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## Chapter 6: Nuclear Structure

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**Chapter 6: Nuclear Structure**Abby Bickley University of Colorado NCSS ‘99 Additional References: Choppin (CLR), Radiochemistry and Nuclear Chemistry, 2nd Edition, Chapter 11 Friedlander (FKMM), Nuclear and Radiochemistry, 3rd Edition, Chapter 10**The Nucleus**• As chemist’s what do we already know about the nucleus of an atom? • Composed of protons and neutrons • Carries an electric charge equivalent to the number of protons & atomic number of the element • Protons and neutrons within nucleus held together by the strong force • Any model of nuclear structure must account for both Coulombic repulsion of protons and Strong force attraction between nucleons**Empirical Observations**• Chart of the nuclides: • 275 stable nuclei • 60% even-even • 40% even-odd or odd-even • Only 5 stable odd-odd nuclei 21H, 63Li, 105B, 147N, 5023Va (could have large t1/2) • Nuclei with an even number of protons have a large number of stable isotopes Even # protons Odd # protons 50Sn:10 (isotopes) 47Ag: 2 (isotopes) 48Cd: 8 51Sb:2 52Te: 8 45Rh:1 49In:1 53I: 1 • Roughly equal numbers of stable even-odd and odd-even nuclei**Implications for Nuclear Models I**• Proton-proton and neutron-neutron pairing must result in energy stabilization of bound state nuclei • Pairing of protons with protons and neutrons with neutrons results in the same degree of stabilization • Pairing of protons with neutrons does not occur (nor translate into stabilization)**Chart of the nuclides**• Light elements: N/Z = 1 • Heavy elements: N/Z 1.6 • Implies simple pairing not sufficient for stability • Neutron Rich: (N>Z) • N>Z: nucleus will - decay to stability • N>>Z: neutron drip line • Proton Rich: (N<Z) • N<Z: nucleus will + decay or electron capture to achieve stability • N<<Z: proton drip line (very rare) Friedlander, “Nuclear and Radiochemistry, 3rd Edition, 1981.**Implications for Nuclear Models II**4. Pairing not sufficient to achieve stability Why? Coulomb repulsion of protons grows with Z2: Nuclear attractive force must compensate all stable nuclei with Z > 20 contain more neutrons than protons Eq. 1**General Nuclear Properties I**8.9 • Binding energy per nucleon approximately constant for all stable nuclei 7.4 Friedlander, “Nuclear and Radiochemistry, 3rd Edition, 1981.**General Nuclear Properties II**• Nuclear radius is proportional to the cube root of the mass r = r0 A1/3 Eq. 2 • Experimental studies show ~uniform distribution of the charge and mass throughout the volume of the nucleus dl = skin thickness Rl = Half density radius Friedlander, “Nuclear and Radiochemistry, 3rd Edition, 1981.**Liquid Drop Model (1935)**• Treats nucleus as a statistical assembly of neutrons and protons with an effective surface tension - similar to a drop of liquid • Rationale: • Volume of nucleus number of nucleons • Implies nuclear matter is incompressible • Binding energy of nucleus number of nucleons • Implies nuclear force must have a saturation character, ie each nucleon only interacts with nearest neighbors • Mathematical Representation: • Treats binding energy as sum of volume, surface and Coulomb energies: Eq. 3**Volume Energy**Surface Energy Liquid Drop Model Components I • Volume Energy: • Binding energy of nucleus number of nucleons • Correction factor accounts for symmetry energy (for a given A the binding energy due to only nuclear forces is greatest for nuclei with equal numbers of protons and neutrons) • Surface Energy: • Nucleon at surface are unsaturated reduce binding energy surface area • Surface-to-volume ratio decreases with increasing nuclear size term is less important for large nuclei c1 = 15.677 MeV, c2 = 18.56 MeV, c3 = 0.717 MeV, c4 = 1.211 MeV, k = 1.79**Liquid Drop Model Components II**• Coulomb Energy: • Electrostatic energy due to Coulomb repulsion between protons • Correction factor accounts for diffuse boundary of nucleus (accounts for skin thickness of nucleus) • Pairing Energy: • Accounts for added stability due to nucleon pairing • Even-even: = +11/A1/2 • Even-odd & odd-even: = 0 • Odd-odd: = -11/A1/2 Coulomb Energy Pairing Energy c1 = 15.677 MeV, c2 = 18.56 MeV, c3 = 0.717 MeV, c4 = 1.211 MeV, k = 1.79**Problem 1**• Using the binding energy equation for the liquid drop model, calculate the binding energy per nucleon for 15N and 148Gd. • Compare these results with those obtained by calculating the binding energy per nucleon from the atomic mass and the masses of the constituent nucleons.**Problem 1: Answers**• 15N = 6.87 MeV/nucleon • 148Gd = 8.88 MeV/nucleon • 15N = 7.699 MeV/nucleon • 148Gd = 8.25 MeV/nucleon**Mass Parabolas**Eq. 4 • Represent mass of atom as difference between sum of constituents and total binding energy: • Substitute binding energy equation for EB and group terms by power of Z: • For a given number of nucleons (A) f1, f2 and f3 are constants • Functional form represents a mass-energy parabola • Single parabola for odd A nuclei ( = 0) • Double parabola for even A nuclei ( = ±11/A1/2) Eq. 5**Mass Parabolas Example 1A = 75 or 157**• Parabola Vertex: • ZA=[-f2/ 2f1] Eq. 6 • Minimum mass & Maximum EB • Used to find mass and EB difference between isobars • Nuclear charge of minimum mass is derivative of Eq. 5 => not necessarily integral • Comparison of Z = 75 and Z = 157 • Valley of stability broadens with increasing A • For a given value of odd-A only one stable nuclide exists • In odd-A isobaric decay chains the -decay energy increases monotonically Friedlander, “Nuclear and Radiochemistry, 3rd Edition, 1981.**Mass Parabolas Example 2A = 156**• Eq. 5 results in two mass parabola for a given even value of A • For a given value of even-A their exist 2 (or 3) stable nuclides • In this figure both 156Gd and 156Dy are stable • In even-A isobaric decay chains the -decay energies alternate between small and large values • This model successfully reproduces experimentally observed energy levels • BUT……. Friedlander, “Nuclear and Radiochemistry, 3rd Edition, 1981.**Problem 2**• Find the nuclear charge (ZA) corresponding to the maximum binding energy for: A = 157, 156 and 75 • To which isotopes do these values correspond? • Compare your results with the mass parabolas on slides 15 & 16.**Problem 2: Answers**• Find the nuclear charge (ZA) corresponding to the maximum binding energy for: A = 157, 156 and 75 ZA = 64.69, 64.32, 33.13 • To which isotopes do these values most closely correspond? 15765Tb, 15664Gd, 7533As • Compare your results with the mass parabolas on slides 15 & 16.**Magic Numbers**• Nuclides with “magic numbers” of protons and/or neutrons exhibit an unusual degree of stability • 2, 8, 20, 28, 50, 82, 126 • Suggestive of closed shells as observed in atomic orbitals • Analogous to noble gases • Much empirical evidence was amassed before a model capable of explaining this phenomenon was proposed • Result = Shell Model Friedlander, “Nuclear and Radiochemistry, 3rd Edition, 1981.**Atomic Orbitals History**• Plum Pudding Model: (Thomson, 1897) • Each atom has an integral number of electrons whose charge is exactly balanced by a jelly-like fluid of positive charge • Nuclear Model: (Rutherford, 1911) • Electrons arranged around a small massive core of protons and neutrons* (added later) • Planetary Atomic Orbitals: (Bohr, 1913) • Assume electrons move in a circular orbit of a given radius around a fixed nucleus • Assume quantized energy levels to account for observed atomic spectra • Fails for multi-electron systems • Schrodinger Equation: (1925) • Express electron as a probability distribution in the form of a standing wave function**3s**2p 2s Atomic Orbitals • Schrodinger equation solution reveals quantum numbers • n = principal, describes energy level • l = angular momentum, 0n-1 (s,p,d,f,g,h…) • m = magnetic, - l l, describes behavior of atom in external B field • ms = spin, -1/2 or 1/2 • Pauli Exclusion Principle: e-’s are fermions no two e-’s can have the same set of quantum numbers • Hund’s Rule: when electrons are added to orbitals of equal energy a single electron enters each orbital before a second enters any orbital; the spins remain parallel if possible. • Example: C = 1s22s22px12py1 1s**Shell Structure of NucleusHistorical Evolution**• Throughout 1930’s and early 1940’s evidence of deviation from liquid drop model accumulates • 1949: Mayer & Jenson • Independently propose single-particle orbits • Long mean free path of nucleons within nucleus supports model of independent movement of nucleons • Using harmonic oscillator model can fill first three levels before results deviate from experiment (2,8,20 only) • Include spin-orbit coupling to account for magic numbers • Orbital angular momentum (l) and nucleon spin (±1/2) interact • Total angular momentum must be considered • (l+1/2) state lies at significantly lower energy than (l-1/2) state • Large energy gaps appear above 28, 50, 82 & 126**Single Particle Shell Model**• Assumes nucleons are distributed in a series of discrete energy levels that satisfy quantum mechanics (analogous to atomic electrons) • As each energy level is filled a closed shell forms • Protons and neutrons fill shells and energy levels independently • Mainly applicable to ground state nuclei • Only considers motion of individual nucleons**Shell Model and Magic Numbers**• Magic numbers represent closed shells • Elements in periodic table exhibit trends in chemical properties based on number of valence electrons (Noble gases:2,10,18,36..) • Nuclear properties also vary periodically based on outer shell nucleons**Pairing**• Just as electrons tend to pair up to form a stable bond, so do like-nucleons; pairing results in increased stability • Even-Z and even=N nuclides are the most abundant stable nuclides in nature (165/275) • From 15O to 35Cl all odd-Z elements have one stable isotope while all even-Z elements have three • The heaviest stable natural nuclide is 20983Bi (N=126) • The stable end product of all naturally occurring radioactive series of elements is Pb with Z=82**Shell Model Evidence - Abundances**• The most abundantly occurring nuclides in the universe (terrestrial and cosmogenic) have a magic number of protons and/or neutrons • Large fluctuations in natural abundances of elements below 19F are attributed to their use in thermonuclear reactions in the prestellar stage**Shell Model Evidence - Stable Isotopes**Stable Isobars • The number of stable isotopes of a given element is a reflection of the relative stability of that element. Plot of number of isotopes vs N shows peaks at • N = 20, 28, 50, 82 • A similar effect is observed as a function of Z # of Isobars # of Neutrons**Shell Model Evidence - Alpha Decay**• Shell Model predictions: • Nuclides with 128 neutrons => • short half life • Emit energetic • Nuclides with 126 neutrons => • Long half life • Emit low energy **Shell Model Evidence - Beta Decay**• If product contains a magic number of protons or neutrons the half-life will be short and the energy of the emitted will be high N = 19 N = 20 N = 21 Z = 21 Z = 20 Z = 19**Shell Model Evidence - Neutrons**• Neutrons do not experience Coulomb barrier even thermal neutrons (low kinetic energy) can penetrate the nucleus • Inside nucleus neutron experiences attractive strong force and becomes bound • To escape the nucleus a neutron’s KE must be greater than or equal to the nuclear potential at the surface of the nucleus • Observation: the absorption cross section for 1.0MeV neutrons is much lower for nuclides containing 20, 50, 82, 126 neutrons compared to those containing 19, 49, 81, 125 neutrons**Shell Model Evidence - Energy**• The energy needed to extract the last neutron from a nucleus is much higher if it happens to be a magic number neutron • Energy needed to remove a neutron • 126th neutron from 208Pb = 7.38 MeV • 127th neutron from 209Pb = 3.87 MeV**Shell Model Evidence - Nucleon Interactions**• Every nucleon is assumed to move in its own orbit independent of the other nucleons, but governed by a common potential due to the interaction of all of the nucleons • Implication: in ground state nucleus nucleon-nucleon interactions are negligible • Implication: mean free path of ground state nucleon is approximately equal to the nuclear diameter • Experimental data does not support this conclusion!!!**Shell Model Evidence - Nucleon Interactions**• Scattering experiments show frequent elastic collisions • Implication: mean free path << nuclear radius • Explanation: Pauli exclusion principle prohibits more than two protons or neutrons from occupying the same orbit (protons and neutrons are fermions) • Why Pauli?: nucleon-nucleon collisions result in momentum transfer between the participants BUT all lower energy quantum states are filled occurrence forbidden • Severely limits nucleon-nucleon collision rate**Nuclear Potential Well**• Nucleon orbit = nucleon quantum state • Similar to quantum state of valence electron BUT • Nucleon “feels” total effect of interactions of all nucleons • Implication: nuclear potential is the same for all nucleons • Strong Force • All nucleons (regardless of their electrical charge) attract one another • Attractive force is short range and falls rapidly to zero outside of the nuclear boundary (~1 fm)**Nuclear Potential - Protons**• Protons do experience a Coulomb barrier a proton must have kinetic energy equal or greater than ECoul to penetrate the nucleus • If Eproton< ECoul proton will back scatter • Inside nucleus proton experiences attractive strong force and becomes bound • To escape the nucleus a proton’s kinetic energy must be greater than or equal to ECoul (in the absence of quantum tunneling)**Nuclear Potential - Neutrons**• Neutrons do not experience Coulomb barrier even thermal neutrons (low kinetic energy) can penetrate the nucleus • Inside nucleus neutron experiences attractive strong force and becomes bound • To escape the nucleus a neutron’s kinetic energy must be greater than or equal to the nuclear potential at the surface of the nucleus**Nuclear Potential Well**Depth of well represents binding energy**Nuclear Potential Functions**• Square Well Potential • Harmonic Oscillator Potential • Woods-Saxon • Exponential Potential • Gaussian Potential • Yukawa Potential: Note: R = nuclear radius r = distance from center of nucleus**Nuclear Potential Functions**Exact shape of well is uncertain and depends on mathematical function assumed for the interaction Yukawa Exponential Gaussian Square Well**Neutron vs Proton Potential Wells**Coulomb repulsion prevents potential well from being as deep for protons as for neutrons**Quantized Energy Levels**• Schrodinger Equation: developed to find wave functions and energies of molecules; also can be applied to the nucleus • Choose functional form of nuclear potential well and solve Schrodinger Equation: H = E • Wave equation allows only certain energy states defined by quantum numbers • n = principal quantum number, related to total energy of the system • l = azimuthal (radial) quantum number, related to rotational motion of nucleus • ms = spin quantum number, intrinsic rotation of a body around its own axis**Angular Momentum**• Associated with the rotational motion of an object • Like linear motion, rotational motion also has an associated momentum • Orbital angular momentum: pl = mvrr • Spin angular momentum: ps = Irot • A vector quantity always has a distinct orientation in space**Magnetic Quantum Effects - Spin**• A rotating charge gives rise to a magnetic moment (s). • Electrons and protons can be conceptualized as small magnets • Neutrons have internal charge structure and can also be treated as magnets • In the absence of a B-field magnets are disoriented in space (can point any direction) • In the presence of a B-field the electron, proton and neutron spins are oriented in specific directions based upon quantum mechanical rules**Spin Angular Momentum**No External B-field Applied External B-field { Project spin angular momentum onto the field axes Allowed values are units of hbar ps(z) = hbar ms**Spin Angular Momentum**• Quantum mechanics requires that the spin angular momentum of electrons, protons and neutrons must have the magnitude • s is the spin quantum number • For protons and neutrons (just like for electrons) spin is always 1/2**Magnetic Quantum Effects - Orbitals**• The orbital movement of an atomic electron or a nucleon gives rise to another magnetic moment (l) • This magnetic moment also interacts with an external B-field in a similar manner to the spin magnetic moment • Quantum mechanics governs how the orbital plane may be oriented in relation to the external field • The orbital angular momentum vector (pl) can only be oriented such that its projection onto the z-axis (field axis) has values pl (z) = hbar ml • Where ml = magnetic orbital quantum number ml = -l, -l+1, -l+2….0…. l-2, l-1, l**B-field axis**pl pl(z) 3 h 2 h 1 h 0 h -1 h -2 h -3 h Orientations of ml in a magnetic field Orbital Angular Momentum • Project orbital angular momentum onto the field axes • Allowed values are units of hbar • pl(z) = hbar ml**Orbital Angular Momentum**• Quantum mechanics requires that the orbital angular momentum of electrons, protons and neutrons must have the magnitude • l is the orbital quantum number • Allowed values of l: • Nucleons: 0 l • Electrons: 0 l< (n-1) • For nucleons (but not electrons) l can exceed n**Orbital Angular Momentum**• The numerical values of the orbital angular momentum quantum number (l) are designated by the familiar spectroscopic notation • Remember: l can only have positive integral values (including 0)