# Solution to Exercise 2.1-6 "Enthalpy difference for the limiting cases of Schottky or Frenkel Defects"

| Calculate the ratio of the concentration of Schottky to Frenkel defect as a funtion of the enthalpy difference
| The equations for the concentrations of the point defects in the "mixed" case are
cV(C) | = cS = exp – | HS 2kT | · | æ ç è | 1 + | N' N | ·exp | HS – HFP kT | ö ÷ ø | 1/2 | = exp – | HS 2kT | · K
cV(A) | = exp – | HS 2kT | · | æ ç è | 1 + | N' N | ·exp | HS – HFP kT | ö ÷ ø | – 1/2 | = exp – | HS 2kT | · K–1
ci(C) | = cFP = | N' N | · exp | Hs 2kT | · exp – | HFP kT | · | æ ç è | 1 + | N' N | ·exp | HS – HFP kT | ö ÷ ø | – 1/2 | = | N' N | · exp | Hs 2kT | · exp – | HFP kT | · K–1
|  | Note that cV(C) or ci(C) is, by definition, identical to the concentration cS or cFP of Schottky or Frenkel defects, respectively. If you have problems with this, refer to the link.
|  | We abbreviated the root of the expression in square brackets by K for writing efficiency.
| The ratio cS /cFP is easy to obtain. The K's cancel, we are left with
cS cFP | = | N N' | · exp – | (HS – HFP) kT | = | N N' | · exp – | D H kT
--- | --- | --- | --- | --- | --- | --- | --- | ---
|  | That is - of course - what we should have expected. The concentrations of Schottky and Frenkel defects are independent of each other and their relation could have been derived straight from the basic equations defining their equilibrium concentrations.

| Show in in particular, how large the difference must be if 90% or 99% of the defects are to be of one kind.
| We want to evaluate the equation for cS/cFP = 0,011 or = 0,001 (prevalence of Frenkel defects) and cS/cFP = 90 or = 99 (prevalence of Schottky defects).
|  | For the difference D H of the formation enthalpies as defined above we obtain
D H | = – | kT | · | æ ç è | ln | N' N | + ln | cS cFP | ö ÷ ø
|  | We have to define a value for N/N' ; we simply take this relation to be 1 or 0,1 as limiting cases.
| Values are easily obtained, we arrange them in a little table
cS cFP
| 99 | 90 | 10 | 0,1 | 0,011 | 0,010
D H [eV] |
N N' | = 1
--- | ---
| –0,115 | –0,112 | –0,058 | 0,058 | 0,112 | 0,115
N' N | = 10
--- | ---
| –0,172 | –0,169 | –0,115 | –0,0004 | 0,054 | 0,057
| Discuss the result.
| We have two interesting results:
|  | If the formation enthalpies of the two defect kinds differ by just about 1/10 of an eV, we are fully justified to consider that only one defect kind is present.
|  | The pre-exponential factor N'/ N, which describes the differences in the basic geometry for interstitials relative to vacancies, accounts at most for about 1/20 of an eV if expressed in enthalpy differences.

With frame

Exercise 2.1-8 Quick Questions

Point Defects in Ionic Crystals

Exercise 2.1-6: Enthalpy difference for the limiting cases of Schottky or Frenkel Defects

© H. Föll (Defects - Script)