# Solution to Exercise 2.1-5 "Do the Math for Mixed Point Defects"

 For obvious reasons some of the symbols deviate a little from the symbols used in the text; e.g. we have hFP instead of HFP. We start with the system of equations that came from the mass action law We start with the calculation of cv(C): Inserting the first and the second equation into the third equation yields: That was the first equation for cV(C). Next we calculate ci(C). Start with the third equation and eliminate cv(A) using the second. We have the final result after a series of mathematical manipulations: That was the third equation. Next we calculate cV(A). Start with the third equation and eliminate ci(C) using the first, we obtain That's it. Nothing to it. ;-) Well, not exactly. I myself certainly cannot solve problems like this without making some dumb mistakes in breaking down the math. Almost everybody does. However, I usually notice that I made a stupid mistake because the result just can't be true. And I can, if I really employ myself, get the right result eventually - because I did some exercises like this before. And that is why you should do it, too!. As a last comment we may note that solving equations coming from the mass action law can become rather tedious very quickly - compare the example in the link, which is about as simple as it could be.

Now we look at the limiting cases of pure Schottky or pure Frenkel disorder.
For pure Frenkel disorder we must have hFP << hS, and cV(A) = 0.
For pure Schottky disorder we must have hFP >> hS, and ci(C) = 0.
For the first case - pure Frenkel disorder - just look at the expression
1  +  N N' æ ç è ö ÷ ø
For hS >> hFP, the exponential in this case is positive which means
N
N'
· exp  hS  –  hFP
kT
>>  1
So you may neglect the 1 in the above expression and replace the whole square root by
N
N'
· exp  hS  –  hFP
2kT
This gives for ci(C)
ci(C)  =  N
N'
·
 exp  hS  –  2hFP kT æ ç è ö ÷ ø

 exp  hS  –    hFP kT æ ç è ö ÷ ø
=  N
N'
· exp –   hFP
kT
This is the result as as it should be.
With this we immediately obtain
cV(C)  =  N
N'
· exp – hFP
2kT

cV(A)  =  0
This is so because
N
N'
· exp  hS  –  hFP
kT
>>  1
Contrariwise, if hS << hFP, 1 + N/N' · exp[(hShFP)/kT] » 1 obtains.
Because hS – 2hFP is a large negative number we get
ci(C)  =   N
N'
· exp  hS  –  2hFP
2kT
»  0
The expressions for cV(C) and cV(A) immediately reduce to the proper equation
cV(C)  = cV(A)  = exp – hS
2kT

Chemical Examples for Mass Action Law Applications

Exercise 2.1-5: Do the Math

Exercise 2.2-1: Properties of Johnson Complexes

© H. Föll (Defects - Script)