Quantum Physics

944 Chapter 29 Nuclear PhysicsIf the binding energy of a nucleus were zero, the nucleus would separate into its constituent protons and neutronswithout the addition of any energy; that is, it would spontaneously break apart.Exercise 29.23Calculate the binding energy of 2He.Answer7.718 MeVApplying Physics 29.1Figure 29.4 shows a graph of the amount of energyrequired to remove a nucleon from the nucleus. Thefigure indicates that an approximately constant amountof energy is necessary to remove a nucleon above A 40,whereas we saw in Chapter 28 that widely varyingamounts of energy are required to remove an electronfrom the atom. What accounts for this difference?Explanation In the case of Figure 29.4, the approximatelyconstant value of the nuclear binding energy is aresult of the short-range nature of the nuclear force. Agiven nucleon interacts only with its few nearest neighbors,rather than with all of the nucleons in the nucleus.Thus, no matter how many nucleons are present in thenucleus, pulling any one nucleon out involves separatingIt’s interesting to examine a plot of binding energy per nucleon, E b /A, as afunction of mass number for various stable nuclei (Fig. 29.4). Except for thelighter nuclei, the average binding energy per nucleon is about 8 MeV. Note thatthe curve peaks in the vicinity of A 60, which means that nuclei with mass numbersgreater or less than 60 are not as strongly bound as those near the middle ofthe periodic table. As we’ll see later, this fact allows energy to be released in fissionand fusion reactions. The curve is slowly varying for A 40, which suggests thatthe nuclear force saturates. In other words, a particular nucleon can interact withonly a limited number of other nucleons, which can be viewed as the “nearestneighbors” in the close-packed structure illustrated in Figure 29.2.Binding Nucleons and Electronsit only from its nearest neighbors. The energy to do this,therefore, is approximately independent of how manynucleons are present. For the clearest comparison withthe electron, think of averaging the energies required tostrip all of the electrons out of a particular atom, fromthe outermost valence electron to the innermost K-shellelectron. This average increases steeply with increasingatomic number. The electrical force binding the electronsto the nucleus in an atom is a long-range force. Anelectron in an atom interacts with all the protons in thenucleus. When the nuclear charge increases, there is astronger attraction between the nucleus and the electrons.Therefore, as the nuclear charge increases, moreenergy is necessary to remove an average electron.Region of greatest stability4 He12 C20 Ne62 Ni208 Pb97235 ClGe8723 Na19 F56 Fe98 Mo107 Ag127 I 159 Tb 197Au226 Ra238 UFigure 29.4 Binding energy pernucleon versus the mass number Afor nuclei that are along the line ofstability shown in Figure 29.3. Somerepresentative nuclei appear as bluedots with labels. (Nuclei to the rightof 208 Pb are unstable. The curverepresents the binding energy forthe most stable isotopes.)Binding energy perparticle, MeV6543210014 N116 B Li9 Be2 H20 406080100 120 140Mass number A160180200220240