# Do nucleons have a lower energy state when bound in a nucleus?

• I
Summary:
I thought that nucleons were usually stable in a nucleus because they were in a lower energy state than when free as individuals.
But when I look at the definition of binding energy that doesnt make seem to make sense. It looks as though they had more energy when they were together and when they were separated that energy turned to mass (the mass defect)? Am I looking at this right?

I also dont understand this explanation on wikipedia..I would have thought energy could decrease and mass could gain if energy was converted to mass?
https://en.wikipedia.org/wiki/Binding_energy
"A bound system is typically at a lower energy level than its unbound constituents. According to relativity theory, a ΔE decrease in the total energy of a system is accompanied by a decrease ΔM in the total mass, where ΔM⋅c2=ΔE."

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BvU
Homework Helper
Summary:: I thought that nucleons were usually stable in a nucleus because they were in a lower energy state than when free as individuals.
Correct
But when I look at the definition of binding energy that doesnt make seem to make sense. It looks as though they had more energy when they were together and when they were separated that energy turned to mass (the mass defect)? Am I looking at this right?
No. Together (for small nuclei) is a lower energy state (ergo: lower mass). Energy is needed to separate the nucleons.
I also dont understand this explanation on wikipedia..I would have thought energy could decrease and mass could gain if energy was converted to mass?
Adding energy can add mass, a transition to a lower mass state releases energy in the form of kinetic energy.
https://en.wikipedia.org/wiki/Binding_energy
"A bound system is typically at a lower energy level than its unbound constituents. According to relativity theory, a ΔE decrease in the total energy of a system is accompanied by a decrease ΔM in the total mass, where ΔM⋅c2=ΔE."
Impeccable

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Orodruin
Staff Emeritus
Homework Helper
Gold Member
Summary:: I thought that nucleons were usually stable in a nucleus because they were in a lower energy state than when free as individuals.

I would have thought energy could decrease and mass could gain if energy was converted to mass?
Common misconception. Energy is never "converted to mass". Mass is a form of energy. In a reaction, other forms of energy may be redistributed into mass energy or vice versa.

What makes mass a form of energy rather than energy a form of mass?

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vanhees71
Gold Member
The structure of the Poincare group ;-)).

BvU
Homework Helper
Haha, and can you now try an explanation (at a slightly lower level) in understandable english ?

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Astronuc
Staff Emeritus
What makes mass a form of energy rather than energy a form of mass?
Note that photons have energy, but no mass.

The simplest group of nucleons is the deuteron formed by the combination of p + n. Then they combine, they emit a gamma ray (binding energy) of about 2.22452 MeV. One can do the calculation based on rest masses. To dissociate the d into to p and n, one needs a gamma ray of about 2.225 MeV or more. One has to consider recoil of the deuteron.

• ohwilleke and vanhees71
Ok so is it true to say that binding energy is the energy released during a nuclear reaction?

Astronuc
Staff Emeritus
Ok so is it true to say that binding energy is the energy released during a nuclear reaction?
Yes. Binding energy is the energy released, e.g., gamma emission following the absorption of a neutron by a nucleus, or the energy added to a system to remove a neutron, or dissociate a nucleus. We talk about the binding energy of the last neutron, i.e., what energy is needed to remove a neutron. If one wanted to completely dissociate a nuclear into its constituent nucleons (p, n), it would take a lot of energy.

In the case of removing a neutron, one would find a gamma ray of sufficient energy to impose on a nucleus. The reaction is known as the photoneutron effect.

In other nuclear reactions, e.g., fusion, say of d + t -> α + n, the resulting energy is manifest in the kinetic energy of the α and n. Both 5He and 5Li are unstable nuclei, so one cannot form either by adding a neutron or proton to 4He.

In the case of 9Be, if one removes the neutron to form 8Be, the 8Be spontaneously dissociates into two α particles.

In nuclear reactions, we consider the Q-value, which is the difference between the mass of reactants and the products, which if positive means some net energy in the form of kinetic energy of the products is produced.
http://hyperphysics.phy-astr.gsu.edu/hbase/NucEne/nucbin.html#c1

Thank you its starting to make more sense. Im still a little bit confused however.
Binding energy is released when a nucleus separates into its components, and it was energy not mass when it was stored in the nucleus? It was energy capable of doing work against any force which tried to tear the nucleus apart?
If you could be successful in separating the nucleons then the binding energy would be stored in the nucleons as a mass gain ? Or is it the energy you externally supplied to tear them apart that gets stored as the extra mass in the nucleons?
During a nuclear reaction however, the binding energy doesnt go into the separated nucleons as extra mass, it directly becomes other types of energy (eg heat)?

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Orodruin
Staff Emeritus
Homework Helper
Gold Member
Binding energy is released when a nucleus separates into its components, and it was energy not mass when it was stored in the nucleus?
No. Binding energy is the energy you need to add to separate a nucleus into components. A bound state has a lower total energy than the sum of its components.

• vanhees71
OK so binding energy is not the energy that is released, but it is equal to the energy that is released?

mfb
Mentor
Same thing. That amount of energy gets released when the nucleus forms.

vanhees71
Gold Member
Perhaps to make the issue clear, it's a good idea to have a quantitative example. The most simple analogous case is the electromagnetic binding of an electron and a proton in the hydrogen atom. By definition the total energy is 0 when both the electron with charge ##-e## and the proton with charge ##+e## are placed "at infinity". Now we calculate the energy to bring the proton at the origin of our coordinate system. Since there's nothing it could interact with, this energy is 0. Next we bring the electron close to the electron (at some finite distance of course). Since the Coulomb force between proton and electron is attractive, the energy of the system gets lower by the amount ##\Delta E=-e^2/(4 \pi r)##, where ##r## is the distance between proton and electron. To get a stable atom you have to think in terms of quantum mechanics and energy eigenstates. The correct order of magnitude for the binding energy in this case is of course the ionization energy of ##\Delta E=-13.6 \; \text{eV}##. Thu hydrogen atom's mass thus is ##m_{\text{H}}=m_{\text{p}}+m_{\text{e}}-13.6 \text{eV}/c^2<m_{\text{p}}+m_{\text{e}}##. Of course in this case due to the small coupling constant of electrodynamics this "mass defect" is very small compared to ##m_{\text{p}}=938 \; \text{MeV}## and ##m_{\text{p}}=0.51 \; \text{MeV}##, but for the strong force binding protons and neutrons together to nucleons the mass defects are not negligible.

• Astronuc
Astronuc
Staff Emeritus
Binding energy is released when a nucleus separates into its components, and it was energy not mass when it was stored in the nucleus? It was energy capable of doing work against any force which tried to tear the nucleus apart?
By any chance, is one referring to a fission reaction? It would help to know the context of one's questions, and to which nuclear reactions one is referring. It would also help to know to what text one is referring in order to understand the source of one's confusion.

Drakkith
Staff Emeritus
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