Before starting discussions on various radiation damage models, let us first under­stand the concept of displacement threshold. While describing the primary damage events, it is essential to develop a clear understanding of the displacement energy or displacement threshold (denoted by Ed), which is defined as the minimum energy that must be transferred to a lattice atom in order for it to be dislodged from its lattice site. Generally, average displacement energy of 25 eV is used. How­ever, the fixed value of 25 eV is only an average of all the possible displacement energies calculated along different crystallographic directions in a given material. This value also agrees well with the experimentally measured displacement energy values. The specific values of displacement energy depend on the nature of the momentum transfer, trajectory of the knock-ons, crystallographic structure, and thermal energy of the atoms. It has been noted that higher melting point metals tend to have higher displacement energies (Figure 3.3). Even though all elements do not follow the trend as there are many other factors that influence displacement

o I— .—- ,—- .—- ,—- .—- ,—- .—-

750 1500 2250 3000 3750

Melting Temperature (K)

Figure 3.3 The variation of displacement energy as a function of melting temperature of metals.

energy, it is clear from Figure 3.3 that the refractory metals (such as Mo and W) tend to have much higher displacement energies compared to those of lower melt­ing point ones. It is most possibly related to the stronger binding energy present in higher melting point metals.

As regards to the lower limiting value of the displacement energy, it is close to the energy needed to produce Frenkel pairs (3-6 eV). If the energy transferred by the knock-on atom to the struck lattice atom is less than the displacement energy, the atom will not dislodge from its site. Rather it will vibrate around an equilibrium position, transfer the energy through the neighboring lattice atoms, and eventually dissipate as heat. In order to calculate the displacement energy, it is essential to know the description of the interatomic potential fields since this is the energy bar­rier that the struck atom needs to surmount to eject successfully from its regular lattice site. A simple example is shown below.

■ Example 3.1

Calculate the displacement energy along (110) direction (i. e., a face-centered position to an adjacent face-centered position) in an FCC metal unit cell (as shown in Figure 3.4a) in which the interatomic potential is given by a simple repulsion potential as described below:

V(r) = — U + У2 k(req — r)2 at r < req, (3.1a)

V(r) — 0 at r > req, (3.1b)

where U is the binding energy of atom (energy per atomic bond), k is a force constant indicative of the repulsive portion of the potential, req is the equili­brium atom separation, and r is the general separation distance. The force constant is k«0 and U values can be taken as 60 and 1 eV, respectively. Note that a0 is the lattice constant of the crystal.

Solution

The atom at position “M” after being dislodged from the site needs to follow the direction of the arrow (parallel to (110) direction) to the position “N.” During the trajectory motion of the atom M, it needs to go through a poten­tial field created by the four-atom barrier (1-1-1-1). The center of this four — atom barrier represents the saddle point (at O) along the trajectory of M atom with the maximum potential energy; the variation of potential energy as a function of position is shown in Figure 3.4b.

The energy of a single atom in the FCC crystal can then be given by Eeq = — 12 U, considering that an FCC atom is surrounded by 12 equidistant near­est neighbors (due to the coordination number of FCC crystal being 12). In turn, we can say that when M atom is at the equilibrium site, the energy of M atom is given by as shown above.

Now when the M atom moves to the center of four-atom barrier, it achieves the highest potential energy following the potential described in Eq. (3.1a). Therefore, the potential energy at the saddle point “O” is given by

E* — 4V(r)=4[- U + =, fe(req — r)2].

The displacement energy (Ed) is given by

Ed — E* — Eeq — 4[— U + =, k(req — r)2] — (-12 U)

— 8 U + 2k(r eq — r)2. (3.2)

Now we need to express req and r in terms of lattice constant (a0). This can be accomplished from the geometrical relations in the FCC unit cell, as shown in Figure 3.4a. In this case, req is the minimum distance from M to O. A simple geometric construction can show that the distance can be
determined by calculating half the hypotenuse of a triangle with other two sides being a/2 and a/2.

Hence, we get

and r (the impact parameter), the distance of each atom-1 of the four-atom barrier from the point O, is given by

r (i. e., O1) = 2 у/(Я0)2 + = y^. (3.4)

Now we take req and r relations from Eqs (3.3) and (3.4), respectively, and use them in Eq. (3.2). Thus, we obtain

Ed along (110}=8U + 2k(req — r)2 = 8U + 2k~ Jh^ = 8U + 2 x k

(0.213)2a2 = (8U) + (2 x 0.045)(k a2) = (8 x 1eV) + (0.09 x 60eV) = ~13.4eV.

As additional exercises, determine the displacement energy along (100} and (111} in FCC yourself using the same potential expression given in Eq. (3.1). It is very clear that accurate knowledge of interatomic potentials is quite important in the accuracy of the calculated displacement energy values.

Special Note: The use of displacement energy in ceramics is similar but bit complicated because of the presence of multiple atomic species (cations and anions). Table 3.1 summarizes some displacement energy values of cations and anions in some well-known ceramics.

In case of multicomponent ceramics, an effective displacement energy (Edff) is often used to calculate the extent of radiation damage. This is given by

where Si is the stoichiometric fraction and Ed is the displacement energy of the ith atomic species. Taking into account the nature of Coulombic interactions in

Table 3.1 A summary of displacement energies in some ceramic materials. Material Threshold displacement energy (eV) Al2O3 eA1 ~ 20, EO = 50 MgO EM = 55, EO = 55 ZnO EZn ~ 50, EO = 55 UO2 EU = 40, EO = 20

collision cascades, the scaling parameter is more appropriately expressed as (SiZi2)/Ai, where Z; and A; are the atomic number and atomic mass of the ith species.

3.2