Collective model. Large quadrupole moments nucleus as a collective

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1 Collective model Large quadrupole moments nucleus as a collective body (Liquid drop model). Interactions between outer nucleons and closed shells cause permanent deformation. Single-particle state calculated in a non-spherical potential ti complicated. Spacing between energy levels depends on size of distortion. ti Doubly magic 1 st excited state away from GS. Near closure single-particle l states. t Further away from closure collective motion of the core excited states. t 1

2 Collective model A net nuclear potential due to filled core shells exists. Collective model combines both liquid drop model and shell model. Two major types of collective motion: Rotations: t ti Rotation ti of a deformed d shape. Vibrations: Surface oscillations.

3 Collective model Rotational motion observed for non-spherical nuclei. Deformed nuclei are mainly 150 < A < 190 and A > 0. Ellipsoid of surface: [ 1 β ( θ, ] R ( θ, φ) = r + Y φ β = 4 π R 3 5 r A A 0 ) 0 Deformation parameter. 1 3 R av Difference between semimajor and semiminor axes. HW 8 Problems 5.11 and 5.1 in Krane. Discuss effect on quadrupole moment. 3 β > 0 β < 0

4 Collective model Symmetry axis 1 E = gω l = g ω l h E = = I ( I g g GS (even-even) even) 0 + Symmetry only even I + E (0 ) = E( E (4 + + ) ) 0 / g ) + 1) = 6( h / g ) = 91.4keV h / g = 15.3keV = 0( h 4

5 Collective model HW 9 compare measured energies of the states of the ground state rotational band to the calculations. Rigid body or liquid drop? Intermediate Short range and saturation ti of nuclear force. 5

6 Collective model HW 9 9 (continued) parity E Spin measured E/E(+ ) I(I + 1)/6 (kev) Higher angular momentum centrifugal stretching 1 higher moment of inertia lower energy than expected additional evidence for lack of rigidity Er 6

7 Odd-A 1 E = gω + Enucleon Collective model 7

8 Collective model Average shape Instantaneous shape 8

9 Collective model R(t) = R + α (t) Y av λµ λµ Instantaneous coordinate r 0 A 1/3 λ λ µ = λ Amplitude (θ,φ) Spherical harmonics α = λµ α λ, -µ Symmetry λ = 0 monopole λ = 1 λ = λ = 3 dipole quadrupole octupole. 9

10 Collective model λ = 0 R(t) = R avr +αα 00 Y 00 monopole R(t) = R avr + α1µ Y1µ ( θ, φ ) = = R R avr avr 1 µ = 1 + α + α Y Y α θ λ = 1 dipole 10 Y 10 + α 1, -1 Y 1, -1 Both monopole and dipole excitations require high energy. 10

11 Collective model R(t) = R avr + α µ Yµ ( θ, φ ) = R avr µ = + α = R + α avr 0 Y Y 0 + α 1 Y 1 + α 0 Y 0 + α,-1 Y λ = quadrupole,-1 + α, - Y, - Quantization of quadrupole vibration is called a quadrupole phonon. A phonon carries two units of angular momentum and even parity (-1 ). This mode is dominant. For most even-even nuclei, a low lying state with J π = + exists. Octupole phonon. 11

12 Collective model Triplet 0 +, +, 4 + l = 4 µ = +4, +3, +, +1, 0, -1, -, -3, -4 l = µ =+, +1, 0, -1, - l = 0 µ =

13 Collective model Two-phonon triplet at twice the HW 30 energy of the single phonon state. Krane

14 Nuclear Reactions X(a,b)Y First in 1919 by Rutherford: 4 He + 14 N 17 O+ 1 H 14 N(α,p) 17 O Incident particle may: change direction, lose energy, completely be absorbed by the target Target may: transmute, t recoil b = γ Radiative Capture. If B.E. permits fission i (comparable masses). Different exit channels a + X Y 1 + b 1 Y +b Y 3 + b 3. 14

15 Nuclear Reactions Recoil nucleus Y could be unstable β or γ emission. One should think about: Reaction dynamics and conservation laws i.e. conditions necessary for the reaction to be energetically possible. Reaction mechanism and theories which explain the reaction. Reaction cross section i.e. rate or probability. 15

16 Nuclear Reactions Conservation Laws Charge, Baryon number, total energy, linear momentum, angular momentum, parity, (isospin??). a p b b θ p a X φ py Y m c m c = T T = i f f i +ve Q-value exoergic reaction. -ve Q-valueQ endoergic reaction. T + T = Q + b +ve Q-value reaction possible if T a 0. -ve Q-value reaction not possible if T a 0. (Is T a > Q sufficient?). Conservation of momentum Y T a Q 16

17 Nuclear Reactions Conservation of momentum. We usually do not detect Y. Show that: T b = m a m T b a cosθ ± m a m T b a cos m θ + ( m + m HW 31 The threshold h energy (for Y T b a ): (the condition occurs for θ = 0º). my + mb T Th = Q my + mb ma +ve Q-value reaction possible if T a 0. Coulomb barriers.!!! -ve Q-value reaction possible if T a > T Th. Y + m b )[ m Y Q + ( m Y m a ) T a ] 17

18 Nuclear Reactions HW (continued) The double valued situation occurs between T Th and the upper limit T a\. m T \ = QQ Y a m m Y a Double-valued in a forward cone. cos θmax = ( m + m )[ m Q + ( m m ) T ] Y b Y m a m T b a Y a a 18

19 Nuclear Reactions Sample 19

20 Nuclear Reactions 0

21 Nuclear Reactions 1

22 Nuclear Reactions

23 Nuclear Reactions Q If the reaction reaches excited states of Y ex = mx c + mac ( my c + Eex ) mbc = Q0 58 Ni(α,p) 61 Cu E ex even less. less proton energy Highest proton energy See Figures 11.4 in Krane 3

24 Nuclear Reactions Neutron scattering Inelastic Q = -E ex (=-E*). Elastic Q = 0. HW 3 Discuss the elastic and inelastic scattering of neutrons using the relations you derived in HW 31. 4

25 Nuclear Reactions Categorization of Nuclear Reactions According to: bombarding particle, bombarding energy, target, reaction product, reaction mechanism. Bombarding particle: Charged particle reactions. [ (p,n) (p,α) (α,γ) heavy ion reactions ]. Neutron reactions. [ (n,γ) (n,p).. ]. Photonuclear reactions. [ (γ,n) (γ,p) ]. Electron induced reactions. Bombarding energy: Thermal. Epithermal. Slow.? Neutrons. Fast. Low energy charged particles. High energy charged particles. 5

26 Nuclear Reactions Targets: Light nuclei (A < 40). Medium weight nuclei (40 < A < 150). Heavy nuclei (A > 150). Reaction products: Scattering. Radiative capture. Fission and fusion. Spallation... Reaction mechanism: Direct reactions. Elastic 14 N(p,p) 14 N Inelastic 14 N(p,p / ) 14 N* Compound nucleus reactions. More in what follows. What is a transfer reaction.????? 6

27 Reaction Cross Section(s) (Introduction) Probability. Projectile a will more probably hit target X if area is larger. Classically: σ = π(r a + R X ). Classical σ =??? (in b) 1 H+ 1 H, 1 H+ 38 U, 38 U+ 38 U Quantum mechanically: σ = π D. HW 33 m a + m X h h D = = m CM X maea µ ax EaX Coulomb and centrifugal barriers energy dependence of σ. Nature of force: Strong: 15 N(p,α) 1 C σ = 0.5 b at E p = MeV. Electromagnetic: 3 He(α,γ) 7 Be σ = 10-6 b at E α = MeV. Weak: p(p,e + ν)d σ = 10-0 b at E p = MeV. Experimental challenges to measure low X-sections.. 7

28 Reaction Cross Section(s) (Introduction) I a θ,φ Detector for particle b dω cm dσ = b particles / s dr I a b N Typical nucleus (R=6 fm): geometrical πr 1 b. Typical σ: <µb to>10 6 b. 8

29 Reaction Cross Section(s) (Introduction) Many different quantities are called cross section. Krane Table 11.1 Units! dω = sinθθ d θd φ σ = dσ dω = Differential cross section σ(θ,φ) or σ(θ ) or cross section!! π sinθdθ π dσ d φ dω dω 0 0 σ t for all b particles. Angular distribution dr b = dσ = dω dσ de r( θ, φ) r( θ, φ) 4πI N a dω 4 π Doubly differential d σ de b dω Energy state in Y 9

30 Coulomb Scattering V = 0 T = ½mv a o l = mv o b b v o Elastic or inelastic. Elastic Rutherford scattering. 1 1 At any distance: v min r min 1 zze mv o = mv + 4πε o r V T a 1 = 4πε = 0 o zze d d No dependence on φ 30

31 Coulomb Scattering 31

32 Coulomb Scattering b θ < b > θ db dθ n target nuclei / cm 3 x target thickness (thin). nx target nuclei / cm HW 34 Show that and hence dσ ( θ ) dω b = zze = 4πε o d θ cot 1 4T 1 sin Rutherford cross section a 4 θ b f df = ( nx ) π b = ( nx ) π bdb 3

33 Coulomb Scattering Study Fig (a,b,c,d) in Krane See also Fig in Krane. HW 35 Show that the fraction of incident alpha particles scattered at backward angles from a µm gold foil is 7.48x

34 Coulomb Scattering Elastic Rutherford scattering. Inelastic Coulomb excitation. See the corresponding alpha spectrum of Fig in Krane. 34

35 Coulomb Scattering 35

36 Nuclear Scattering Elastic or inelastic. Analogous to diffraction. Alternating maxima and minima. First maximum at λ θ h R λ = R = p R o A 1 3 Minimum not at zero (sharp edge of the nucleus??) Clear for neutrons. Protons? High energy, large angles. Why? Inelastic Excited states, energy, X-section and spin-parity. 36

37 Compound Nucleus Reactions E CM a Q CN Direct CN decays Time. Energy. Two-step reaction. CN forgets how it was formed. Decay of CN depends on statistical factors that are functions of E x, J. Low energy projectile, medium or heavy target. 37

38 Compound Nucleus Reactions 38

39 Compound Nucleus Reactions Consider p + 63 Cu at E CM p = 0 MeV. Calculate E CM +[m( p 63 Cu) + m(p) m( 64 Zn)]c. Divide by 64 available energy per nucleon << 8 MeV. Multiple collisions long time statistical distribution of energy small chance for a nucleon to get enough energy Evaporation. Higher incident id energy more particles evaporate. See also Fig in Krane. 39

40 Direct Reactions Random collisions nearly isotropic angular distribution. Direct reaction component strong angular dependence. See also Fig in Krane. 40

41 Direct Reactions Peripheral collision with surface nucleon. 1 MeV incident nucleon D?? more likely to interact with the nucleus CN reaction. 0 MeV incident nucleon D?? peripheral collision Direct reaction. CN and Direct (D) processes can happen at the same incident particle energy. Distinguished by: D (10 - s) CN ( s). [Consider a 0 MeV deuteron on A=50 target nucleus]. Angular distribution. 41

42 Direct Reactions (d,n) stripping (transfer) reactions can go through both processes. (d,p) stripping (transfer) reactions prefer D rather than CN; protons do not easily evaporate (Coulomb). [(p,d) is a pickup reaction]. What about (α,n) transfer reactions? HW 36 Show that for a (d,p) reaction taking place on the surface of a 90 Zr nucleus, and with 5 MeV deuterons, the angular momentum transfer can be approximated by l = 8sin(θ/), where θ is the angle the outgoing proton makes with the incident id deuteron direction. (Derive a general formula first). J π ( 90 Zr )=0 + l gs Fig in Krane. J( 91 Zr) = l ± ½,π = (-1) l Optical model, DWBA, Shell model, Spectroscopic Factor. θ 0º 14.4º 9º 44º dσ dω meas = S dσ dω calc 4

43 Neutron-induced Reactions X(n,b)Y( σ n D Y + b H II C C H I X + n 1 1 E v Γ b b( (Q+E n n) Γ n (E n ) v P n ln ( E ) n Probability to penetrate the potential barrier For thermal neutrons Γ b (Q) constant P o (E thermal ) = 1 Q >> E n P >o (E thermal )=0 Non-resonant σ E ) n n ( E ) (Saed Dababneh). 1 v 43

44 Neutron-induced Reactions 44

45 (Saed Dababneh). 45

46 Neutron-induced Reactions n-tof CERN 46

47 (Saed Dababneh). 47

48 Neutron-induced Reactions n_tof CERN 48

49 Neutron-induced Reactions 49

50 Charged Particle Reactions What is the Gamow Peak? Nuclear Radius 50

51 Charged Particle Reactions Electron Screening 51

52 Charged Particle Reactions e = 1.44x10-1 kev.m Tunneling probability: P HW 37 e πη Gamow factor η = Z 1 Z hv e Sommerfeld parameter In numerical units: πη = 31.9Z1Z E µ(u) µ ( ( kev CM ) L+ 1 For γ-ray emission: Multipolarity Γ L ( E γ ) = α LE γ Γ Dipole E γ ) = α E 3 ( 1 γ 5

53 Charged Particle Reactions σ ( E) e πη σ ( E) σ ( E) = πd 1 πη E e 1 E S( E) Nuclear (or astrophysical) S-factor 53

54 Charged Particle Reactions E C =?? 54

55 Resonance Reactions Projectile Projectile Target Q-value Target Q-value Q + E R = E r E γ = E + Q - E ex Non-resonant Resonant Capture Capture σ γ Y H a + γ (all energies) X σ γ (selected energies with large X-section) f H γ Er Er H CN a + E + X σ γ Γ a Γ b 55

56 Resonance Reactions E t CN particle emission E E > spacing between virtual states continuum. Lower part larger spacing isolated resonances. 56

57 Resonance Reactions HW 38 Entrance 38 Channel a + X J π Excited State In the 19 F(p,αγ) reaction: Compound b + Y The Q-value is 8.??? MeV. Nucleus C* The Q-value for the formation of the C.N. is 1.??? MeV. For a proton resonance at 668 kev in the lab system, the corresponding energy level in the C.N. is at 13.??? MeV. If for this resonance the observed gamma energy is 6.13 MeV, what is the corresponding alpha particle energy? If for this resonance there has been no gamma emission observed, what would then be the alpha particle energy? E x Exit Channel 57

58 Resonance Reactions L = l h = bp = b = ld D b h D π π l max l + 1 l σ = b b = ( l + 1 ), max + HW 40 πdπ D ( b ) = µ ( u) E ( πd CM ( kev J + 1 σ max = πd ax (1 + δ (J + 1)(J + 1) a X ω ) ax ) 58

59 Resonance Reactions J a +J X +l= J (-1) l π(j a ) π(j X ) = π(j) (-1) l = π(j) Natural parity. Entrance Channel a + X J π Excited State Compound Nucleus C* E x Exit Channel b + Y 59

60 Resonance Reactions Entrance Channel a + X σ σ J π E x a+x Y+b Q>0 Excited State Exit Channel Compound b + Y Nucleus C* J + 1 a + X Y + b Q > 0 b + Y X + a Q < 0 Inverse Reaction ax = πd ax (1 + δax ) + II I + ( J a + 1)( J X + 1) QM Statistical Identical Nature of force(s). Factor (ω) particles Time-reversal invariance. J + 1 by = πd by (1 + δ by ) a + X H I C C H II b + ( J b + 1)( JY + 1) Y b H C HW 39 C σ σ ax by H a =?? 60 X Y

61 (Saed Dababneh). 61

62 Resonance Reactions Damped Oscillator Oscillator strength response f ( ω ) δ ω o + ( ( ) δ = 1 Damping 0 factor t eigenfrequency σ ( E) ( E E Γ R a Γ ) b + ( Γ ) Γt o = h 6

63 Resonance Reactions σ ( E) = πd ax J + 1 Γ (1 ) ( + ax J + 1)( J + 1) δ ( a X E E R Breit-Wigner formula All quantities in CM system Only for isolated resonances. σ σ R e Γ Γ a a Γ Γ σ R Γb = σ e Γ a b a Reaction Elastic scattering a Γ ) Γ = Γ b + ( ) + Usually Γ a >> Γ b. Γ a Γ b HW 41 When does σ R take its maximum value? 63

64 Resonance Reactions What is the Resonance Strength? HW 4 What is its significance? In what units is it measured? J J + 1 Γa Γb ωγ = (1 + δ ax ) (J a + 1)(J X + 1) Γ Cro oss sect tion ωγ Γ a E C Charged particle radiative capture (a,γ) (What about neutrons?) ωγ Γ γ Energy 64

65 14 N(p,γ) HW Resonance Reactions HW 43 Q =?? E C =?? E C.M. R =.0 MeV Formation via s-wave protons, Take J = ½, Γ p = 0.1 MeV, dipole radiation E γ = 9.3 MeV, Γ γ = 1 ev. Show that ωγ = 0.33 ev. If same resonance but at E R = 10 kev Γ p =?? E γ =?? Γ γ (dipole) =?? Show that ωγ = 3.3x103x10-3 ev. Huge challenge to experimentalists 65

66 α-transfer reactions Angular distribution Resonance J π Estimated Energy (kev) ωγ (µev) Experimental upper limit < 1.7 µev 18 O(α, γ) Ne 66

67 Neutron Resonance Reactions 67

68 Neutron Activation Analysis (Z,A)( ) + n (Z, A+1) ) β - γ (β-delayed γ-ray) (Z+1, A+1) 68

69 Neutron Attenuation Neutrons Target Thickness x di I = σ t ndx I = I o e σ nx t Similar to γ-attenuation. Why? 69

70 Neutron Moderation Show that, after elastic scattering the ratio between the final neutron energy E \ and its initial energy E is given by: \ CM E A + 1+ Acosθ = E ( A +1) \ E A 1 For a head-on collision: = E min A + 1 After n s-wave collisions: where ζ = ln E E \ = 1+ av ln \ E n ( A 1) A = ln ln E A 1 A + 1 nζζ HW 44 70

71 Neutron Moderation HW 44 (continued) How many collisions are needed to thermalize a MeV neutron if the moderator was: 1 H H 4 He 1 C 38 U Discuss the effect of the thermal motion of the moderator atoms. 71