During fusion, for instance hydrogen fuse to form a helium nucleus which has lower mass than the sum of its constituents.... Why is it so?

I have read that energy is required to break apart the nucleus to its nuclei hence energy is required but here energy is also required to bring the two nuclei together so why is energy released ? Also i have read that helium nucleus has more negative potential energy than hydrogen nuclei ..... Why ?

Please answer in detail with examples and no (if possible) complex mathematical terms.....

  • 1
    $\begingroup$ Similar question to physics.stackexchange.com/q/667124/313823 $\endgroup$
    – RC_23
    Commented Jun 2, 2022 at 12:42
  • $\begingroup$ Nope this doesn't answer my question $\endgroup$ Commented Jun 2, 2022 at 13:05
  • 3
    $\begingroup$ You might consider this answer which discusses the necessity of including energy stored in fields when computing the total mass-energy of a system. $\endgroup$
    – rob
    Commented Jun 2, 2022 at 13:40

5 Answers 5


Binding energy, and (via $E=mc^2$) mass defect, are not unique to nuclear systems. The definition of a “bound system” is that its components cannot be separated from each other without some energy input.

For example, you are (I assume) gravitationally bound to the Earth. If you wanted to leave the Earth, you would have supply some amount of energy. A small amount of energy might get you a transition between two different bound states. For example, you might metabolize a little bit of your breakfast to drive a transition from “downstairs” to “upstairs.” If you wanted to become gravitationally unbound from the Earth, like the Juno spacecraft did en route to Jupiter, your breakfast is not enough energy to get you there; you would instead need to fill a giant tube with slow-release explosive, and then point the correct end of the tube in the correct direction.

If a object were to fall towards Earth from deep space, there are two major classes of outcomes. The unbound object might miss the Earth at its closest approach, in which case it follows some hyperbolic orbit and eventually escapes back to infinity with all of its initial kinetic energy. Or it might strike the Earth, and release its kinetic energy as heat, light, sound, dead dinosaurs, and so on. After such a collision, the fragments of the meteorite are gravitationally bound to the Earth.

You also have binding energies in electronic systems. For example, to separate the electron from the proton in a hydrogen atom, you have to supply at least 13.6 electron-volts of energy from somewhere. You can supply somewhat less energy than this and drive an internal transition in the hydrogen atom.

The only difference between gravitational binding energies, electronic binding energies, and nuclear binding energies is the scale of the binding energy relative to the masses of the particles involved. Most nuclear binding energies are roughly 0.1% of the mass of the nucleus, which means you have to care about the binding energy if you are doing mass spectroscopy with three significant figures. (The deuteron, mass 2 GeV, has binding energy 2 MeV; fission of uranium, mass 235 GeV, releases about 180 MeV.) Compare with the hydrogen atom, mass 1 GeV, which has electronic binding energy 13 eV: a part-per-billion correction, smaller than typical uncertainties in measurements which are sensitive to mass directly.

  • 1
    $\begingroup$ But how is this related to the cause of energy release in nuclear fusion? Could you explain @rob $\endgroup$ Commented Jun 3, 2022 at 15:39
  • 1
    $\begingroup$ I already mentioned deuterium formation, which is the simplest case of nuclear fusion. $\endgroup$
    – rob
    Commented Jun 3, 2022 at 15:42
  • $\begingroup$ Op is asking about how does mass defect is converted into energy. $\endgroup$ Commented Jun 3, 2022 at 16:32
  • 1
    $\begingroup$ @JohnDoty It’s not the relativistic mass. A better statement would have been $m_\text{bound}= \gamma_1 m_1 + \gamma_2 m_2 + \cdots + (U_\text{grav}+U_\text{EM}+U_\text{nuclear})/c^2$, with this argument for assigning negative values to the interaction energies. But that seemed a little over-the-top for the pedagogical level of this asker. $\endgroup$
    – rob
    Commented Jun 3, 2022 at 17:27
  • 1
    $\begingroup$ @Davius See this answer for an argument that the negative energy stored in an attractive field is exactly as real as the positive energy associated with motion. In general relativity, it is made quite clear that all contributions to the energy density contribute to gravitation. $\endgroup$
    – rob
    Commented Jul 20, 2022 at 14:31

During fusion, you have to put energy into the constituent nuclei in order to surmount the coulomb repulsion of the positively charged nuclei. You could imagine that this energy you put into the system (in the form of photons, which are the exchange particles of the electromagnetic interaction) first increases the mass. It is kind of like an activation energy.

Only as soon as the nuclei have been brought together close enough so that the short ranged strong interaction starts attracting them, you will start gaining back energy. You can imagine that this decreases mass again (according to Einstein's mass energy relation).

Depending on how much you have to put in and how much you gain back, the total energy balance will be positive or negative, i.e. you have a total gain or a total loss of mass-energy. The analysis of the involved energies is the subject of nuclear models. (e.g. the liquid drop model)

The described energy barrier is an obstacle in both directions. So you (temporarily) need some activation energy, no matter if you want to fuse or you want to split nuclei.

By the way, this is hardly any different from situations in chemistry. For the synthesis of ammonia from nitrogen and hydrogen you formally need 2253 kJ/mol to split them into the atoms, and you gain 2346 kJ/mol during the formation of ammonia, resulting in an energy gain of a miserable 93 kJ/mol. This extremely high activation energy is the reason why ammonia does not form spontaneously at ambient pressure, and why a nitrogen/hydrogen mixture is not particularly useful as a combustible. Other than that, the energies involved in chemistry are way too low to be detectable as a mass defect (although it is there in principle).

  • $\begingroup$ But @oliver, won't the energy released be equal to the activation energy and thus there is no overall gain???? Right? $\endgroup$ Commented Jun 4, 2022 at 4:58
  • $\begingroup$ If the final state has higher energy (i.e. the reaction requires energy overall), the activation energy is (by definition) the energy that is finally released from the activated state. By contrast, if the initial state has higher energy (i.e. the reaction causes overall energy gain), the activation energy is the energy that is initially needed. There could also be the case that there is no overall gain (if the intial and final state are at the same energy), but that would just be a coincidence. $\endgroup$
    – oliver
    Commented Jun 4, 2022 at 5:50

Another answer here:

Nuclear binding energy is basically the energy required to dismantle a nucleus into free unbound neutrons and protons. It is the energy equivalent of the mass defect, the difference between the mass number of a nucleus and its measured mass. Nuclear binding energy is derived from the residual strong force or nuclear force which again is mediated by 3 types of mesons.

Nuclear binding energy can be determined once the mass defect is calculated, usually by converting mass to energy by applying $E = mc²$. When this energy is calculated which is of joules for a nucleus, you can scale it into per-mole quantities and per-nucleon. You need to multiply by Avogadro’s number to convert into joules/mole and divide by the number of nucleons to convert to joules per nucleon.

Nuclear binding energy is also applied to situations where the nucleus splits into fragments that consist of more than one nucleon wherein, the binding energies of the fragments can be either negative or positive based on the position of the parent nucleus on the nuclear binding energy curve. When heavy nuclei split or if the new binding energy is known when the light nuclei fuse, either of these processes results in the liberation of binding energy.

The nuclear binding energy holds a significant difference between the nucleus actual mass and its expected mass depending on the sum of the masses of isolated components.

Since energy and mass are related based on the following equation:

$E = mc²$

Where c is the speed of light. In nuclei, the binding energy is so high that it holds a considerable amount of mass.

The actual mass is less than the sum of individual masses of the constituent neutrons and protons in every situation because energy is ejected when the nucleus is created. This energy consists of mass which is ejected from the total mass of the original components and called a mass defect. This mass is missing in the final nucleus and describes the energy liberated when the nucleus is made.

Mass defect is determined as the difference between the atomic mass observed ($m_0$) and expected by the combined masses of its protons ($m_p$, every proton has a mass of 1.00728 AMU) and neutrons ($m_n$, 1.00867 AMU).

$M_d = (m_n + m_p) - m_0$


Generally in nuclear fusion more nucleons take part in it than nuclear fission. So, more mass is converted into energy in nuclear fusion than fission.

Hope this helps.

  • $\begingroup$ Ummm... It is helped to an extent but can you tell why does mass defect exactly occur??? What causes the mass of the final nucleus to be smaller then the initial Nuclei???? Why is this related to binding energy???? Please help ... $\endgroup$ Commented Jun 3, 2022 at 9:24
  • 2
    $\begingroup$ Mass and energy are the same thing of different forms. When nuclear fusion occurs, the anti particles of nucleons brake and build up a thousands of photons. The particles brake due to the energy required to fuse the nucleis. Photons are the energy particles of electromagnetic waves. The energy produced by the photons is called the binding energy. Because, some particles of the initial nuclei are converted into energy, the mass of the final nuclei is less than the initial nuclei. $\endgroup$ Commented Jun 3, 2022 at 14:47
  • $\begingroup$ @Nuclearfusion Have you any doubt left in your mind? $\endgroup$ Commented Jun 3, 2022 at 16:35
  • $\begingroup$ I still don't understand how does so much energy gets released during nuclear fusion? How is it related to mass defect and binding energy? Does negative potential energy play any role in it? Why is the helium nucleus at a lower energy state than hydrogen? $\endgroup$ Commented Jun 4, 2022 at 4:56
  • 1
    $\begingroup$ It means that mass and energy are two faces of a coin. A body's mass is its mass + energy contained in it. We can convert mass to energy by multiplying the speed. Of light squared $\endgroup$ Commented Jun 4, 2022 at 7:40

With increased binding energy, the rest mass of the products of the reaction is less than the rest mass of the reactants of the reaction, and due to conservation of energy the difference appears as an increase in the kinetic energy of the products compared to the reactants. The decrease in rest mass is due the nucleons in the products being more tightly bound (in lower energy states inside the nucleus) compared to the nucleons in the reactants. The energy "released" as kinetic energy is given by $E = mc^2$. Fission of heavy nuclei as well as fusion of light nuclei result in a decrease in rest mass. Such reactions are exothermic; reactions that increase rest mass are endothermic.

A similar change in rest mass occurs in chemical reactions, but the magnitude is very much less than for nuclear reactions. This change in energy for a chemical reaction is called the energy of formation or the enthalpy of formation in classical thermodynamics; we now know it is a change in rest mass.

Update based on OP follow-up comment/question

The question is “why does binding energy cause mass defect?”.

For a given nucleus, its total binding energy is the energy required to separate the nucleus into its individual nucleons. Energy is required to separate the nucleons because the strong nuclear force inside the nucleus attract the nucleons to each other. A crude classical physics analogy is two masses connected by an elastic spring with the spring not stretched initially; energy is required to pull the masses apart to overcome the force of the stretched spring tending to pull the masses toward one another.

Binding energy is a measure of how tightly the nucleons (protons and neutrons) of a nucleus are bound together. If the binding energy is increased the nucleons are more tightly bound, or in other words the nucleus is in a less energetic internal state; the nucleus has less internal energy. A graph of the average binding energy per nucleon versus the mass number for various nuclei is called a curve of binging energy; a figure below is this curve. The reason for the difference in the average binding energy per nucleon as a function of the nucleus mass number and atomic number has been addressed using models by Brueckner et. al. A simpler model is the liquid-drop model that empirically explains the curve based on: the short range strong nuclear force, an assumption that the strong nuclear force for one nucleon saturates (does not extend) beyond about four adjacent nucleons, and the Coulombic repulsion of protons. Details of these models are complex and are available online and in standard nuclear physics textbooks.

The curve of binding energy indicates that either the fusion of light nuclei or the fission of a heavy nucleus due to a nuclear reaction can result in an increase in the average binging energy per nucleon. An increase in binding energy is a decrease in internal energy. Applying Einstein’s $E = mc^2$ where $E$ is internal energy, $m$ is rest mass, and $c$ the speed of light, the decrease in internal energy is a decrease in rest mass.

Here is a simple example of a change in internal energy. First let’s use a macroscopic viewpoint. Consider a solid that is heated; using the first law of thermodynamics the increase in internal energy of the solid is equal to the heat added. The additional insight due to Einstein is that this increase in internal energy is an increase in the rest mass of the solid. Now, let’s use a microscopic viewpoint. For the solid that is heated, the temperature increases due to the increased kinetic energy of the atoms in the solid. That is, looking inside the solid what we identified as increase in internal energy using the macro viewpoint- not looking inside the solid- is an increase in kinetic energy of the atoms using the micro viewpoint.

Clarification added in italics

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.

enter image description here

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. There are some other chat links there. Since this thread somehow got to 75 comments, I’m going to pause further comments here just to make the point that extended discussions don’t belong in comments. $\endgroup$
    – rob
    Commented Jul 20, 2022 at 20:26

I personally think that this is due to the nucleons in the product nucleus being more tightly bound with higher nuclear energy experienced by each nucleon.

  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Jun 12, 2022 at 9:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.