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4 Mean Values and Thermodynamics
We shall consider only Schottky defects that appear in the form of n ion pairs
at the surface of the crystal. We shall assume that an energy E f must be expended
in the formation of each such ion pair within the lattice and in transporting it to
the crystal surface. We shall further assume that the concentration of these defects
is sufficiently small that the creation of any one of them may be treated as an
independent event.
With these assumptions, the number of ways of creating n ion-pair defects
in the ionic crystal is given by the product of the numbers of ways of independently
choosing n positive ions and n negative ions, each from a set of N such ions, i.e.,
=
N !
n!(N − n)!
·
N!
n!(N − n)!
.
The (configurational) entropy S associated with the formation of these n ion-pair
defects will thus be given in terms of as
S(n) = k B ln = k B ln
N!
n!(N − n)!
2
,
or
S(n) = 2k B ln
N!
n!(N − n)!
.
The Helmholtz energy A ≡ U − T S associated with the formation of n such
Schottky defects is then given by A(n) = U(n) − T S(n), in which the internal
energy is given in terms of the energy E f associated with the formation of a single
ion-pair defect as U(n) = nE f , and the configurational entropy, S(n), as
A(n) = U(n) − T S(n)
= nE f − 2k B T [ln N! − ln n! − ln(N − n)!] .
Minimization of the expression for A(n) with respect to n will provide an optimized
value n for the number of Schottky defects. Thus, we must set the derivative of A
with respect to n to zero and solve for the value n of n that minimizes A(n). From
our expression
A(n) = nE f − 2k B T [ln N! − ln n! − ln(N − n)!]
for A(n), we obtain
dA
dn
n=n
= E f − 2k B T [− ln n + ln(N − n)] ≡ 0 .