Basic Nuclear Physics 2. Nuclear Reactions. I(j,k)L. I + j L +k or I( j,k )L. Glatzmaier and Krumholz 7,8 Prialnik 4 Pols 6 Clayton 4

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1 A nuclear reaction turns one nucleus to another. We have already discussed several kinds: Beta decay, electron capture, positron emission Basic Nuclear Physics Nuclear Reactions Glatzmaier and Krumholz 7,8 Prialnik 4 Pols 6 Clayton 4 Alpha decay and fission (technically everything heavier iron is unstable) Neutron or proton drip Here we expand the discussion to include fusion reactions where two (or rarely more) nuclei come together to produce a third, often with the emission of some lighter particles and energy. The generic binary fusion reaction is: I + j L +k or I( j,k )L I(j,k)L PP Cycle I = Target nucleus j = incident particle L = Product nucleus k = outgoing particle or particles If there is no incident particle put the outgoing particles together without a comma E.g., pp; the main reaction sequence powering the sun Here β + = e + p(p,e + ν) H(p,γ ) 3 He( 3 He, p) 4 He γ = heat and light The symbol in the second reaction is is sort of a catch-all for energy that comes out in forms other than neutrinos, electrons, and positrons. Energy comes out as photons as well as kinetic energy (heat).

2 The cross section for the reaction I(j,k)L is number of reactions/nucleus I/sec (I) = number of incident particles j/cm /sec σ thus has units of cm though other units of area are sometimes used (typically "barns" are the unit in nuclear physics; barn = 0 4 cm as in "big as a barn") The reaction rate for the reaction I (j,k) L is given by the product of the number densities of the reactants times their relevant speed and cross section. n j v can loosely be thought of as the flux of particles j on a nucleus of species I n I n j σ I j v This has units of reactions cm 3 s It is more convenient to write things in terms of the Y s previously defined Y I = X I n I = ρn A Y I gm atoms Mole A I cm 3 Mole gm so that the rate becomes (ρn A ) Y I Y j σ I j v and a term in a rate equation decribing the destruction of I might be Here dy I = ρy IY j N A (I)v +... = ρy I Y j λ jk (I) +... Equivalent to denotes a suitable average over energies and angles and the reactants are usually assumed to be in thermal equilibrium. dn I = n I n j σ Ij v +... For a Maxwell-Boltzmann distribution of reactant energies the average of the cross section times velocity is µ (I)v = 4π π kt (I)v = 8 πµ / 3/ 0 kt where µ is the "reduced mass" µ = M I m j M I + m j (v)v 3 e µv /kt dv 3/ 0 (E) E e E/kT de for the reaction I (j, k) L and E = µv is the kinetic energy in the center of mass frame. For T in 0 9 K = GK, in barns ( barn = 0-4 cm ), E 6 in MeV, and k = /.6045 MeV/GK, the thermally averaged rate factor in cm 3 s - is v = 6.97 x 0-4 Â / T 9 3/ (E 6 ) E 6 e.6045 E 6 /T 9 de 6 Â = A A I j A I + A j 0 for the reaction I(j,k)L If you know jk from the lab, or a calculation, just put it in and integrate. But often we don t know the cross section at all, or know it in a limited energy range or only at a few points. How to estimate, interpolate, extrapolate?

3 = p = k The Cross Section Area subtended by a de Broglie wavelength in the c/m system How much the nucleus I+j looks like the target nucleus I with j sitting at its surface. Liklihood of staying inside R once you get there. σ (E) = π P(E) Χ(E, A) geometry penetration nuclear term factor structure (Cla 4-80) probability a flux of particles with energy E at infinity will reach the nuclear surface. Must account for charges and QM reflection. see Clayton Chapter 4 is the de Broglie wavelenth in the c / m system π = π µ v = π µe = 0.656barns  E(MeV) where barn = 0-4 cm is large for a nuclear cross section. µ = M M M + M Â= A A A + A ~ for neutrons and protons on heavy nuclei ~ 4 for α-particles if A I is large, >> nuclear potential U o U 0 = potential energy r o R 3 r o ~0 cm Classical turning radius mv = KE = Z Z e r r min = >> fm r 3 kt = Z Z e r Z Z e 3kT min E ~ 3( )(0 7 ) =. 0 0 cm = 3/ kt ( ) QM Barrier Penetration The classical turning radius is given by energy conservation mv = KE = Z Z e r r classical = Z Z e mv The De Broglie wavelenth of the particle with mass m is, So the ratio λ = p = mv r classical = Z Z e mv λ mv = Z Z e η v Which is close to the factor in exponential in the penetration function P~exp(- r classical / λ) exp π Z Z e v The Sommerfeld parameter note : r classical >> λ Note that as the charges become big or E gets small, P gets very small.

4 For particles with charge, providing X(A,E) does not vary rapidly. with energy (exception to come), i.e., the nucleus is "structureless" σ (E) = π P l X ( A, E) e πη E This motivates the definition of an "S-factor" which should be ~ constant S(E)=σ (E) E exp(πη) η = Z Z I j e v = 0.575Z I Z j  / E  = A A I j E in MeV A I + A j This S-factor should vary slowly with energy. The first order effects of the Coulomb barrier and Compton wavelength have been factored out. For those reactions in which S(E) is a slowly varying function of energy in the range of interest and can be approximated by its value at the energy where the integrand in the reaction rate formula is a maximum, E 0, 8 N A σ v N A πµ σ (E) S(E 0) E exp( πη) / where η(e) = Z I Z j e v kt 3/ S(E o ) exp( E / kt πη(e))de 0 =  / E(MeV ) Z I Z j For accurate calculations we would just enter the energy variation of S(E) and do the integral numerically. However, Pols 6.. shows that exp E kt πη can be replaced to good accuracy by ( ) C exp E E 0 ( Δ / ), i.e. a Gaussian The quantity in the integral looks like

5 Detail: ( ) exp E kt πη C exp E E 0 ( Δ / ) ) Get E 0 from d E exp de kt πη =0 with η =0.575 Â / E(MeV ) Z I Z j =η 0 / E(MeV ) gives E 0 = ( πη 0 kt ) /3 ( ) /3 ( ) /3 MeV = π (0.575Â/ Z I Z j )kt = 0. Z I Z j ÂT 9 ( ) ) Evaluate C at E 0 C exp E E 0 ( Δ / ) = exp E kt πη gives C =exp E 0 kt πη(e 0) = exp 3E 0 kt [Pols 6..] 3) Δ is determined by matching the nd derivatives at E 0 [Pols 6..] Δ = 4 E 0kT 3 / E 0 Detail: Then if τ = 3E 0 kt 8 λ N A µπ / kt 3/ e τ S(E)exp E E 0 Δ / de Then if S(E) is constant in the vicinity of E 0 8 λ = N A µπ 8 = N A µπ / / kt 3/ e τ S(E 0 ) exp E E 0 Δ / de 3/ kt e τ S(E 0 ) but Δ/((kT) 3/ 4 ) = 9 3 πη 0 τ So λ = N A σ v = N A 3µ / π Δ FW 8 9 πη 0 S(E o)τ e τ = S(E 0 ) τ e τ [Pols 6.8] Â Z I Z j (Pols 6..7 with Δ / Δ FW ) Summary We replaced the exponential term with a Gaussian that was analytically integrable. Matching the first and second derivatives of the two functions, the maximum and full wih of the Gaussian were determined The peak is E o, the Gamow Energy. This is the center-o-mass energy where most reactions happen. It is >>kt E o = 0. Z I Z ( j ÂT 9 ) /3 MeV The Gaussian has a full wih at /e times the maximum. This gives the range of interesting energies, e.g., in which the cross section needs to be measured. Δ = 4 ( 3 E okt ) / = 0.37 Z I Z 5 ( j ÂT 9 ) /6 MeV Â= A I A j A I + A j T 9 = T / 0 9 K Example 3 He + 3 He at K E o = 0. Z I Z ( j ÂT 9 ) /3 MeV = 0. ( 9 6 (0.05) ) /3 =0.04 MeV =.4 kev ( ) /6 MeV Δ = 0.37 Z I Z j ÂT = (0.05)5 /6 = 0.0 MeV =. kev The relevant energy range in which we need to know S(E) is ± 6 kev The value of kt for comparison is.3 kev

6 A Gaussian integral is analytic e x dx = π and so N A σ v =  Z I Z j S(E 0 ) τ e τ cm 3 / (Mole s) where S(E 0 ) is measured in MeV barns. If we define λ jk = N A v then a term in the rate equation for species I such as Y j ρλ jk has units Mole gm gm cm 3 cm 3 Mole s = s Note that τ here is τ = 3E 0 kt = 4.48 Z I Z j  T 9 /3 Different people use different conventions for which sometimes do or do not include or N A. This defines mine.% The lifetime of the species I against the reaction I(j,k)L is given by Adelberger et al, RMP, (998( τ jk (I) = Y I dy i jk = ( ρy j λ jk (I)) E.g., the lifetime of 4 N against the reaction 4 N (p,γ ) 5 O is τ pγ ( 4 N)= ( ρy p λ pγ ( 4 N) )

7 λ f (T ) = τ e τ τ = A T /3 df dt = dτ τe τ dt τ dτ e τ dt dτ dt = A 3T = τ 4/3 3T For example, C + C at 8 x 0 8 K τ = 4.48 = /3 T f df dt = T τ e τ (τ e τ )( τ 3T ) = d ln f d lnt = τ 3 f T n n= τ 3 T τ e τ (τ e τ )( τ 3T ) n = = p + p at.5 x 0 7 K τ = = 3.67 n = /3 = 3.89 Resonant reactions The previous discussion was predicated upon the assumption that the nuclear structure factor was slowly varying with energy. This led to an S-factor that was also nearly constant. This turns out to be the case when the excited state structure of the nucleus can be ignored. But suppose I and j come together at the Gamow energy to produce L in an excited state that has just the right spin, angular momentum, and parity to closely resemble I + j. Such reactions are called resonant and the cross section in a narrow energy range can be many of orders of magnitude larger than if the resonant state were not there. See appendix to these notes.% Specific Reactions in Hydrogen and Helium Burning pp chain pp and pp3 chains CNO cycle 3-alpha and C(, ) 6 O

8 Proton-proton reaction (the basic first step for pp,, 3): p( p,e + ν e ) H (+0.4MeV) This cross section is far too small (~0-47 cm at MeV) to measure in the laboratory but it does have a nearly constant, calculable S-factor. Two stages: Temporarily form diproton. The initial wave function is the same as for proton scattering. Initially the diproton must have its protons spins counter alligned because can t have protons in identical states. Unbound. Diproton experiences a weak interaction (with a spin flip) to make deuteron with the neutron and proton alligned. Higher this is more tightly bound than the counter alligned state no weak interaction needed, very fast 4 Li is unbound pp pp,3 PP Cycle Here β + = e + γ = heat and light H. A. Bethe (b 906) Nobel 967

9 Lifetimes against various reactions Reaction Lifetime (years) H(p,e + ) H 7.9 x 0 9 H(p, ) 3 He 4.4 x He( 3 He,p) 4 He.4 x He( 4 He, ) 7 B 9.7 x 0 5 For 50% H, 50% He at a density of 00 g cm -3 and a temperature of 5 million K The time between proton collisions, for a given proton, is about a hundred millionth of a second. Nuclear energy yield How much energy is produced and how fast is it liberated. We could just add up the energies of each reaction and divide that number by the time scale for the slowest reaction. Instead let s solve the general problem (make things easier in the future). We have a set of nuclei {Y i } that is transformed into a new set {Y i + Y i ). Each nucleus has binding energy BE i in MeV. q nuc = N A [ (δy i )(BE i q weak )] erg/gm Here.60 x 0-6 is the conversion factor from MeV (which are the units of BE) to erg. q weak is a correction term that is only non zero for weak interactions. It accounts for neutrinos lost and mass changes when protons and electrons turn to neutrons or vice versa. q weak Example: Hydogen burning Basically important only in hydrogen burning q weak accounts for the neutron proton mass difference, the creation and annihilation of positrons, and energy lost to neutrinos q weak = ΔZ * [( n p mass difference)+ m e c ] q ν = ΔZ * [.94 MeV-0.5 MeV ] q ν = ΔZ * [0.783 MeV ] q ν For example in the pp process 4 protons urn into neutons and protons so ΔZ =. In addition two neutrinos are emitted with an average of 0.6 MeV each, so q = 0.5 MeV ν and q weak =.09MeV a) 00% H 4 He δ Y( H) = BE( H) = 0 From previous page q weak =.09 δ Y( 4 He) = q= = erg g b) 70% H; 30% 4 He 4 He δ Y( H) = 0.7 δy ( 4 He) = q= BE( 4 He)= 8.96MeV ( ) = erg g

10 BE( C) = 9.6 MeV BE( 6 O) = 7.69 BE( ) = 8.96 MeV q weak = 0 A related quantity, the energy generation rate is given by ε nuc = dy i values for helium burning (BE i q weak ) erg g - sec - Additional nuclear physics needed for post-helium burning stages will be covered when we treat the advanced burning stages of massive stars later in the course. For hydrogen burning by the pp process We could include all the reactions and evaluate dy i for each species but it is much easier to just realize that the net result is that every time 4 protons are burned by the reaction p(p,e + ν), one 4 He appears. So dy( 4 He) = dy( H ) = 4 4 ρy ( H )λ pp = ρy ( H )λ pp dy ε nuc = i (BE i q weak ) erg g - s - = ρy ( H )(6.)λ pp = ρy ( H ) λ pp erg g s Previously we showed λ =  Z I Z j S(E 0 ) τ e τ cm 3 / (Mole s) for the pp reaction  = */ (+) = / τ = 4.48 Z I Z j  T 9 /3 = 3.37 T 9 /3 = 33.7 T 6 /3 = 4.48 ii/ T 9 S 0 = kev b = MeV b λ pp = /3 / T 6 e 33.7/T 6 /3 = T 6 /3 e 33.7/T 6 /3 Compare with Clayton 5. /3 Putting it together ε = ρy H T 6 /3 e 33.7/T 6 /3 e.g. ρ =50 g cm 3 T 6 =5 Y H =0.35 ε =( )(50)(0.35) (5) /3 e 3.66 =8.7 erg g s erg g s erg g s Given our previous discussion this can also be written as a power law of the temperature with n = τ 3 τ = 33.7 / (5) /3 =3.66 n= 3.89 So ε pp 8.7 ρ Y H 50g cm T K 3.89 erg g s

11 Reality checks: ) Energy budget for the sun Q = L τ MS ( ) ( erg s )(0 0 yr)( s/yr) 0 5 erg Q nuc ( erg g )(0.)(M ) =( erg g )( 0 3 g) = erg ) Luminosity of the sun L = erg s ε pp i(0.m ) 9erg g s ( 0 3 g)~ 0 33 erg s Could adjust Y H,T,ρ, or fraction burning to get better agreement 3) Lifetime τ MS τ pp = Y H / dy H Y = H ρy H λ pp = ( ρy H T 6 /3 e 33.7/T 6 /3 ) At ρ=50g cm 3 ; T 6 = 5;Y H = 0.7 ( ) ( ) s = 8 billion years τ MS (x0 3 )(50)(0.7)(0.64)(.x0 6 ) = pp and pp3 chains (more important at higher T) In all cases 4 + 4p He + e + ν e Neutrino Energies Species Average energy Maximum energy p+p 0.67 MeV 0.40 MeV 7 Be MeV MeV 0% % 8 B MeV 5 MeV In the sun pp 85% pp 5% pp3 0.0% T central = 5.7 Million K In the case of 8 B and p+p, the energy is shared with a positron hence there is a spread. For 7 Be the electron capture goes to two particular states in 7 Li and the neutrino has only two energies

12 min For temperatures above 8 million K, the CNO cycle dominates energy production 0 min Detail Rate equations: dy( C) = Y( C)Y p ( C)+Y( 5 N)Y p ρ λ pα ( 5 N) dy( 3 N) = Y( C)Y p ( C) Y( 3 N)λ e + ( 3 N) dy( 3 C) = Y( 3 N)λ e + ( 3 N) Y( 3 C)Y p ( 3 C) dy( 4 N) = Y( 3 C)Y p ( 3 C) Y( 4 N)Y p ( 4 N) dy( 5 O) = Y( 4 N)Y p ( 4 C) Y( 5 O)λ e + ( 3 N) dy( 5 N) = Y( 5 O)λ e + ( 3 N) Y( 5 N)Y p ρ λ pα ( 5 N) dy p = Y( C)Y p ( C) Y( 3 C)Y p ( 3 C) Y( 4 N)Y p ( 4 N) Y( 5 N)Y p ρ λ pα ( 5 N) dy α = Y( 5 N)Y p ρ λ pα ( 5 N) The slowest reaction is 4 N(p,γ ) 5 O so for purposes of energy generation the rate equations can be approximately be written: dy p = 4Y( 4 N)Y p ( 4 N) dy α = Y( 4 N)Y p ( 4 N) where Y( 4 N) 4 (X C + X N + X O ) Z 4 One still needs to subtract off the energy carried away by neutrinos and adjust for n - p mass differences q weak = ΔZ * [0.783 MeV ]+ q =()(0.783) +.70 ν = 3.6 MeV λ = What is λ pγ ( 4 N)? The S factor for 4 N(p, ) 5 O is.64 x0-3 MeV barns (including a recent downward revision)  Z I Z j S(E 0 ) τ e τ cm 3 / (Mole s) for the 4 N(p,γ ) reaction  = *4 / (4 +) = 4 /5 λ pγ ( 4 N) = τ = 4.48 Z I Z j  T (4 /5) = 5.9 T 9 /3 = T 9 /3 e 5.9/T 9 /3 /3 = 4.48 i 7 i4 /5 T 9 ( )(5.9) T /3 9 e 5.9/T 9 /3 = for example at T 9 = 0.0 (0 M K) /3

13 dy α = Y( 4 N)Y p ( 4 N) Y Z p 4 (4 N) At T 9 = 0, τ =56.0, n = τ 3 =8 T λ pγ ( 4 N) MK dy α 8 T. 0 7 ρy p Z 0MK dy ε nuc = i (BE i q weak ) erg g - sec - 8 T =.7 ρy p Z 0MK ( ) ε CNO T = 93 ρy p Z 0MK 8 8 erg g - sec - Ratio of CNO energy generation to pp ε CNO T = 93 ρy p Z 0MK 8 erg g - sec - ε pp = ρy p T 6 /3 e 33.7/T 6 /3 T =.36 ρy 9 p 0.0 The ratio = when Z T Y p and if Z= 0.05, Y p = T erg g - sec - 4 = = T 9 = CNO will dominate above about 8 MK CNO Tri-cycle Note nucleosynthetic implications: Synthesis of some 3 C. 4,5 N, 7,8 O, 9 F

14 Helium Burning Helium burning is a two-stage nuclear process in which two alpha-particles temporarily form the ground state of unstable 8 Be *. Occasionally the 8 Be * captures a third alpha-particle before it flies apart. No weak interactions are involved. Helium burning the C(, ) rate Resonance in Gamow window - C is made! unstable No resonance in Gamow window C survives! But some C is converted into O The current value is due to Caughlan and Fowler (988) using mesurements from Sam Austin λ 3α = T 9-3 exp ( / T 9 ) T 9 d lnλ d ln T = T Slight revisions to Unlike most reactions in astrophysics, the temperature dependence here is not determined by barrier penetration but by the Saha equation. In fact, at high temperature (T 9 >.5) the rate saturates and actually begins to decline slowly as the resonance slips out of the Gamow window. Γ γ here dy α =-3ρ Y 3 α λ 3α /6 - Y α Y( C) ρ λ αγ ( C) dy( C) = ρ Y 3 α λ 3α /6 - Y α Y( C) ρ λ αγ ( C) dy( 6 O) = Y α Y( C) ρλ αγ ( C) For Y small or ρ large Helium Burning α 6 C For Y large or ρ small α O For binary reactions, λ N σ v A

15 Energy Generation from Helium Burning BE( 4 He) = 8.96 MeV BE( C)= 9.6 MeV BE( 6 O) = 7.67 MeV ) 3 4 He C q = (δy i )(BE i ) erg/gm (q weak = 0) = Y final ( C) BE( C) Y initial ( 4 He) BE( 4 He) = = erg g ) 8 4 He C O (i.e. = 5.8% C %O) = [Y final ( C) BE( C) +Y final ( 6 O) BE( 6 O) Y initial ( 4 He) BE( 4 He)] = = erg g Energy Generation from Helium Burning ε nuc = dy i (BE i ) erg g - sec - dy α =-3ρ Y 3 α λ 3α /6 ignore C(α,γ ) 6 O dy( C) = ρ Y 3 α λ 3α /6 dy ε nuc = i (BE i ) = ρ Y 3 α λ 3α / (8.96) = ρ X 3 α λ 3α = ρ 3 X α ( T -3 9 exp ( 4.403/ T 9 )) ε 3α = ρ X α 3 T 9-3 exp ( 4.403/ T 9 ) note typo in GK coeff off by 000 In a 5 solar mass star: Extra Material λ αγ ( C) / 3 λ 3α

16 Resonant Reactions The resonances can be broad or narrow Not constant S-factor for resonances (log scale!!!!) ~ constant S-factor for direct capture Rate of reaction through a narrow resonance If the resonance is much broader than the Gamow energy, then treat with the formalism we have already developed by with a slowly varying S-factor. For narrow resonances though, wih << the Gamow energy, another formalism is employed. This would take some time to develop, so here we just give the result. Narrow means: Γ << ΔE In this case, the resonance energy must be near the relevant energy range E to contribute to the stellar reaction rate. pull out front Recall: and < σv >= 8 πµ (kt ) 3/ σ(e)e e E ktde 0 Γ σ(e) = π ω (E)Γ (E) (E E r ) + (Γ(E) / ) Here E r is the energy of the resonance and the s are the partial wihs of the state to break up into I+j, L+k, etc.

17 Then one can carry out the integration analytically (Clayton 4-93) and finds: For the contribution of a single narrow resonance to the stellar reaction rate: Light particle (n, p, α): λ jk N A v N A < σv >=.54 0 (AT 9 ) 3/ ωγ [MeV]e The rate is entirely determined by the resonance strength!ωγ ωγ = J r + Γ Γ (J j +)(J I +) Γ In general the reactions during the proton-proton cycles are non-resonant but the reactions for the CNO cycle and more advanced burning stages are resonant..605 E r [MeV] T 9 cm 3 s mole Y I Y p (I); Y I Y n ρ λ nγ (I); Y I Y p ρ λ pα (I) Heavy Ion 6 Y α 3 λ 3α ; ρy ( C)λ, ; ρy ( 6 O)λ 6,6 Weak interaction (beta decay, electron capture, positron emission) Y I λ β (I); Y I λ ec (I); Y I λ β (I) λ = ln τ / Example of solving a rate equation E.g., pp hydrogen burning: E.g. the decay of 4 C So dy( 4 C) = Y( 4 C) λ β ( 4 C) Y( 4 C) =Y 0 ( 4 C) exp( λ β ( 4 C) t) τ / = 5730 y =.8 0 sec λ = ln.8 0 = sec p(p,e + ν) H(p,γ ) 3 He( 3 He,p) 4 He dy p = Y ρ λ p pp + Y 3 He dy( H) dy( 3 He) dy( 4 He) = Y ρ λ p pp Y p Y H =Y p Y H ( ) ( H) = Y 3 He ( ) ρ λ 3,3 ( ) ρ λ 3,3 ( ) ( H) ( ) ρ λ 3,3 Y 3 He The rate factors λ contain the temperature dependence of the rate equations. The density dependence has been separated out.