(e) To observe the rings on the film we need
φ
<
π
2
, so
λ
2
a
(
p
2
+
q
2
+
r
2
)
1
/
2
<
sin
π
4
=
2

1
/
2
. For
a
=
2
.
8
×
10

10
and
λ
=
0
.
7
×
10

10
, we obtain the condition
p
2
+
q
2
+
r
2
<
32
.
(12.13)
The permutations of
p
,
q
, and
r
lead to the same difraction rings, so they must be excluded.
In addition, the sum
p
+
q
+
r
must be an even number. The inspection shows that the
number of relevant combinations is 16.
(f) The primitive vectors of the tetragonal structure are
t
1
=
a
2
[
e
1
+
e
2

(
1
+
ε
)
e
3
]
,
t
2
=
a
2
[
e
2
+(
1
+
ε
)
e
3

e
1
]
,
and
t
3
=
a
2
[
e
1
+(
1
+
ε
)
e
3

e
2
]
.
(12.14)
87
The volume of the primitive cell is
V
t
=
t
1
·
t
2
×
t
3
= (
1
+
ε
)
a
3
2
,
so the fundamental vectors of the reciprocal lattice are
˜
t
1
=
2
π
a
(
e
1
+
e
2
)
(12.15)
˜
t
2
=
2
π
a
e
2
+
e
3
1
+
ε
¶
(12.16)
˜
t
3
=
2
π
a
e
1
+
e
3
1
+
ε
¶
.
(12.17)
So a general vector of the reciprocal lattice is
G
t
=
k
˜
t
1
+
l
˜
t
2
+
m
˜
t
3
=
2
π
a
•
(
k
+
m
)
e
1
+(
k
+
l
)
e
2
+(
l
+
m
)
e
3
1
+
ε
‚
≡
2
π
a
•
p
e
1
+
q
e
2
+
r
e
3
1
+
ε
‚
.
Since
G
=
2
π
a
p
p
2
+
q
2
+
l
2
(
1
+
ε
)

2
, the maximum interference condition reads:
2
π
a
s
p
2
+
q
2
+
r
2
(
1
+
ε
)
2
=
4
π
λ
sin
φ
2
.
In the case of the cubic lattice any permutation of the integers
p
,
q
, and
r
would lead to the
formation of the same cone of maximum interference, while in the case of tetragonal lat
tice, the interchange of
p
and
q
leaves the condition of maximum interference unchanged,
while the interchange of
r
with any of
p
or
q
leads to a different angle of maximum inter
ference. Beside this, the degeneracy is lifted for any other combination of
p
,
q
, and
r
and
p
0
,
q
0
, and
r
0
, so that
p
2
+
q
2
+
l
2
= (
p
0
)
2
+(
q
0
)
2
+(
l
0
)
2
, while
l
6
=
l
0
.
Problem 12.2:
(a) The von Koch curve is constructed by dividing each side of equilateral triangle by 3 and
replacing the central part by another equilateral triangle, see Fig. 12.4. Find fractal dimen
sion of the von Koch curve.
(b) Consider a square lattice
N
×
N
of atoms with isotropic scattering formfactor.
88
CHAPTER 12. TESTS
Figure 12.4:
1. Assume that a plane wave has a front parallel to the crystal plane, so the incident
angle
α
=
0.
Calculate the scattering intensity
I
as a function of the scattering angle,
θ
, for
ka
=
π
(Bragg conditions) and for
N
=
5, 10, and 20.
Plot the dependence in polar coordinates.
Show that the maximal scattering intensity is proportional to the total number of
atoms squared.
2. Repeat the calculation for
ka
=
1
.
5
π
and
α
=
0.
Discuss the difference between the scattering curves.
3. Find the incident angle satisfying the Bragg conditions at
ka
=
1
.
5
π
and repeat cal
culations of the item 1.
How the Bragg condition manifest itself?
Solution 12.2:
(a) We have the recursive formula for the perimeter
P
(
‘/
3
) =
4
3
P
(
‘
)
.
Assuming
P
(
‘
)
∝
‘

(
D

1
)
we get
D
=
log4
log3
≈
1
.
26
.
This value is close to experimental one (1.3) for Norwegian coast.
(b) For an incident wave with
α
=
0, the difference of the optical path corresponding to the
adjacent atom to the right is

a
sin
θ
.
89
Figure 12.5: Sketch of the angles.
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 Fall '13
 Cubic crystal system, Reciprocal lattice, Lattice points