Lecture 15: Nuclear Reactions RELEVANCE: Basic Research reaction mechanisms structure exotic nuclei Applications: analytical tool neutron activation analysis astrophysics origin of elements space radiation effects satellites, Mars landing materials science T c (high) superconductors medical therapy tumor treatment nuclear power 25% of U.S. electricity DEFINITION: A collision between two nuclei that produces a change in the nuclear composition and/or energy state of the colliding species ; i.e., 2 nd order kinetics 1. Nomenclature A. Constituents 1. Projectile: nucleus that is accelerated (v > 0) ; 0 2 10 12 ev a. neutrons reactors b. light ions A 4: AGS (NY), FNAL (IL) c. heavy ions A > 4 U: ANL (IL), MSU (MI), TAMU (TX), RHIC (NY) d. electrons (and photons) SLAC (CA), CEBAF (VA), MIT (MA) e. exotic beams:, K, p : FNAL f. radioactive beams: MSU, ANL (IL) HHJRF (TN) 2. Target: fixed (v = 0), usually But now have colliding beams, RHIC, FNAL
3. Products ANYTHING PERMITTED BY CONSERVATION LAWS. Note: since projectile kinetic energy can be converted into mass, only limit on mass-energy is beam energy. B. Notation: Target [projectile, light product(s)] heavy product 7 e.g., 3 Li + 208 82 Pb 215 At + : 212 At + 3n : 208 Pb( 7 Li, ) 215 At 208 Pb( 7 Li, 3n) 212 At 197 Au + 18 C : 208 Pb( 7 Li, 18 C) 197 Au fission : 208 Pb( 7 Li, f) C. Energetics: Q-values revisited 1. Q = (reactants) (products) = available energy 2. Example: 208 Pb( 7 Li, ) reaction: Q = ( 208 Pb) + ( 7 Li) ( 215 At) (, ) Q = 21.759 + 14.908 (1.263) 0 Q = 5.589 MeV ENDOTHERMIC Accelerate 7 Li to 5.589 MeV, no reaction. WHY? II. Energetic Conditions for Reaction A. Energetic Threshold: E th & Excitation Energy: E* 1. DEFINITION: E th is the minimum projectile energy necessary to satisfy mass-energy and momentum conservation (i.e., compensate for Q)
2. Derivation: a. Apply conservation Laws projectile + target composite nucleus products Mass-energy: E p + p + t = CN + E CN + E* E p = projectile E T = 0 kinetic energy E* = Excitation energy; When E* = 0 E p = E th E 2 p ; 2m p 2 2mE Linear Momentum: pp + pt = pcn 2M p E p 0 2M E CN CN OR M p E p = M CN E CN A p E p = A CN E CN b. Combining E p + p + t = CN + (A p /A CN )E p + E* E p (A p /A CN )E p = CN p T + E* E p (1 A p /A CN) = E* [p + T CN ] A A E CN p p A = E* Q CN E p (A T /A CN ) = E* Q IF E* = 0 (minimum), then E p = E th and E th = (A CN /A T )(Q) IF E* > E th E* = (A T /A CN ) E p + Q
c. Example: 208 Pb( 7 Li, ) 215 At ; Q = 5.589 MeV E th = Q A A CN T = (5.589 MeV)(215/108) = 5.777 MeV E CN = E th E p = 5.777 5.589 = 0.188 MeV i.e., of total of 5.777 MeV, 5.589 MeV goes into mass (M = E/c 2 ) and 0.188 goes into kinetic energy of the composite nucleus. d. Example: E p = 40.0 MeV, 208 E* = 40.0 MeV 5.589 = 33.1 MeV 215 This energy is converted into heat Fermi Gas model: T E*/A 1 MeV 10 10 K (kt) E* 0 e. Accelerate to 5.777 MeV Still no reaction. WHY? B. Coulomb Barrier Charged Projectiles Only 1. Nucleus-Nucleus Charge Repulsion V coul = (Z e)(z e) d ZZe R +R p T p T R d 2. Potential Energy Surface Nuclear Reactions begin to occur when tails of nuclear matter distributions overlap P T P T Bottom line: r 0 for reactions is greater than for potential well. (r 0 ~ 1.60 fm) (r 0 ~ 1.44 fm) p P 2 T V 0 T ; somewhat fuzzy
3. Net Result V 1.44 Z Z MeV fm 0.90 Z Z p T r ( A A ) ( A A ) cm p T Coul = 0 p 1/3 T 1/3 p 1/3 T 1/3 MeV r 0 = 1.60 fm 4. Momentum Conservation CN must carry off some kinetic energy; same correction as for Mass-Energy Conservation (Q-value) V lab Coul A A CN T V cm Coul 5. Example 208 Pb( 7 Li, ) 215 At V lab Coul 215 208 (82)(3)(0.9) 1/3 (208 7 13 / ) = 29.1 MeV NOW THINGS HAPPEN 6. Diffuse nuclear surface fuzzes precision of Coulomb barrier E th is an EXACT condition V Coul is an APPROXIMATE condition C. Centrifugal Barrier NOT A MINIMUM CONDITION Alpha Scattering Geometry The scattering of the alpha particle by the central repulsive Coulomb force leads to a hyperbolic trajectory. From the scattering angle and momentum, one can calculate the impact parameter and closest
approach to the target nucleus. Rotational energy = ( ) 2I 1 2 where l mvb and I = r 2 where is the reduced mass of the system. The rotational energy is not available for reaction! D. 1. Summary of Energetic Factors E p E th : Mass-Energy Conservation: ABSOLUTE CONDITION 1 st law 2. E p V C l lab Coul : Charge Repulsion Constraint BARRIER PENTRATION Probability low below thiss energy 3. 4. E p E rot : No constraint since = 0 is always possible. IN GENERAL V Coul > E th ; except for very light nuclei and neutral projectiles, e.g., neutronss
III. Reaction Probability: The Second-Order Rate Law Probability = cross section [SAME FOR CHEMICAL Rx] A. Schematic Picture 1. ON-OFF nature of nuclear force suggests a simple geometric model: Touching Spheres Model IF projectile and target touch, REACTION ; b R p + R T IF projectile and target don't touch, NO REACTION; b > R p + R T 2. Bottom Line: Probability is proportional to cross-sectional area Area = (R P + R T ) 2 = b T R = r 0 2 (A p 1/3 + A T 1/3 ) 2 r 0 1.60 fm TOTAL REACTION CROSS SECTION 3. Unit: 1 barn = 1 10 24 cm 2 = 1b
B. Definitions 1. Sequential Process COLLISION STAGE DECAY STAGE (entrance channel) (exit channel) ~ 10 21 10 23 Mass-Energy s ~ 10 15 10 21 s = REACTION CROSS SECTION Equilibration (a,b)= PRODUCTION CROSS SECTION Collision Probability Probability of forming a given product (many may be possible) 2. R = (a,b) i.e., Sum of all possible production 's equals the total reaction 3. Example 40-MeV 4 He + 232 Th 236 U* products (, pn) = 0.010 b 234 91Pa (, 2n) = 0.080 b 234 U (, 3n) = 0.245 b 233 U (, 4n) = 0.125 b 232 U (, f) = 1.190 b fission R = 1.650 b = (, x)
C. Cross Section Measurements: Nuclear Reaction Rates 1. Review of Biomolecular Rate Law A + B C + D [Elementary Reaction only kind for nuclei] Rate = d [A] = k [A][B] dt i.e., a second-order rate process a. [A][B] factor: Collision Probability Defines collision geometry ; e.g., two gases or two liquids, molecular beam + gas, etc b. k = rate constant = probability of reaction IF collision occurs. k = f (H a, T, structure, etc.) 2. Nuclear Case: Charged-Particle-Induced Reactions k = ; [A][B] = n p n T (projectile nuclei target nuclei); definition of n is geometry dependent R = Rate = Number of Product nuclei/time (a, b) OR Number of Reactions/time total = cross section must be given 3. Thick Target: General Case For thick targets, nuclear reactions remove a significant fraction of the beam: I n p = number of projectiles/time I I 0 I = rate of reaction = di/dx x = target thickness I 0 0 x R = (di/dx) = n t I dx, where n t = N 0 /(gat. wt.) i.e., number density (N/cm 3 ) I = I 0 e nx I di x Beer-Lambert Law = -n dx I t 0 I 0
4. Thin Target Case GEOMETRY: Beam + infinitely thin target [thin = no shadowed nuclei] Beam Target = fixed (i.e., v = 0); Area of target > Area of beam N p N T Collision Probability: (N p /t)(n T /area) Reaction probability: = f(q, V c, I, etc.) Thin target Result: Rate = I(n t x) a. I : Projectile current A Z q X A Z X Z I e (I electrons/projectile nucleus) Projectile Beam Target Faraday Cup To count projectiles, measure current i collected in Faraday cup I = i / particle charge q I is measured in Amps, since charged particles 1A = 6.28 10 18 e /s q = Z p e (ion charge) e.g., 10nA 12 C beam (stripped of electrons) 9 1nA 6.28x10 e / s 10 10nA 6.28x10 e / s However, each 12 C has a charge of +6. # 12 C/s = 6.28x10 6e / 10 12 e C / s 1.05x10 10 12 C/s = I 2 b. (n t x) = # t arg etatoms / cm = density x thickness c. TOTAL RATE R = Number of Reactions/unit time R = I(n t x)
266 5. Problem: What is the production rate of 106 Sg if a 100 g/cm 2 target of 248 96Cm is bombarded with a 1.0 A beam of 22 Ne ions? ( 22 Ne, 4n) = 1.0 nb R = I(n t x) = (1.0 10 24 cm 2 )(10 9 ) = 1.0 10 33 cm 2 (n t x) = (100 10 6 )g (6.02 10 23 atoms/mole) = 2.43 10 17 atoms/cm 2 248 g/mole cm 2 I = 1.0 A 6.28 10 12 e /A / (10 e /Ne) = 6.28 10 11 Ne/s R = (1.0 10 33 cm 2 2.4310 ) 2 cm R = 1.5 10 4 /s = 0.54/hr 17 (6.28 10 11 /s). ( 22 Ne, f) = 2.5 b R = 7.6 10 5 /s NOTE: 2 fragments/fission i.e., humongous fission background