Fundamentals of Materials Science and Engineering: An Integrated Approach, 3e

GTBL042-03 GTBL042-Callister-v2 September 6, 2007 15:33

72 • Chapter 3 / Structures of Metals and Ceramics

Solution

Since the plane passes through the selected originO, a new origin must be

chosen at the corner of an adjacent unit cell, taken asO′and shown in sketch

(b). This plane is parallel to thexaxis, and the intercept may be taken as

∞a.Theyandzaxes intersections, referenced to the new originO′, are –b

andc/2, respectively. Thus, in terms of the lattice parametersa,b, andc, these

intersections are∞,−1, and^12. The reciprocals of these numbers are 0, –1, and

2; and since all are integers, no further reduction is necessary. Finally, enclosure

in parentheses yields (012).

These steps are briefly summarized below:

xyz

Intercepts ∞a −bc/2

Intercepts (in terms of lattice parameters) ∞− (^112)
Reciprocals 0 − 12
Reductions (unnecessary)
Enclosure (012)
z
x
y
z
x
y
z
x
y
(b)
(c)
(a)
O
(001) Plane referenced to
the origin at point O
(111) Plane referenced to
the origin at point O
(110) Plane referenced to the
origin at point O
Other equivalent
(001) planes
Other equivalent
(111) planes
Other equivalent
(110) planes
O
O
Figure 3.25
Representations of a
series each of (a)
(001), (b) (110), and
(c) (111)
crystallographic
planes.