I have read from my textbook about binding energy but it indicates two completely different ideas.I am listing them below:

Definition (1):

"Binding Energy : An atomic nucleus is a stable structure.Inside it,the protons and neutrons are bond together by means of strong attractive nuclear forces.Thus a definite amount o work is required to be done to breakup the nucleus into its constituent particles and to place them at infinite distance from one another.This work gives a measure of the binding energy of the nucleus."

This definition gives the idea that binding energy is needed to overcome the Nuclear force between nucleons.

Definition (2):

"It is seen that the mass of a stable nucleus is always less than the sum of the masses of the constituent protons and neutrons in their free state.This mass difference is called 'mass defect which accounts for the $\Delta E_b$' energy released when a certain number of neutrons and protons are brought together to form a nucleus of a certain charge and mass.

This gives the idea that binding energy is given to nucleus then this energy reflects as extra mass of protons and nucleus.

So what actually is binding energy.Another definition tells it is the Energy required to break up nucleus. This suggests Binding energy should account for both mass defect and Nuclear force potential energy.Then why the formula of binding energy only accounts for mass defect ?

i.e.$E_b=Δm\times C^2$

And as far as I know from graph (below) Nuclear potential energy is not negligible.enter image description here

Edit From the given answers till now,what i get is idea that binding energy is required to break up nucleus into constituent particles.enter image description here So i applied conservation of energy to find this binding energy. Equation that i get is $$E_b+(m-Δm)*C²+NFP=m*C²$$ (NFP Is net nuclear force potential among all nucleons & m is mass of individual protons and neutrons when they are in free state and Δm is mass defect) rearranging we get


And as NFP is negative and very large in magnitude(ie:- NFP "-100MeV" was between two nucleons[data from graph] so for a smaller nucleus also NFP should be" $-100*(N choose 2)$) Please help in clarifying this Edit.

Please help me make out what actually binding energy is.

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    $\begingroup$ Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. $\endgroup$
    – ACuriousMind
    Apr 17 '20 at 15:06
  • $\begingroup$ As your final diagram shows, (energy of the nucleus) + (binding energy) = (energy of the unbound nucleons). How is that not clear? $\endgroup$
    – PM 2Ring
    Apr 18 '20 at 5:45

As is stated in one of the texts above you start with a nucleus and then split the nucleus into individual neutrons and protons (the constituent parts of the nucleus).
The work done to split up the nucleus into its constituent parts is the binding energy of the nucleus.
In the reverse process if the individual neutrons and protons are brought together and form the nucleus the amount of energy released in that process is equal to the binding energy of the nucleus.

It is found that the mass of the nucleus is smaller that the total mass of the individual neutrons and protons which make up the nucleus.
The difference between these two masses is called the mass defect, ie the nucleus is deficient in some mass as compared with the sum of the masses of the individual particles which make up the nucleus.

If the binding energy is $E_{\rm b}$ nad the mass defect is $\Delta m$ then the two are related vis Einstein's equation $E_{\rm b} = \Delta m \,c^2$ where $c$ is the speed of light.

So to break up a nucleus into its constituent parts the minimum amount of energy input into the nucleus is the binding energy and at the end of the process the process the total mass of the constituent parts increase by an amount equal to the mass defect.


I think you know that two protons will repel each other, according to Coulomb's law. And the coulomb force of repulsion between two protons in a nucleus is approximately $250 N$ if they are separated by a distance of one fermi, which is very large considering the small mass of the proton.

So, as you might well know, Strong Nuclear force, for a short-range, manages to keep the nucleus together. Now, if you want to break the nucleus into its constituents, you need to do a certain amount of work- which is the binding energy.

Now, think about the nucleus before it was formed. All its constituents would have been separate entities. If you wished to make them into a single body, for the sake of spontaneity (my use of this word may not be entirely correct), the end product must possess lower energy than the sum of energies of independent entities.

This is rather counter-intuitive if you don't have an idea about Strong Nuclear forces, as there would be an increase in the energy of the system, by Coulomb's law.

But what actually happens is that the mass of the nucleus is less than the mass of its constituents taken one at a time. So, from Prof. Einstein's mass-energy equivalence relation, we can say that the energy contained in this mass was released during the formation of the Nucleus and this difference in mass is called the mass defect.

So, in order to break the Nucleus, you need to remove (or rather give back) what was holding them together-so, quite naturally, the binding energy is the same as the energy released from mass defect


As a response to your edited question, I'm including an analogy. Let us assume that you have a ball initially placed on the top of a shelf of height $h$. Now, if it slips and falls down, it will lose its internal energy in the form of kinetic energy and you will have the relation $$mgh+\frac{1}{2}mv^2=mgh_{2}$$ (essentially a special case of third equation of motion) where $h_2$ is its present height. If you want to lift it back to the position it initially was at, you need to supply energy equal to the energy lost as kinetic energy. $$E=\frac{1}{2}mv^2$$ You might have noticed that we don't consider $GFP$ (Gravitational Force Potential) in the equation for energy that has to be supplied.

In a similar fashion, if the initial energy of the nucleus (before formation) is $E$, and the particles lose a mass $\Delta m$, then the energy of the nucleus will be $$E_{nucleus}=E_i-\Delta mc^2$$

Now to bring it back to the initial stage (The $E_i$ already includes the $NFP$ you're talking about, in the same way the $GFP$ is included in the analogy above ) Now, to take the nucleus back to its initial position, (with energy $E_i$) you need to supply it the binding energy. $$E_{nucleus}+E_b=E_i$$ rearranging, $$E_{nucleus}=E_i-E_b$$

Comparing this equation with that of formation of the nucleus, we have $$E_b=\Delta mc^2$$

  • $\begingroup$ So are you saying that nucleons are bound together by binding energy due to mass defect and not because of nuclear force?? $\endgroup$
    – sarthak
    Apr 17 '20 at 16:00
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    $\begingroup$ So, they are held together because of the forces- that is right. They prefer that because they can be at lower energy. $\endgroup$
    – Elendil
    Apr 17 '20 at 16:08
  • $\begingroup$ I understood your answer but i did not got the point that i asked why we are neglecting nuclear force potential in binding energy.Please consider explaining it. $\endgroup$
    – sarthak
    Apr 17 '20 at 18:05
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    $\begingroup$ @sarthak, ok, as you wish- let $E_n=E_i-\Delta mc^2 + NFP.$ Then, the nucleus is gaining NFP while being formed. This means that it music lose the NFP (conservative field) when it is broken into nucleons. So, the next equation will read, $E_n+ E_b-NFP=E_i$. So, you will get $E_b=∆mc^2$ $\endgroup$
    – Elendil
    Apr 19 '20 at 2:24
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    $\begingroup$ We can ignore NFP, not because it is small, but because it is a state dependant function. During formation, it decreases (-ve) during destruction, it increases, and the net would cancel out $\endgroup$
    – Elendil
    Apr 19 '20 at 2:29

Binding energy is the minimum energy required to bring the system out of stable equilibrium. So in your context it is the energy required to break the nucleus into its constituents. It’s the value of the minima in the graph.

The reason why this value only accounts for mass defect is because once you have provided energy equal to the mass defect, the system has enough energy to break into its individual components. If any less is provided, the total energy of the system isn’t enough to break it because the total energy is still less than the sum of the mass energy of the individual components.


enter image description here Basic Defination of Binding energy

The Binding energy of a nucleus is defined as the energy required to break up a nucleus into constituent Protons and Neutrons and to such a distance that they may not interact with each other.

Cause of mass defect Mass defect is diffrent for diffrent element.This suggests us that there is no such conclusion can be made that a proton or neutron will lose a particular mass.The mass defect created purely depends on structure of nuclei and the structure of nuclei determines drop in nuclear force potential this gives us a hint that mass defect some how depends on drop in Nuclear force potential.So conclusion is that Mass defect is some how correlated to loss in nuclear force potentialHence we can write that $$ΔmC²=0-NFP..........(1)$$ (NFP is nuclear force potential when system of nucleons are bound as nucleon and has negative value )

Formula for Binding energy Now applying energy conservation we get


(We should not consider Nuclear Force potential as this same thing is accounted by mass defect(from above conclusion) .Hence considering NFP is like double counting) $$E_b=ΔmC².............(2)$$ This equation suggests us the defination 2 given that is binding energy is given to overcome the mass defect. From equation (1) & (2) We can say $$E_b=-NFP$$ Which implies $$E_b+NFP=0$$ As 0 is final Nuclear force potential (ie NFP_f) $$E_b=NFP_f-NFP=ΔNFP$$ $$E_b=ΔNFP..........(3)$$ This defination suggests us that Binding energy is given to overcome Nuclear force of attraction.In this case Binding energy is the measure of how strong is the nuclear force of attraction.

Both of defination are correct.They are just like opposite faces of a coin. (For better accuracy we can include EFP(Electrostatic force potential))

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    $\begingroup$ It is nice that you have attempted to reason south things, but I feel that you have read your Binding energy value wrong. To my knowledge, binding energy is in itself determined by mass defect measurement as opposed to actual application of energy to break a nucleus (which is a lot less accurate, for energy is not used entirely for the purpose of breaking the nucleus) $\endgroup$
    – Elendil
    Apr 18 '20 at 13:40
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    $\begingroup$ Well,yes. I'm happy that you realized that the Binding Energy takes care of NFP. Anyway, upvotes and reputation are not what matters. When I get some free time (i.e. when I don't have to teach or learn (both)) I spend it here. The drive in PhysicsSE is teaching. $\endgroup$
    – Elendil
    Apr 19 '20 at 15:05
  • $\begingroup$ @Krishna Are you a teacher?? Atlast thank you for your guidance $\endgroup$
    – sarthak
    Apr 19 '20 at 15:57
  • 1
    $\begingroup$ No, I'm not anything close to being a teacher. But I've been holding classes for students-like tution- for free. It keeps me busy, nevertheless. Not much, I might be just slightly older than you (an year, probably) $\endgroup$
    – Elendil
    Apr 19 '20 at 18:29

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