HISTORICAL NOTE
Zone RefiningWilliam G. Pfann
A
Thirty-five years ago this July, Bell Labs
r e s e a r c h e r William G. P f a n n first described the process of zone refining a
simple, elegant way to make materials
very pure or having a desired level of

121
2) The Ge is a fine black powder. It is melted (T m = 936 C) and solidified into a
polycrystalline ingot which has a shiny silver appearance.
3) Common impurities (also found in Si) are P, As, Sb, B, A, Ga, In.
4) Zone refining: The basic principle is

41
(at least for the low-level states, convince yourself from the graphical solution drawn
to proper scale not as in McKelvey)
that the k x s are regularly spaced at intervals (to very good accuracy) of
2 3 4 5
,
,
,
kx = ,
L L L L L
:
L
.
within this se

101
As the E-field makes the e trace out A B C C B A . we have:
A B
Normal acceleration in the positive direction
v k t is increasing
B C
The effective mass < 0, the external field appears to decelerate the e, but it is
really the effect of the atom poten

71
these equations may be written in the matrix mode form, with c a vector =
(M) (c)
then the above equations are:
(
c k1
c k1 G1
)
= 0
This is a standard problem in linear algebra: If there is to be a solution where the vector c 0,
then we must have: d

131
W) But in finding out which sets are most likely i.e., with Q (or n Q) the largest, we must
make sure that (11) and (12) are obeyed while we maximize Q. This will shift the N j s away
N
from the value N j
which occurs when energy does not matter.
i
X

161
low concentration
high concentration
n (x) large
n (x) small
1
1
n ( x ) moving to the right
n ( x ) moving to the left
2
2
x = 0
net flow crossing x = 0 is greater from the left (high c) than from the right (low c)
so
J (x = 0) > 0
concentration di

21
3K T
B
me
vr ms
T
and
=
then
ne 2
me
T
T
1
1
2
2
(n does not charge much with temperature for metals)
1
1
2
2
but the experiment shows T, not T
1
2
One obvious way to fix things would be to take a more realistic distribution of velocities,

31
me
dvy
dt
0 e Ey ev x B
me
vy
(2)
x-equation:
me
d vx
m
0 e Ex e v y B e v x
dt
put vy = 0 in (2), (3)
(3) v x
(3)
(remember, vy =0 in the steady-state)
e Ex
v drift
me
(4)
A) This is the same result as before for vdrift (Drude), i.e., the magnet

61
then the electric potential from the charge (Ze) located at x = Ln
V n (x) =
( Ze) e
x nL
(Coulomb potential)
Z e2
x nL
the total potential:
n = 0, 1, 2, . is
V (x) =
clearly, if x x + m L
n
m = 0, 1, 2
a translation by a lattice vector:
V (x + m L)

141
note: (EG/2), not EG
so
n, p ~ T
3
2
e E G / 2 K BT
measure using Hall effect
A) The electrons and holes behave like free electrons and free anti-electrons with effective
masses me, mh. When collisions are added, we get Drude-like results.
in an E -fi

1
OhmsLaw
(Georg Ohm, 1787-1854)
L
wire
cross-sectional area A
Ohm measured the electric current I passing through a given piece of material for different
applied voltages V and found V I for a wide range of materials.
i.e.,
Ohms Law
V = IR
R = resistance

51
z (free e/atom)
n
Cu
1
8.47 x 10 /cm
Fe
2
17.0 x 10 /cm
Au 1
5.90 x 10 /cm
F
vF
22
3
7.00 eV
1.57 x 10 cm/s
8
22
3
11.1 eV
1.98 x 10 cm/s
22
3
5.53 eV
1.40 x 10 cm/s
8
8
Note: The average kinetic energy/particle in a classical (no Pauli principle) elec

111
dv h
2
eE
E h k h d t
mh
E e (k e )
now
d 2 E e k e
d 2 E h k h
d kh d kh
d k e2
ke ke
so
m h (k L ) = m e (k e )
(4)
E) Equation (4) above says that the m effective for holes at the top of the band is > 0 (i.e.,
normal). So, the behavior of the

11
5) A reasonable guess (but much too simple, we shall see) made by Drude was that the es
collide with fixed, heavy atoms which are randomly placed (not crystalline).
(If he thought metals were crystals, he might have thought the e could move freely thro

151
so
Probability () = B (T) constant
and at equilibrium
or
np
A (T) np = B (T) constant
B T
constan t F T
A T
A) So going back:
which is the law of mass action for this case
n p N C N V e E G / K BT
K T
4 B 2
2
3
(=
m e m h 3 2
F (T)
e E G / K

81
so
G1
x
G1 ~ sin
2
2
G1
~ sin
x,
2
so G1 are really the same, only one distinct state
2
So there are N states in band 1 which can hold 2N electrons. (+ spin, spin in each spatial
state)
G) This is also true for the other bands and in 2 and 3 d

91
d
v
dk
k
independent
of k
= group velocity
i.e.,
vphase = vgroup
Non-dispersive
(k0) (k0) k0 = v k0 v k0 = 0
then
also, relation (1) above is then exact
(x, t) = ( x v t, 0)
and
undistorted wave packet travels with velocity v
exact
If material is