Structure of Atomic Nuclei Introduction In our investigation of the electronic properties (chemical, physical, etc.) of atoms, the nucleus is regarded as a positively charged point particle that embodies the great majority of the mass of the atom. The Coulomb interaction between the nucleus and the atomic electrons holds the atom together. The structure of the nucleus plays almost no role in atomic properties because nuclear excitation energies are several orders of magnitude higher than atomic or electronic excitation energies. 1 Hence, in normal atomic interactions (e.g., chemical reactions), the nucleus is always in the ground state. However, we do know, based upon experiments conducted by Rutherford, Chadwick, and others, that the nucleus does have structure. Our question is: how are the structure of the nucleus and the properties of the nucleus related? Nuclear Constituents Nuclei are made up of protons (charge + e) and neutrons (no charge). Hydrogen is unique in that its nucleus has a single proton and no neutrons. The sum of the number of protons (atomic number Z) and the number of neutrons (neutron number N) is called the mass number ( A) of the nuclear species or nuclide : Z+N = A. Protons and neutrons are collectively called nucleons. Each A nuclide is symbolized by its chemical symbol (X) along with the values of A and Z: Z X. 4 8 14 Examples: 2 He , 4 Be, 6 C. A [The symbol in the textbook is ZNX , but this is redundant.] Isotopes are nuclides that have the same Z, but different N (and therefore different A). Since the number of electrons in a neutral atom (i.e., Z) determines the chemical behavior of an element, isotopes have the same chemical symbol. 4 3 12 14 Examples: 2 He , 2 He ; 6 C, 6 C. Isobars are nuclides that have the same mass number A, but different Z (and therefore different N). 2 14 14 Examples: 7 N , 6 C Nuclear Sizes and Shapes Like atoms, nuclei have no precisely defined size because of the extended nature of the wave functions of nuclear constituents. However, one can define an “average” radius. From scattering experiments, we know that nuclear radii have values in the range 2 – 8 fermi (10 -15 m). 1 MeV’s versus eV’s 2 Isotones have the same N: 7 B, 6 Li. 4 3 1 Experiments also show that nuclei are for the most part spherical. Departures from sphericity are usually slight (either elongation or flattening). Another interesting feature is that the density of the nucleons, i.e., the number of the nucleons per unit volume, is virtually independent of the mass number of the nucleus. (See diagram above, taken from Krane’s Modern Physics .) Hence, to good approximation, A = constant, 4 3 3 π R i.e., 1/3 RRA= 0 , (Empirical) (9.1) where R0 is a constant of proportionality. The value of R0 must be determined experimentally. Its value depends on how the nuclear radius R is defined. One such definition is the distance (from the center of the nucleus) at which the nuclear density falls to one-half of its value at the center of the nucleus. Whatever the definition for R, R0 usually lies in the range 1.0 – 1.5 fermis. Because the nucleus occupies such a small volume and contains virtually all the mass of the atom, the mass densities of nuclei are immense. 12 ≈ Example: 6 C; R 2.7 fm, A = 12 12 u ρ = = 2.4 ×10 17 kg/m 3! 4 15 3 3 π (2.7× 10 m) 3 3 For comparison purposes, ρwater = 10 kg/m To measure nuclear sizes, one can do scattering experiments in which charged nucleons 3 are scattered from the nucleus to be investigated. Departures from the Rutherford formula indicate that the nucleus is being impinged upon. 4 Why is there a deviation from the Rutherford formula when the nucleus is approached closely? The deviation suggests that when nucleons are in close proximity, i.e., within ~2 fermis of each other, another interaction dominates the electromagnetic Coulomb interaction. This interaction is the strong nuclear interaction . 3 E.g., protons or alpha particles (which contain protons) 4 Distance of closest approach < nuclear radius 2 The Strong Nuclear Force It is easy to see that another interaction between U(r) nucleons (apart from the Coulomb interaction) must exist inside the nucleus. The Coulomb interaction causes a repulsive interaction among the protons in the nucleus. Since neutrons have no charge, they are unaffected by the Coulomb interaction. Why then does a nucleus remain bound? The reason is that for 1 2 nucleon-nucleon distances of up to about 2 fm, the strong nuclear force dominates the Coulomb force. r (fm) At typical inter-nucleon distances in the nucleus, the strong force is attractive and therefore binds the -100 MeV nuclear constituents together. [Note that the attractive gravitational force is about 36 orders of magnitude too weak to do the job.] Experiments have shown that the strong force has the following properties (see sketch of nucleon-nucleon potential energy vs. separation distance above): 1. It is extremely short-ranged: For nucleon-nucleon separations r > ~ 3 fm, the strong force is negligibly small. 5 2. For r ≤ 1 fm, the strong force is strongly repulsive. The constancy of nucleon densities is evidence for this repulsive interaction. (If the interaction were purely attractive, nuclear densities at the centers of large nuclei would be significantly larger than those for small nuclei.) 3. The strong force acts equally between any pair of nucleons. Thus, the strong interactions for n- n, n-p, and p-p are the same for a given nuclear quantum state. The strong nuclear force is therefore charge-independent. 4. The strong interaction affects only certain types of particles called hadrons (e.g., neutrons, protons, pions). Non-hadrons e.g., electrons, are not acted upon by the strong nuclear force. Thus, when high-energy electrons are used to bombard the nucleus, the characteristics of the scattering are not influenced by the strong force. [Electron diffraction is a useful tool for determining nuclear sizes.] [Show Fig. 9.3 p. 235, Krane.] Nuclear Masses and Binding Energies We saw in chapter 3 that the mass of a nucleus is less than the total mass of its constituent nucleons because energy is liberated when individual nucleons bind together to form a nucleus. This energy, called the binding energy, decreases the mass of the nucleus relative to that of the separated nucleons, according to mass-energy equivalence. Thus, 5 The constancy of nuclear density supports this, as it indicates that each nucleon only interacts with its nearest neighbors and not with all other nucleons. 3 2 E = (Nm + Zm − m )c , (9.2) b n p A X,nuc Z where m is the mass of the A nucleus. Since it is the masses of atoms that are usually A X,nuc Z X Z A measured, we add the mass of Z electrons to the protons and to the Z X nucleus to obtain: 2 Eb=(). Nm n + Zm1 − m A c (9.3) 1H Z X,atom Note that we have neglected the binding energy of the electrons in the atom under consideration. However, since nuclear binding energies are several orders of magnitude greater than electronic binding energies, it is an excellent approximation to neglect electronic binding energies. 56 Example: Find the binding energy and the binding energy per nucleon for 26 Fe. Solution: Z = 26, N = 30. mn =1.0086654 u. m1 = 1.007825 u, and m56 = 55.934939 u 1 H 26 Fe,atom Eb = (30 ×1.008665 u + 26 ×1.007825 u – 55.934939 u) × 931.5 MeV / u = 492.3 MeV, where we have used the equality c2 = 931.5 MeV / u. 56 [Note that binding energy for innermost electrons in 26 Fe is tens of keV.] 56 Binding energy per nucleon for 26 Fe = Eb /A = 492.3 MeV / 56 = 8.791 MeV/nucleon. 238 A similar calculation for 92 U gives Eb = 1802 MeV and Eb / A = 7.571 MeV per nucleon. 238 Hence, 92 U has the greater binding energy as would be expected, but it takes more energy per 56 nucleon to break up the 26 Fe nucleus. The diagram [p. 237 Krane or TZ&D] shows the trend in binding energy per nucleon for all the 1 elements. The lightest nuclei have the smallest binding energies per nucleon ( 1 H has a binding energy of zero). This is because the number of nearest-neighbors is smaller than the "saturation" value, i.e., the maximum number consistent with the short range of the strong force. The binding 56 energy per nucleon rapidly climbs, peaks at 26 Fe, and then slowly declines. The binding energy per nucleon chart explains the release of energy in nuclear fission and nuclear fusion. If a very massive nucleus is split into two fragments having comparable masses, then each fragment will have a larger Eb/A value than the original. Thus, energy is released in this process. The process of splitting a massive nucleus with attendant release of energy is called nuclear fission . To more clearly see that energy is released, we write the nuclear reaction: X → C + D (X splits into C and D). Let us consider the binding energies. Total energy is conserved, so −EEEb,X= − b ,C− b ,D + energy. (9.4) Note that the binding-energy terms are negative (the binding energy Eb itself is defined to be positive) because the nuclei are bound states , i.e., energy must be supplied to separate the nuclei into individual nucleons having no kinetic or potential energy.