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Unformatted text preview: 材材材材材材
Fundamental of Materials
Prof: Tian Min Bo
Prof: Tian
Tel: 62795426 材 62772851
Email: tmb@mail.tsinghua.edu.cn
tmb@mail.tsinghua.edu.cn
Department of Material Science and Engineering
Tsinghua University. Beijing 100084 Lesson five §2.5 Indices of crystal planes and directions
§2.5
Ⅰ.What’s crystal planes and directions? The atomic planes and
The
directions passing
through the crystal are
called (crystal) planes
and directions
respectively.
respectively. Ⅱ. Plane indices 1. Steps to determinate the plane indices:
1. Establish a set of coordinate axes Find the intercepts of the planes to be
indexed on a, b and c axes (x, y, z).
c
z a x y b Take the reciprocals of the intercepts 1/x, 1/y,
1/z. Clear fractions but do not reduce to lowest
integers. Enclose them in parentheses, (h k l) Example: 1/2,1,2/3
1/2,1,2/3 2,1,3/2 (423) Plane indices referred to three axes a, b and c
Plane
and
are also called Miller Indices.
are Several important aspects of the Miller indices for
planes should be noted: Planes and their negatives are identical. (020) = (0
Therefore .20) Planes and their multiples are not identical. In cubic systems, a direction that has the same indices as a plane is perpendicular to that plane. 2. The important planes in cubic crystals
2. (001)
(110) (111)
(112) 3. A family of planes consists of equivalent
planes so far as the atom arrangement is
concerned.
concerned.
{110} = (110) + ( 1 10) + (101) +
(10 1 ) + (011) + (0 1 1)
Total: 6 {111} = (111) + ( 1 11) + (1 1 1) +
(11 1 )
Total: 4 {112} = (112) + ( 1 12) + (1 1 2) + (112 ) +
(121) + ( 1 21) + (1 21) + (12 1 ) +
(211) + ( 2 11) + (2 1 1) + (21 1 ) Total: 12 {123} = (123) + ( 1 23) + (12 3) + (12 3) + (132) +
( 1 32) + (1 3 2) + (132) + (231) + ( 2 31) +
(2 3 1) + (23 1 ) + (213) + ( 213) + (2 1 3) +
(21 3) + (312) + ( 3 12) + (3 1 2) + (312) +
(321) + ( 3 21) + (321) + (32 1 ) Total: 4×3 ！ =24 Ⅲ. Direction Indices 1. Derivation for the crystallographic direction
a.
b.
c.
d. As the first above, set the origin on the direction
to be indexed.
Find the coordinates of another point on the
direction in questions.
Reduce to three smallest integers: u, v, w.
Enclose in square brackets [u v w]. 2. The important direction in cubic crystals:
2.
<100> : crystal axes
<110> : face diagonal
<111> : body diagonal
<112> : apices to opposite facecenters
3. Family of directions consists of
Family
crystallographically equivalent directions,
denoted <u v w>
denoted
e.g. < 100 >= [100] + [010] + [001] +
[ 1 00] + [0 1 0] + [00 1 ] §2.6 Hexagonal axes for hexagonal
crystals
crystals Ⅰ. Why choose fouraxis system?
Four indices has been devised for hexagonal
unit cells because of the unique symmetry of the
unique
system.
system. c
(1 1 0)
(100) b
a
[100] [110] Ⅱ. Plane indices (hkil)
It can be proved: i ≡ 材 (h 材 k) (100) → (10 1 0)
(1 1 0) → (1 1 00) Important planes :
Important planes :
c (10 1 2) (0001) a3 (11 20) a2
a1
(10 1 1) (10 1 0) Ⅲ. Direction indices [ u v t w ]
Direction To make the indices unique, an additional
To
condition is imposed.  Let t 材材 (u 材 v)
condition
[0001] [ 1 011] Important directions [01 1 0]
[ 2 1 1 0] Transformation of indices
Transformation
Transformation of 3 to 4 indices, or vice versa. Suppose we
have a vector, whose 3 indices [u v w], and 4 indices [u v t w].
have
We have Since L = ua1 + va2 + ta3 + wc = Ua1 + Va2 + Wc a3 = − (a1 + a2 )
t = − (u + v) ∴ ua1 + va 2 +(u + v)(a1 + a2 ) + wc = Ua1 + Va2 + Wc (2u + v)a1 + (u + 2v)a 2 +wc = Ua1 + Va2 + Wc ∴ U = 2u + v V = 2v + u u = 1 3 (2U − V )
v = 1 3 (2V − U )
or:
or:
w =W
t = −(u + v) W =w For example:
For
2
u=
3 [100] : 1
v=−
3
1
t=−
3 w=0 ⇒ [2 1 1 0] Examples and Discussions
1. Quick way for indexing the directions in cubic
1.
crystals:
crystals: The value of a direction depends on its feature
while the sign on direction.
while 2. The coordinate origin can be set arbitrarily
2.
(for example on apices, bodycenter, face(for
centers etc.), but never on plane in
centers
questions, otherwise the intercepts would be
0,0,0 .
0,0,0
3. The coordinate system can be transferred
arbitrarily, but rotation is forbidden.
arbitrarily,
c c´ (11 1 ) b ( 1 1 1) a
a´ b´ 4. The atomic arrangement and planar density of the
4.
important direction in cubic crystal.
important
plane indices BCC FCC atomic
atomic
planar density
arrangement
arrangement planar density a 1
4= 1
a2
a2 1
4× +1
2
4
=2
a2
a 2a 1
4× +1
1.4
4
=2
a
2a 2 {100} a a {110} {111} 2a 2a
2a 4× 1
6 = 0.58
a2
32
a
2 3× a
a a
2a
2a 2a 2a 1
1
4× + 2×
4
2 = 1 .4
a2
2a 2
1
1
3× + 3×
6
2 = 2.3
a2
32
a
2 5. The atomic arrangement and linear density of the
5.
important direction in cubic crystal.
important
linear indices <100> <110> <111> BCC
atomic
arrangement FCC linear density
1
2=1
a
a 2× a 2a 3a 1
2 = 0 .7
a
2a 2× 1
2× +1
1.16
2
=
a
3a atomic
arrangement linear density
1
2=1
a
a 2× a 2a 3a 1
2× +1
1.4
2
=
a
2a
1
2 = 0.58
a
3a 2× Exercise
Exercise
1. Calculate the planar density and planar packing
fraction for the (010) and (020) planes cubic
polonium, which has a lattice parameter of 0.334nm.
Solution atom per face 1
planar density (010) =
=
area of face 0.334 2
= 8.96 ×1014 atoms/cm 2
area of atoms per face
packing fraction (010) =
area of face
1× (πr 2 )
=
= 0.79
2
( 2r ) However, no atoms are centered on the (020) planes.
There fore, the planar density and the planar packing
fraction are both zero. Thank you !
Thank you !
5 ...
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 Spring '11
 MingjieZhao

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