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Nuclear Reaction Rates 9 – Nuclear reaction rates introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 1 9.1 – Nuclear decays: Neutron decay By increasing the neutron number, the gain in binding energy of a nucleus due to the volume term is exceeded by the loss due to the growing asymmetry term. After that, no more neutrons can be bound, the neutron drip line is reached beyond which, neutron decay occurs: (Z, N) à (Z, N - 1) + n (9.1) Q-value: Qn = m(Z, N) - m(Z, N-1) - mn (9.2) Neutron Separation Energy Sn Sn(Z, N) = m(Z, N-1) + mn - m(Z, N) = -Qn for n-decay (9.3) Neutron drip line: Sn = 0 Beyond the drip line Sn < 0 neutron unbound Gamma-decay • When a reaction places a nucleus in an excited state, it drops to the lowest energy through gamma emission • Excited states and decays in nuclei are similar to atoms: (a) they conserve angular momentum and parity, (b) the photon has spin = 1 and parity = -1 • For orbital with angular momentum L, parity is given by P= (-1)L • To first order, the photon carries away the energy from an electric dipole transition in nuclei. But in nuclei it is easier to have higher order terms in than in atoms. introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 2 Proton decay Proton Separation Energy Sp: Sp(Z,N) = m(Z-1,N) + mp - m(Z,N) (9.4) Proton drip line: Sp = 0 beyond the drip line: Sp < 0 the nuclei are proton unbound α-decay α-decay is the emission of an a particle ( = 4He nucleus) The Q-value for a decay is given by Q "m(Z,A) m(Z 2,A 4) m $c2 α = # − − − − α % (9.5) Decay lifetimes Nuclear lifetimes for decay through all the modes described so far depend not only on the Q-values, but also on the conservation of other physical quantities, such as angular momentum. Typical decay times for similar lifetimes are (a) 10-10 s for beta decay, (b) 10-16 s for gamma-decay and (c) up to many years for alpha- decay. introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 3 Fission Very heavy nuclei can fission into two parts (Q > 0) This happens because for large nuclei the surface energy is less important – large deformations are then possible. With a small amount of additional energy the nucleus can be deformed sufficiently so that Coulomb repulsion wins over nucleon-nucleon attraction and the nucleus fissions. This will be possible if Q "m(Z,A) 2m(Z / 2,A / 2)$c2 0 (9.6) f = # − % > Ma ss Num ber A Using the LDM formula, Eq. 80 100 120 140 160 20 (7.5) we can shown that Qα > 0 this happens Nuc lear C harge Yield in Fission of 234U for A > 90. 15 ) % ( Symmetric fission (fission into ) Z 10 two nearly equal size fragments) ( Y d as well as asymmetric fission are l e i likely to occur. Y 5 0 Example from 25 30 35 40 45 50 55 60 65 Moller et al., Nature 409 (2001) 485 Proton Num b e r Z introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 4 4 9.2 - Observations 1. Most nuclei found in nature are the stable and not too heavy to undergo alpha decay or fission) 2. There are many more unstable nuclei that can exist for short times after their creation. 3. Alpha particle is favored building blocks of nuclei because of their comparably low Coulomb barrier when fused with other nuclei high binding energy per nucleon This explains abundance peaks at nuclei composed of multiples of alpha particles in solar system abundance distribution 4. The binding energy per nucleon has a maximum in the Iron-Nickel region The iron peak in solar abundance distribution implies that some fraction of matter was brought into equilibrium, probably during supernovae explosions. introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 5 Observations – Solar abundances introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 6 9.3 - Nuclear reactions in stars The plasma within a star is a mix of fully ionized projectiles and target nuclei at a temperature T. Using the definition of cross section in Eq. (7.24), we get that for particles with a given relative velocity v and within a volume V with projectile number density ni, and target number density nj: λ = σ n v i (9.7) R = σ nivn jV 3 so for reaction rate per second and cm is (9.8) r = nin jσ v This is proportional to the number of i-j pairs in the volume. However, when the species are identical, i.i., i =j, one has to divide by 2 to avoid double counting 1 r = n n σ v 1 i j (9.9) +δij introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 7 Reaction rates at temperature T Assuming Local Thermal Equilibrium LTE the velocity distribution of particles in a plasma follow a Maxwell-Boltzmann velocity distribution: 3/ 2 mv2 ⎛ m ⎞ − 2 2kT (9.10) with Φ(v)dv =1 (9.11) Φ(v) = 4π ⎜ ⎟ v e ∫ ⎝ 2π kT ⎠ where Φ(v) is the probability to find a particle with a velocity between v and v + dv. This distribution is easily deduced from the Boltzmann distribution f(E) = A exp(-E/kT) by replacing E = mv2/2 and adjusting the constant A to conform with the max at normalization condition E = kT given by Eq. (9.11). introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 8 Reaction rates at temperature T In terms of the Maxwell-Boltzmann distribution, with the mass m replaced by the reduced mass m of the i and j particles, i.e., m m 3/ 2 µ v2 i j − µ = (9.12) ⎛ µ ⎞ 2 2kT (9.13) m m Φ(v) = 4π ⎜ ⎟ v e i + j ⎜ ⎟ ⎝ 2π kT ⎠ In the stellar environments, the reaction rates given in Eq. (9.9) has to be averaged over velocities, 1 r = nin j < σ v > (9.14) 1+δij using over the distribution Φ(v), 1 r n n (v) (v)vdv (9.15) = i j ∫ σ Φ 1+δij introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 9 Reaction rates in terms of abundances 3 In terms of units of stellar reaction rate given by NA<σv> in cm /s/g, the reaction rate per second and per cm3, is 1 2 2 (9.16) r = YiYjρ NA < σ v > 1+δij Sometimes, one uses another notation for reactions per second and target nucleus j, by dividing the above equation by YjρNA, i.e., 1 One usually call this by the λ = Yiρ NA < σ v > (9.17) stellar reaction rate of a 1+δij specific reaction As an application, lets assume that the species i and j are destructed by j capturing the projectile i in the form i + j à k. Then dn j = −n λ = −n Y ρ N < σ v > dt j j i A (9.18) dn k = n λ dt k introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 10 Reaction rates in terms of abundances Since the dnj/dt is proportional to –nj, the solution of the equation is a decaying exponential. And since the species k is built from the destruction of of j, its time dependence should be proportional to [1 – nj(t)]. In other words, n (t) = n e−λ t j 0 j (9.19) −λ t n k (t) = n0k (1− e ) In terms of the abundances: Y (t) = Y e−λ t j 0 j (9.20) −λ t Yk (t) = Y0k (1− e ) The lifetime for the destruction of j is 1 1 τ = = (9.21) λ Yjρ NA < σ v > After a time equal to a few multiples of τ, the species j is almost completely destroyed while k is almost completely built up from i and j capture. One often uses the name “half-life”, τ1/2, instead of lifetime, where τ1/2 = τ ln2. introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 11 9.4 - Energy generation Lets consider the energy generation by nuclear reaction sin stars. As we know the energy released is given by the Q-values ⎛ ⎞ Q c2 ⎜ m m ⎟ (9.22) = ⎜ ∑ i − ∑ j ⎟ ⎝ initial nuclei i final nuclei j ⎠ where m are the respective masses of the particles. The energy generated per gram and second by a reaction i + j is then rQ 1 2 ε = = Q YiYjρNA < σ v > (9.23) ρ 1+δij We can now define the reaction flow as the abundance of nuclei converted in time T from species j to k via a specific reaction as T !dY $ T F = dt# j & = λ(t)Y (t)dt ∫ # dt & ∫ j (9.24) 0 " %via specific reaction 0 introduc)on to Astrophysics, C. Bertulani, Texas A&M-Commerce 12 Branching Consider possibilities as in the figure on the side in terms k of several possible reaction rates λi i j The total destruction rate for the nucleus is m λ = ∑λi (9.25) i And its total lifetime is given by 1 1 τ = = (9.26) λ ∑λi i The reaction flow branching into reaction i, bi is the fraction of destructive flow through reaction i. (or the fraction of nuclei destroyed via reaction i) λi bi = (9.27) ∑λ j j introduc)on to Astrophysics, C.
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