Ion Implantation and Synthesis of Materials - Studium

80 7 Displacements and Radiation DamageThere are two possibilities for the second statement in (7.1). Consider a PKAborn with an energy between E d and 2E d . Now consider the sequence of events asthe PKA undergoes a collision with a lattice atom. If the energy transferred by thePKA to the lattice atom is greater than E d , but less than 2E d , the lattice atom willbe displaced, but the initial PKA is left with energy less than E d . In this situationthe struck atom moves off its lattice site, but the PKA falls into the vacated site,dissipating its remaining kinetic energy as heat. This process represents areplacement collision. Alternatively, if the PKA transfers less than E d to the latticeatom, the struck atom will not be displaced, leaving the PKA as the only displacedatom with insufficient energy left to displace another lattice atom. In either of theabove two possibilities, the PKA collision results in only one moving atom, whichhas an energy less than the original PKA. Therefore, a PKA with kinetic energybetween E d and 2E d produces only one displaced atom.We must now determine the functional form of the damage function 〈N d (E)〉 forE > 2E d . This can be accomplished by calculating the average energy recoilenergy, 〈T〉 produced by a PKA of energy E. From the definition of an averagevalue, the mean recoil energy is given by∫TM1 TM E( , )d d0 ∫ 0TM2〈 T〉 = TP E T T = T T =where we have again used the hard-sphere probability function. Since theminimum energy needed to produce one displacement is E d , the average numberof displacements produced by a mean recoil energy of 〈T〉 is simply 〈T〉/E d orE/2E d .Kinchin and Pease defined a critical energy, E c , to accommodate electronenergy losses by the PKA. Values of E c are taken as M 2 keV, for exampleE c (Cu) = 64 keV. For PKAs generated with energy greater than E c , the number ofdisplacements is assumed to beEc〈 Nd( E) 〉 = (for E > Ec)2Ed(7.2)The total Kinchin–Pease PKA damage function now can be constructed⎧ 0 (if E < Ed)⎪ 1 (if Ed< E < 2 Ed)〈 Nd( E)〉 = ⎨⎪ E/2 Ed(if2 Ed< E < Ec)⎪⎩Ec/2 Ed(if E > Ec)(7.3)and is shown in Fig. 7.2.