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27 - 725Materials for Nuclear Energy

Radiation Damage:The Radiation Damage Event

Tony (A.D.) Rollett, Carnegie Mellon University

Spring 2019Updated: 22nd Jan. 2019

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Lecture Objective• The objective of this lecture is to give students the basic

tools needed to understand what happens in a Radiation Damage Event.

• The topics follow the chapter order in Was, except that some chapters are omitted for lack of time.

• The more applied topics correspond to characteristic issues in some of the materials commonly used in nuclear reactors.

• For example, stress corrosion cracking is a particular issue for the nickel and stainless steel alloys.

• For example, creep and swelling are issues for both the uranium oxide used as fuel and the zirconium alloys used as cladding.

Basics

• Atomic and microscopic changes in structure lead to macroscopic changes in properties (swelling, embrittlement, etc.).

• The scientific discipline of radiation damage research involves revealing mechanisms responsible for atomic and microstructural changes produced by irradiation and predicting macroscopic property changes.

• The key concept is that of displacements per atom (dpa) as being more indicative of damage than particle fluence.

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Outline1. Radiation Damage Event2. Atom Displacement3. Damage Cascade4. Point Defect Formation & Diffusion5. Loss of, Sinks for Point Defects6. Radiation Induced Segregation7. Radiation Induced Dislocations8. Radiation Induced Voids, Bubbles9. Radiation Effects on Phase Stability10. Ion Bombardment, Implantation11. Surrogates for Neutron Irradiation12. Radiation Effects on Hardening, Deformation13. Radiation Effects on Fracture Toughness, Embrittlement14. Radiation Effects on Creep and Growth15. Radiation Effects on Corrosion: emphasis on Stress Corrosion

Cracking (SCC)

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Nomenclature• Flux: number of particles crossing unit area per unit time.

Generally the particle of interest here is the neutron. Electrons and photons are also interesting, sometimes. For pressure vessels, the fluxes are typically of order 1010

0n/cm2/s at the inner surface (cf. 1014 0n/cm2/s at the baffle). Minimizing the flux that actually reaches the wall is important for maximizing reactor life.

• Fluence: integrate flux over time and the result is fluence, which is the total irradiation dose. For pressure vessels, the doses are typically of order 0n/1019/cm2.

• Displacements per atom, dpa: properties correlate better with dpa than with fluence.

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Radiation Damage Event

• The key components of a radiation damage event are as follows.– Interaction between an (energetic) incident particle (or photon) and an

atom in the material– Transfer of kinetic energy from the incident particle to the atom

(and/or nuclear reaction), possibly giving rise to a primary knock-on atom (PKA).

– Displacement of the affected atom from its site (given sufficient energy).

– Passage of the displaced atom through the surrounding lattice, with possible creation of additional displaced atoms.

– Generation of a displacement cascade from the set of point defects created by the original PKA.

– Cessation of motion of PKA, leaving behind point defects.• Analogy: creation, storage of dislocations during plastic

deformation.

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From: T. Allen, U Wisconsin

Simple Picture: Incoming Particle

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From: T. Allen, U Wisconsin

Simple Picture: Damage Event

Primary Knock-on Atom ≡ PKA

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From: T. Allen, U Wisconsin

Simple Picture: secondary, tertiary knock-on

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From: T. Allen, U Wisconsin

Simple Picture: Damage Cascade

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From: T. Allen, U Wisconsin

Simple Picture: Defect Formation

vacancy

interstitial

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From: T. Allen, U Wisconsin

Displacement Damage

� Primary Damage – 10-11 seconds� Migration of Defects – seconds� Property Changes – hours, days, months

Displacement per atom (dpa)=average number of displacements of each lattice atom;this is generally much larger than the number of input ions/particles, especially for energetic neutrons.

Note: impossible to observe experimentally; only studied via computer simulation (molecular dynamics)

Primary Damage• Vacancy• Interstitialpka

Vacancy

Interstitial

Focusson

LCSReplacement collision

Radiation Damage Effects• 1 ms = 10-3 s; 1 µs = 10-6 s; 1 ns = 10-9 s; 1 ps = 10-12 s;• The typical time involved in a damage event is extremely short ~

10-11 seconds (10 ps). Migration of defects via thermally activated mechanisms such as diffusion require (much) longer times.

• Changes in atom positions that occur during a damage event are classified as radiation damage effects.

• The main objective of the next section is to quantify the transfer of energy from incoming particles and the atoms in a material.

• The focus will be on neutron-nucleus interactions, which dominates in nuclear reactors. Neutrons are neutral so we can neglect (electrostatic) interactions with the electrons and focus on the nucleus.

• Was’s book also deals with ion bombardment of materials where the charged particles (ions) interact strongly with the electrons, at least at low energies. Ion bombardment is a useful surrogate for neutron irradiation and is also useful as a surface treatment.

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Fluence versus dpa

• Greenwood showed [J. Nucl. Mater. 216 29 (1994)] that displacements per atom, dpa, is the best measure of radiation damage. Compare the spread in yield strength increase as a function of fluence, versus the unified curve that results from using dpa.

• For engineering purposes, however, although dpa are used for stainless steels, it is more common to use fluence in assessing pressure vessels.

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dpa – displacements per atom• Experience has shown that the number of

displacements per atom (dpa) is the most useful measure of damage.

• Therefore we need to calculate the number of displacements per unit volume, per unit time, R. This depends on the density of atoms, N, the range of incident energies from Emin to Emax, the particle flux (as a function of energy, Ei), f(Ei), the displacement cross section (as a function of energy), sD(Ei):

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Was’s notation for min. and max. energies isrespectively.

Energy Spectrum16

http://nuclearpowertraining.tpub.com/h1019v1/css/h1019v

1_138.htm

NEUTRON FLUX SPECTRUM DOE-HDBK-1019/1-93 Reactor Theory (Neutron Characteristics)

NP-02 Page 34 Rev. 0

Figure 3 Comparison of Neutron Flux Spectra for Thermal and Fast Breeder Reactor

Thermal and Fast Breeder Reactor Neutron Spectra

The spectrum of neutron energies produced by fission varies significantly from the energyspectrum, or flux, existing in a reactor at a given time. Figure 3 illustrates the difference inneutron flux spectra between a thermal reactor and a fast breeder reactor. The energydistribution of neutrons from fission is essentially the same for both reactors, so the differencesin the curve shapes may be attributed to the neutron moderation or slowing down effects.

No attempt is made to thermalize or slow down neutrons in the fast breeder reactor (liquid metalcooled); therefore, an insignificant number of neutrons exist in the thermal range. For thethermal reactor (water moderated), the spectrum of neutrons in the fast region (> 0.1 MeV) hasa shape similar to that for the spectrum of neutrons emitted by the fission process.

In the thermal reactor, the flux in the intermediate energy region (1 eV to 0.1 MeV) hasapproximately a 1/E dependence. That is, if the energy (E) is halved, the flux doubles. This1/E dependence is caused by the slowing down process, where elastic collisions remove aconstant fraction of the neutron energy per collision (on the average), independent of energy;thus, the neutron loses larger amounts of energy per collision at higher energies than at lowerenergies. The fact that the neutrons lose a constant fraction of energy per collision causes theneutrons to tend to "pile up" at lower energies, that is, a greater number of neutrons exist at thelower energies as a result of this behavior.

• The figure presents a typical energy spectrum from two different types of reactors, showing the broad range of energies that is present in an operating reactor.

• For the current fleet of civilian reactors, use the “thermal” curve

Capture Cross Sections• The nuclear capture cross section, scapture, for neutrons is the effective area within

which a neutron passes in order to be captured by an atomic nucleus. It is a probabilistic value dependent on the nature and energy of the particle as well as the nature of the capturing nucleus. Nuclear capture cross section is often measured in barns (yes, a humorous reference to barn doors);(1 barn = 10–24 cm2 = 10–28 m2).

• Example of Boron:– The element Boron has two stable isotopes, 10B and 11B. They differ by one neutron.– The neutron capture cross section of B-10, with 5 neutrons, is 3835 barns (this means it has high

probability of capturing another neutron).– The neutron capture cross section of B-11, with 6 neutrons, is 0.0055 barns (low probability).

• Any "Macroscopic cross section”, S, is the effective cross-sectional area per unit volume of material for whatever process is of interest (capture, scattering, fission etc.); hence, it is equal to the product of the number of atoms per unit volume, N, and their (microscopic) cross sections (written as s). The atom number density is equal to the product of Avogadro’s Number and density, divided by atomic weight, N=NA*r/M.

• Thus, the macroscopic capture cross section is the sum of the various weighted capture cross sections. The unit of measure for S is cm2/cm3 or reciprocal cm (cm–1). It is often measured in "capture units" or "sigma units." A capture unit (c.u.) = 10–3 cm–1. S = Ns.

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Fission Cross Section18

Thermal neutrons

Neutron Cross-section for Fehttp://www.nndc.bnl.gov/exfor/servlet/E4sMakeE4

1 of 1 3/27/13 7:24 AM

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Note that the (total) cross-section for Fe is about x10 smaller than the fission cross-section for U-235 at 1eV (10-6 MeV).

Cross-sections from ENDF• Searching on “ENDF BNL” should show you a link to the ENDF

database at Brookhaven Natl. Lab.• This allows you to look up cross-section data for a huge variety of

nuclear reactions.• Check the help pages for what is reasonable to do.• The main page asks you to choose the nuclide (multiple nuclides

can be requested), the reaction type, and the type of cross section.• Reactions include (e.g.) neutron-gamma (n-g), neutron-fission (n-f),

neutron-elastic (n-el), neutron-inelastic (n-inl).• If you choose an element with many nuclides along with a broad

range of reactions, the website may complain that you have selected too many records.

• In general, it is reasonable to select just one set of records that cover the cross-section types that you are interested in (e.g., one each of n-f, n-el, n-inl, n-g, n-tot).

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Types of cross sections• There are, obviously, many types of cross sections. We

can define cross sections for fission cross section, sf, radiative capture cross section, sr, absorption cross section, sa and scattering cross section, ss.

• An important example of a cross-section is the fission cross section for the (thermal) neutron fission (n,2n)reaction for U-235, sf = 577 barns. Given a density of 18.7gm/ccwhich translates to N=4.8 1028 m-3, then the macroscopic cross section, S, is about 28cm-1. Clearly this will depend on the enrichment of the uranium and its chemical form (metal vs. oxide, e.g.).

• The other important cross section is the scattering cross section because this is central to quantifying radiation damage. The displacement cross section is a subset of the scattering cross section.

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Capture Cross Sections: 1• What is the physical meaning of a capture cross section? • The approximate radius of a nucleus as a function of its mass, MT, is given by:

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• For a lithium-6 nucleus, with mass 6.01522amu, this works out to be 2.27.10-5 Å.

• Compare this with the cross section of 941barnsfor the (n,a) reaction whereby the nucleus captures a thermal neutron and promptly emits an alpha particle. This is a fairly large capture cross section.

• If we take the circle-equivalent radius corresponding to the cross section, we find a radius of 173.10-5 Å, which is astonishingly larger than the actual physical size of the nucleus. Hence the physical mechanisms of interaction are more complicated than just hard sphere collisions!

Displacement Cross-sections

We can expand the displacement cross-section in terms of the energy transferred from the particle to an atom, T, and the scattering cross-section, ss. The number of displacements per PKA is n(T).

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Then we expand the equation for the displacement rate as follows, where s is the cross section associated with energy transfer in collisions.

The overall equation for the displacement rate:

Rd = N

Z Emax

Emin

�(Ei)�D(Ei)dEi

Rd = N

Z Emax

Emin

Z Tmax

Tmin

�(Ei) �(Ei, T ) ⌫(T ) dEidT

Energy Transferred• A critical quantity is the ratio of masses of the

incident particle, m, and the atom struck, M. We define g as follows.

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• The relationship between the energy transferred, T, and the incident particle energy and the scattering angle in the center-of-mass-frame, f, is as follows.

• This then makes it easy to see that the maximum energy transferred is in a head-on collision (f =180�).

• Now let’s see where this comes from …

Note how for neutrons, m=1, the efficiency of energy transfer drops off steeply as the mass of the struck atom increases. Hence, light elements are most effective as moderators

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T. Allen, Wisconsin

Binary Collisions

M1 M2v10

v2f

v1fq1

q2

Laboratory Frame of Reference-elastic collision

Momentum (vector) balance: Energy (scalar) balance

€

M1v102 = M1v1 f

2 +M2v2 f2

x

yConservation of momentum

and energy applies

Section 1.1.1 in Was

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Center of Mass (COM) System• It turns out, however, to be much easier to do the analysis

in the center-of-mass frame of reference, denoted by the suffix cm.

Why work in center-of-mass frame of reference?If external force on system of particles is zero, the center of mass velocity (in the lab. frame) is constant:

even for inelastic collisions, which most real collisions are.

• When we transform the coordinates to the center of mass reference frame, vcm=0. This simplifies the analysis.

€

Fcm = ma = mcmdvcmdt

= 0

vcm = constant

T. Allen, Wisconsin

Definitions27

• Masses in both frames are obviously invariant:M1 and M2 are taken to be the neutron and nucleus masses, thus mand M, respectively.

• Velocities in the laboratory frame are designated as v• Initial velocities in the laboratory frame are designated as v10 or v20

• Scattering angles in the laboratory frame for the particle and PKA are , q1 and q2, respectively

• Velocities in the center-of-mass (COM) frame are designated as u , thus u10 or u20, with angles f1 and f2.

• Momentum is Mv, or Mu• The primary knock-on atom is known as the PKA• Kinetic energy KE is Mv2, or Mu2

• Kinetic energy of the incoming neutron is Ei (as in incident energy) and the PKA has T, or recoil energy

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Center-of-Mass (COM) System

M1 M2

M2

M1

j1

j2

u10

u2F

u1F

u20€

xcm M1 +M2( ) = x1M1 + x2M2vcm M1 +M2( ) = v1M1 + v2M2

€

u10 = v10 − vcm = v10 1−M1

M1 +M2

#

$ %

&

' ( = v10

M2M1 +M2

#

$ %

&

' (

u20 = vcm = v10M1

M1 +M2

#

$ %

&

' (

€

vcm =M1

M1 +M2

"

# $

%

& ' v10

In Lab frame, if v2 = 0

COM definition (Lab frame)

Initial velocities, u, in the COM frame:

T. Allen, Wisconsin

For this discussion,

M1=neutron, m, and M2=nucleus

Position:Momentum:

COM characteristics29

• The net linear momentum in CM coordinates is zero (at alltimes).• In the absence of external forces (an inertial frame-of-reference), the velocity of the center-of-mass is constant (equal to vcm, relative to the lab frame-of-reference).• In a two-body system, the center of mass Ocm must lie on a line adjoining the two interacting particles at all times.• In a two-body system, the velocity vectors for the interacting particles, m1 and m2, u1 and u2, respectively, are anti-parallel at all times.• In a two-body system, the magnitudes of the velocities of the particles in CM coordinates remain unchanged by the collision (i.e., the scattering interaction).

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Collision description

€

Initial Condition : M1u10 = M2u20

In the COM frame, the magnitude of the velocity does not change during the collision

€

Incoming Particle : u10 = u1 f = v10 − vcmStruck Particle : u20 = u2 f = vcm€

//x :M1u1 f cos ϕ1( ) = M2u2 f cos ϕ2( )//y :M1u1 f sin ϕ1( ) = M2u2 f sin ϕ2( )

€

ϕ1 = ϕ2

T. Allen, Wisconsin

x

y

Resolve the velocities on each axis after the collision:

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Collision descriptionm

u1f

u2fv2f

v1f

q1q2

M

vcmj1

M is the PKA

m is the neutron

€

v2f2 = vcm

2 + u2f2 − 2vcmu2f cosϕ1

Goal: determine energy of PKA, T, as a function of neutron energy Ei

Law of cosines:

Define T=PKA kinetic energy, or, recoil energy

Define Ei=neutron kinetic energy

T. Allen, Wisconsin

Laboratory Frame of Reference-elastic collision

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Collision description

€

vcm =m

m + M

"

# $ $

%

& ' ' v10

€

v2 f2 = vcm

2 + u2 f2 − 2vcmu2 f cosϕ1

€

T =12M v2 f

2

€

v2 f2 = 2vcm

2 1− cosϕ1( )Recall

€

u2f = vcm

Recall

Recall

€

T =12

4mMm + M( )2

1− cosϕ( )Ei

T. Allen, Wisconsin

vcm = v10M1

M1 +M2Based on:

Note that we still have the scattering angle, f, in the COM frame

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Collision description

Define (as before):

€

γ =4m M

m + M( )2

€

T =12γ 1− cosϕ( )Ei

gives

€

Tmax = γEi

Collisions between similar size masses lead to the greatest PKA energy; again, this explains why light elements are more effective moderators than heavy ones.

T. Allen, Wisconsin

So for head-on collision with f = 180°:

Elastic Scattering: 2• We need to know how much energy is transferred to an atom as a

function of the energy of the incident particle. Starting with the total elastic scattering cross section, s(Ei) is given in the COM frame by the following*.

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• If we assume, as is reasonable, that the (elastic) scattering cross-section is independent of the scattering angle, then the integral is very simple.

• One can also show that there is a simple relationship between the scattering cross section written in terms of energy transferred versus scattering angle:

* See derivation in Was, Eqs. 1.15-1.18, p9.

Eq. 1.19b

Eq. 1.20

Eq. 1.19

�S(Ei,�) = �S(Ei)/4⇡

Elastic Scattering: 3Combining these last two equations yields the following, which shows that the probability of transferring a given amount of energy T is the same, regardless of the value of T.

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Thus the average energy transferred is just the average of the max and min energies, but if the minimum energy is negligible then the average is about half the maximum, which is just given by the head-on collision case.

Examples for neutrons with 1 MeV: 1 MeV n on C : g = 0.28, <T> = 0.14 MeV 1 MeV n on Fe : g = 0.069, <T> = 0.035 MeV = 35 keV = 35,000 eV! 1 MeV n on U : g = 0.017, <T> = 0.009 MeV

Eq. 1.21

Eq. 1.22

Inelastic Scattering• Inelastic scattering is exactly what the name

suggests – a particle interacts with a nucleus but there is a loss of kinetic energy.

• The typical event is capture of a neutron by a nucleus, forming an excited state [nucleus] that subsequently emits a neutron (but with a different energy) and one or more gamma rays.

• There are two significant types of reactions: (n,2n) and (n,g). The first is the fission reaction characteristic of 235U and 239Pu. The second is particularly important to damage in the reactor pressure vessel where the gamma flux is comparable to the (fast) neutron flux.

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Inelastic Scattering: 237

Was gives an equation (Eq. 1.44) that summarizes how the capture cross section varies as a function of the input neutron energy, Ei, where E0 is the energy associated with the main absorption resonance, and G is the neutron width of the compound nucleus. The equation below gives the cross section for gamma production, which is important in pressure vessels (because the gamma flux in the wall is comparable to the neutron flux). However, Was does not provide values for the neutron width of typical elements, so one would need to search in the literature for data (e.g. ref. 6, Evans, The Atomic Nucleus (1956)).

CYLINDRICAL REACTOR PRESSURE VESSELS 5

6

m GAMMA ENERGY = 2 MeV £ o_ 5

O

>. 4 O 5 MeV

,.z. ~ 3

u ~ 2

A i i i i i

0 1 2 3 4 S b

RELAXATION LENGTH OF GAMMA PENETRATION

Fig. 4. The effect of gamma energy on the energy absorption buildup factor at various relaxation lengths.

1.0

Z 0 . 8 O

Z 0.6 O

0 . 4 >

0.2

THERMAL NEUTRON -- -- --

G~MMA RAYS 2 MeV

~ G~,MMA RAYS 7 M e V O ' ~ O " ~ O

REFERENCE HEAT f t 3 _ h r

I I I I I 5 10 IS 20 25 PRESSURE VESSEL WALL THICKNESS (in.)

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Fig. 5. Relative heat generation rate varies with pressure vessel wall thickness.

4.1. Thermal neutrons The rate of volumetric heat generation due to the capture of thermal neutrons and the subsequent absorption

of secondary gamma rays in the pressure vessel can be given by eqs. (1), (5) and (6). Here,

Effect of Gamma Radiation- Ma (1969) made estimates of the energy deposition rates in reactor pressure vessel walls and demonstrated that neutron and gamma heating are nearly equivalent.- Note that the interaction of gammas with atoms is complicated, with three different mechanisms, depending on the energy range. From to low to high energy: photoelectric, Compton and pair production.- Kwon & Motta (2000) showed that the displacement cross section for gammas in iron reaches several barns above ~7 MeV (which is the average energy of gammas reaching the wall). Therefore the damage rate from gammas is significant.

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Ma, Ann. Nucl. Eng. (1969)

Ion-Atom Interactions39

• There are approximately 3 classes of ions whose interactions with atoms we need to consider.

1. Light energetic ions with Ei> 1 MeV.

2. Highly energetic heavy ions, such as fission fragments, Ei ~ 100 MeV.

3. Lower energy heavy ions produced by an accelerator, or that recoil from earlier, higher energy collisions, < 1 MeV.

The graph (Fig. 1.12 in Was) of distance of closest approach, r, (as a ratio of the screening radius of the atom, a) versus recoil energy, T, shows examples of each of the 3 types of ion in copper, along with the different interatomic potential that should be used to characterize the interaction.

Ion-Atom Collisions• We can also summarize the types of

reactions in the following table.

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Summary• Capture Cross Sections are an important piece of information about interactions

between incoming particles and atoms. They are measured in barns, which is of the order of the cross-sectional area of a nucleus, although the values vary considerably. Macroscopic cross sections are also used, with units of cm-1.

• In a collision between an incoming particle (e.g. neutron) and an atom, the maximum energy that can be transferred depends on the ratio of the masses. The larger the atom, the smaller the energy transfer from a light particle such as a neutron.

• We are particularly interested in nuclear reactions that absorb a neutron and then give out either 2 neutrons, or that give out a gamma ray.

• In Light Water Reactors where the moderator and the coolant are the same, this results in approximately 5 Mev per fission deposited directly into the coolant by neutron thermalization alone [Doster]; most of the available 190 MeV is deposited into the fuel.

• In terms of radiation damage, displacements per atom (dpa) are a much better measure of (potential) damage than particle flux (the rate) or fluence (the total dose).

• Accordingly, we are particularly interested in measuring or predicting the damage rate in a given material, under a given irradiation.

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Supplemental Slides• Graphs from Lau

Fission of 235U43