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Lecture 5 Basic Nuclear Physics – 3 Nuclear Cross Sections And

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Lecture 5 Basic Nuclear Physics – 3 Nuclear Cross Sections And = 1 cm2 Lecture 5 Basic Nuclear Physics – 3 j Nuclear Cross Sections Flux per cm2 I = n v and Reaction Rates j Total area of target nuclei per cm3 = n σ I I Reaction rate per cm3 assuming no blocking = n v n σ j I I The reaction rate for the two reactants, I and j as in e.g., I (j,k) L is then: nI n j σ I j v which has units reactions cm-3 s-1 For example, a term in the rate equation It is more convenient to write things for 12C during the CNO cycle might look like in terms of the mole fractions, X Y = I n = ρN Y I A I A I gm atoms ⎛ Mole⎞ 12 I ⎛ ⎞ ⎛ ⎞ dY( C) 12 12 ⎜ 3 ⎟ ⎜ ⎟ ⎜ ⎟ = − ρY( C)Y N σ ( C)v +... so that the rate becomes ⎝ cm ⎠ ⎝ Mole ⎠ ⎝ gm ⎠ dt p A pγ 2 (ρN A ) YI Yj σ I j v and a term in a rate equation decribing the destruction of I might be 12 13 for the reaction C(p, ) N γ Equivalent to dYI = − ρYIYj N A σ I jv +.... dn dt I = − n n σ v + ... dt I j Ij Here denotes a suitable average over energies and angles and the reactants are usually assumed to be in thermal equilibrium. The thermalization time is short compared with the nuclear timescale. The cross section for reaction is defined in the usual way: number reactions/nucleus /second σ = number incident particles/cm2 / second Center of mass system – that coordinate system in which the total initial momenta of the reactants is zero. clearly has units of area (cm2 ) σ For a Maxwell-Boltzmann distribution of reactant energies The energy implied by the motion of the center of mass is not available to cause reactions. Replace mass by the "reduced mass" The average of the cross section times velocity is 1/2 3/2 ⎛ 8 ⎞ ⎛ 1 ⎞ ∞ M M σ v = σ (E) E e− E/kT dE µ = 1 2 I j ⎜ ⎟ ⎜ ⎟ ∫ I j ⎝ πµ ⎠ ⎝ kT ⎠ 0 M1 + M2 1/2 1/2 ⎛ 2E ⎞ ⎛ 1⎞ ⎛ 2 ⎞ v = dv = dE where µ is the "reduced mass" ⎝⎜ m ⎠⎟ ⎝⎜ 2⎠⎟ ⎝⎜ mE ⎠⎟ 3/2 1/2 Read Clayton – Chapter 4.1 M m 3 ⎛ 2E ⎞ 1 ⎛ 2 ⎞ I j σ v dv = σ dE µ = ⎝⎜ m ⎠⎟ 2 ⎝⎜ mE ⎠⎟ M + m I j 2 = σE dE m2 for the reaction I (j, k) L 9 -24 2 The Cross Section For T in 10 K = 1 GK, in barns (1 barn = 10 cm ), E6 in MeV, and k = 1/11.6045 MeV/GK, the thermally averaged rate factor in cm3 s-1 is Area subtended by a de Broglie wavelength in the c/m system. How much the nucleus I+j looks -14 ∞ like the target nucleus I with j 6.197 x 10 11.6045 E /T 1 Characteristic quantum − 6 9 sitting at its surface. Liklihood σ v = σ (E ) E e dE = = mechanical dimension jk ˆ 1/2 3/ 2 ∫ jk 6 6 6 of staying inside R once you get A T9 0 p k of the system there. A A ˆ I j A = for the reaction I(j,k)L 2 A + A I j σ(E) = π ρPl(E) Χ(E, A) geometry penetration nuclear term factor structure If you know jk from the lab, or a calculation, in the center of mass (Cla 4-180) frame, just put it in and integrate. The end probability a flux of particles with energy E The actual form of may be very complicated and depends upon the at infinity will reach the presence or absence of resonances and the reaction mechanism. nuclear surface. Must account In general, however, it is of the form … for charges and QM reflection. see Clayton Chapter 4 where is the de Broglie wavelenth in the c/m system 2 2 Consider just the barrier penetration part (R < r < infinity) π π 0.656barns π2 = = = 2v2 2 E ˆ µ µ A E(MeV) Clayton p. 319ff shows that Schroedinger's -24 2 where 1 barn = 10 cm is large for a nuclear cross section. equation for two interacting particles in a radial Note that generally E(MeV) < 1 and > Rnucleus but potential is given by (Cla 4-122) [see also our Lec 4] is much smaller than the interparticle spacing. χl (r) m Ψ(r, θ,φ) = Y (θ,φ) * r l potential where χ(r) satisfies 2 M1M 2 Z Z e µ = V (r)= I j r > R M M 2 2 2 1 + 2 ⎡- d l(l +1) ⎤ r ⎢ 2 + 2 +V (r) − E⎥ χl (r)=0 A A 2µ dr 2µr ˆ 1 2 ⎣ ⎦ V (r)= V r < R A= ~ 1 for neutrons and protons nuc A1 + A2 (Clayton 4-103) ~ 4 for α-particles if AI is large for interacting particles with both charge and angular momentum. The angular momentum term represents the known eigenvalues of the operator L2 in a spherical potential *The 1/r cancels the r 2 when integrating Ψ*Ψ over solid angles (e.g. Clayton 4-114). It is not part of the potential dependent barrier penetration calculation. To solve ⎡-2 d 2 l(l +1)2 ⎤ + +V (r) − E χ (r)=0 Classically, centrifugal force goes like ⎢ 2 2 ⎥ l ⎣ 2µ dr 2µr ⎦ 2 2 2 2 2 mv m v R L divide by E and substitute for V(r) for r >R Fc = = 3 = 3 R mR mR 2 2 2 2 ⎡ − d l(l +1) ZI Z je ⎤ One can associate a centrifugal potential with this, ⎢ 2 + 2 + −1⎥ χ(r) =0 2 E dr 2 r E rE ⎣⎢ µ µ ⎦⎥ −L2 F dR = Change of radius variable. Substitute for r ∫ c 2mR2 2µE 2µE 2µE Expressing things in the center of mass system and ρ = r dρ → dr d 2ρ → d 2r 2 2 2 taking the usual QM eigenvaluens for the operator L2 and for Coulomb interaction chain rule one has 2 ZI Z je 2E 2 η = v = −l(l +1) v µ 2 2µR to obtain ρ and η are dimensionless numbers ⎡ −d 2 l(l +1) 2η ⎤ ⎢ 2 + 2 + −1⎥ χl (ρ)=0 ⎣ dρ ρ ρ ⎦ multiply by -1 d 2χ 2η l(l +1) + (1− − ) χ = 0 dρ 2 ρ ρ 2 This is the solution for ρPl gives the probability of barrier penetration to the nuclear R < r < has solutions (Abromowitz and Stegun 14.1.1) ∞ radius R with angular momentum l. In general, http://people.math.sfu.ca/~cbm/aands/ ρ χ = C1 Fl (η,ρ) + C2 Gl (η,ρ) C1 = 1 C2 = i ρPl = 2 2 where Fl is the regular Coulomb function Fl (η,ρ) +Gl (η,ρ) where F and G, the regular and irregular Coulomb functions and G is the irregular Coulomb function are the solutions of the differential equation and the constants e.g., Illiadis 2.162 l come from applying the boundary conditions See Abramowitcz and Stegun, Handbook of Mathematical Functions, p. 537 The barrier penetration function P is then given by l These are functions of the dimensionless variables 2 ( ) 2 2 2 χl ∞ Fl (ρ = ∞)+Gl (ρ = ∞) 1 Z Z e P I j l = 2 = 2 2 = 2 2 = 0.1575Z Z Aˆ / E F (η,ρ)+G (η,ρ) F (η,ρ)+G (η,ρ) η = I j contains all the charge dependence χl (R) l l l l v 2µE Cla 4-115 ˆ contains all the radius dependence ρ = 2 R=0.2187 AE Rfm For the one electron atom with Ze2 The “1” in the numerator corresponds to a a potential , one obtains the r purely outgoing wave at infinity from a same solution but the radial component decaying state. is Laguerre polynomials. Z Z e2 Physical meaning of η= I j v On the other hand, The classical turning radius, r , is given by 0 = = = 2µE R p µv ⎛ 1 2 ⎞ ρ = R = 2(µ)⎜ µv ⎟ 2 2 ⎝ 2 ⎠ 1 Z Z e µv2 = I j 2 r0 The de Broglie wavelength on the other hand is is just the size of the nucleus measured in de Broglie wavelengths. 2 2ZI Z je 2 r = =η = 2η = = o µv2 µv p µv This enters in, even when the angular momentum and charges are zero, because an abrupt change in potential r at the nuclear surface leads to reflection of the wave Hence η= 0 nb., both η and ρ are dimensionless. 2 function. The probability of finding the particle inside of its classical turning radius decreases exponentially with this ratio. Z Z e2 For low interaction energy, (2η>>ρ, i.e., E << I j ) R and Z ≠0, ρP has the interesting limit j l There exist other interesting limits for ρPl , for example when η is small - as for neutrons where it is 0 ⎡ 2l(l +1) ⎤ ρ ∝E1/2 ρ P = ρ ρP ≈ 2ηρ exp⎢−2πη + 4 2ηρ − ⎥ Abramowitz and Stegun, 0 l 3 ⎣⎢ 2ηρ ⎦⎥ 14.6.7 ρ ρ << 1 for cases of interest ρ P1= 2 where 1+ ρ for neutron capture 1/2 independent of E and 5 2 0.2625 Z Z AˆR ! ρ ηρ = ( I j fm ) ρP = 2 9 + 3ρ2 + ρ4 which is independent of energy but depends on nuclear size. This implies that for l = 0 neutrons 2 the cross section will go as 1/v. ZIZ j e Note: η= = 0 v rapid decrease with smaller energy and increasing charge(η ↑ ) 2µE 1/ 2 ρ = R=0.2187 AˆE R rapid decrease with increasing angular momentum 2 E −1/ 2 2 fm 2 Z Z e i.e., π ρ P0 ∝ ∝ E I j ˆ E η = =0.1575ZI Z j A / E The leading order term for l = 0 proportional to v 2µE For low energy neutron induced reactions, the ρ = R=0.2187 AˆE R 2 fm cross section times velocity, i.e., the reaction rate ρP ∝ exp −2πη l ( ) term, is approximately a constant For particles with charge, providing X(A,E) does not vary rapidly.
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