The total scattering factor is given as
S
=

A
(
N
)

2
·
B
(
N
)

2
.
Performing the summation of the geometric series we get

A
(
N
)

2
=
•
sin
[(
π
N
/
2
)
sin
θ
]
sin
[(
π
/
2
)
sin
θ
]
‚
2
,

B
(
N
)

2
=
•
sin
(
π
N
cos
θ
/
2
)
sin
(
π
cos
θ
/
2
)
‚
2
.
Substituting these expression and assuming the Bragg condition
ka
=
π
we obtain the
graphs shown in Fig. 12.6. Similar curves for
ka
=
3
π
/
2 are shown in Fig. 12.7.
Figure 3.5: Plots of
S
(
N
,
θ
)
/
N
4
for
ka
=
π
;
N
=
5 (red curve), 10 (blue curve), and 20 (green
curve)
Problem 3.4:
Consider a random 1dimensional chain along a straight line, the probability of
finding the nearest neighbor at distance
x
being
P
(
x
) =
1
√
2
πσ
e

(
x

a
)
2
/
2
σ
.
(3.1)
Assuming that
σ
¿
a
2
and the atoms are pointlike find the density distribution function,
g
(
x
)
.
Plot
g
(
x
)
for
σ
=
0
.
1 and
σ
=
0
.
3 and discuss the results.
Hints:
(i) Recall the rules of calculating probability of a sum of random quantities; (ii) Have in
mind that the Fourier transform of the Gaussian probability is
P
(
k
) =
e

k
2
σ
/
2
.
transform
36
CHAPTER 3. FRACTALS AND DISORDERED SYSTEMS
Figure 3.6: Plots of
S
(
N
,
θ
)
/
N
4
for
ka
=
3
π
/
2;
N
=
5 (red curve), 10 (blue curve), and 20 (green
curve)
Solution 3.4:
Let us define the distance between the initial and
n
th atom as
x
=
na
+
U
n
,
n
∑
j
=
1
u
j
.
The probability of the the sum of of displacement are given by the general prescription
W
(
U
n
)
=
n
∏
j
=
1
Z
du
j
P
(
u
j
)
δ
ˆ
U
n

n
∑
j
=
1
u
j
!
=
∏
j
Z
dk
j
2
π
P
(
k
j
)
Z
du
j
e
ik
j
u
j
Z
dk
2
π
e
ik
(
U
n

∑
u
j
)
=
Z
dk
2
π
e
ikU
n
[
P
(
k
)]
n
=
Z
dk
2
π
e
ikU
n
e

k
2
σ
n
/
2
.
Here
P
(
k
)
is Fourier transform of
P
(
u
)
. We have also used the hints and the propeety
Z
due
i
(
k
1

k
2
)
=
2
πδ
(
k
1

k
2
)
.
Now we have to find the distribution
g
(
x
)
, which is just
g
(
x
) =
Z
du
0
P
(
u
0
)
W
(
x

u
0
) =
Z
dk
2
π
P
(
k
)
W
(
k
)
.
37
From this we get,
g
n
(
x
) =
Z
dk
2
π
e

kx
e

k
2
σ
(
n
+
1
)
/
2
=
1
p
2
πσ
(
n
+
1
)
e

(
x

na
)
2
/
2
σ
(
n
+
1
)
.
Since for small dispersion
σ
the Gaussian spikes do not overlap, the final answer can be expressed
as
g
(
x
) =
∞
∑
n
=
1
1
p
2
πσ
(
n
+
1
)
exp

(
x

na
)
2
2
σ
(
n
+
1
)
¶
.
The plots are shown in Fig. 3.7
Figure 3.7: Plot of the distribution function for
σ
=
0
.
1 (red curve) and
σ
=
0
.
03 (green curve).
38
CHAPTER 3. FRACTALS AND DISORDERED SYSTEMS
Chapter 4
Defects and Diffusion
Quick access: 1 2 3 4 5
Problem 4.1:
Austenite:
Austenite is a type of steel with less than 1% carbon.
Call this
concentration
C
0
. It is desirable to increase the carbon content near the surface to give the steel
a harder surface. This can be achieved by keeping the Austenite in a mixture of CO and CO
2
where the reaction 2CO
→
CO
2
+ C
aust
will produce a given concentration
C
s
on the surface.
This concentration will then diffuse into the steel.
(a) Show that the expression
C
(
x
,
t
) =
C
s

(
C
s

C
0
)
erf
x
2
√
Dt
¶
(4.1)
satisfies the diffusion equation
∂
C
∂
t
=
D
∂
2
C
∂
x
2
,
(4.2)
where erf
(
y
)
≡
2
√
π
R
y
0
e

t
2
dt
. Sketch the function
C
(
x
,
t
)
at different times.
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 Cubic crystal system, Reciprocal lattice, Lattice points