Lecture 4: Nuclear Energy Generation

Transcription

1 Lecture 4: Nuclear Energy Generation Literature: Prialnik chapter 4.1 & 4.2!" 1

2 a) Some properties of atomic nuclei Let: Z = atomic number = # of protons in nucleus A = atomic mass number = # of nucleons N = A-Z = # of neutrons Isotopes: nuclei with same Z, different A Isobars: nuclei with same A, different Z Isotones: nuclei with same N, different Z A < 40: A ~ 2Z A > 40: A > 2Z Empirically: nuclear radius: R A 1/3 cm nuclear density ~ independent of A ~ g/cm 3 Binding energy of nucleus: BE = (m p Z + m N N M nucl )c 2 2 (m H Z + m N N M atom )c 2

3 Average binding energy BE per nucleon: f = BE A 3

4 Average binding energy cont. Liquid drop model leads to semi-empirical expression: BE = a 1 A a 2 A 2/3 (Z 1) 2 ([A / 2] Z) 2 a 3 a A 1/3 4 δ o A a i (MeV) a 1 : average BE per nucleon in ideal case 15.7 a 2 : surface tension: surface 17.8 a 3 : Coulomb repulsion of protons: # pairs/r 0.7 a 4 : asymmetry energy 94.5 The first three terms speak for themselves. The a 4 term reflects that when the additional neutrons (or protons for that matter) are added to the nucleus, the Pauli exclusion principle forces the excess neutrons to be in a higher energy level. The 5 th term is the pairwise interaction term and its sign depends on A, Z, & N (see next slide) 4

5 Asymmetry energy δ o = 12 MeV Z, N, and A even 3/4 A δ o = 0 A odd δ o = 12 MeV Z and N odd, & A even 3/4 A Even-even nuclei: stable pairs with opposing spins Odd-odd nuclei relatively unstable; often converted to even-even nuclei via β-disintegration (ΔZ=±1) Nuclei with even Z more stable than those with odd Z Stable nuclei with even A have even Z, except 2 H, 6 Li, 10 B, 14 N Especially stable nuclei: Z and/or N = 2, 8, 20, 28, 50, 82, 126, 5 (magic numbers)

6 Magic numbers: nuclear burning Pattern recognition is a key aspect of science (& mankind). For this lecture on the nuclear energy of stars, the key species are: 4 2 He, 12 6 C, 16 8 O, Ne, Mg, Si,! Fe 8 Be The magic numbers are all a-nuclei up to 32 S (except for: ) 4 6

7 Magic numbers: nuclear physics Nuclear physics looks at magic numbers slightly differently: The liquid drop model is a good guide on the overall behavior of the average binding energy per nucleon. However, detailed structures are not accounted for. There is a set of magic numbers and nuclei with that number of protons or neutrons are more stable than expected. Indeed, double magic numbers are even more stable. This pattern of magic numbers is also obvious when considering eg., elemental abundances, the additional binding energy of a single neutron, neutron absorption cross sections, or the nuclear quadrupole moment. Magic numbers when Z and/or N = 2, 8, 20, 28, 50, 82, 126 Magic Numbers: 50 Sn has 10 stable isotopes, 49 In & 51 Sb have only 2 Double magic numbers: 2 4 He, 8 16 O, Ca, Ca, 7

8 Magic numbers and number of isotopes/isotones & abundance 8

9 Magic numbers &neutron binding energy 9

10 Magic numbers & neutron cross section 10

11 Magic numbers & nuclear quadrupole moment 11

12 Interlude: magic numbers and the electronic structure of atoms This was first interpreted in terms of the Bohr model a quantum interpretation of the classical motion of electrons bound by Coulomb forces to a nucleus and with Sommerfeld s modifications led to the valence shell model for atoms 12

13 Patterns can lead to deep insight: Schrödinger equation for a particle in a spherically symmetric central potential leads to three quantum numbers: n, l, m which are degenerate with respect to l: and we know this from electronic structure studies: n 1 2 ( 2l +1) = 2n 2 0 or magic numbers: 2, 10, 28, if shells fill sequentially Actual magic numbers are 2, 10, 18, 36, 54,... as magic numbers correspond to filled subshells 13

14 14

15 Magic numbers & nuclear potential Magic numbers provide insight in the nuclear potential. For a 3-D harmonic oscillator the energies are specified by the oscillator quantum number, n, and the degeneracy is then 2 n +1 ( )( n + 2) with magic numbers : n +1 = k +1 2, 8, 20, 40, 70, 112,... k 0 ( ) ( )( n + 2) / 2 A square well potential yields the magic numbers: 2, 8, 18, 20, 34, Adding a non-central force (involving spin-orbit coupling) yields the right magic numbers (Woods-Saxon potential) 15

16 Single particle shell model with oscillator potential (actually Woods-Saxon potential) with spin 16 orbit interaction

17 Nuclear potential well For energies involved, see assignment b) Bohr picture of nuclear reactions Reaction: a+x Y+b shorthand: X(a,b)Y Bohr: reaction proceeds in two steps: a+x C* Y+b 1. Formation of compound nucleus C* 2. Disintegration of C* into particular products ( channels ) X + a elastic scattering a + X C* X*+ a Y + b inelastic scattering particle emission : C + γ radiative capture 17 (always required: conservation of energy, angular momentum, parity)

18 Bohr: if lifetime of C* >> time required for nucleon to traverse nucleus (t ~ sec at v~0.1c) then: assume steps 1 and 2 are independent Mode of disintegration of C* depends on its global properties (energy, angular momentum, parity), and not on specific way it was formed 18

19 Two rules of thumb: 1. Particle emission: If excitation energy of C* sufficiently large fission in ~ equal parts otherwise: emission of n, p, α (in that order of likelihood) 2. Radiative capture If excitation energy of C* too small for particle emission γ -emission (very unlikely if particle emission possible) 19

20 Q (MeV) Example: 1 H + 9 Be 10 B* 8 Be + 2 H 2 4 He + 2 H Li + 4 He B + γ 6.49 (rhs arranged in order of likelihood) Each mode has a mean life time for decay, t i, and thus an energy width, Γ i τ i =! Probability to decay through channel i is then P i = Γ i Γ, with Γ= Γ i i Note that the compound nucleus can also decay back to the entrance channel. 20 Γ i

21 c) Cross sections Reaction: a + X C* Y + b Average kinetic energy in the core of a star is much less (~10-3 ) than Coulomb energy barrier (see assignment). Reaction will still occur because of QM tunneling. Consider the incoming particle as a wave characterized by its angular momentum,, and its corresponding impact parameter, b, mvb = l". Semiclassically, the cross section is then, π (" mv) 2 l+1 ( ) 2 l 2 = ( π#2 2l+1) (1 barn=10 24 cm 2 ) 0.6 A(amu)E(MeV ) barn with # the reduced de Broglie wavelength ( = " / p = " / 2mE ) 15 N(p,a) 12 C, s=0.5 barn (strong nuclear force) 3 He(a,g) 7 Be, s=10-6 barn (electromagnetic interaction) p(p,e + g)d s=10-20 barn (weak force) 21

22 C,i) Non-resonant reaction Cross section: σ (a, b) = (2 +1)πλ 2 g P (a) G(b) With: g = statistical factor, G(b) = branching ratio (relative probability that C* decays into b) and P l (a) = tunneling probability ( a) exp 2πη P l 2πη = 2 " r 1 r 0 2m V(r) E dr = 2(2mE)1/2 " r 1 r 0 r 1 r 1 1/2 dr with r 1 = Z 1 Z 2 e2 E = 2Z 1 Z 2 e2 mv 2 the classical turning point 2πη = 4Z Z 1 2 e2 "v P i a ( ) exp 2πZ 1 Z 2 e 2 "v arccos r 0 r 1 r 0 r ( 1 1 r 0 r ) 1 = exp b 22 E1/2 2πZ Z 1 2 e2 "v

23 Aside: Heuristically, the wave function in the classically forbidden zone decreases exponentially: ψ exp (r 1 r)! with r = Z Z e2 /E and! the reduced de Broglie wavelength (see assignment) P ψ 2 = exp 2r 1! = exp 4η Close to the expression derived before: 2πη 31.3Z 1 Z ( 2 A E) 1/2, here A is reduced mass in m u and E is in kev Strongly dependent on mass and charge of nuclei involved 23

24 Cross section: σ (a, b) (2l +1) λ 2 P l (a) σ (E) S(E) E exp[ b E1/2 ] Here S(E) is defined as the astrophysical S factor, which is a weakly varying function of E. As measurements are done at energies well above stellar energies, S(E) is determined through extrapolation. 24

25 C,ii) Resonant reactions Cross sections can be greatly enhanced when the compound nucleus has an (unbound) excited state coinciding with the projectile energy (subject to selection rules). P( E)dE = Γ 2π ( E E p ) 2 + Γ 2 ( ) 2 de where Γ=! τ = Γ p + Γ n + Γ α + Γ β + Γ γ + E R =E p +Q Γ a is the probability to decay back to the entrance channel and to a specific exit channel. The cross section is then: Γ b E is center of mass energy σ ( E) = π! 2 ( E p )g Γ a Γ b ( E E p ) 2 + Γ 2 ( ) 2 25 Breit-Wigner formula

26 Two examples Proton capture by 12 C Non-resonant Extrapolation of laboratory data to stellar energy region reliable C(p, γ ) 13 N Resonant Extrapolation tricky

27 d) Reactions involving neutrons Reactions involving neutrons do not have Coulomb barriers but may have an angular momentum barrier. Classically, L = pd with p momentum, d impact parameter QM: L = l( l +1)! with l = 0 s-wave, l =1 p-wave, etc Classically, energy, E, of particle with angular momentum, L, E = L 2mr 2 QM: E l = l( l +1)! 2 / 2µr 2 E l =12l l +1! " ( ) / A 1 A 2 ( A 1 + A 2 ) ( ) ( ) A 1 2/3 + A 2 2/3 # $ MeV 27

28 s-wave neutron capture has no Coulomb or angular momentum barrier and dominates at low temperature. But note, the transmission probability at the nuclear (step) potential is proportional to E and hence the cross section will scale with 1/ E or e.g., 1/v and <σv> is constant. σ v = 2 π v th σ th v th σ th with v th = 2kT µ Thermal cross section: σ th 28

29 7 Li(n,g) 8 Li: note 1/v behavior and resonance Important for Big Bang nucleosynthesis as well as a model system for 7 Be(p,g) 8 B,which is important for the Solar neutrino 29 problem

30 Neutron capture with higher angular momentum Penetration scales with: P l ( E) E l+1/2 and σ ( E) E l 1/2 Experimental data compared to p-wave direct radiative capture model 30

31 e) Thermonuclear Reaction Rate What is: # of nuclear reactions X(a,b)Y per cm 3 per sec resulting from thermally induced collisions between a and X, as function of ρ and T? Energy release per reaction known Let: a subscript 1, X subscript 2 ( ) ε ρ,t,x i total # of reactions X(a,b)Y per cm 3 per sec is given by: r 12 = n 1 n 2 σ v 12 With συ 12 = π ( kt ) 3/2 m 0 Eσ $ 12(E) exp E ' % & kt ( ) de Note 1: above result general, valid for any interaction Note 2: if reaction endothermic, take σ (υ) = 0 for υ < υ min 31

32 Energy production Let: Q = energy release in 1 reaction X(a,b)Y Then: Q ε 12 = r ρ = n Q an X συ ρ = N 2 0 ρ X a A a For identical particles: ε 11 = 1 2 N 2 ρ X 0 A 2 συ 11 Q 11 X X A X συ 12 Q 12 Abundance change given by: dx i dt = r A m ijkl i u or more general: dx i dt = A m r i u i x + x y 32 r y i

33 f) Calculation of For low energies, far from resonances: Take S=S 0 (constant): συ = 2 1 # 2 & % ( π (kt ) 3/2 $ m ' 1/2 S 0 Integrand is sharply peaked: Gamow peak Occurs at E=E 0 >> kt, with:! E 0 = m $ # & " 2 % 1/3! π Z a Z X e 2 kt # "! $ & % 2/3 =1.22Z a 2/3 Z X 2/3 A 1/3 T 6 2/3 kev σ v σ (E) = S E exp[ 2πη] * exp E kt η -, / + E. de η = π 2m Z Z a X e2 0 (reduced mass!) 33 Proton-proton reaction

34 Examples 1. H-burning via proton-proton cycle ( 10b) Z a = Z X = A a = A X =1, T=15x10 6 K E 0 = 6 kev 2. H-burning via CNO-cycle Z a = A a =1, Z X ~ 7, A X ~15, T=15x10 6 K Evaluation of remaining integral Write: Expand f(e) around maximum at E=E 0 : E 0 = 27 kev For most relevant thermonuclear reactions (10 7 < T < 10 8 K) we have 5 < E 0 < 100 kev: stellar energy region I = 0 exp[ f (E)]dE f (E) = E kt b E f (E) = τ τ " E % $ 1' 4 # E 0 & 2 + τ = 3E 0 kt 1 T 1/3 Retain only first 2 terms in expansion approximate Gamov 34 peak by a Gaussian of proper height and width

35 Use: ξ = 1 " E % $ 1' τ I = exp[ f (E)]dE = 2 2 # E 0 & 3 kt τ e τ exp) * ξ 2 +, dξ τ Integrand peaks at ξ = 0, and has width << τ replace lower 2 τ limit by, which shows that I kt π τe and hence 3 συ = 4 # 2 & % ( 3 $ m ' 1/2 1 (kt ) S 0τ 1/2 exp[ τ ] τ 2 exp[ τ ] % Z az X 1/2 $ A Þ Reactions in which at least one of the two partners is a proton have much larger cross section than when both Z a and Z X >1 # & ( ' 1/3 T 2/3 exp( Z 2/3 a Z 2/3 X A 1/3 T 1/3 ) Interpolation formula Write: # T συ T = συ T0 % $ T 0 & ( ' ν ν = ln συ lnt T0 Then: ε = ε 0 ρt ν ν = 1 ( 3 τ 0 2) ε τ 0 >> 2 is a very steep function of T, since 35 for 10 7 < T < 10 8 K

36 Resonant reaction It is straightforward to show that for a narrow resonance at energy E r : & συ = n(e)σ (E)υ de n(e r )υ r σ (E)dE = ( 2π ' m 0 In this case we obtain: ε = ε 0 ρt ν ν = E r kt g) Electron shielding Maxwell-Boltzmann distribution and use of Breit-Wigner formula Coulomb repulsion crucial for thermonuclear reaction rate Shielding of nuclear charge by surrounding free electrons Potential of point charge r D : Debye-Hückel length r D 2 = kt 4π e 2 n 0 Ze r now replaced by Ze r exp r / r D Z n = n e + n i, χ = µ i (Z i +1)X i µ i 36 ) + * 3/2 2 ω ς Γ a i ( E r )Γ b Γ A i 1 ( kt ) exp. E 1 r 3/2 / 0 kt 2 3 [ ]

37 Decrease of repulsive barrier increases P 0 so: συ f συ Weak screening: E D = Z az X e 2 [ ] 1.1 r D << kt f = exp E D / kt Strong screening: complicated expression for E D Interpolation formula # ρ & f συ = f 0 συ 0 % ( $ ' ρ 0 λ # T % $ T 0 & ( ' ν ν = 1 3 (τ 0 2) E D kt λ =1+ E D 3kT ρ For very large and low T pycnonuclear reactions 37