Chapter 2 solutions
___________________________________________________________________________________
2.2
a.
Apply the equation for temperature dependence of resistivity,
ρ
(
T
) =
ρ
o
[1 +
α
o
(
T

T
o
)]. We have the
temperature coefficient of resistivity,
α
o
, at
T
o
where
T
o
is the reference temperature. The two given
reference temperatures are 0
°
C or 25
°
C, depending on choice. Taking
T
o
= 0
°
C + 273 = 273 K,
ρ
(40
°
C + 273 = 233 K) =
ρ
o
[1 +
α
o
(233 K
−
273 K)]
ρ
(25
°
C + 273 = 298 K) =
ρ
o
[1 +
α
o
(298 K
−
273 K)]
Divide the above two equations to eliminate
ρ
o
,
ρ
(40
°
C)/
ρ
(25
°
C) = [1 +
α
o
(40 K)] / [1 +
α
o
(25 K)]
Next, substitute the given values
ρ
(25
°
C) = 2.72
×
10
8
Ω
m and
α
o
= 4.29
×
10
3
K
1
to obtain
K)]
)(25
K
10
(4.29
+
[1
K)]
)(40
K
10
(4.29
+
[1
m)
10
(2.72
=
C)
(40
1

3

1
3
8

×
×
Ω
×
°
ρ
= 2.03
×
10
8
Ω
m
b.
In
ρ
(
T
) =
ρ
o
[1 +
α
o
(
T
−
T
o
)] we have
α
o
at
T
o
where
T
o
is the reference temperature, for example, 0
°
C
or 25
°
C depending on choice. We will choose
T
o
to be first at 0
°
C = 273 K and then at 40
°
C = 233 K
so that
ρ
(40
°
C) =
ρ
(0
°
C)[1 +
α
o
(233K
─
273K)]
and
ρ
(0
°
C) =
ρ
(40
°
C)[1 +
α

40
(273K
─
233K)]
Multiply and simplify the two equations above to obtain
[1 +
α
o
(233 K
−
273 K)][1 +
α
40
(273 K
−
233 K)] = 1
or
[1
−
40
α
o
][1 + 40
α
40
] = 1
Rearranging,
α
40
= (1 / [1
−
40
α
o
]
−
1)(1 / 40)
∴
α
40
=
α
o
/ [1
−
40
α
o
]
i.e.
α
40
= (4.29
×
10
3
K
1
) / [1
−
(40 K)(4.29
×
10
3
K
1
)] =
5.18
×
10
3
K
1
Alternatively, consider,
ρ
(25
°
C) =
ρ
(40
°
C)[1 +
α
40
(298 K
−
233 K)] so that
α
40
= [
ρ
(25
°
C)
−
ρ
(40
°
C)] / [
ρ
(40
°
C)(65 K)]
∴
α
40
= [2.72
×
10
8
Ω
m
−
2.03
×
10
8
Ω
m] / [(2.03
×
10
8
Ω
m)(65 K)]
∴
α
40
=
5.23
×
10
3
K
1
c.
We know that 1/
ρ
=
σ
=
en
μ
where
σ
is the electrical conductivity,
e
is the electron charge, and
μ
is
the electron drift mobility. We also know that
μ
=
e
τ
/
m
e
, where
τ
is the mean free time between
electron collisions and
m
e
is the electron mass. Therefore,
1/
ρ
=
e
2
n
τ
/
m
e
2.1
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