COURSE TITLE MODULE 2 CRYSTAL PLANES AND MILLER INDICES

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COURSE TITLE: X-RAY CRYSTALLOGRAPHY MODULE 2 CRYSTAL PLANES AND MILLER INDICES This module is aimed at 1. Understand the concept of a crystal plane; 2. Be able to determine the Miller indices of a plane 3. To understand how lattice planes and their Miller indices can assist to comprehend other concepts in materials science. 4. To be able to calculate the interplanar distance in unit cells Crystal planes Crystal planes are defined as some imaginary planes inside a crystal in which large concentration of atoms are present. Inside the crystal, there exists certain directions along which large concentration of atoms exists. These directions are called crystal directions. Crystal planes is an important concept used in powder diffraction and crystallography in general. One can imagine a crystal being sub-divided into smaller component units (Fig. 1); crystallographers use, depending on context,

Transcript of COURSE TITLE MODULE 2 CRYSTAL PLANES AND MILLER INDICES

Page 1: COURSE TITLE MODULE 2 CRYSTAL PLANES AND MILLER INDICES

COURSE TITLE: X-RAY CRYSTALLOGRAPHY

MODULE 2

CRYSTAL PLANES AND MILLER INDICES

This module is aimed at

1. Understand the concept of a crystal plane;

2. Be able to determine the Miller indices of a plane

3. To understand how lattice planes and their Miller indices can assist to

comprehend other concepts in materials science.

4. To be able to calculate the interplanar distance in unit cells

Crystal planes

Crystal planes are defined as some imaginary planes inside a crystal in which large

concentration of atoms are present. Inside the crystal, there exists certain

directions along which large concentration of atoms exists. These directions are

called crystal directions.

Crystal planes is an important concept used in powder diffraction and

crystallography in general. One can imagine a crystal being sub-divided into

smaller component units (Fig. 1); crystallographers use, depending on context,

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two alternative sub-divisions: one is the unit cell, the crystal building block, and

the other components are sets of planes also known as diffracting planes,

reflecting planes, Bragg planes, crystal planes or hkl planes. A set of such planes

consists of parallel evenly spaced planes which are extended to exactly fill the

entire crystal; each plane is at equal distance, d (the inter-planar spacing), from its

neighbouring plane. There can be an infinite number of such types of planes,

which cover every region of space, and indeed every atom, within the crystal: this

is one of the properties that makes their concept useful.

Hauy observed that when a crystal was cleared, the corresponding faces of the

different fragments were equivalent. In order to describe these crystal faces,

Hauy proposed the law of Rational intercept. It states that it is always possible to

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find a set of axes that can be used to describe the crysral face in terms of intecept

along the axes.

Miller indices are small integers which describes the orientation of a plane and

they are reciprocals of intercepts. They were developed by William Hallowes

Miller. Miller indices define directional and planar orientation within a crystal

lattice. The indices may refer to a specific crystal face, a direction, a set of faces,

or a set of directions. Indices that refer to a crystal plane are enclosed in

parentheses (hkl), indices that refer to a set of symmetrically equivalent planes or

family of planes are enclosed in braces (curly brackets) {hkl}, indices that

represent a direction are enclosed in square brackets [hkl], and indices that

represent a set of equivalent directions or family of directions are enclosed in

angle brackets ˂hkl˃.

Crystal plans pass through lattice points and are parallel to the crystal faces.

They are also described by the three whole numbers h, k, l which may be

positive or negative. Parallel planes may be viewed as cutting a unit length

of each axis of the unit cell into an integral number of equal parts: the a axis

into h equal parts, the b axis into k equal parts and the c axis into l parts.

Since the parallel planes are exactly alike one generally considers the planes

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nearest to the origin passing through the set of points closest to the origin.

The next plane parallel to this will pass through the second nearest set of

points to the origin and so forth. The main point is that each plane passé

through the lattice points of the crystal. Such a set of planes will have the

interplanar spacing d equal to the distance of the 1st plane from the origin.

If the intercept is at infinity (that is the plane is parallel to the axes), the

reciprocal is zero. For emphasis, zero means the axis is parallel to the

plane. If a plane cuts an axis on the negative side of the origin, the

corresponding index is negative and is indicated with a minus sign over the

index [i.e hkl) (h bar).

For instance, consider the Figure below

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The arrangement of atoms in the lattice is such that the plane can intercept these

axes at 3 on the a axis, 2 on the b axis and 2 on the c axis. The reciprocal of the

intercepts are

. It the fractions are cleared, the Miller indices are (2 3 3).

Procedure for obtaining Miller Planes

1. Determine the intercept.

2. Find the reciprocal of the intercept.

3. Clear fractions

The above diagram is in two dimension. The Miller Indices are [4 2]

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Example 1.

Academic resource centre.

https://web.iit.edu/sites/web/files/departments/academic-affairs/academic-

resource-center/pdfs/Miller_Indices.pdf

The origin above is the intersection of the three axes. The planes can be infinite

in number. The first plane is a distance a ( or 1a) away from the origin. The second

and the third planes are a distance 2a and 3a away from the origin respectively.

The 1st, 2nd and 3rd planes intersect the x axis at point a, 2a and 3a. The planes are

parallel to the y and z axes. Therefore to calculate their Miller indices;

1st plane:

Intercepts : ,,1

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Reciprocal of the intercept: 1 0 0

Miller Indices; (1 0 0)

The 2nd and 3rd planes have the same Miller indices.

Intercepts : ,,2

Reciprocal of the intercept:

Clear fraction: 1 0 0

Miller Indices; (1 0 0)

Example 2:

Consider the diagram above.

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The intercept of the pink plane is ,,1 . The intercepts of the yellow and the

green planes are ,1, and 1,,

For the pink plane:

Intercepts : ,,1

Reciprocal of the intercept:

1.

1,

1

1= 1 0 0

Miller Indices: (1 0 0)

For the yellow plane:

Intercepts : ,1,

Reciprocal of the intercept:

1.

1

1,

1= 0 1 0

Miller Indices: (0 1 0)

For the green plane:

Intercepts: 1,,

Reciprocal of the intercept: 1

1.

1,

1

= 0 0 1

Miller Indices: (0 0 1)

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Example 3

Determine the Miller indices of the following intercepts by crystal planes:

1. .2

1,

3

2 2. 1.

5

3,

7

2 3. 2,

5

1,

5

2

Solution:

1. Intercepts : .2

1,

3

2

Reciprocal of the intercept:

1.

1

2,

2

3

Clear fraction: 3 4 0

Miller Indices; (3 4 0)

2. Intercepts : 1.5

3,

7

2

Reciprocal of the intercept: 1.3

5,

2

7

Clear fraction: 21 10 6

Miller Indices: (21 10 6)

3. Intercepts : 2,5

1,

3

2

Reciprocal of the intercept: 2

1,5,

2

3

Clear fraction: 3 10 1

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Miller Indices: (3 10 1)

Uses of Miller Indices

They are useful in understanding many phenomena in materials science,

such as

i. Explaining the shapes of single crystals,

ii. Explaining the form of some materials' microstructure,

iii. Used to interpret x. ray diffraction patterns

iv. Used to explain the movement of a dislocation which may determine

the mechanical properties of the material

Calculation of the distance between planes

The distance between planes is written as d. Specific formular can be

worked out relating d to a, b, c and the Miller Indices of the planes. Three

simple cases include

Cubic cell: 2

222

2

1

a

lkh

dhkl

(a=b=c)

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Tetragonal Cell: 2

2

2

22

2

1

c

l

a

kh

dhkl

Orthorhombic cell: 2

2

2

2

2

2

2

1

c

l

b

k

a

h

dhkl

Example: Calculate the separation of a. the [123] planes and b. the [246]

planes of an orthorombic unit cell given that a = 0.82 nm, b = 0.94 nm and

c = 0.75 nm. (Atkins and de Paula, 2008)

Solution

a. The [123] plane

Using 2

2

2

2

2

2

2

1

c

l

b

k

a

h

dhkl

h = 1, k = 2, l = 3

2

2

2

2

2

2

2

12375.0

3

94.0

2

82.0

11

d

= 22.014 nm-2

Hence, d123 = 0.21 nm

b. The [246] plane

h = 2, k = 4, l = 6

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2

2

2

2

2

2

2

24675.0

6

94.0

4

82.0

21

d

d246 = 0.11 nm

In general, the separation of [nh, nk, nl] planes is n times smaller than

the separation of the [hkl] planes.

From the example above d246 for[246] planes is half that of d123 for [123]

planes.

For further reading:

1. Bames P, Csoka T, Jacques S.(2006). A little more on crystal planes. Birkbeck

College, London

2. https://www.doitpoms.ac.uk/tlplib/miller_indices/printall.php

3. http://chemistry.bd.psu.edu/jircitano/Miller.html

4. Academic resource centre.

https://web.iit.edu/sites/web/files/departments/academic-

affairs/academic-resource-center/pdfs/Miller_Indices.pdf

5. Atkins P. and de Paula J. 2008. Physical Chemistry. 9th

Ed. p697