# Why is energy released when nucleons come together to form a nucleus?

Binding energy is defined as the energy required to split a given nucleus into it's individual protons and neutrons. But why when nucleons are combined they release energy? What is the origin of this energy? If the mass is converting to energy, then what exactly causes/drives this mass to convert to energy. Kindly explain in detail.

The binding energy is a negative number for stable nucleons, which means that you have to pay energy to split the nucleon into separate protons and neutrons.

For example let's consider Helium. When the two protons and neutrons come close together in space strong interaction starts to do its job and creates a bound state of the four particles. This bound state is the Helium nucleon. Of course, in reality the creation on nucleons is more complicated, but for now this simple approach is enough. When this bound state is created some energy gets "released". This "released" energy is not necessarily released in terms of photons, it can become the kinetic energy of the new particle (if momentum conservation does not exclude this). Now if you want to split this Helium nucleon you have to pay energy to separate the particles from each other. The energy you have to pay to separate the particles is exactly the absolute value(!) of the binding energy.

The source of this binding energy is of course in the interaction between the particles. If two strongly interacting particles gets near each other, strong interaction starts to kick in and it results in an attractive force between the particles. This attractive force is so strong that it wins over the Coulomb repulsion of protons for example. So essentially in first approximation we can say that when you want to separate the particles, you have to work against the attractive force of strong interaction.

In reality the whole story is of course much more complicated. The binding energy can be calculated by the so called Semi-empirical mass formula (see Wiki) which gives the binding energy of many nucleon to a high precision.

Finally, mass $$\textbf{does not}$$ convert into energy. It is not driven by anything. The relation $$E=m c^2$$ merely states that the energy of a particle is related to its mass, but this mass is not the $$\textit{rest mass}$$ of the particle, but rather the kinetic mass, which depends on the velocity of the particle. The difference between the total mass of the initial particles and the mass of the nucleon they form is just a statement that they have different energies. In case of Helium the mass difference tells you that it is energetically favorable for the protons and neutrons to form a Helium nucleon and the change of the total mass is related to the binding energy by Einstein's expression

$$$$E_{\mathrm{binding}} = \Delta m c^2$$$$ where $$\Delta m$$ is the mass difference between the final and initial total masses.

• When this bound state is created some energy gets "released''. My main question was that what causes this and why this happens? Why is the helium nucleus at a lower energy state? An excellent answer otherwise 👏. Kindly elaborate on the above questions. Dec 24, 2021 at 8:43
• @Hark zltn.guba just gave you the answer : "strong interaction" is such that the helium nucleus has a lower anergy than two free protons and two free neutrons. "Why" ? You might just as well ask why two water molecules have lower energy than one oxygen molecule and two hydrogen molecules. Or any other chemical or nuclear reaction. Dec 27, 2021 at 21:45
• In a reaction where binding energy is increased, rest mass is converted into kinetic energy; the total energy and total relativistic mass do not change. Please see my answer Dec 27, 2021 at 22:02
• Mass is not converted to energy. Saying that rest mass is converted into kinetic energy is equivalent to saying that potential energy is converted into kinetic energy. Dec 29, 2021 at 13:18

Rest mass is converted to kinetic energy when the binding energy is increased in a reaction (nuclear or chemical). The total energy (kinetic plus rest mass) does not change, hence the relativistic mass (a term longer in current use) does not change. Fission and fusion reactions increase the binding energy and therefore result in large increases in kinetic energy. For the details please see What is the relationship between binding energy, energy released, and mass defect? on this exchange. Also, https://physics.stackexchange.com/questions/598694/can-mass-energy-equivalence-be-used-to-measure-absolute-internal-energy… may help.

You asked @zltn.guba When this bound state is created some energy gets "released''. My main question was that what causes this and why this happens? Here is simple analogy. When you cool a body at rest its internal energy decreases. Since energy is related to mass, here $$E = m_0c^2$$, where $$m_0$$ is the rest mass, the rest mass of the body decreases. Similarly, when a reaction occurs in which the products of the reaction are in a less energetic state than the reactants, the kinetic energy of the products is increased over that of the reactants. By a less energetic state I mean the electrons (chemical reaction) or nucleons (nuclear reaction) are more tightly bound so the atoms or nuclei of the products have less internal energy than the internal energy of the reactants, and this decrease in internal energy is a decrease in rest mass. In a reaction the total energy $$E$$ of the products equals that of the reactants. For the reaction, energy is equivalent to mass through $$E = mc^2 = T + m_0c^2$$, where $$m$$ is the relativistic mass and $$T$$ is the relativistic kinetic energy; for an exothermic reaction $$T$$ of the products is greater than $$T$$ of the reactants and $$m_0$$ of the products is less than $$m_0$$ of the reactants. The decrease in rest mass is a decrease in internal energy. The decrease in rest mass is a result of the electrons or nucleons of the products being in a less energetic state (more tightly bound) than the reactants, and energy would be required to restore the internal energy of the products to the internal energy of the reactants. Binding energy is a measure of how tightly bound are the electrons/nucleons and an increase in binding energy is a decrease in rest mass.

Older texts use $$E = mc^2$$ where $$m$$ is the relativistic mass as distinct from rest mass $$m_o$$, so in the reaction the total energy and the relativistic mass do not change (but rest mass is converted to kinetic energy). Now the term relativistic mass is not widely used. In current use, $$m$$ means rest mass and the relationship is written $$E = T + mc^2$$ where $$m$$ means rest mass and $$T$$ is the kinetic energy which is in general the relativistic kinetic energy. In a reaction the total energy of the reactants and products is constant, but for an exothermic reaction $$T$$ of the products is greater than $$T$$ of the reactants and $$m$$ (rest mass) of the products is less than $$m$$ of the reactants. See the related questions on this exchange noted in the first paragraph, and/or see a good basic nuclear physics test such as Meyerhof, Elements of Nuclear Physics.

In any exothermic reaction (chemical or nuclear), rest mass is converted to kinetic energy. But the decrease in rest mass is much smaller for a chemical than for a nuclear reaction. Conversely, in any endothermic reaction kinetic energy is converted to rest mass.

• Do you mean that the reactant nuclei are less tightly bound than the product nucleus and hence the "internal energy" of the product nucleus is less than the reacting nuclei and hence the resultant less energy is released in the form of energy of nucleus fusion? Dec 28, 2021 at 7:04
• And here what do you exactly mean by internal energy? Dec 28, 2021 at 7:04
• Regarding your first comment, yes that is the idea. You reduce the internal energy in an exothermic reaction and since energy is mass you reduce the rest mass, and that reduction appears as increased kinetic energy. See a nuclear physics text such as the one I mentioned for the complete discussion of the energy balance in a reaction. Dec 28, 2021 at 13:01
• Regarding your second comment. Internal energy is the total energy within system of molecules, or the sum of the molecular kinetic energy and potential energy in a substance. You choose to not look internally into the system but look at it from the outside and account for all the energy within as internal to the system. The macroscopic kinetic and potential energies don’t matter for internal energy – if you move the whole closed system or change its gravitational potential energy, the internal energy remains the same. See a thermodynamics text, such as one by Sonntag and Van Wylen. Dec 28, 2021 at 13:03
• But why does being tightly bound change the internal energy? Dec 28, 2021 at 15:45

Consider the formation of a helium nucleus from deuterium ions.

In order for this reaction to occur, the electrostatic repulsion between the two electrically positive protons has to be overcome. This is a large amount of energy, and to overcome it, the deuterium nuclei have to be moving very fast towards each other.

Now let's say all the conditions are perfect and the fusion reaction takes place. The resulting helium atom is practically at rest compared to the two original deuteriums. All of that kinetic energy has to go somewhere. It will be released through various forms of radiation, notably gamma and/or x-rays and energetic neutrons (there is a potential reaction that does not release neutrons, but its rare).

So that's where the energy comes from - you provided it in order for the reaction to occur in the first place. It is basically the leftover bits when the reaction is complete in largely the same basic way that chemical reactions can produce heat when you re-arrange electrons.

• The binding energy of helium, due to the so-called "strong interaction", or "strong force" is orders of magnitude larger than the kinetic energy needed to overcome the electrostatic repulsion. Dec 27, 2021 at 21:39
• Is this hypothesis scientifically proven ? Dec 28, 2021 at 8:09