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I have found from this link http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin.html that: Nuclei are made up of protons and neutron, but the mass of a nucleus is always less than the sum of the individual masses of the protons and neutrons which constitute it. The difference is a measure of the nuclear binding energy which holds the nucleus together. This binding energy can be calculated from the Einstein relationship: Nuclear binding energy $= Δmc^2$.

Now from the above it seems that the nuclear binding energy should be positive to compensate the energy difference between between the mass of the protons and neutron and the mass of a nucleus. But my question is how the binding energy can be positive if this binds the nucleons by attractive force?

I have also gone through this The "binding energy" of bonded particles adds mass? , but didn't get my question resolved.

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    $\begingroup$ Sign conventions. There are more than one, and you have to be careful when comparing two sources that they are using the same one. You'll notice that David Z is using a different one from hyperphysics in his answer below. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Feb 22, 2014 at 19:13

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Let's take the simple system of a deuterium nucleus, that is the bound state of a proton and neutron.

It's tempting to think of the binding energy as something that has to be added to a proton and neutron to glue them together into a deuteron, and therefore that the deuteron must weigh more than the proton and neutron because it has had something extra added to it. However this is the exact opposite of what actually happens.

Suppose we start with a proton and a neutron at a large separation, and we let them go and allow them to attract each other. As they move towards each other their potential energy decreases, so because of energy conservation their kinetic energy increases. They accelerate towards each other in just the same way you accelerate towards the Earth if you jump out of a window. This means that when the proton and neutron meet they are moving very rapidly towards each other, and they just flash past each other and coast back out to a large separation. This is not a bound state. To form a deuteron you have to take energy out of the system so that when the proton and neutron meet they are stationary with respect to each other. Then they can bond to each other to form a deuteron.

And this is the key point. To form a bound state you need to take an amount of energy out that is equal to the kinetic energy gained as the two particles collide. This energy is the binding energy, which is 2.2 MeV for a deuteron. If you take the energy 2.2 MeV and convert it to a mass using Einstein's equation $E = mc^2$ you get a mass of $3.97 \times 10^{-30}$ kg, and if you compare the mass of a deuteron with the mass of a proton + neutron you find the deuteron is indeed $3.97 \times 10^{-30}$ kg lighter.

Just for completeness let's look at this the other way round. Suppose we start with a deuteron and we want to separate it into a proton and neutron. Because the two particles in a deuteron attract each other we need to do work to pull them apart, which means we need to add energy to the system. To pull them completely apart we need to add an amount of energy equal to the binding energy of 2.2 MeV. That means our separated proton and neutron have 2.2 MeV more energy than the deuteron, so the total mass of the separated particles has increased by 2.2 MeV and they are $3.97 \times 10^{-30}$ kg heavier than the deuteron.

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  • $\begingroup$ Though my original question is resolved from the two answers. But I am not able to apply the same concept that you have told in the case of quarks forming a proton or neutron. m(u)=2.3 MeV , m(d)=4.8 MeV , then how come mass of protron is 931 MeV if you want to bring analogy with "nucleons forming nucleus"? $\endgroup$
    – user22180
    Commented Feb 23, 2014 at 8:28
  • $\begingroup$ @user22180: this also confused me when I first encountered it, and I'm still not sure I fully understand what's going on. The difference between a nucleus and a proton is that the quarks in a proton are confined and to separate them to infinity would take an infinite energy. Unlike a nucleus you cannot even theoretically start at large separation and consider what happens as you bring the quarks together. $\endgroup$
    – John Rennie
    Commented Feb 23, 2014 at 8:45
  • $\begingroup$ Why do the proton and neutron attract each other so strongly? $\endgroup$
    – J Thomas
    Commented Jul 1, 2022 at 11:59
  • $\begingroup$ @JohnRennie can you please explain then that why do some atoms and subatomic particles have positive binding energy while the others have a negative binding energy? $\endgroup$
    – Asad
    Commented May 30, 2023 at 0:33
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The binding energy doesn't compensate for the difference between the mass of the nucleons and that of the nucleus, it is that difference.

$$m_\text{protons and neutrons}c^2 + \text{binding energy} = m_\text{nucleus}c^2$$

Since the mass of the nucleus is less than the mass of the protons and neutrons, the binding energy must be negative.

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  • $\begingroup$ how does this actually work. For the mass $M$ of a system of non-interacting particles $M^2=\left(\sum_{i}E_i\right)^2 - \left(\sum_{i}p_i\right)^2$. So which part exactly changes? Is $\left(\sum_{i}E_i\right)^2$ getting smaller in the case of interaction? $\endgroup$
    – Alexander Cska
    Commented Jan 20, 2020 at 8:53
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  • Binding energy is the energy a nucleus doesn't have, not an energy it does have. The nucleus lost mass in its formation from its components ( and due to mass-energy equivalence) and therefore it equivalently lost this energy. The nucleus doesn't posses this "binding energy" so in fact its a terrible name, its an amount of energy it simply doses not possess.
  • The binding energy of the nucleus is intrinsically negative, in the same way that my total energy within the gravitational field is negative, so I am bound to that field. I am trapped.
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It's just a definition. The binding energy is defined as the energy it takes to take a system apart. The advantage is that you avoid writing minus signs all the time.

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