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Homework #9
12.1
(a)
Compute the electrical conductivity of a 7.0mm (0.28in.) diameter cylindrical silicon specimen 57 mm
(2.25 in.) long in which a current of 0.25 A passes in an axial direction.
A voltage of 24 V is measured across two
probes that are separated by 45 mm (1.75 in.).
(b)
Compute the resistance over the entire 57 mm (2.25 in.) of the specimen.
This problem calls for us to compute the electrical conductivity and resistance of a silicon specimen.
Solution
(a)
We use Equations 12.3 and 12.4 for the conductivity, as
σ
=
1
ρ
=
Il
VA
=
Il
V
π
d
2
2
Here
d
is the specimen diameter.
Now, incorporating values for the several parameters provided in the problem
statement, leads to
σ
=
(0.25 A)
(
45
×
10
−
3
m
)
(24 V)(
π
)
7.0
×
10
−
3
m
2
2
= 12.2 (
Ω
 m)
1
(b)
The resistance,
R
, may be computed using Equations 12.2 and 12.4, as
R
=
l
σ
A
=
l
σπ
d
2
2
=
57
×
10
−
3
m
12.2 (
Ω −
m)
−
1
[
]
(
π
)
7.0
×
10
−
3
m
2
2
= 121.4
Ω
12.6
(a)
Calculate the number of free electrons per cubic meter for silver, assuming that there are 1.3 free
electrons per silver atom.
The electrical conductivity and density for Ag are 6.8 × 10
7
(Ω
m)
–1
and 10.5 g/cm
3
,
respectively.
(b) Now compute the electron mobility for Ag.
Solution
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View Full Document(a)
This portion of the problem asks that we calculate, for silver, the number of free electrons per cubic
meter (
n
) given that there are 1.3 free electrons per silver atom, that the electrical conductivity is 6.8
×
10
7
(
Ω
m)
1
,
and that the density
(
ρ
Ag
'
) is 10.5 g/cm
3
.
(Note:
in this discussion, the density of silver is represented by
ρ
Ag
'
in
order to avoid confusion with resistivity which is designated by
ρ
.)
Since
n
= 1.3
N
Ag
, and
N
Ag
is defined in
Equation 5.2 (and using the atomic weight of Ag found inside the front cover—viz 107.87 g/mol), then
n
= 1.3
N
Ag
= 1.3
ρ
Ag
'
N
A
A
Ag
= 1.3
(
10.5 g/cm
3
)(
6.02
×
10
23
atoms /mol
)
107.87 g/mol
= 7.62
×
10
22
cm
3
= 7.62
×
10
28
m
3
(b)
Now we are asked to compute the electron mobility,
µ
e
.
Using Equation 12.8
µ
e
=
σ
n

e

=
6.8
×
10
7
(
Ω −
m)
−
1
(
7.62
×
10
28
m
−
3
)(
1.602
×
10
−
19
C
)
= 5.57
×
10
3
m
2
/V s
12.7
(a)
Using the data in Figure 12.8, determine the values of ρ
0
and a from Equation 12.10 for pure copper.
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 Summer '08
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