11795 Ch3-Miller Indices Examples

7/28/2019 11795 Ch3-Miller Indices Examples

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Determine the Miller indices of directions A, B, and C.

Miller Indices, Directions

(c) 2003 Brooks/Cole Publishing /

Thomson Learning


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SOLUTION

DirectionA

1. Two points are 1, 0, 0, and 0, 0, 0

2. 1, 0, 0, -0, 0, 0 = 1, 0, 0

3. No fractions to clear or integers to reduce

4. [100]

Direction B

1. Two points are 1, 1, 1 and 0, 0, 0

2. 1, 1, 1, -0, 0, 0 = 1, 1, 1

3. No fractions to clear or integers to reduce

4. [111]

Direction C

1. Two points are 0, 0, 1 and 1/2, 1, 02. 0, 0, 1 -1/2, 1, 0 = -1/2, -1, 1

3. 2(-1/2, -1, 1) = -1, -2, 2

2]21[.4


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For some crystal structures, several

nonparallel directions with different

indices are crystallographically

equivalent; this means that atomspacing along each direction is the

same.

3

Families of Directions


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If the plane passes thru origin, either:

Construct another plane, or

Create a new origin

Then, for each axis, decide whether plane

intersects or parallels the axis. Algorithm for Miller indices

1. Read off intercepts of plane with axes interms ofa, b, c

2. Take reciprocals of intercepts

3. Reduce to smallest integer values

4. Enclose in parentheses, no commas.4

Crystallographic Planes


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Crystallographic Planesz

x

ya b

c

4. Miller Indices (110)

example a b cz

x

ya b

c


1. Intercepts 1 1

2. Reciprocals 1/1 1/1 1/

1 1 03. Reduction 1 1 0

1. Intercepts 1/2

2. Reciprocals 1/ 1/ 1/

2 0 03. Reduction 2 0 0

example a b c


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Crystallographic Planes

z

x

ya b

c


example1. Intercepts 1/2 1 3/4

a b c

2. Reciprocals 1/ 1/1 1/

2 1 4/3

3. Reduction 6 3 4


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Determine the Miller indices of planes A, B, and C.

Miller Indices -Planes

(c) 2003 Brooks/Cole Publishing /

Thomson Learning

Crystallographic planesand intercepts.


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SOLUTION

PlaneA

1. x= 1, y= 1, z= 1

2.1/x= 1, 1/y= 1,1 /z= 1

3. No fractions to clear

4. (111)

Plane B

1. The plane never intercepts the zaxis, so x= 1, y= 2, andz=

2.1/x= 1, 1/y=1/2, 1/z= 0

3. Clear fractions:1/x= 2, 1/y= 1, 1/z = 0

4. (210)

Plane C

1. We must move the origin, since the plane passes through

0, 0, 0. Lets move the origin one lattice parameter in the y-direction. Then, x= , y= 1, and z =

2.1/x= 0, 1/y= 1, 1/z = 0

3. No fractions to clear.

)010(.4


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Drawing Direction and Plane

Draw (a) the direction and (b) the plane in acubic unit cell.

1]2[1 10)2(


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SOLUTION

a. Because we know that we will need to move in thenegative y-direction, locate the origin at 0, +1, 0. The

tail of the direction will be located at this new origin. Asecond point on the direction can be determined bymoving +1 in the x-direction, 2 in the y-direction, and+1 in the z direction.

b. To draw in the plane, first take reciprocals ofthe indices to obtain the intercepts, that is:

x= 1/-2 = -1/2 y= 1/1 = 1 z= 1/0 =

Since the x-intercept is in a negative direction, and wewant to draw the plane within the unit cell, move the

origin +1 in the x-direction to 1, 0, 0. Then locate the x-intercept at 1/2 and the y-intercept at +1. The plane willbe parallel to the z-axis.

10]2[


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Family of Planes

Planes that are crystallographicallyequivalent have the same atomic packing.

Also, in cubic systems only, planes having

the same indices, regardless of order andsign, are equivalent.

Ex: {111}

= (111), (111), (111), (111), (111), (111), (111), (111)

(001)(010), (100), (010),(001),Ex: {100} = (100),

_ __ __ _ __ _ _ __


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FCC Unit Cell with (110) plane

12


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BCC Unit Cell with (110) plane

13


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SUMMARY

Crystallographic points, directions and planes are

specified in terms of indexing schemes.

Materials can be single crystals orpolycrystalline.

Material properties generally vary with single

crystal orientation (anisotropic), but are generallynon-directional (isotropic) in polycrystals with

randomly oriented grains.

Some materials can have more than one crystal

structure. This is referred to as polymorphism (or

allotropy).

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11795 Ch3-Miller Indices Examples