Solutions to exercises
271
P
e1
1
O 2
1 5
σ
σ
=
#
−
°
where 1
σ ° is a constant.
2 Oxygen defect: TiO 2(1−x)
K g2 = [V O
::
]
2 # n
4 # P O 2
By using the electroneutrality equation, we obtain
K g2 = 4
1
# n
6 # P O 2
or
n A P
2 O 2
1 6
=
−
with
A
(4K )
const.
2
g 2
1 6
=
=
and
A F u P
e2
2
e
O 2
1 6
σ =
#
#
#
−
P
e2
2
O 2
1 6
σ
σ
=
#
−
°
where 2
σ ° is a constant.
2. a. Experimental equation log σ e = f (log P O 2 ) at 1 000 °C
2 By exploiting the results shown in figure 89, we obtain in the domain
10
−12
≤ P O 2 ≤ 10
−4
bar an equation of the type
log
log ° log P O 2
1 4
σ
σ
=
+
−
2 Likewise, in the domain 10
−12
≤ P O 2 ≤ 10
−20
bar, we obtain an equation of the type
log
log ° log P O 2
1 6
σ
σ
=
+
−
Remark – The preceding reasoning associates the total conductivity to
the electronic conductivity of the semiconductor. This is completely
justified because of the differences in concentration and, especially, in
the mobility of the ionic and electronic carriers.
b. Referring to question 1(c), we notice that the majority point defect is the
oxygen vacancy V O
::
compensated by the electronic species e ′ .
3. a. We consider the equilibrium with gaseous oxygen in the following reaction:
Ti
#
Ti + 2O
#
O m O 2(g) + Ti i
3• + 3e ′
(3)