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CHAPTER
27
The Microscopic Theory of Electrical Conduction
1* ·
In the classical model of conduction, the electron loses energy on average in a collision because it loses the
drift velocity it had picked up since the last collision. Where does this energy appear?
The energy lost by the electrons in collision with the ions of the crystal lattice appears as Joule heat (
I
2
R
).
2
·
A measure of the density of the freeelectron gas in a metal is the distance
r
s
, which is defined as the radius
of the sphere whose volume equals the volume per conduction electron. (
a
) Show that
r
s
= (3/4
p
n
)
1/3
, where
n
is the freeelectron number density. (
b
) Calculate
r
s
for copper in nanometers.
(
a
) The volume occupied by one electron is 1/
n
= (4/3)
p
r
s
3
. Thus,
r
s
= (3/4
p
n
)
1/3
.
(
b
)
n
= 8.47
×
10
28
m
–3
(See Table 271). Evaluate
r
s
r
s
= 1.41
×
10
–10
m = 0.141 nm
3
·
(
a
) Given a mean free path
?
= 0.4 nm and a mean speed
v
av
= 1.17
×
10
5
m/s for the current flow in copper
at a temperature of 300 K, calculate the classical value for the resistivity
?
of copper. (
b
) The classical model
suggests that the mean free path is temperature independent and that
v
av
depends on temperature. From this
model, what would
?
be at 100 K?
(
a
) Use Equ. 277
(
b
)
v
av
∝
T
1/2
?
=
10
4
)
10
(1.6
10
8.47
10
1.17
10
9
10
2
19
28
5
31



×
×
×
×
×
×
×
×
O
.
m
= 0.123
μ
O
.
m
?
100
= (0.123
μ
O
.
m)(100/300)
1/2
= 0.071
μ
O
.
m
4
·
Calculate the number density of free electrons in (
a
) Ag (
?
= 10.5 g/cm
3
) and (
b
) Au (
?
= 19.3 g/cm
3
),
assuming one free electron per atom, and compare your results with the values listed in Table 271.
(
a
), (
b
)
n
=
?
N
A
/
M
;
?
= density,
M
= molar mass
(
a
)
n
Ag
= 5.86
×
10
22
el./cm
3
; (
b
)
n
Au
= 5.90
×
10
22
el./cm
3
.
The results agree with Table 271
5* ·
The density of aluminum is 2.7 g/cm
3
. How many free electrons are present per aluminum atom?
n
e
= electrons/atom =
nM
/
?
N
A
n
e
=
10
6.02
2.7
26.98
10
18.1
23
22
×
×
×
×
= 3.00 electrons/atom
6
·
The density of tin is 7.3 g/cm
3
. How many free electrons are present per tin atom?
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View Full Document Chapter 27
The Microscopic Theory of Electrical Conduction
See Problem 5
n
e
=
10
6.02
7.3
118.7
10
14.8
23
22
×
×
×
×
= 4.00 electrons/atom
7
·
Calculate the Fermi temperature for (
a
) Al, (
b
) K, and (
c
) Sn.
(
a
), (
b
), (
c
)
T
F
=
E
F
/
k
;
k
= 8.625
×
10
–5
eV/K
(
a
)
T
F
= (11.7 eV)/
k
= 1.36
×
10
5
K; (
b
)
T
F
= 2.45
×
10
4
K
(
c
)
T
F
= 1.18
×
10
5
K
8
·
What is the speed of a conduction electron whose energy is equal to the Fermi energy
E
F
for (
a
) Na, (
b
)
Au, and (
c
) Sn?
(
a
), (
b
), (
c
)
m
/
E
2
=
e
F
F
u
; use Table 271 for
E
F
(
a
)
u
F
= 1.07
×
10
6
m/s;
(
b
)
u
F
= 1.39
×
10
6
m/s;
(
c
)
u
F
= 1.89
×
10
6
m/s
9* ·
Calculate the Fermi energy for (
a
) Al, (
b
) K, and (
c
) Sn using the number densities given in Table 271.
(
a
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This note was uploaded on 05/01/2008 for the course PHYS 13/14 taught by Professor Millan/thorstensen during the Spring '08 term at Dartmouth.
 Spring '08
 MILLAN/THORSTENSEN
 Physics, Energy

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