Page 102 - Materials Science and Engineering An Introduction

74 • Chapter 3 / The Structure of Crystalline Solids

EXAMPLE PROBLEM 3.10

Determination of Directional Indices for a Hexagonal Unit Cell
Determine the indices (four-index system) for the direction z
shown in the accompanying figure.

Solution
The first thing we need to do is determine U, V, and W indices
for the vector referenced to the three-axis scheme represented in
the sketch; this is possible using Equations 3.13a through 3.13c.
Because the vector passes through the origin, a 1 = a 2 = 0a and
z = 0c. Furthermore, from the sketch, coordinates for the vector
head are as follows: c
a 2
a 1 = 0a
a 2 = -a
c
z = a
2
a a 1
Because the denominator in z is 2, we assume that n 2.
Therefore,
a 1 - a 1 0a - 0a
U = na b = 2a b = 0
a a
a 2 - a 2 -a - 0a
V = na b = 2a b = -2
a a
z - z c/2 - 0c
W = na b = 2a b = 1
c c
This direction is represented by enclosing the above indices in brackets—namely, [021].
Now it becomes necessary to convert these indices into an index set referenced to the four-
axis scheme. This requires the use of Equations 3.11a–3.11d. For this [021] direction,

U = 0 V = -2 W = 1
and
1 1 2
u = (2U - V) = [(2)(0) - (-2)] =
3 3 3
1 1 4
y = (2V - U) = [(2)(-2) - 0] = -
3 3 3
2 4 2
t = -(u + y) = - a - b =
3 3 3
w = W = 1
Multiplication of the preceding indices by 3 reduces them to the lowest set, which yields values
for u, y, t, and w of 2, 4, 2, and 3, respectively. Hence, the direction vector shown in the figure
is [2423].