Paul 10.10

1. 1
 
 
5
a Given a mean free path 0.4 and
a mean speed 1.17 10 for the
current flow in copper at a temperature of
300 , calculate the classical value for the
resistivity of copper. b The classical
mod
nm
m
v X
s
K




el suggests that the mean free path is
temperature independent and that
depends on temperature. From this model,
what would be at 100 ?
v
K

Solution

2. 2

3. 3
2
We know that mean free path is equal
to the product of the average speed
& the average time between collisions
, called the relaxation time.
1
=
where is the number of ions per unit
volum
a
a
v
v
n r
n


 


2
2
e & is the ion radius. In terms
of , the resistivity & conductivity are
given by
= & (1)
e
e
r
m v ne
ne m v


 

   

4. 4
31
5
28 3
19
Using
9.11 10 ,
1.17 10 ,
8.47 10 / ,
1.60 10 ,
0.4
in eq. (1), we get,
e
m X kg
m
v X
s
n X electrons m
e X C
nm









5. 5
2
31 5
28
3
19 2 9
3
7 7
2
2 2
2 2 2 3 2
=
9.11 10 1.17 10
8.47 10
(1.60 10 ) 0.4 10
.
1.23 10 1.23 10 .
. . .
[ ]
. . .
e
m v
ne
m
X kgX X
s
electrons
X X
m
X C X X m
kg m
X X m
sC
kg m J s J kg m
s C C s A s A



 
 

  
    

6. 6
23
31
10
2
10
2
4
( )
8
We know that
8 1.38 10 100
3.14 9.11 10
0.386 10
0.386 10
3860000000
62128.9 6.2 10
e
b
kT
v
m
J
X X X K
K
X X kg
J
X
kg
m
X KgX
s
kg
m
s
m m
X
s s








 

7. 7
2
31 4
28
3
19 2 9
3
8 8
2
2 2
2 2 2 3 2
=
9.11 10 6.2 10
8.47 10
(1.60 10 ) 0.4 10
.
6.5 10 6.5 10 .
. . .
[ ]
. . .
e
m v
ne
m
X kgX X
s
electrons
X X
m
X C X X m
kg m
X X m
sC
kg m J s J kg m
s C C s A s A



 
 

  
    