Nuclear Binding Energy

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5. NUCLEAR REACTIONS (ZG: P5-7 to P5-9, P5-12, 16-1D; CO: 10.3) Binding energy of nucleus with Z protons and N neutrons is: Q(Z, N) = [ZM p + NM n M(Z, N)] c 2. } {{ } mass defect Nuclear Binding Energy Energy release: 4 H 4 He 6.3 10 14 J kg 1 = 0.007 c 2 ( 56 H 56 Fe 7.6 10 14 J kg 1 = 0.0084 c 2 ( = 0.007) = 0.0084) H burning already releases most of the available nuclear binding energy. 5.1 Nuclear reaction rates: (ZG: P5-7) 1 + 2 1,2 + Energy Reaction rate is proportional to: 1. number density n 1 of particles 1 2. number density n 2 of particles 2 v σ 3. frequency of collisions depends on relative velocity v of colliding particles r 1+2 = n 1 n 2 (v)v 4. probability P p (v) for penetrating Coulomb barrier (Gamow factor) P p (v) exp[ (4π 2 Z 1 Z 2 e 2 /hv)] v t

MeV Coulomb Barrier Z 1Z 2 e 2 l (l+1) h 2 V = + 4 πε r 2 µ r 2 kev 0 Reaction Rate 5. define cross-section factor S(E): = [S(E)/E] Pp(E) depends on the details of the nuclear interactions insensitive to particle energy or velocity (nonresonant case) S(E) is typically a slowly varying function evaluation requires laboratory data except in p-p case (theoretical cross section) 6. particle velocity distribution (Maxwellian). D(T, v) (v 2 /T 3/2 ) exp[ (m H A v 2 /2kT)] where A = A 1 A 2 (A 1 + A 2 ) 1 is the reduced mass. The overall reaction rate per unit volume is: Setting n 1 = ( R 12 = 0 n 1n 2 v[s(e)/e P p (v)] D(T, v) dv /m1) X1, n2 = ( /m 2 ) X 2 and = 3E 0 /kt = 3{2π 4 e 4 m H Z 2 1Z 2 2A /(h 2 kt)} 1/3 e E kt kt = 1 3 kev Product (magnified) E = 15 30 kev 0 = 4 10 kev 1/2 be e R 12 = B 2 (X 1 X 2 /A 1 A 2 ) 2 exp( )/(A Z1Z2) where B is a constant depending on the details of the nuclear interaction (from the S(E) factor) Low temperature: is large; exponential term leads to small reaction rate. Increasing temperature: reaction rate increases rapidly through exponential term. High temperature: 2 starts to dominate and rate falls again. (In practice, we are mainly concerned with temperatures at which there is a rising trend in the reaction rate.)

(1) Reaction rate decreases as Z 1 and Z 2 increase. Hence, at low temperatures, reactions involving low Z nuclei are favoured. (2) Reaction rates need only be significant over times 10 9 years. 5.2 HYDROGEN BURNING 5.2.1 PPI chain: (ZG: P5-7, 16-1D) 1) 2) 3) 1 H + 1 H 2 D + e + + 2 D + 1 H 3 He + + 1.44 MeV + 5.49 MeV 3 He + 3 He 4 He + 1 H + 1 H + 12.85 MeV for each conversion of 4 H 4 He, reactions (1) and (2) have to occur twice, reaction (3) once the neutrino in (1) carries away 0.26 MeV leaving 26.2 MeV to contribute to the luminosity reaction (1) is a weak interaction bottleneck of the reaction chain Typical reaction times for T = 3 10 7 K are (1) 14 10 9 yr (2) 6 s (3) 10 6 yr (these depend also on, X 1 and X 2 ). Deuterium is burned up very rapidly. If 4 He is sufficiently abundant, two further chains can occur: PPII chain: 3a) 4a) 5a) PPIII chain: 4b) 5b) 6b) 3 He + 4 He 7 Be + 7 Be + e 7 Li + + 1.59 MeV + 0.86 MeV 7 Li + 1 H 4 He + 4 He + 17.35 MeV 7 Be + 1 H 8 B + 8 B 8 Be + e + + + 0.14 MeV 8 Be 4 He + 4 He + 18.07 MeV In both PPII and PPIII, a 4 He atom acts as a catalyst to the conversion of 3 He + 1 H 4 He +. E total is the same in each case but the energy carried away by the neutrino is different. All three PP chains operate simultaneously in a H burning star containing significant 4 He: details of the cycle depend on density, temperature and composition.

5.2.2 THE CNO CYCLE (ZG: P5-9; 16-1D) (T < 10 8 K) Carbon, nitrogen and oxygen serve as catalysts for the conversion of H to He 12 C + 1 H 13 N + 13 N 13 C + e + + 13 C + 1 H 14 N + 14 N + 1 H 15 O + 15 O 15 N + e + + 15 N + 1 H 12 C + 4 He The seed nuclei are believed to be predominantly 12 C and 16 O: these are the main products of He burning, a later stage of nucleosynthesis. cycle timescale: is determined by the slowest reaction ( 14 N + 1 H) Approach to equilibrium in the CNO cycle is determined by the second slowest reaction ( 12 C + 1 H) in equilibrium 12C 12C = 13 C 13 C = 14N 14N = 15 N 15 N (where are reaction rates and 13 C, etc. number densities) most of the CNO seed elements are converted into 14 N Observational evidence for CNO cycle: 1. In some red giants 13 C/ 12 C 1/5 (terrestrial ratio 1/90) 2. Some stars with extremely nitrogen-rich compositions have been discovered

4He + 4He + 5.3 Energy generation from H burning (CO: 10.3) Using experimental or extrapolated reaction rates, it is possible to calculate (T) for the various chains. PP X2 H CNO XHXCNO Energy generation occurs by PP chain at T 5 10 6 K. High-mass stars have higher T c (CNO cycle dominant) than low-mass stars (pp chain) Analytical fits to the energy generation rate: PP 1X 2 H T4; CNO 2X H X CNO T20. 5.4 Other Reactions Involving Light Elements (Supplementary) Both the PP chain and the CNO cycle involve weak interactions. First reaction of PP chain involves two steps 2 D + e + + 1 H + 1 H 2 He 1 H + 1 H In the CNO cycle, high nuclear charges slow the reaction rate. D, Li, Be and B burn at lower temperatures than H, because all can burn without -decays and with Z < 6. 2 D + 1 H 3 He + 6 Li + 1 H 4 He + 3 He 7 Li + 1 H 8 Be + 9 Be + 1 H 6 Li + 4 He 10 B + 1 H 7 Be + 4 He 11 B + 1 H 4 He + 4 He + 4 He + 7 Be is destroyed as in the PP chain These elements always have low abundances and play no major role for nuclear burning they take place at T 10 6 10 7 K they are largely destroyed, including in the surface layers, because convection occurs during pre-mainsequence contraction.

5.5 HELIUM BURNING (ZG: P5-12; 16-1D) When H is exhausted in central regions, further gravitational contraction will occur leading to a rise in T c, (provided material remains perfect gas) Problem with He burning: no stable nuclei at A = 8; no chains of light particle reactions bridging gap between 4 He and 12 C (next most abundant nucleus). Yet 12 C and 16 O are equivalent to 3 and 4 particles. Perhaps many body interactions might be involved? These would only occur fast enough if resonant. Triple reaction: 4He + 4He + 4He 12C + Ground state of 8 Be has = 2.6 10 16 s = 2.5 ev Time for two s to scatter off each other: t scatt 2d/v 2 10 15 /2 10 5 10 20 sec A small concentration of 8 Be builds up in 4 He gas until rate of break-up = rate of formation. At T = 10 8 10 9. K and = 108 kg m 3, n(8be)/n(4he) This is sufficient to allow: 8 Be + 4 He 12 C + The overall reaction rate would still not be fast enough unless this reaction were also resonant at stellar temperatures. An s-wave resonance requires 12 C to have a 0 + state with energy E 0 + 2 E 0 where E 0 = 146(T 10 8 ) 2/3 kev and 2 E 0 = 164(T 10 8 ) 5/6 kev. Such an excited state is found to lie at a resonance energy E res = 278 kev above the combined mass of 8 Be + 4 He. - Best available estimates of partial widths are: = 8.3 ev; = (2.8 ± 0.5)10 3 ev. Thus resonant state breaks up almost every time. Equilibrium concentration of 12 C and the energy generation rate can be calculated. At T 10 8 K 3 3X 3 He 2 T 30. energy generation in He core strongly concentrated towards regions of highest T other important He-burning reactions: 12 C + 13 C + 14 N + 16 O + 18 O + 20 Ne + 16O + 16O + n 18O + e+ + 20Ne + 22Ne + 24Mg + in some phases of stellar evolution and outside the core, these can be the dominant He-burning reactions in a stellar core supported by electron degeneracy, the onset of He burning is believed to be accompanied by an explosive reaction THE HELIUM FLASH once He is used up in the central regions, further contraction and heating may occur, leading to additional nuclear reactions e.g. carbon burning by the time that H and He have been burnt most of the possible energy release from fusion reactions has occurred