College Physics

Science and Technology

Binding Energy

Tác giả: OpenStaxCollege

The more tightly bound a system is, the stronger the forces that hold it together and the greater the energy required to pull it apart. We can therefore learn about nuclear forces by examining how tightly bound the nuclei are. We define the binding energy (BE) of a nucleus to be the energy required to completely disassemble it into separate protons and neutrons. We can determine the BE of a nucleus from its rest mass. The two are connected through Einstein’s famous relationship E=(Δm)c2 size 12{E= \( Δm \) c rSup { size 8{2} } } {}. A bound system has a smaller mass than its separate constituents; the more tightly the nucleons are bound together, the smaller the mass of the nucleus.

Imagine pulling a nuclide apart as illustrated in [link]. Work done to overcome the nuclear forces holding the nucleus together puts energy into the system. By definition, the energy input equals the binding energy BE. The pieces are at rest when separated, and so the energy put into them increases their total rest mass compared with what it was when they were glued together as a nucleus. That mass increase is thus Δm=BE/c2 size 12{Δm="BE"/c rSup { size 8{2} } } {}. This difference in mass is known as mass defect. It implies that the mass of the nucleus is less than the sum of the masses of its constituent protons and neutrons. A nuclide AX size 12{"" lSup { size 8{A} } X} {} has Z size 12{Z} {} protons and N size 12{N} {} neutrons, so that the difference in mass is

Δm=(Zmp+Nmn)mtot. size 12{Δm= \( ital "Zm" rSub { size 8{p} } + ital "Nm" rSub { size 8{n} } \) - m rSub { size 8{"tot"} } } {}


BE=(Δm)c2=[(Zmp+Nmn)mtot]c2, size 12{"BE"= \( Δm \) c rSup { size 8{2} } = \[ \( ital "Zm" rSub { size 8{p} } + ital "Nm" rSub { size 8{n} } \) - m rSub { size 8{"tot"} } \] c rSup { size 8{2} } } {}

where mtot size 12{m rSub { size 8{"tot"} } } {} is the mass of the nuclide AX, mp is the mass of a proton, and mn is the mass of a neutron. Traditionally, we deal with the masses of neutral atoms. To get atomic masses into the last equation, we first add Z size 12{Z} {} electrons to mtot size 12{m rSub { size 8{"tot"} } } {}, which gives mAX size 12{m left ("" lSup { size 8{A} } X right )} {}, the atomic mass of the nuclide. We then add Z size 12{Z} {} electrons to the Z size 12{Z} {} protons, which gives Zm1H size 12{ ital "Zm" left ("" lSup { size 8{1} } H right )} {}, or Z size 12{Z} {} times the mass of a hydrogen atom. Thus the binding energy of a nuclide AX size 12{"" lSup { size 8{A} } X} {} is

BE = { [ Zm ( 1 H ) + Nm n ] m ( A X )} c 2 .

The atomic masses can be found in Appendix A, most conveniently expressed in unified atomic mass units u (1u=931.5MeV/c2 size 12{1" u"="931" "." 5 "MeV"/c rSup { size 8{2} } } {}). BE is thus calculated from known atomic masses.

Work done to pull a nucleus apart into its constituent protons and neutrons increases the mass of the system. The work to disassemble the nucleus equals its binding energy BE. A bound system has less mass than the sum of its parts, especially noticeable in the nuclei, where forces and energies are very large.

What patterns and insights are gained from an examination of the binding energy of various nuclides? First, we find that BE is approximately proportional to the number of nucleons A in any nucleus. About twice as much energy is needed to pull apart a nucleus like 24Mg compared with pulling apart 12C, for example. To help us look at other effects, we divide BE by A size 12{A} {} and consider the binding energy per nucleon, BE/A size 12{ {"BE"} slash {A} } {}. The graph of BE/A size 12{ {"BE"} slash {A} } {} in [link] reveals some very interesting aspects of nuclei. We see that the binding energy per nucleon averages about 8 MeV, but is lower for both the lightest and heaviest nuclei. This overall trend, in which nuclei with A size 12{A} {} equal to about 60 have the greatest BE/A size 12{ {"BE"} slash {A} } {} and are thus the most tightly bound, is due to the combined characteristics of the attractive nuclear forces and the repulsive Coulomb force. It is especially important to note two things—the strong nuclear force is about 100 times stronger than the Coulomb force, and the nuclear forces are shorter in range compared to the Coulomb force. So, for low-mass nuclei, the nuclear attraction dominates and each added nucleon forms bonds with all others, causing progressively heavier nuclei to have progressively greater values of BE/A size 12{ {"BE"} slash {A} } {}. This continues up to A60 size 12{A approx "60"} {}, roughly corresponding to the mass number of iron. Beyond that, new nucleons added to a nucleus will be too far from some others to feel their nuclear attraction. Added protons, however, feel the repulsion of all other protons, since the Coulomb force is longer in range. Coulomb repulsion grows for progressively heavier nuclei, but nuclear attraction remains about the same, and so BE/A size 12{ {"BE"} slash {A} } {} becomes smaller. This is why stable nuclei heavier than A40 size 12{A approx "40"} {} have more neutrons than protons. Coulomb repulsion is reduced by having more neutrons to keep the protons farther apart (see [link]).

A graph of average binding energy per nucleon, BE/A size 12{ {"BE"} slash {A} } {}, for stable nuclei. The most tightly bound nuclei are those with A size 12{A} {} near 60, where the attractive nuclear force has its greatest effect. At higher A size 12{A} {} s, the Coulomb repulsion progressively reduces the binding energy per nucleon, because the nuclear force is short ranged. The spikes on the curve are very tightly bound nuclides and indicate shell closures.

The nuclear force is attractive and stronger than the Coulomb force, but it is short ranged. In low-mass nuclei, each nucleon feels the nuclear attraction of all others. In larger nuclei, the range of the nuclear force, shown for a single nucleon, is smaller than the size of the nucleus, but the Coulomb repulsion from all protons reaches all others. If the nucleus is large enough, the Coulomb repulsion can add to overcome the nuclear attraction.

There are some noticeable spikes on the BE/A size 12{ {"BE"} slash {A} } {} graph, which represent particularly tightly bound nuclei. These spikes reveal further details of nuclear forces, such as confirming that closed-shell nuclei (those with magic numbers of protons or neutrons or both) are more tightly bound. The spikes also indicate that some nuclei with even numbers for Z size 12{Z} {} and N size 12{N} {}, and with Z=N size 12{Z=N} {}, are exceptionally tightly bound. This finding can be correlated with some of the cosmic abundances of the elements. The most common elements in the universe, as determined by observations of atomic spectra from outer space, are hydrogen, followed by 4He, with much smaller amounts of 12C and other elements. It should be noted that the heavier elements are created in supernova explosions, while the lighter ones are produced by nuclear fusion during the normal life cycles of stars, as will be discussed in subsequent chapters. The most common elements have the most tightly bound nuclei. It is also no accident that one of the most tightly bound light nuclei is 4He, emitted in α decay.

What Is BE/A size 12{ {"BE"} slash {A} } {} for an Alpha Particle?

Calculate the binding energy per nucleon of 4He size 12{"" lSup { size 8{4} } "He"} {}, the α size 12{α} {} particle.


To find BE/A, we first find BE using the Equation BE={[Zm(1H)+Nmn]m(AX)}c2 and then divide by A. This is straightforward once we have looked up the appropriate atomic masses in Appendix A.


The binding energy for a nucleus is given by the equation


For 4He size 12{"" lSup { size 8{4} } "He"} {}, we have Z=N=2 size 12{Z=N=2} {}; thus,


Appendix A gives these masses as m(4He)=4.002602 u, m(1H)=1.007825 u, and mn=1.008665 u size 12{m rSub { size 8{n} } =0 "." "008665"`" u"} {}. Thus,

BE=(0.030378 u)c2. size 12{"BE"= \( 0 "." "030378 u" \) c rSup { size 8{2} } } {}

Noting that 1 u=931.5 MeV/c2 size 12{"1u"="931" "." "5 MeV/"c rSup { size 8{2} } } {}, we find

BE=(0.030378)(931.5 MeV/c2)c2=28.3 MeV. size 12{"BE"= \( 0 "." "030378" \) \( "931" "." "5 MeV/"c rSup { size 8{2} } \) c rSup { size 8{2} } ="28" "." 3" MeV"} {}

Since A=4 size 12{A=4} {}, we see that BE/A size 12{ {"BE"} slash {A} } {} is this number divided by 4, or

BE/A=7.07 MeV/nucleon. size 12{"BE"/A=7 "." "07"" MeV/nucleon"} {}


This is a large binding energy per nucleon compared with those for other low-mass nuclei, which have BE/A3 MeV/nucleon. This indicates that 4He is tightly bound compared with its neighbors on the chart of the nuclides. You can see the spike representing this value of BE/A for 4He on the graph in [link]. This is why 4He is stable. Since 4He is tightly bound, it has less mass than other A=4 nuclei and, therefore, cannot spontaneously decay into them. The large binding energy also helps to explain why some nuclei undergo α decay. Smaller mass in the decay products can mean energy release, and such decays can be spontaneous. Further, it can happen that two protons and two neutrons in a nucleus can randomly find themselves together, experience the exceptionally large nuclear force that binds this combination, and act as a 4He unit within the nucleus, at least for a while. In some cases, the 4He escapes, and α decay has then taken place.

There is more to be learned from nuclear binding energies. The general trend in BE/A size 12{"BE"/A} {} is fundamental to energy production in stars, and to fusion and fission energy sources on Earth, for example. This is one of the applications of nuclear physics covered in Medical Applications of Nuclear Physics. The abundance of elements on Earth, in stars, and in the universe as a whole is related to the binding energy of nuclei and has implications for the continued expansion of the universe.

Problem-Solving Strategies

For Reaction And Binding Energies and Activity Calculations in Nuclear Physics

  1. Identify exactly what needs to be determined in the problem (identify the unknowns). This will allow you to decide whether the energy of a decay or nuclear reaction is involved, for example, or whether the problem is primarily concerned with activity (rate of decay).
  2. Make a list of what is given or can be inferred from the problem as stated (identify the knowns).
  3. For reaction and binding-energy problems, we use atomic rather than nuclear masses. Since the masses of neutral atoms are used, you must count the number of electrons involved. If these do not balance (such as in β+ size 12{β rSup { size 8{+{}} } } {} decay), then an energy adjustment of 0.511 MeV per electron must be made. Also note that atomic masses may not be given in a problem; they can be found in tables.
  4. For problems involving activity, the relationship of activity to half-life, and the number of nuclei given in the equation R=0.693Nt1/2 size 12{R= { {1 "." "693"N} over {t rSub { size 8{1/2} } } } } {} can be very useful. Owing to the fact that number of nuclei is involved, you will also need to be familiar with moles and Avogadro’s number.
  5. Perform the desired calculation; keep careful track of plus and minus signs as well as powers of 10.
  6. Check the answer to see if it is reasonable: Does it make sense? Compare your results with worked examples and other information in the text. (Heeding the advice in Step 5 will also help you to be certain of your result.) You must understand the problem conceptually to be able to determine whether the numerical result is reasonable.

Section Summary

  • The binding energy (BE) of a nucleus is the energy needed to separate it into individual protons and neutrons. In terms of atomic masses,
    where m1H size 12{m left ("" lSup { size 8{1} } H right )} {} is the mass of a hydrogen atom, mAX is the atomic mass of the nuclide, and mn is the mass of a neutron. Patterns in the binding energy per nucleon, BE/A, reveal details of the nuclear force. The larger the BE/A size 12{"BE"/A} {}, the more stable the nucleus.

Conceptual Questions

Why is the number of neutrons greater than the number of protons in stable nuclei having A greater than about 40, and why is this effect more pronounced for the heaviest nuclei?

Problems & Exercises

2H is a loosely bound isotope of hydrogen. Called deuterium or heavy hydrogen, it is stable but relatively rare—it is 0.015% of natural hydrogen. Note that deuterium has Z=N size 12{Z=N} {}, which should tend to make it more tightly bound, but both are odd numbers. Calculate BE/A, the binding energy per nucleon, for 2H and compare it with the approximate value obtained from the graph in [link].

1.112 MeV, consistent with graph

56Fe is among the most tightly bound of all nuclides. It is more than 90% of natural iron. Note that 56Fe has even numbers of both protons and neutrons. Calculate BE/A, the binding energy per nucleon, for 56Fe and compare it with the approximate value obtained from the graph in [link].

209Bi is the heaviest stable nuclide, and its BE/A is low compared with medium-mass nuclides. Calculate BE/A, the binding energy per nucleon, for 209Bi and compare it with the approximate value obtained from the graph in [link].

7.848 MeV, consistent with graph

(a) Calculate BE/A for 235U, the rarer of the two most common uranium isotopes. (b) Calculate BE/A for 238U. (Most of uranium is 238U.) Note that 238U has even numbers of both protons and neutrons. Is the BE/A of 238U significantly different from that of 235U ?

(a) Calculate BE/A for 12C. Stable and relatively tightly bound, this nuclide is most of natural carbon. (b) Calculate BE/A for 14C. Is the difference in BE/A between 12C and 14C significant? One is stable and common, and the other is unstable and rare.

(a) 7.680 MeV, consistent with graph

(b) 7.520 MeV, consistent with graph. Not significantly different from value for 12C, but sufficiently lower to allow decay into another nuclide that is more tightly bound.

The fact that BE/A is greatest for A near 60 implies that the range of the nuclear force is about the diameter of such nuclides. (a) Calculate the diameter of an A=60 nucleus. (b) Compare BE/A for 58Ni and 90Sr. The first is one of the most tightly bound nuclides, while the second is larger and less tightly bound.

The purpose of this problem is to show in three ways that the binding energy of the electron in a hydrogen atom is negligible compared with the masses of the proton and electron. (a) Calculate the mass equivalent in u of the 13.6-eV binding energy of an electron in a hydrogen atom, and compare this with the mass of the hydrogen atom obtained from Appendix A. (b) Subtract the mass of the proton given in [link] from the mass of the hydrogen atom given in Appendix A. You will find the difference is equal to the electron’s mass to three digits, implying the binding energy is small in comparison. (c) Take the ratio of the binding energy of the electron (13.6 eV) to the energy equivalent of the electron’s mass (0.511 MeV). (d) Discuss how your answers confirm the stated purpose of this problem.

(a) 1.46×108u vs. 1.007825 u for 1H

(b) 0.000549 u

(c) 2.66×105 size 12{2 "." "66" times "10" rSup { size 8{ - 5} } } {}

Unreasonable Results

A particle physicist discovers a neutral particle with a mass of 2.02733 u that he assumes is two neutrons bound together. (a) Find the binding energy. (b) What is unreasonable about this result? (c) What assumptions are unreasonable or inconsistent?

(a) –9.315 MeV

(b) The negative binding energy implies an unbound system.

(c) This assumption that it is two bound neutrons is incorrect.