317. Binding EnergyLearning Objectives
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 Imagine pulling a nuclide apart as illustrated in Figure 1. 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 $\text{\Delta}m=\text{BE}/{c}^{2}$. 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 ${}^{A}\text{X}$ has $Z$ protons and $N$ neutrons, so that the difference in mass is $$\mathrm{\Delta}m=({\text{Zm}}_{p}+{\text{Nm}}_{n}){m}_{\text{tot}}\text{.}$$Thus, $$\text{BE}=(\mathrm{\Delta}m){c}^{2}=[({\text{Zm}}_{p}+{\text{Nm}}_{n}){m}_{\text{tot}}]{c}^{2}\text{,}$$where ${m}_{\text{tot}}$ is the mass of the nuclide ${}^{A}\text{X}$, ${m}_{p}$ is the mass of a proton, and ${m}_{n}$ 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 $Z$ electrons to ${m}_{\text{tot}}$, which gives $m\left({}^{A}\text{X}\right)$, the atomic mass of the nuclide. We then add $Z$ electrons to the $Z$ protons, which gives $\text{Zm}\left({}^{1}\text{H}\right)$, or $Z$ times the mass of a hydrogen atom. Thus the binding energy of a nuclide ${}^{A}\text{X}$ is $$\text{BE}=\{[\text{Zm}({}^{1}\text{H})+{\text{Nm}}_{n}]m({}^{A}X)\}{c}^{2}.$$The atomic masses can be found in Appendix A, most conveniently expressed in unified atomic mass units u ($1\phantom{\rule{0.25em}{0ex}}\text{u}=\text{931}\text{.}5\phantom{\rule{0.25em}{0ex}}\text{MeV}/{c}^{2}$). BE is thus calculated from known atomic masses. Things Great and SmallNuclear Decay Helps Explain Earth’s Hot Interior A puzzle created by radioactive dating of rocks is resolved by radioactive heating of Earth’s interior. This intriguing story is another example of how smallscale physics can explain largescale phenomena. Radioactive dating plays a role in determining the approximate age of the Earth. The oldest rocks on Earth solidified about $3\text{.}5\times {\text{10}}^{9}$ years ago—a number determined by uranium238 dating. These rocks could only have solidified once the surface of the Earth had cooled sufficiently. The temperature of the Earth at formation can be estimated based on gravitational potential energy of the assemblage of pieces being converted to thermal energy. Using heat transfer concepts discussed in Thermodynamics it is then possible to calculate how long it would take for the surface to cool to rockformation temperatures. The result is about ${\text{10}}^{9}$ years. The first rocks formed have been solid for $3\text{.}5\times {\text{10}}^{9}$ years, so that the age of the Earth is approximately $4\text{.}5\times {\text{10}}^{9}$ years. There is a large body of other types of evidence (both Earthbound and solar system characteristics are used) that supports this age. The puzzle is that, given its age and initial temperature, the center of the Earth should be much cooler than it is today (see Figure 2). We know from seismic waves produced by earthquakes that parts of the interior of the Earth are liquid. Shear or transverse waves cannot travel through a liquid and are not transmitted through the Earth’s core. Yet compression or longitudinal waves can pass through a liquid and do go through the core. From this information, the temperature of the interior can be estimated. As noticed, the interior should have cooled more from its initial temperature in the $4\text{.}5\times {\text{10}}^{9}$ years since its formation. In fact, it should have taken no more than about ${\text{10}}^{9}$ years to cool to its present temperature. What is keeping it hot? The answer seems to be radioactive decay of primordial elements that were part of the material that formed the Earth (see the blowup in Figure 2). Nuclides such as ${}^{\text{238}}\text{U}$ and ${}^{\text{40}}\text{K}$ have halflives similar to or longer than the age of the Earth, and their decay still contributes energy to the interior. Some of the primordial radioactive nuclides have unstable decay products that also release energy— ${}^{\text{238}}\text{U}$ has a long decay chain of these. Further, there were more of these primordial radioactive nuclides early in the life of the Earth, and thus the activity and energy contributed were greater then (perhaps by an order of magnitude). The amount of power created by these decays per cubic meter is very small. However, since a huge volume of material lies deep below the surface, this relatively small amount of energy cannot escape quickly. The power produced near the surface has much less distance to go to escape and has a negligible effect on surface temperatures. A final effect of this trapped radiation merits mention. Alpha decay produces helium nuclei, which form helium atoms when they are stopped and capture electrons. Most of the helium on Earth is obtained from wells and is produced in this manner. Any helium in the atmosphere will escape in geologically short times because of its high thermal velocity. 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 ${}^{\text{24}}\text{Mg}$ compared with pulling apart ${}^{\text{12}}\text{C}$, for example. To help us look at other effects, we divide BE by $A$ and consider the There are some noticeable spikes on the $\text{BE}/A$ graph, which represent particularly tightly bound nuclei. These spikes reveal further details of nuclear forces, such as confirming that closedshell 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 $Z$ and $N$, and with $Z=N$, 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 ${}^{4}\text{He}$, with much smaller amounts of ${}^{\text{12}}\text{C}$ 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 ${}^{4}\text{He}$, emitted in $\alpha $ decay. Example 1: What Is $\text{BE}/A$ for an Alpha Particle?Calculate the binding energy per nucleon of ${}^{4}\text{He}$, the $\alpha $ particle. Strategy To find $\text{BE}/A$, we first find BE using the Equation $\text{BE}=\{[\text{Zm}({}^{1}\text{H})+{\text{Nm}}_{n}]m({}^{A}\text{X})\}{c}^{2}$ and then divide by $A$. This is straightforward once we have looked up the appropriate atomic masses in Appendix A. Solution The binding energy for a nucleus is given by the equation $$\text{BE}=\{[\text{Zm}({}^{1}\text{H})+{\text{Nm}}_{n}]m({}^{A}\text{X})\}{c}^{2}\text{.}$$For ${}^{4}\text{He}$, we have $Z=N=2$; thus, $$\text{BE}=\{[2m({}^{1}\text{H})+{2m}_{n}]m({}^{4}\text{He})\}{c}^{2}\text{.}$$Appendix A gives these masses as $m({}^{4}\text{He})=\text{4.002602 u}$, $m({}^{1}\text{H})=\text{1.007825 u}$, and ${m}_{n}=\text{1.008665 u}$. Thus, $$\text{BE}=(0\text{.}\text{030378 u}){c}^{2}\text{.}$$Noting that $\text{1 u}=\text{931}\text{.}\text{5 MeV/}{c}^{2}$, we find $$\text{BE}=(\text{0.030378})(\text{931}\text{.}\text{5 MeV/}{c}^{2}){c}^{2}=\mathrm{28.3\; MeV}\text{.}$$Since $A=4$, we see that $\text{BE}/A$ is this number divided by 4, or $$\text{BE}/A=\text{7.07 MeV/nucleon}\text{.}$$Discussion This is a large binding energy per nucleon compared with those for other lowmass nuclei, which have $\text{BE}/A\approx \text{3 MeV/nucleon}$. This indicates that ${}^{4}\text{He}$ is tightly bound compared with its neighbors on the chart of the nuclides. You can see the spike representing this value of $\text{BE}/A$ for ${}^{4}\text{He}$ on the graph in Figure 3. This is why ${}^{4}\text{He}$ is stable. Since ${}^{4}\text{He}$ 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 $\alpha $ 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 ${}^{4}\text{He}$ unit within the nucleus, at least for a while. In some cases, the ${}^{4}\text{He}$ escapes, and $\alpha $ decay has then taken place. There is more to be learned from nuclear binding energies. The general trend in $\text{BE}/A$ 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. ProblemSolving StrategiesFor Reaction And Binding Energies and Activity Calculations in Nuclear Physics
PhET Explorations: Nuclear FissionStart a chain reaction, or introduce nonradioactive isotopes to prevent one. Control energy production in a nuclear reactor! Section Summary
Conceptual QuestionsExercise 1Why 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 & ExercisesExercise 1${}^{2}\text{H}$ 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 $Z=N$, which should tend to make it more tightly bound, but both are odd numbers. Calculate $\mathrm{BE/}A$, the binding energy per nucleon, for ${}^{2}\text{H}$ and compare it with the approximate value obtained from the graph in Figure 3. Show/Hide Solution Solution1.112 MeV, consistent with graph Exercise 2${}^{\text{56}}\text{Fe}$ is among the most tightly bound of all nuclides. It is more than 90% of natural iron. Note that ${}^{\text{56}}\text{Fe}$ has even numbers of both protons and neutrons. Calculate $\mathrm{BE/}A$, the binding energy per nucleon, for ${}^{\text{56}}\text{Fe}$ and compare it with the approximate value obtained from the graph in Figure 3. Exercise 3${}^{\text{209}}\text{Bi}$ is the heaviest stable nuclide, and its $\text{BE}/A$ is low compared with mediummass nuclides. Calculate $\mathrm{BE/}A$, the binding energy per nucleon, for ${}^{\text{209}}\text{Bi}$ and compare it with the approximate value obtained from the graph in Figure 3. Show/Hide Solution Solution7.848 MeV, consistent with graph Exercise 4(a) Calculate $\text{BE}/A$ for ${}^{\text{235}}\text{U}$, the rarer of the two most common uranium isotopes. (b) Calculate $\text{BE}/A$ for ${}^{\text{238}}\text{U}$. (Most of uranium is ${}^{\text{238}}\text{U}$.) Note that ${}^{\text{238}}\text{U}$ has even numbers of both protons and neutrons. Is the $\text{BE}/A$ of ${}^{\text{238}}\text{U}$ significantly different from that of ${}^{\text{235}}\text{U}$ ? Exercise 5(a) Calculate $\text{BE}/A$ for ${}^{\text{12}}\text{C}$. Stable and relatively tightly bound, this nuclide is most of natural carbon. (b) Calculate $\text{BE}/A$ for ${}^{\text{14}}\text{C}$. Is the difference in $\text{BE}/A$ between ${}^{\text{12}}\text{C}$ and ${}^{\text{14}}\text{C}$ significant? One is stable and common, and the other is unstable and rare. Show/Hide Solution Solution(a) 7.680 MeV, consistent with graph (b) 7.520 MeV, consistent with graph. Not significantly different from value for ${}^{\text{12}}\text{C}$, but sufficiently lower to allow decay into another nuclide that is more tightly bound. Exercise 6The fact that $\text{BE}/A$ 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 $A=\text{60}$ nucleus. (b) Compare $\text{BE}/A$ for ${}^{\text{58}}\text{Ni}$ and ${}^{\text{90}}\text{Sr}$. The first is one of the most tightly bound nuclides, while the second is larger and less tightly bound. Exercise 7The 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 13.6eV 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 Section 314 Table 1 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. Show/Hide Solution Solution(a) $1\text{.}\text{46}\times {\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\mathrm{u}$ vs. 1.007825 u for ${}^{1}\text{H}$ (b) 0.000549 u (c) $2\text{.}\text{66}\times {\text{10}}^{5}$ Exercise 8Unreasonable 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? Show/Hide Solution Solution(a) $\mathrm{\u20139.315\; MeV}$ (b) The negative binding energy implies an unbound system. (c) This assumption that it is two bound neutrons is incorrect.
