1. 1
The electrical
conductivity of copper at
room temperature is
5.96 × 107
Ω−1
m−1
.
Evaluate the average
distance between
electron scatterings.
How many lattice
spacings does this
amount to?
Helping Tools
No.1
OR
4. 4
Solution
2
2
We know that conductivity
of free electron gas is given as:
(1)
Re-arranging eq.(1),
where is the number density of
conduction
we
electrons, is the
electron charg
get
(2
e,
)
is
n
e
e
m
n
m
ne
= − − − −
= − − − − −
the average
time between electron-ion collisions,
and is the electronic mass.
m
5. 5
7 1 1 31
28 19
3
Using
=5.96X10 , 9.11 10 ,
8.47 10 , & 1.60 10 ,
in eq(2), we get
m m X kg
atoms
n X e X C
m
− − −
−
=
= =
7 1 1 31
28 3 19 2
7 38 2
28 31 2
45 2
59 2
14 2 1 2
14 2 1 2 2
14 2
5.96X10 9.11 10
=
8.47 10 (1.60X10 C)
5.96 10 9.11 10
8.47 10 10 1.6 1.6
54.29 10 .
21.68 10
2.50 10
2.5 10 1
2.5 10
m X X kg
X m X
X X X kgXm
X X X X XC
X Kg m
X XC
X kgm C
X kgm XAV XA s
X kgm X
− − −
− −
− − −
− − − −
−
=
=
=
= −
= 1 1 2
14 1 1
14 1
14 1
14
2.5 10
2.5 10
2.5 10
2.5 10
A V s
X NmA XAW
X NmW
X JXsJ
X s
− − −
− − −
− −
− −
−
=
=
=
=
6. 6
14 6
8 9
Now,
2.5 10 1.57 10
3.925 10 39.25 10
39.2
where is the mean free path
of the electrons & is the
Fermi velocity.
In comparison with 0.256nm,
this amounts to 153 lattice spacings.
F
F
m
l v X sX X
s
X m X m
nm
l
v
−
− −
= =
= =
=
