1 Lecture 3 Nuclear Decay modes, Nuclear Sizes, shapes, and the Liquid drop model Introduction to Decay modes (continued) Gamma Decay Electromagnetic radiation corresponding to transition of nucleus from a higher excited state to a lower excited state. A Z 110 * A E E Z m 110 47 Ag 47 Ag ; t 1/2 = 249.8 d The m in the superscript by the mass number denotes a metastable state. Notice that in gamma decay the atomic number of the parent and the atomic number of the daughter are unchanged. Sometimes a superscript g is written to denote the ground state. Three modes of gamma decay: a) Pure gamma emission b) Internal conversion (IC) c) Pair production (PP) A Z E a) Pure gamma emission : The gamma rays emitted by a nucleus in the gamma decay process are monoenergetic for each transition between energy levels. The gamma energies typically range from 2 kev to 7 MeV. Obviously energy must be conserved and momentum must be conserved. Thus, a small recoil energy/momentum is imparted to the daughter nucleus (can generally be ignored). Nevertheless, this energy is small so the energy of the gamma is very close to the energy of the transition. b) Internal conversion (IC): The excited nucleus de-excites by transferring its energy to an orbital electron. This energy exceeds the binding energy of that electron so consequently that electron is ejected. * A Z E ICelectrons Xrays The IC electrons are mono-energetic. Their kinetic energy is equal to the energy of the transition minus the binding energy of the electron. Internal conversion and pure gamma decay are competing processes! Because IC decay results in a vacancy in an atomic orbital in the daughter nucleus, the electrons in the daughter nucleus shuffle down resulting in the emission of X-rays and Auger electrons. As IC and pure gamma decay compete we can define the internal conversion coefficient ( ): decay ; ICdecay
2 c) Pair Production : For nuclear transitions with energies greater than 1.022 MeV, it is possible for the decay energy to directly produce an electron-positron pair which is then ejected from the nucleus. Recall E=mc 2 where m= 2m e. The total kinetic energy of the pair is equal to the difference between the transition energy and 1.022 MeV (which is 2m e ). e.g 16 m 16 8O 8 O ; E trans = 6.05 MeV ; t 1/2 = 7 x 10-11 s Fission Spontaneous fission is the process in which a heavy nucleus breaks into two smaller nuclei. From the curve of <BE> we understand why this is energetically favorable. 98 151 Cf Sr Nd 2n 252 98 38 60 The resulting heavy nuclides following fission are called fission products. Notice that a couple of neutrons are also produced. Fission can also be induced by bombarding a heavy target nucleus with neutrons or charged particles thus exciting it. Obviously this process is not spontaneous! e.g. 1 140 94 U n Ba Kr 2n 235 92 56 36 Remember conservation of mass number and atomic number (charge). Nuclear Sizes and Shapes Size Measurements Perform relative to a standard length. Atoms and Nuclei microscopic: measurement: meter stick? Criterion of applicability: size of object Perform scattering experiments (fog analogy) 1. Metric: Photon or Particle Wave Lengths: DeBroglie Wave Length photon hc 1.24 10 3 fm E E (MeV) h p h mv 28.7 fm A E (MeV) e.g. Rutherford Scattering Experiment NUCLEAR ATOM
3 Relative Dimentions probe Atoms(10 8 cm) Nuclei (<10 12 cm) photon ~ 10 kev (v c) ~ 100 MeV (v c) electron ~ 100 ev (v ~ 0.1c) ~ 100 MeV (v c) nucleon ~ 0.1 ev (v ~ 10 5 cm/s) ~ 10 MeV (v ~ 0.1 c) RESULT: Nuclear Sizes(r) and Densities ( ) Density Distributions: (r) Product of Scattering Experiments Nuclear Size Density Figure
4 2. Nucleus of A Nucleons a. First Approximation: Uniform Density Sphere Volume V A V nucleon = (4/3) R 3 ; V nucleon constant b. Liquid Drop Analogy: Assume all nucleons uniformly distributed for V then, R = [(3A V nucleon )/4 ] 1/3 = r 0 A 1/3 ; r 0 is the nuclear radius constant. r 0 1.2 1.4 fm (ALWAYS GIVEN) c. Calculate the radius of 216 Po ; r0 = 1.40 fm 84 R = r 0 A 1/3 = (1.40 fm)(216) 1/3 = 8.40 fm = 8.40 10 13 cm 3. Radii of Real Nuclei: Second Order Approximation a. Electron Scattering Results: Fig 2-4 SLAC: 21 GeV e s ; ~ 0.1 fm CONCLUSION: Central uniform density and diffuse surface (cloudy crystal ball)
5 b. Woods-Saxon Shape (Fermi function) (r) = 0 (r R 1/2 )/d OPTICAL MODEL 1 e 0 = central density ( ~ 2 10 14 g/cm 3 ) R 1/2 = half-density radius ; R 1/2 = r 0 A 1/3, r 0 = 1.07 fm i.e., radius at which = 0 /2 d = diffuseness ; distance over which (r) decreases from 0.90 0 to 0.10 d~ 2.4 fm
6 Nuclear Shapes 1. Spherical Near N or Z = 2, 8, 20, 28, 50, 82 and 126 (neutrons) MAGIC NUMBERS 2. Spheroidal : For nucleon numbers midway between magic number a. Prolate: a > b = c rugby ball b. Oblate: a < b = c discus c b a 3. Exotic Shapes Octupole (pear-shaped) ; fission (dumbell) ; Scissors
7 Figure 3 1 Motion of the charged particles (protons) within a nucleus represents a current just as the electrons moving through a wire do. Consequently a magnetic field is generated by the nucleus. This field is static (unchanging) and is called the magnetic moment of the nucleus. This magnetic moment tells us about the shape of the nucleus.
8 Nuclear Models Philosophy and Difficulties Nuclear Force no analytical expression Many-Body Problem no mathematical solution computer approximations MODELS Macroscopic Properties Energetics, Sizes, Shapes Assumes all nucleons are alike (except charge) MicroscopicModels Spins, Quantum States, Magic Numbers Assumes all nucleons are different Unified Model Liquid Drop Model (Neils Bohr- 1940 s) Assumption: Justification The nucleus is a charged, nonpolar liquid drop, Chemical analogy: a cluster of Xe and Xe + atoms held together by Van der Waals attractions 1. Nuclear Behavior: Similarities to liquid drop a. Force is short-ranged ; i.e., sharp boundary at surface b. Force is saturated; i.e. all nucleons in bulk of the liquid are bound equally, independent of radius c. Nucleus is incompressible at low temperatures accounts for uniform density and constant <BE> d. Surface Tension Surface nucleons lose binding ; spherical 2. Differences a. Few Particles ; A 270 vs 10 23 b. Protons carry charge c. Two types of particles d. Result: microscopic properties exert significant influence and modify simple results.
9 B. Contributions to the Total Binding Energy of a Charged Liquid Drop 1. Attractive Forces STRONG NUCLEAR FORCE: EQUAL AMONG ALL NUCLEONS Rationale: <BE> curve (Fig. 2.1) <BE> ~ 8 MeV = TBE/A TBE = C 1 A = E v, where E v is called the volume term NOTE: TBE Volume = 4/3 R 3 and C 1 is a constant related to strength of nuclear force THIS IS THE PRIMARY ATTRACTIVE TERM neutron stars) (~ only term in 2. Loss of Binding Energy Nucleus is a small system and also contains charged particles; two components (n & p) Major loss terms: subtract from attractive term a. Surface Tension: Surface Energy -- E s Nucleons on surface lose binding relative to bulk Surface Energy E s : DECREASES TOTAL BINDING ENERGY E s = 4 R 2, where = surface-tension constant E s = 4 (r 0 A 1/3 ) 2, and 4 R 2 is the surface area of a sphere E s = C 2 A 2/3, ; small nuclei lose binding to greater extent than large nuclei (A 2/3 /A) b. Electric Charge: Coulomb Energy -- E c Protons are charged and repel one another DECREASES BINDING Coulomb's Law: E c = 3 5 sphere 2 2 Ze R ; Electrostatic energy of a charged
10 E c = C 3 Z 2 /A 1/3 Nuclei with large atomic numbers lose binding (Z 2 ) E s, and E c ARE MAJOR SOURCES OF BE LOSS c. Two-Component Liquid: Symmetry Energy E sym Raoult's Law: Minimum energy of a two-component solution with non-polar attractive forces occurs when mole fractions are equal (e.g. C 8 H 18 and C 10 H 22 ) p A o p o A p o B p o A p o B Symmetry Energy: 0 0.5 1 X B i.e., minimum E loss when N = Z E sym = C (N Z) 4 2 A 2 d. Diffuse Surface Correction E diff = C 5 Z 2 /A ; small ; we'll ignore e. Pairing Effect Small but systematic differences depending on even-odd character of nucleus Nucleons of the same type prefer to exist in pairs (n-n, p-p) (anti-hund's rule) WHY? Empirical evidence: stable nuclei even Z. even N even Z, odd N odd Z, even N odd Z, odd N (e-e) (e-o) (o-e) (o-o) Number 157 55 50 4 ( 2 H, 6 Li, 10 B & 14 N)
11 Pairing Energy: E p = C 6 /A 1/2, where = + 1 e-e 0 e-o & o-e 1 o-o 4. Additional Terms i.e., extra binding for e-e and extra loss for o-o C 1 to C 6 give fundamental information about nuclear matter; Equations with up to 250 parameters have been added (based on ~2500 pieces of data). However, fit improvement is small and physical significance of C i 's is modified. D. Semi empirical Mass Equation: Nuclear Equation of State (T=0) (half theory/half data) Equation summarizes the macroscopic properties of nuclei in their ground states (T=0). Useful for predictive purposes, searching for new isotopes and elements, and astrophysics 1. Simplest Form TBE = C 1 A C 2 A 2/3 C 3 Z 2 /A 1/3 C 4 (N-Z) 2 /A 2 + C 6 /A 1/2 <BE> = C 1 C 2 A 1/3 C 3 Z 2 /A 4/3 C 4 (N-Z) 2 /A 3 + C 6 /A 3/2 Nuclear Attraction Surface Loss Charge Loss Asymmetry Loss Pairing Gain/Loss
12 2. Constants C C i a. Derive from ~ 2500 known nuclear masses with only C 1 CC 6 terms get 5.0 MeV accuracy (0.005u) b. Deviations: Evidence for shell structure and sharp changes Nuclear Shell Model proposed independently by Maria Goeppert Mayer and Johannes H. D. Jenson in 1949. (Nobel Prize in Nuclear Physics to Goeppert Mayer in 1963).
13 c. Physical significance C 1 strength of nuclear force C 2 nuclear shapes and compressibility C 3 nuclear radius d. With ~250 parameters, get 0.1 MeV accuracy (0.0001 u) 3. Interpretation a. 56 Fe Most Stable Differentiate <BE> with respect to Z and A; set to zero; gives minimum at 56 Fe REASON: Competition between surface-energy losses (favors large A) and Coulomb losses (favors low Z) balance at 56 Fe REMEMBER THIS b. Loss of binding at low A / Increase in Q value for fusion Large surface energy loss at low A since most nucleons are on surface c. Loss of binding at high A / Increase in Q value for fission Coulomb repulsion competes with nuclear force d. N/Z ratio > 1 for heavy nuclei e. Fine Structure: Pairing 4. Problem: Which is more stable: 124 Sn or 200 Hg? 50 80 Major Terms: Surface Energy small for both (1/A 1/3 ), favors 200 Hg slightly. Coulomb Energy large for 200 Hg (80 2 /50 2 ), favors 124 Sn. Minor Terms: Symmetry Energy 200 Hg : (N-Z) 2 = 40 2 80
14 Pairing Energy 124 Sn : (N-Z) 2 = 34 2 ; favors 124 Sn 50 No effect, both e-e.