the nuclei close.

  • $\begingroup$ The nucleons each gain kinetic energy (KE) as the force accelerates them toward each other I would not say it this way. The nucleons gain potential energy (electrostatic) from being closer together, then also lower their potential energy (strong nuclear) when close enough. The decrease in strong nuclear PE is translated to increase in KE of the products (neutrons, beta, neutrinos, etc) $\endgroup$
    – RC_23
    Commented Jul 26, 2022 at 15:39
  • $\begingroup$ So what do you hold the cause for the energy release during nuclear fusion? $\endgroup$ Commented Jul 28, 2022 at 7:25
  • $\begingroup$ @RC_23 , Why is a tremendous amount of energy released and what is the relation of binding energy here ? $\endgroup$ Commented Jul 28, 2022 at 7:26
  • $\begingroup$ Take protons fusing into deuterium for example. Deuterium is a much lower energy configuration for the system than being two free protons. That difference in energy is carried away by a position and neutrino as their kinetic energy. No different than hydrogen and oxygen reacting to form water, releasing energy. $\endgroup$
    – RC_23
    Commented Jul 29, 2022 at 2:33
  • $\begingroup$ @RC_23 , That much is known to me , what I am actually asking is why does the deuterium have a lower energy state and a detailed explanation or an answer at best. Thanks $\endgroup$ Commented Jul 29, 2022 at 10:41

2 Answers 2


Part of what I actually provided in answer to your earlier question at Why does binding energy cause mass defect? is below. What you say above does not address the major increase in kinetic energy which is due to decrease in rest mass (increase in binding energy). The above just addresses the part of the process before the actual nuclear reaction, and that is not responsible for most of the increase in kinetic energy. What you provide above is probably based on our lengthy follow-up discussions where you were asking about how the nuclei come together to fuse in the first place. Perhaps other responses can help explain this better.

Part of earlier response follows

Now, let’s consider a nuclear fusion reaction. Let the system be two nuclei that undergo an exothermic reaction: $ a + b -> c$ where $a$ and $b$ are the reactants and $c$ is the product of the reaction. That is, we “move up the curve of binding energy” in this reaction; the reaction increases the average binding energy per nucleon. We observe that the total kinetic energy of $c$ is greater than the total kinetic energy $a$ and $b$. From the macro viewpoint, no heat or work was added to the system (defined here as all the nuclei inside a system boundary) so according to thermodynamics the total internal energy of the system does not change. However, the kinetic energy of the products is greater than that of the reactants, so inside the system the internal energy of the separate nuclei decreases to account for this increased kinetic energy. Before nuclear reactions were discovered and before Einstein, classical thermodynamics observed exothermic chemical reactions and attributed the required decrease in internal energy of the interacting constituents to “heat (or enthalpy) of formation”; for example, see one of the texts on ‘thermodynamics by Sonntag and Van Wylen. We now know that this classical “heat of formation” used to explain the decrease in internal energy of the constituents is a decrease in rest mass. So, for both a chemical and a nuclear exothermic reaction, the increase in kinetic energy of the products compared to the reactants is accompanied by a decrease in rest mass of the products compared to the reactants. For our fusion example, $c$ has less rest mass than the sum of the rest masses of $a$ and $b$.

For an exothermic nuclear reaction, once the reaction occurs the internal forces (nuclear and Coulombic) reconfigure the nucleons in an overall lower energy state. But, some initial kinetic energy of $a$ and $b$ is required to fuse since we are fusing charged particles and the Coulombic repulsion of the two positively charged nuclei must be overcome before $a$ and $b$ are sufficiently close for the short range strong nuclear force to become dominant. It is sort of like a ball on a high shelf would like to move to the ground due to gravity but some initial energy is required to bump the ball off the shelf.

For a fission reaction using a charge-neutral neutron, for some target nuclei- $235 U$ for example- a very low energy neutron can cause fission. But, you have to cause the free neutron to be released (e.g., from a previous fission) to be available to cause fission

The micro viewpoint for fusion (and fission) is addressed by details in models of nuclear binding mentioned earlier. These models propose explanations for why the binding energy of $c$ is greater than the sum of the binding energies of $a$ and $b$; that is, the nucleons are reconfigured in the reaction to be in overall less energetic states inside the nucleus, reflected on the macro scale as a decrease in internal energy and therefore a decrease in rest mass.

  • $\begingroup$ I couldn't understand the relation of binding energy here and what role it plays here and as to why the KE of the resultant system increases .... $\endgroup$ Commented Jul 27, 2022 at 12:06
  • $\begingroup$ I hope others can offer you a better explanation. $\endgroup$
    – John Darby
    Commented Jul 27, 2022 at 14:04
  • $\begingroup$ I Dont think there is anybody active here..... :( $\endgroup$ Commented Jul 30, 2022 at 12:11
  • $\begingroup$ @JohnDarby, can you help me further in understanding this concept ...? $\endgroup$ Commented Aug 4, 2022 at 11:30
  • $\begingroup$ Am the same person but different account , from the lengthy discussions one..... $\endgroup$ Commented Aug 4, 2022 at 11:30

At first it's important to know what actually nuclear fission and nuclear fusion is.

Nuclear fission is a reaction in which the nucleus of an atom splits into two or more smaller nuclei.

Nuclear fusion is a reaction in which two or more atomic nuclei are combined to form one or more different nuclei or subatomic particles (protons or neutrons).

We need to provide an amount of energy to cause nuclear fission. The minimum amount of energy to completely disassemble an atomic nucleus to its nucleons is called nuclear binding energy. As energy is provided to the nuclei in this process, the mass of each nucleons increases (because mass and energy are equivalent). The relation of binding energy and the total increased mass of all nucleons can be given by Einstein's mass-energy equivalence equation: $$E = ∆mc^{2}$$ Where $E$ is binding energy, $∆m$ is the total increased mass and $c$ is speed of light.

(N. B. The above isn't true for unstable nucleus like the nucleus of Uranium-235. Mass is decreased during the process of fission in these cases due to their different structure of nucleus)

So, when we give the required energy to two or more nucleus to come closer so that the nuclear force comes into play and bound them together, their increased mass or energy must be lower to hold them together. This is why the product nucleus comes into a lower internal energy state and the decrease in internal lower energy state is translated to the external energy including kinetic energy of its products (the fused nucleus, neutrons, beta neutrons etc.) . The amount of released energy is the same of their binding energy. Because energy is released there is a decrease in the rest mass of the product. By using the decreased mass the released energy can be calculated by $E = ∆mc^{2}$ where $∆m$ is decreased mass.

Now why is the kinetic energy of the product nuclei is greater than the reactant nuclei?

The classical kinetic energy equation is $$E_K = \frac {1}{2}mv^{2}$$ Now the mass is decreased in the nuclear fusion but acc. to mass-energy equivalence a huge amount of energy produced for which the velocity of the product nuclei is greatly increased. So the overall kinetic energy increases.

  • $\begingroup$ But why the increased mass or energy must be lowered to hold them together $\endgroup$ Commented Oct 15, 2022 at 8:04
  • $\begingroup$ Suppose there are gas inside a container. According to thermodynamics the entropy of the gas must be lower for the molecules come closer and then the molecules form liquid and then solid and now the molecules have a lower internal energy state. But the molecules have released energy to its nearby environment and also the intermolecular attraction is increased between the molecules. Just similar incident happens in the case of fusion. $\endgroup$ Commented Oct 15, 2022 at 10:06
  • $\begingroup$ Two nucleus are separate because they have a lot of energy and when they comes closer due to the nuclear force they are bound together and meanwhile their internal energy decreases and external energy increases. Then the increased external energy radiated away. $\endgroup$ Commented Oct 15, 2022 at 11:02
  • $\begingroup$ Because a lot of energy was needed to separate the nucleus into its nucleons, a lot of energy released when the process is reversed. $\endgroup$ Commented Oct 15, 2022 at 11:04

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