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
 ·   ( 1 +   N'
N
 · exp   HS  –  HFP
kT
) 1/2  = exp – HS
2kT
 · K

  cV(A)  = exp – HS
2kT
 ·   ( 1 +   N'
N
 · exp   HS  –  HFP
kT
) – 1/2  = exp – HS
2kT
· K–1

  ci(C)  = cFP =  N'
N
  ·  exp  Hs
2kT
 ·  exp – HFP
kT
 ·  ( 1 +   N'
N
 · exp   HS  –  HFP
kT
) – 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 abbreviate 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 –   (HSHFP)
kT 
 =  N
N'
  ·  exp –   ΔH
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.
 
Discuss the result. 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 ΔH of the formation enthalpies as defined above we obtain
 
ΔH  = – kT  ·  ( ln   N'
N
  +  ln   cS
cFP
)
 
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
ΔH [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
 
We have an interesting result: If the formation enthalpies of the two defect kinds differ by just about 1/10 of an eV, we are fully justified to consider one defect kind only.
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.
 

To index To index

go to Point Defects in Ionic Crystals

go to Formation Enthalpies and Entropies for Frenkel and Schottky Defects

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

© H. Föll