44
Derive simplified expressions for
F
2
for diamond, including the rules governing observed
reflections.
This crystal is cubic and contains 8 carbon atoms per unit cell, located in the
following positions:
0,0,0
½
,
½
,0
½
,0,
½
, 0,
½
,
½
¼
,
¼
,
¼
¾
,
¾
,
¼
¾
,
¼
,
¾
¼
,
¾
,
¾
Solution
Begin by summing amplitudes and phases according to the structure factor equation (411).
Note
that because there are 8 atoms in the unit cell, the summation must contain 8 terms.
Simplify,
Now by trialanderror determine those values of
h
,
k
, and
l
that make this expression zero.
Recall that e
n
π
i
= e

n
π
i
=
(1)
n
where
n
is any integer, and that 0 is an even integer.
F
2
= 0 for mixed indices (even and odd),
F
2
= 0 for (
h
+
k
+
l
) = odd multiple of 2.
Finally, determine the rules governing observed reﬂections.
F
2
= 64
f
2
for (
h+k+l
) = even multiple of 2, and
F
2
= 32
f
2
for (
h+k+l
) = odd.
F
=
f
exp(2
π
i
[0 + 0 + 0]) +
f
exp(2
π
i
[
h/
2 +
k/
2 + 0])
+
f
exp(2
π
i
[
h/
2 + 0 +
l/
2]) +
f
exp(2
π
i
[0 +
k/
2 +
l/
2])
+
f
exp(2
π
i
[
h/
4 +
k/
4 +
l/
4]) +
f
exp(2
π
i
[3
h/
4 + 3
k/
4 +
l/
4])
+
f
exp(2
π
i
[3
h/
4 +
k/
4 + 3
l/
4]) +
f
exp(2
π
i
[
h/
4 + 3
k/
4 + 3
l/
4])
F
=
f
(1 + exp(
π
i
[
h
+
k
] + exp(
π
i
[
h
+
l
] + exp(
π
i
[
k
+
l
])
+
f
(exp(
π
i
[
h
+
k
+
l
2
]) + exp(
π
i
[
3
h
+ 3
k
+
l
2
])
+ exp(
π
i
[
3
h
+
k
+ 3
l
2
]) + exp(
π
i
[
h
+ 3
k
+ 3
l
2
]))
PROBLEM 44
B.D. Cullity and S.R. Stock, Elements of XRay Diffraction, 3
rd
Ed., Prentice Hall, (2001)
MSE 104
Materials Characterization
Professor R. Gronsky
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