A simple model for calculating the atomic displacements is due to Kinchin and Pease [3], known as the Kinchin-Pease model. Before discussing this model, let us consider a case of collision between a high-energy neutron of mass M1 and a lattice atom of mass M2 (Figure 3.5). At certain energy ranges, neutron-nucleus interac­tion can be described by an elastic scattering process, where both kinetic energy and momentum of the particles are conserved before and after the collision. In the majority of the events like this, binary (two-body) collision is an appropriate approx­imation. A neutron has no electrical charge (i. e., neutral particle), and hence does not get perturbed by electrical fields exerted by the nucleus or electrons.

Let E1 be the energy of the neutron and due to the collision, the atom is scattered with energy T transferred from the neutron. There exists a threshold energy Ed for the atom to be displaced that is related to the binding energy of the atom. Thus, the struck atom will be displaced only if the transferred energy T is at least equal to or greater than this threshold value. As we have noted before, the displacement threshold energy is very small on the order of ~25 eV, while the transferred energy might be in keV or MeV range. Thus, there is a very high probability that the struck atom will get displaced and become a PKA. This PKA with reasonably high energy will thus interact with other atoms and displace them from their positions leading to secondary, tertiary, and other higher order knock-on atoms.

In the collision process between the incoming high-energy particle and the atom being displaced, if the knocked-on atom flies away with energy, T the incoming neutron will have energy E1 — T — Ed, at a scattering angle of в. In general, how­ever, the energy consumed in the scattering process is assumed negligible and thus the scattered neutron will have energy equal to E1 — T. Of course, the neutron

Figure 3.5 Various characteristics of an elastic collision event are shown in (a) the laboratory coordinates and (b) in the center-of-mass coordinates.

with energy [E1 — T] and the PKA with energy Tproduce other knock-on atoms. A simple calculation can show the likelihood of neutron-nucleus interaction. If we consider a neutron scattering cross section (a) as 1 b (i. e., 10—24 cm) and the num­ber density of atoms (N) in a material as 0.85 x 1023 cm—3 (typical of alpha-Fe), the distance between collisions, that is, the mean free path of the neutron (calculated by 1/Na) will be several centimeters (in this case, ~11.8 cm).

Analytical derivation based on the assumption of the kinetic energy and momen­tum conservation for an isotropic elastic scattering event between a neutron and an atom nucleus gives the following relations:

where Л — (4M1M2 )/(M1 + M2)2 and En is a generalized form of average incident neutron energy and is essentially equal to E10.

From Eq. (3.8), it is clear that the value of Tdepends on the scattering angle, the energy of the incident particle and the masses of the collided particles.

We can see that the maximum energy that can be transferred is given for the angle of scattering of 180°. In other words, the minimum possible value of cos(e), that is, —1, gives the transferred energy the highest value (Tmax), which is then given by the following relation:

T LE

max — L n

Of course, the minimum energy (Tmin) that can be transferred would be zero given a в value of0°.

Table 3.2 A summary of maximum percentage energy transfer that takes place from a 2 MeV neutron to different nuclides in the event of an isotropic elastic scattering (mass of neutron = 1.008665 amu). Nuclides Nuclide mass M2 (amu) , 4MnM2 L =——————— 2 (Mn + M2 )2 Tmax — AEn (MeV) %Energy approximately transferred H1 1.007825 1.0000 2.0000 100 H2 2.014102 0.8894 1.7787 89 C12 12.000000 0.2861 0.5722 28.6 Si12 27.976927 0.1344 0.2687 13.4 Au197 196.96654 0.0203 0.0406 2.0

However, the neutron-nucleus elastic scattering event could lead to glancing col­lisions. In that case, the use of average recoil energy (Tavg) is more appropriate:

Tavg = AEn • (3.10)

Table 3.2 lists the maximum percentage energy that can be transferred to differ­ent nuclides in the event of an elastic collision between a fast neutron (kinetic energy of 2 MeV) and the nuclides. One important observation from this table could be that the neutron transfers more energy to lighter nuclei. Thus, nuclides with low mass numbers generally act as good moderators. As we have discussed in brief, the materials often used are light water (H2O), heavy water (D2O), and graphite (C).

Even though earlier in the chapter we assumed that neutrons are scattered through every possible direction (Isotropic Scattering), in reality they tend to scatter preferentially in the forward directions, and hence the scattering behavior is aniso­tropic. That is why Eqs (3.9) and (3.10) may need to be corrected using a correction factor (f). The value off varies with the nuclide type; for example, for Be it is 0.56, whereas for C and Cu they are 0.84 and 0.6, respectively.

■ Example 3.2

A beryllium (Be) reflector is exposed to a neutron of 1 MeV energy. Let us assume the masses of the neutron and the Be atom are 1 and 9.01 amu, respectively. (a) Calculate the maximum and average energies transferred from the neutron to the beryllium atom based on isotropic elastic scattering event. (b) Determine the average energy transferred if the anisotropic scat­tering occurs (f— 0.56).

Solution

From Eq. (3.8), we know that Tmax = AEn, where Л = (4M1M2)/ (M1 + M2 )2.

a) In order to find out Tmax, we need to first calculate Л. Given M1 = 1 amu and M2 = 9.01 amu,

(4)(1)(9.01) (1 + 9.01)2

Hence, Tmax = (0.36) x (1 MeV) = 0.36 MeV.

Tavg = -^ = 0.18 MeV.

b) For anisotropic elastic scattering event, Taniso(avg) = f (LEn/2) = (0.56 x 0.18 MeV) = ~0.1 MeV.

Now, let us try to get an expression of the displacement damage rate (Rd) defined as the number of displacements per unit volume (cm3) per second. Rd is basically proportional to the number of target atoms per cm3, N (i. e., number density of atoms), and the displacement cross section od(En) for neutrons with energy En, and the neutron flux W(En), and can generally be written as

Rd = N ■ Od(En)- W(En). (3.11)

The important unit of radiation damage, displacements per atom (or dpa), can now be defined as

for neutrons with energies varying from 0 to 1.

Note that the neutrons with En < Ed/L do not generate any displaced atoms since Tshould be >Ed (i. e., the maximum transferable energy given by LEn > Ed to pro­duce a knock-on atom).

The displacement cross section is a function of the sum of the number of atomic displacements, n(T), produced by PKAs with energies T from Ed (the minimum needed) to Tmax(= LEn). The interaction probability is given by the differential energy transfer cross section, od(En, T)dE, for producing a PKA with energy (T, dT) due to the interaction with a neutron of energy En.

Therefore, we can write

Г LEn

Od(En)= On(En, T)■ n(T)dT. (3.13)

Ed

Note that the minimum energy of the PKA to produce displacements is Ed (Л = 1) since scattering between like atoms and the maximum energy of PKA is Tmax = LEn.

The Kinchin-Pease model is the simplest of all radiation damage models. It gives an estimate of the number of displacements produced by a PKA. The model is based on certain assumptions:

1) The cascade is created due to a sequence of binary (two-body) elastic collisions and the potential is based on the hard sphere model approximation.

2) Atomic displacements occur only when T> Ed (i. e., no quantum effects).

3) No energy is passed to the lattice during the collision phase.

4) Energy loss by electronic stopping is given by a cutoff energy known as elec­tronic cutoff energy (Ec). If the PKA energy is more than Ec, no additional displacements occur until electron energy losses reduce the PKA energy to Ec. For all energies less than Ec, electronic stopping is ignored and only atomic collisions take place.

5) Atomic arrangement is random (i. e., the effect of crystal structure is neglected).

6)

No annihilation of defects is assumed.

For the derivation of the K-P model, readers are referred to Refs [4, 5]. Here, we present the end result of the K-P model derivation:

According to Eq. (3.14), the number of displaced atoms can be plotted as a func­tion of PKA energy, as shown in Figure 3.6. The relation (3.14a) is self-explanatory from the very definition of Ed. That is, if T is less than Ed, we expect no atom dis­placement (i. e., n(T) = 0). However, for the relation (3.14b), when Tis greater than Ed but less than 2Ed, there could be two possibilities. One scenario could be that the struck atom does not get energy Ed from the PKA and stays in the same place. Second scenario could be that the struck atom gets more than Ed energy to be dis­lodged from the lattice site. But the PKA now with less than Ed energy falls back to

the original site of the lattice atom. So, in either case, only one displacement takes place. When the PKA energy is greater than 2Ed but less than Ec, the number of displacements increases monotonically with PKA energy until it reaches the elec­tronic cutoff energy (Ec). Beyond that, the number of displacements does not change. Even though in the simple K-P model it was assumed that all energy losses of the PKA go toward elastic collisions, with the increasing PKA energy, a greater fraction of energy will be lost into electronic excitation and ionization. In this case, we need to recognize that PKAs are basically fast ions. So, the number of displace­ments calculated at higher PKA energy would not be all consumed in creating the displacement damage. There are several modifications to the simple K-P model depending on the relaxation of the various assumptions.

Now getting back to the expression of Rd developed in Eq. (3.11) and combining with Eq. (3.13), we get the following expression:

From the theory of isotropic elastic scattering and hard sphere approximation, we can get a relation between double differential energy transfer cross section and scattering cross section:

where elastic collision cross section weakly depends on the neutron energy (En). Thus, it can be shown that

and

where Ф is the total integrated neutron flux given by

Ф = / W(En)dEn.

J0

(3.20)

■ Example 3.3

Consider 0.5 MeV neutrons (ael = 3 b) with a flux of 5 x 1015ncm-2s-1 interacting with alpha-iron (atomic mass number 56, lattice constant 0.287 nm) target. Find Rd.

Solution

Given are En = 0.5 MeV, ael = 3 b, Ed = 24 eV, Ф = 5 x 1015 n cm-2 s-1, and a0 = 0.287 nm.

From Eq. (3.17), we have

We have values of all the terms in the above equation except N and Л.

For Fe, we can calculate the number density of atoms (N) in the target material. There are two ways to do it.

1) If the material is a crystalline solid and we know its crystal structure (here, alpha-Fe that is BCC), we first find out the volume (Vcell) of the unit cell of alpha-iron from its lattice parameter (a0 = 0.287 nm). We know that the effective number of atoms (q) in a BCC unit cell is 2. So, N = q/Vcell = 2/a03 = 2/(0.287 nm)3 = 0.85 x 1023 atoms per cm3.

2) Another way to calculate N can be applied to all state of matter. This has been explained in Chapter 1. This needs knowledge of density and atomic mass of the material.

For Л, we use

Hence,

(0.85 x 1023 cm-3)(0.069)(3 x 10-24 cm2) , , w 2

Rd =±—————— 4(24 eV) —————- “ ^0’5 x 106 eV (5 x 1015 n cm-2 s^1)

=~ 4.5 x 1017 displacements cm-3 s_1.

Now if we divide Rd just by N, we obtain ~5 x 10 6dpas x.

We can also find the dpa value by multiplying the above number by the days of neutron exposure (say, 30 days), we get a dpa of (~5 x 10-6 dpa s-1). (30 x 24 x 3600s) = ~13.0.

Furthermore, if we wish to calculate the number of displaced atoms per neutron, we can use the following expression:

ЛЕn (0.069)(0.5 x 106eV)

4Ed = 4(24 eV)

3.3

Summary

In this chapter, we focused our attention on understanding the concepts of primary radiation damage involving neutron-nucleus collision and subsequent PKA and lattice atoms interaction leading to the formation of displacement cascades. The displacement threshold concept is elucidated with a simple example. The value of displacement energy is generally about 25 eV. Then, the K-P model is introduced. This gives the number of atomic displacements produced by a PKA. Finally, the methods of calculating the damage displace­ment rate and dpa are described. The field of radiation damage is vast, but discussion on all these aspects is out of the scope of this chapter. Interested readers may refer to texts listed in Bibliography.

Problems

1.1 a) If a copper target (mass number 64) is exposed to monoenergetic neutrons

of flux 2 x 1015 n cm-2 s-1 for 6 days continuously, show that the number of atomic displacements per atom (dpa) is 2.856 (Ed = 22 eV, ael = 2 b).

b) What will be the enhancement in self-diffusion in copper at 600 °C if only vacancy survived for million displacements (due to in situ annealing)? Given: Coordination number of copper = 12, lattice constant = ~0.22 nm, vacancy formation energy = 20kcalmol-1, and vacancy migration energy = 18 kcal mol-1.

1.2 a) Using the simple Kinchin-Pease model, evaluate the number of atomic

displacements per atom (dpa) of iron due to a monoenergetic neutron (2MeV) flux of 3 x 1015ncm-2s-1 for 1 year (assume isotropic elastic scattering ael = 3 b and Ed = 40 eV. Iron is BCC with A = 56 and q = 7.9 g cm-3).

b) Calculate the dpa for neutrons of energy 0.25 MeV and compare with the above.

1.3 a) Calculate the average energy of primary knock-ons for Zr when struck with 2MeV neutrons (Zr: (HCP) A — 91, a — 3.23A, c — 5.147A, Ev — 30kcal mol_1, Ed — 59kcalmol_1, sel — 2 b, Ed — 25eV).

b) If in a zirconium target exposed to high-energy neutrons, PKAs were pro­duced with 240 keV, calculate the number ofatoms displaced due to a PKA (assume Kinchin-Pease model).

c) Where do the displaced atoms go to and what is the primary defect created by these atomic displacements?

d) Some of these zirconium specimens were exposed to high-energy neutrons and dpa was calculated tobe2.5, while only one vacancy survived for billion displacements. Calculate the vacancy concentration in the irradiated sample at 500 °C.

e) Evaluate the %increase in lattice diffusivity ofthe irradiated material.