254
5 – Applications
c. The mobilities will be equal when
T
e
T e
6 10
3 6
.
kT
eV
kT
eV
3
0 4
1
–
–
#
=
−
or
.
ln
T
k
eV
6 10
0 6
3
#
=
#
^
h
.
.
.
.
ln
T
8 314
6 10
0 6 1 6 10
6 02 10
3
19
23
#
#
=
−
#
#
#
#
^
h
T
K
799
=
d. 2 Given that the conductivity of particle i is given by
z F u [i]
i
i
2 2 i
σ =
u
the total conductivity of SrCl 2 under the conditions of question (c) is
t
V
Cl
Cl
i
σ
σ
σ
=
+
:
l
F u
[V ] [Cl ]
t
2 Cl
Cl
i
i
σ =
+
:
l
l
u ^
h
By neglecting the electronic conductivity, the electroneutrality equation takes the form [Cl ] [V ]
K
i
C l
A F
=
=
:
l
The final expression for the total conductivity is thus
2F u
K
t
2 Cl
AF
i
σ =
l
u
2 Numerical evaluation gives
2 (96 480) 799
36
e
3 10 e
t
2
2
kT
1eV
kT
0.8 eV
σ =
−
− −
#
#
#
#
2 (96 480) 799
36
e
3 10 e
t
2
2
1.38 10
799
1.6 10
1.38 10
799
0.8 1.6 10
23
19
23
19
σ =
−
−−
#
#
#
#
#
#
#
#
#
#
#
−
−
−
−
1.14 10 S cm
t
4
1
σ =
−
−
#
2. a. The reaction for dissolution of KCl in SrCl 2 is
KCl $ K ′
Sr + Cl
#
Cl + V
•
Cl
It leads to an increase in the concentration of chlorine vacancies V
•
Cl .
b. In figure 85(b),
2 the B ′ C ′ domain corresponding to the smallest slope represents the
extrinsic domain; that is, the domain where the concentration of
charge carriers that give rise to electrical conduction is fixed by foreign species. In this case, this involves essentially the dopant KCl.
5 – Applications
c. The mobilities will be equal when
T
e
T e
6 10
3 6
.
kT
eV
kT
eV
3
0 4
1
–
–
#
=
−
or
.
ln
T
k
eV
6 10
0 6
3
#
=
#
^
h
.
.
.
.
ln
T
8 314
6 10
0 6 1 6 10
6 02 10
3
19
23
#
#
=
−
#
#
#
#
^
h
T
K
799
=
d. 2 Given that the conductivity of particle i is given by
z F u [i]
i
i
2 2 i
σ =
u
the total conductivity of SrCl 2 under the conditions of question (c) is
t
V
Cl
Cl
i
σ
σ
σ
=
+
:
l
F u
[V ] [Cl ]
t
2 Cl
Cl
i
i
σ =
+
:
l
l
u ^
h
By neglecting the electronic conductivity, the electroneutrality equation takes the form [Cl ] [V ]
K
i
C l
A F
=
=
:
l
The final expression for the total conductivity is thus
2F u
K
t
2 Cl
AF
i
σ =
l
u
2 Numerical evaluation gives
2 (96 480) 799
36
e
3 10 e
t
2
2
kT
1eV
kT
0.8 eV
σ =
−
− −
#
#
#
#
2 (96 480) 799
36
e
3 10 e
t
2
2
1.38 10
799
1.6 10
1.38 10
799
0.8 1.6 10
23
19
23
19
σ =
−
−−
#
#
#
#
#
#
#
#
#
#
#
−
−
−
−
1.14 10 S cm
t
4
1
σ =
−
−
#
2. a. The reaction for dissolution of KCl in SrCl 2 is
KCl $ K ′
Sr + Cl
#
Cl + V
•
Cl
It leads to an increase in the concentration of chlorine vacancies V
•
Cl .
b. In figure 85(b),
2 the B ′ C ′ domain corresponding to the smallest slope represents the
extrinsic domain; that is, the domain where the concentration of
charge carriers that give rise to electrical conduction is fixed by foreign species. In this case, this involves essentially the dopant KCl.