Definition of cross sections 1 - Moodle

Cross sectionsAlain Hébertalain.hebert@polymtl.caInstitut de génie nucléaireÉcole Polytechnique de MontréalENE6101: Week 4 Cross sections – 1/37

Content (week 4) 1Definition of cross sectionsFormation of a compound nucleusselection rulesresonancesPorter and Thomas distributionWigner distributionThe single level Breit-Wigner (SLBW) formularesonance parameters of 232 Th235 U, 238 U and 239 Pu cross sectionslow-energy variation of cross sectionsENE6101: Week 4 Cross sections – 2/37

Definition of cross sections 11. The trajectory of a neutron in a material is a straight line which can be interrupted by anuclear interaction with a nucleus of the material.2. The neutron-nucleus collision can result in a variety of nuclear reactions (such as theelastic scattering reaction).3. The concept of cross section is used to describe the probability of each type of nuclearreaction. The probability for a neutron located at r and moving in a material at velocityV n to undergo a nuclear reaction in a differential element of trajectory ds isindependent of the past history of the neutron and is proportional to ds.Let’s consider a one-speed and parallel beam of neutrons of intensity I neutrons per unitsurface and unit time. The beam hits perpendicularly a target of width ds. The targetcontains a unique type of nuclide with a number density of N nuclei per unit volume.I (cm-2 s-1)N (cm-3)ENE6101: Week 4 Cross sections – 3/37ds

Definition of cross sections 2The number density of nucleus in the target is given by(1)N = ρA 0Mwhere ρ is the density (g/cm 3 ), M is the atomic mass of one nuclide (u) and A 0 is theAvogadro number defined as 6.022094×10 23 u/g.The intensity I if the beam is obtained from(2)I = V R nwhere V R is the relative velocity of the neutrons with respect to the target and n is thenumber density of the neutrons (cm −3 ) in the beam. If the target is at 0K, V R = V n .The surfacic reaction rate dR x is defined as the number of nuclear reactions of type x perunit time and unit surface of the target. It is experimentally found that dR x is proportional tothe number density N, to the intensity of the beam I, and to the width of the target, at thelimit of zero width. The microscopic cross section σ x is defined as the proportionality factor:(3)dR x = σ x N I ds .ENE6101: Week 4 Cross sections – 4/37

Definition of cross sections 3The microscopic cross section must have the dimension of a surface to make Eq. (3)dimensionally consistent. They are generally expressed in barn (b), with 1 b = 10 −24 cm 2 . Itis also common to define the macroscopic cross section Σ x to group all the characteristics ofthe target in a single value. It is defined as Σ x = N σ x .If the material of the target is an homogeneous mixture of different types i of nuclides, theresulting macroscopic cross section is(4)Σ x = ∑ iN i σ x,i .Moreover, we define the total macroscopic cross section as the sum of cross sections fromall nuclear reactions. We write(5)Σ = ∑ xΣ x .ENE6101: Week 4 Cross sections – 5/37

Definition of cross sections 4I 0 (cm-2 s-1)sLet’s consider now a one-speed and parallel beam of neutrons of intensity I 0 neutrons perunit surface and unit time. The beam hits perpendicularly a slab of finite width. The collisionor total reaction rate dR is the number of collisions of neutrons in an elemental width dslocated at distance s, per unit time and unit surface of the slab. Each time a neutron collidein ds, it is removed from the uncollided beam. The reaction rate is writtends(6)dR = −dI(s) = I(s)Σdswhere I(s) is the intensity of the uncollided beam after a distance s in the slab. Integratingthis equation between 0 and s, we obtain(7)I(s) = I 0 e −Σs .ENE6101: Week 4 Cross sections – 6/37

Definition of cross sections 5The ratio I(s)/I 0 is the probability for the neutron to be uncollided after a distance s. Theprobability of an interaction for this neutron in the following elemental width ds is Σds. TheprobabilityP(s)ds for a neutron in I 0 to collide in ds is therefore(8)P(s)ds = Σe −Σs ds .This equation is useful to define the mean free path λ of neutrons in an infinite slab, animportant quantity in reactor physics. The mean free path is the average trajectory length ofthe neutrons in an infinite and homogeneous material. This value is obtained from equation(9)λ =∫ ∞0dssP(s) = 1 Σ .ENE6101: Week 4 Cross sections – 7/37

Definition of cross sections 61. Each type of nuclear reaction is characterized by a specific microscopic cross sectionσ x which is also function of the type of nuclide and on the velocity or kinetic energy ofthe neutron. However, the potential cross section is independent of neutron energy.2. An elastic neutron-nucleus collision is characterized by an elastic scattering crosssection σ e . This reaction includes both potential and resonant elastic collisions.3. The symbol Q represents the energy produced by a nuclear reaction in the form ofkinetic energy. In a threshold reactions, Q is negative and the nuclear reaction canoccur only if e exc ≥ −Q. If a scattering reaction has a threshold energy −Q, thecompound nucleus is left excited after the collision and decay with gamma rayemission of energy −Q (short half life). This is the inelastic scattering cross section(σ in ).The scattering cross section σ s is written σ s = σ e +σ in + ∑ x≥2 σ n,xn.σtotalσsσe σin σn,2n σ n,3n σfσγσασpscattering(n,xn)absorptiontransmutationENE6101: Week 4 Cross sections – 8/37

Definition of cross sections 7In a similar way, we may define cross sections for reactions involving the absorption of theincident neutron. A representation of this hierarchy is given in figure.Nuclear reactions in reactor physics are often represented with the following notation:(n,n):(n,n’):(n,γ):elastic scattering,inelastic scattering,the radiative capture cross section σ γ(n,f): the fission cross section σ f ,(n,α), (n,p):(n,2n):transmutation cross sections σ α (an α particle is emitted)or σ p (a proton is emitted)n-2n reaction, etc.All these reactions, with the exception of the potential scattering, involve the formation of acompound nucleus and are characterized by cross sections that may exhibit high variationwith neutron energy. This phenomena will be studied in the next section.The total cross section is the sum of all existing cross sections.ENE6101: Week 4 Cross sections – 9/37

Formation of a compound nucleus 1Any nuclear reaction with formation of a compound nucleus are proceeding in twosuccessive steps:1. The incident neutron is absorbed by the target nucleus A ZX to form an excited state ofnucleus A+1ZX called the compound nucleus. During the formation process, theavailable kinetic energy of the incident neutron e exc and the binding energy of thisadditional neutron S n (A+1,Z) are distributed among all the nucleons of thecompound nucleus. The binding energy is computed in term of the mass default using(10)S n (A+1,Z) = [M(A,Z)+m n −M(A+1,Z)]c 2where M(A,Z) is the mass of isotope A Z X expressed in u, m n = 1.008665u is themass of the neutron and c 2 = 931.5MeV/u is the square of the light velocity in void.2. The compound nucleus decay with a half life between 10 −22 and 10 −14 second,without reminding how it was formed. The neutronic reactions are not the only onesthat can produce a compound nucleus; a photo-nuclear interaction (with an incident γray) can produce the same effect. This decay can occur following a number of decaychannels, as represented in table. Each decay channel involves the production ofsecondary rays and/or particles which can emerge in ground or excited state.ENE6101: Week 4 Cross sections – 10/37

Formation of a compound nucleus 2Type of reactionNumber of decay channels1○ elastic scattering 1 channel2○ inelastic scattering 1 channel if e exc ≥ −Q3○ radiative capture very high number of channels4○ fission ≃ 2 or 3 channels if e exc +S n ≥ e fThe compound nucleus model consists to write the interaction as a two-stage reaction,(11)AZ X +1 0 n → A+1Z X∗ →where the asterisk indicates that the compound nucleus is in an excited state. The first arrowdenotes the formation stage and the second, the decay stage. The energy-level diagram forthis reaction is shown in figure.⎧⎪⎨⎪⎩1○2○3○4○,ENE6101: Week 4 Cross sections – 11/37

Formation of a compound nucleus 3≈ 1 eV2J π 2(excited level)eexces = –Q14AZXI π 1negative resonances3Sn(A+1,Z)≈ 5 MeV≈ 40 keVA+1Z X (compound nucleus)ENE6101: Week 4 Cross sections – 12/37

Formation of a compound nucleus 4A decay channel is open only if the three following laws are observed: conservation ofenergy, linear and angular momentum of the nuclide-nucleus pair. The first two laws havealready been studied in Week 3. The conservation of angular momentum is written(12)J = I +K +L whereI , K = angular momentum of the target nucleus A ZX in its ground state and of the neutronJ = angular momentum of the compound nucleus A+1ZX in its excited stateL = orbital angular momentum of the nuclide-nucleus pair in the CM.The classical definition of the orbital angular momentum of the nuclide-nucleus pair aboutthe origin of the CM is L = m(r n −r CM )×v n +mA(r A −r CM )×v A .Using relations in week 3 and defining the e z –directed axis as perpendicular to the planewhere the particles are moving, this equation simplifies to(13)L = ∆RV R m 0 e zwhere ∆R is the distance between the lines of definition of vectors v n and v A in the CM andm 0 is the reduced mass of the neutron defined as m 0 = mAA+1 .ENE6101: Week 4 Cross sections – 13/37

Formation of a compound nucleus 5In reactor physics, the principle of conservation of angular momentum involves principles ofquantum mechanics. There is a relation between the orbital angular momentum of theneutron-nucleus pair and the angular momentum quantum number.In quantum mechanics, the modulus of the orbital angular momentum vector L is related tothe angular momentum quantum number l, a positive integer, by the relation(14)L = √ l(l+1)where = 1.054494×10 −34 J·s is the reduced Plank constant. A comparison of theclassical expression of L in Eq. (13) and of the quantum expression in Eq (14) indicates thatl = 0 is likely to occur if V R is low. However in cases where ∆R or V R is high, greater valuesof l become possible. Note that high values of ∆R is only possible when the target nucleushas a large radius, a characteristics of heavy nuclides. The following nomenclature isuniversally accepted:l =⎧⎨0; s–wave interaction,1; p–wave interaction,⎩2; d–wave interaction.In conclusion, s–wave interactions are the most common for low-energy incident neutronsand p– or d–wave interactions will be more probable for heavy target nuclides.ENE6101: Week 4 Cross sections – 14/37

Selection rules 1The parity π is also required. It is a binary quantity (equal to +1 or -1) that characterize theenergy levels of the particles in interaction. The basic information about parity is:Spin (I, J or K)Parity (π)Neutron 1/2 +1Nucleus (ground state):even A and even Z 0 +1even A and odd Z integer and ≠ 0 unknownodd A half integer unknownNotations I π 1 and J π 2 are often used to represent the spin and parity of initial and finallevels, respectively. An energy level of the compound nucleus can be excited if the followingtwo selection rules are fulfill (see ENDF–102):1. the spin J of the excited level in A+1ZX is an element of set{J min ,J min +1,...,J max } where J min = ∣ |I −l|−1∣ and J2 max = l+I + 1 2 ,2. the parity of the excited level in A+1 Z X must obey π 2 = π 1 (−1) l where π 1 and π 2 arethe parity of the target (ground level) and of the compound nucleus (excited level).ENE6101: Week 4 Cross sections – 15/37

Resonances 1Notice that e ∗ = S n (A+1,Z)+e exc is the available excitation energy of the compoundnucleus and is measured on the CM. If the compound nucleus has an excited state at e i thatis close to e ∗ , then one can have resonance condition and the compound nucleus formationcross section will show a peak at the neutron incident energy corresponding to e i . Thesepeaks are at the origin of the resonant cross section phenomena, as depicted in figure. Wealso observe that the resonance peaks for a cross section of isotope A ZX correspond to theexcitation states of isotope A+1Z X located above energy S n(A+1,Z).eexcAZX02.55.07.510.012.5σxENE6101: Week 4 Cross sections – 16/37

Resonances 21. Each excitation level e i has a certain energy width γ i , due to the Heisenberguncertainty principle. The width γ i corresponds to its finite lifetime τ i :(15)γ i = τ i.The finite lifetime is the average lifetime of the compound nucleus at level i. It is thereciprocal of the radioactive decay constant. The smaller the width means the longerthe lifetime of the level. The energy width γ i is defined in the CM.2. The resonance width is related to the formation of the compound nucleus at level i.The probability for it to decay with a nuclear reaction of type x is given in term of thepartial resonance width γ x,i as(16)P x,i = γ x,iγ iso that(17)γ i = ∑ xγ x,i .The decay channel x is one of the nuclear reactions. The partial width correspondingto the scattering reaction is the neutron width γ n,i .ENE6101: Week 4 Cross sections – 17/37

Resonances 33. The peak energy of a resonance may be negative, in the case where a level of thecompound nucleus is close to the state corresponding to the capture of a neutron bythe target but is located below this state (called a negative resonance).4. Intermediate and heavy isotopes are characterized by a high number of resonances. Itbecome possible to apply statistical studies on the resonance widths γ and of theirspacing D, corresponding to a given spin and parity. This statistical treatment is usefulfor the representation of the unresolved resonances in domain 10keV ≤ E ≤ 300keV.5. The neutron width is a function of the incident neutron energy and is written γ n (e exc ).The statistical treatment on the neutron width is applied on the random variable(18)x = γl n〈γ l n〉where the reduced neutron width γn l , a function of the angular momentum quantumnumber l, is defined asγn l = γ n (e exc )(19)v l (e exc ) √ e excwhere v l (e exc ) is a coefficient related to the penetrability of the potential barrier in thetarget nucleus.ENE6101: Week 4 Cross sections – 18/37

Resonances 4The first three values of the penetrability coefficient arev 0 (e exc ) = 1( [R̷λ) 2 ( ]R̷λ) 2−1v 1 (e exc ) = 1+(20)v 2 (e exc ) =( [R̷λ) 4 (9+3 R̷λ) 2 ( ]R̷λ) 4−1+where R is the hard sphere radius of the nucleus. It can be computed in term of the atomicmass ratio A using R = ( 0.123A 1/3 +0.08 ) ×10 −12 cm, and where ̷λ is the reducedwavelength of the neutron, defined by the relation(21)̷λ =m 0 V R=√ 2m0 e excwhere m 0 is the reduced mass of the neutron.ENE6101: Week 4 Cross sections – 19/37

Resonances 5The reduced wavelength can also be written in term of the neutron mass m and of theexcitation energy E exc in the LAB. We write(22)where(23)̷λ = A+1A√2mEexc= ( 0.4552136×10 −9) A+1AE exc = A+1A e exc1√Eexccmis expressed in eV. Using the above definitions, we find that the reduced neutron width γ l ncan be expressed in √ meV.ENE6101: Week 4 Cross sections – 20/37

Porter and Thomas distribution 1P(x,ν)1.5ν = ∞1ν = 16ν = 40.5ν = 2ν = 100 1 2 3xPorter and Thomas have shown that the resonance widths γ γ , γ f and the normalizedneutron width γ l n corresponding to a single spin and parity obey a chi-square statistics:(24)P(x,ν) =ν2G(ν/2)( νx2)ν2−1e−νx/2where x = γ x /〈γ x 〉 is the random variable defined as the ratio of the actual resonance widthdivided by the average resonance width for reaction x. The random variable x = γl n〈γ l n〉 mustbe used in the case of the neutron width.ENE6101: Week 4 Cross sections – 21/37

Porter and Thomas distribution 2The number of degrees of freedom ν in the Porter-Thomas distribution is equal to thenumber of decay channels of the compound nucleus. The average value of x is 1 and itsmost probable value is x p = 1−2/ν if ν ≥ 2. In the case where ν = 1, P(x,1) becomesinfinite at x = 0. The gamma function G(ρ) is defined as(25)G(ρ) =∫ ∞0dte −t t ρ−1so that G(1/2) = √ π, G(1) = 1 and G(ρ) = (ρ−1)G(ρ−1). The Porter-Thomas isdepicted in figure for different values of ν.In the case of a scattering reaction, ν = 1 and the Porter-Thomas distribution simplifies to asimple decreasing exponential. In the case of a radiative capture, the number of decaychannels is very high and P(x,∞) = δ(x−1), the Dirac delta distribution. In this case, γ γwill be almost constant from resonance to resonance.ENE6101: Week 4 Cross sections – 22/37

Wigner distribution 1S(x)0.80.70.60.50.40.30.20.100 1 2 3xThe statistical distribution of the resonance spacings corresponding to a single spin andparity follows the Wigner distribution written as(26)and depicted in figure.S(x) = π 2 xe−π 4 x2where x = D〈D〉ENE6101: Week 4 Cross sections – 23/37

The SLBW formula 1The single level Breit and Wigner (SLBW) formulas result from the application of theSchrödinger equation to a compound nucleus model with a single excitation level. With thismodel, variations in cross sections are related to the width and energy characteristics of thisexcited level.For reactions involving the absorption of the incident neutron, such as radiative capture orfission, the SLBW formula is written in term of the reduced energy variable(27)u = 2 γ 1(e exc −e 1 )as(28)whereσ x (e exc ) = σ 0γ x,1γ 111+u 2γ x,1 = resonance width of the excited level for an absorption reaction of type xγ 1 = total resonance width of the excited levele 1 = energy of the excited level in the CM relative to the ground level of the target nucleus.ENE6101: Week 4 Cross sections – 24/37

The SLBW formula 2The parameter σ 0 is defined as(29)σ 0 = 4π̷λ 2 g Jγ n,1 (e exc )γ 1= 2π2m 0√eexcv l (e exc )g Jγ l n,1γ 1where the expression for γ n,1 (e exc ) and ̷λ 2 were recovered from Eqs. (19) and (21),respectively. In this relation, γ l n,1 is constant. The statistical spin factor g J is defined as(30)g J =2J +1(2I +1)(2K +1)=2J +12(2I +1)since K = 1/2 for the neutron. The excited level of the compound nucleus has a spin Jfunction of the angular momentum quantum numberland consistent with the selection rules.ENE6101: Week 4 Cross sections – 25/37

The SLBW formula 3In the case of the elastic scattering reaction, the cross section is the sum of threecomponents. The first two are the potential and resonant cross section terms. There is alsoan interaction term between the first two components. The interference term arises in thequantum mechanical model when the modulus of the sum of two complex quantities is taken.The SLBW formula corresponding to the elastic scattering cross section is(31)σ e (e exc ) = σ l p +σ 0 sin2φ l( )u1+u 2 +σ γn,1 10 −2sin 2 φ lγ 1 1+u 2 .The potential component σ l p is defined in term of the angular momentum quantum number las(32)σ l p = 4π̷λ 2 (2l+1) sin 2 φ lwhere the integer quantum number l is ≥ 0 and where φ l are the shift factors.ENE6101: Week 4 Cross sections – 26/37

The SLBW formula 4The first values of the shift factor φ l are written asφ 0φ 1= a̷λ= a̷λ −1−tana̷λ(33)φ 2 = a̷λ −tan −1 ̷λ( a̷λ) 2.3−3aWe have used the diffusion radius a different from the hard sphere radius R introducedearlier. In general, the ratio a/̷λ is small with respect to one, and the potential cross sectioncomponent σp 0 for s wave interactions is almost equal to the classical value corresponding toa “billiard-ball" collision:(34)σ 0 p = 4πa2 .ENE6101: Week 4 Cross sections – 27/37

The SLBW formula 5Summation of Eq. (28) with Eqs. (31) corresponding to all absorption types of nuclearreactions leads to the SLBW expression of the total cross section:(35)σ(e exc ) = σ l p +σ 0 sin2φ lu1+u 2 +σ 0 cos2φ l11+u 2 .Equations (28) and (31) are giving the variation of the absorption-type and scattering crosssections with energy e exc in the case of a unique resonance. These variations are depictedin figure.σ e (b)9σγ (b)20σf (b)808.751040208.50 00 0.2 0.4 0.6 0.8 0 0.2 0.4 0.6 0.8 0 0.2 0.4 0.6 0.8E (eV)E (eV)E (eV)Figure depicts a nice illustration of the effect of the interference term in Eq. (31).ENE6101: Week 4 Cross sections – 28/37

The SLBW formula 6This is a very well known shielding issue associated with 56 Fe. The cross section is almostvanishing near 24 keV, so that incident neutrons with this energy are undergoing very fewcollisions in iron.Microscopic total cross section (b)100101100010000100000Energy (eV)ENE6101: Week 4 Cross sections – 29/37

The SLBW formula 71. The SLBW formula can only be applied to the case of an isolated resonance. However,most isotopes feature many resonances at increasing energies. A crude approximationconsists to neglect resonance interactions, so that the cross section of a reactionx ≠ e is written by summing the contributions of I resonances as(36)σ x (e exc ) =I∑i=1σ 0γ x,iγ i11+u 2 .A similar expression can be written for σ e (e exc ).2. In the general case, many levels of the compound nucleus interact together and amultilevel formula is required. In this case, the cross section are obtained by summingthe contributions from each individual resonance, taking care to introduce interactionterms in the sum.3. At energies above ≃ 10 keV, resonances become unresolved and are represented bystatistical parameters. At higher energies, the resonance become tighter up to thepoint where they overlap, leading to the continuum energy domain.ENE6101: Week 4 Cross sections – 30/37

The SLBW formula 8The SLBW equations (28) and (31) are function of the excitation energy e exc in the CM.However, most theoretical developments used in reactor physics are expecting crosssections defined in term of LAB-related quantities. We remember the expression of theexcitation energy E exc in the LAB as(37)E exc = A+1A e exc = 1 2 mV 2 R .In case where the target nuclide is initially at rest, E exc is equal to the initial energy of theneutron in the LAB. If we replace e exc by E exc in Eqs. (??) and (??), these equations remainvalid provided that we redifine LAB-related peak energy and resonance widths as(38)E 1 = A+1A e 1 and Γ x,1 = A+1A γ x,1 .Peak energy and resonance widths are measured and are reported in reference tables asLAB-defined values, similar to those defined in Eqs. (38). The neutron widths are reportedwith the statistical spin factor included, as g J Γ n,1 (E 1 ).ENE6101: Week 4 Cross sections – 31/37

Resonance parameters of 232 Th 1i E i (eV) gΓ n,i (E i ) (meV) Γ γ,i (meV) l1 8.346 0.0003 29.0 12 13.111 0.0002 13 21.783 2.0200 24.5 04 23.439 3.8800 26.6 05 36.926 0.0010 16 38.165 0.0006 17 40.925 0.0006 18 47.001 0.0014 19 49.850 0.0006 110 54.130 0.0011 111 58.780 0.0096 112 59.514 3.9000 23.7 013 64.580 0.0005 114 69.224 43.8000 21.9 0ENE6101: Week 4 Cross sections – 32/37

Uranium-235 total cross section 1Microscopic total cross section (b)1000010001001010.1110100Energy (eV)ENE6101: Week 4 Cross sections – 33/37

Uranium-238 total cross section 1Microscopic total cross section (b)100001000100101110100Energy (eV)ENE6101: Week 4 Cross sections – 34/37

Plutonium-239 total cross section 1Microscopic total cross section (b)1000010001001010.1110100Energy (eV)ENE6101: Week 4 Cross sections – 35/37

Low-energy variation of cross sections 1We consider a resonance located at an energy e x,1 above the thermal energy domain (> 1eV) and study its contribution at thermal energies. We will rewrite the SLBW formulas andtake their limits as e x,1 approaches zero.If R/̷λ

Low-energy variation of cross sections 21. These equations are valid at energies E ≤ 300 keV for a heavy nuclide and atenergies E ≤ 1 MeV for an intermediate nuclide (A ≃ 40).2. We will next consider an isotope without negative resonance and without resonancesin the thermal energy domain. In this case, the energy-variation of the cross sectionsat thermal energies is dictated by the energy variation of ̷λ and Γ n,1 . Theabsorption-type widths, such as Γ γ,1 or Γ f,1 , are constant in energy.3. The neutron width Γ n,1 , on the other hand, varies as √ E for a s wave interaction, asE 3/2 for a p wave interaction and as E 5/2 for a d wave interaction. The squaredreduced wavelength ̷λ 2 varies as 1/E.4. We therefore observe that s wave absorption-type reactions feature a characteristic1/v–dependence. At low energies, the probability of these interactions is directlyproportional to the time the neutron spends within the reach of the nuclear force.5. Assuming s wave interaction, we see that σ e (e exc ) is almost constant at low energyand that the absorption-type reactions with no threshold energy vary as 1/ √ E.ENE6101: Week 4 Cross sections – 37/37