Objectives By the end of this section you should: understand the concept of planes in crystals know that planes are identified by their Miller Index and.

Objectives

By the end of this section you should:• understand the concept of planes in crystals• know that planes are identified by their Miller

Index and their separation, d• be able to calculate Miller Indices for planes• know and be able to use the d-spacing

equation for orthogonal crystals• understand the concept of diffraction in crystals• be able to derive and use Bragg’s law

Lattice Planes and Miller IndicesImagine representing a crystal structure on a grid (lattice) which is a 3D array of points (lattice points). Can imagine dividing the grid into sets of “planes” in different orientations

• All planes in a set are identical• The planes are “imaginary”• The perpendicular distance between pairs of adjacent

planes is the d-spacing

Need to label planes to be able to identify them

Find intercepts on a,b,c: 1/4, 2/3, 1/2

Take reciprocals 4, 3/2, 2

Multiply up to integers: (8 3 4) [if necessary]

Exercise - What is the Miller index of the plane below?

Find intercepts on a,b,c:

Take reciprocals

Multiply up to integers:

Plane perpendicular to y cuts at , 1, (0 1 0) plane

General label is (h k l) which intersects at a/h, b/k, c/l

(hkl) is the MILLER INDEX of that plane (round brackets, no commas).

This diagonal cuts at 1, 1, (1 1 0) plane

NB an index 0 means that the plane is parallel to that axis

Using the same set of axes draw the planes with the following Miller indices:

(0 0 1)

(1 1 1)

Using the same set of axes draw the planes with the following Miller indices:

(0 0 2)

(2 2 2)

NOW THINK!! What does this mean?

Planes - conclusions 1

• Miller indices define the orientation of the plane within the unit cell

• The Miller Index defines a set of planes parallel to one another (remember the unit cell is a subset of the “infinite” crystal

• (002) planes are parallel to (001) planes, and so on

d-spacing formula

For orthogonal crystal systems (i.e. ===90) :- 2

2

2

2

2

2

2 c

l

b

k

a

h

d

1

For cubic crystals (special case of orthogonal) a=b=c :- 2

222

2 a

lkh

d

1

e.g. for (1 0 0) d = a(2 0 0) d = a/2(1 1 0) d = a/2 etc.

A tetragonal crystal has a=4.7 Å, c=3.4 Å. Calculate the separation of the:

(1 0 0)

(0 0 1)

(1 1 1) planes

A cubic crystal has a=5.2 Å (=0.52nm). Calculate the d-spacing of the (1 1 0) plane

]ba[c

l

a

kh

d

12

2

2

22

2

Question 2 in handout:

If a = b = c = 8 Å, find d-spacings for planes with Miller indices (1 2 3)

Calculate the d-spacings for the same planes in a crystal with unit cell a = b = 7 Å, c = 9 Å.

Calculate the d-spacings for the same planes in a crystal with unit cell a = 7 Å, b = 8 Å, c = 9 Å.

X-ray Diffraction

Diffraction - an optical grating

XY

1

2

a

Coherent incident light Diffracted light

Path difference XY between diffracted beams 1 and 2:

sin = XY/a

XY = a sin

For 1 and 2 to be in phase and give constructive interference, XY = , 2, 3, 4…..n

so a sin = n where n is the order of diffraction

Consequences: maximum value of for diffraction

sin = 1 a =

Realistically, sin <1 a >

So separation must be same order as, but greater than, wavelength of light.

Thus for diffraction from crystals:

Interatomic distances 0.1 - 2 Å

so = 0.1 - 2 Å

X-rays, electrons, neutrons suitable

Diffraction from crystals

XY

Z

d

Incident radiation “Reflected” radiation

Transmitted radiation

1

2

?

X-ray Tube

Detector

Beam 2 lags beam 1 by XYZ = 2d sin

so 2d sin = n Bragg’s Law

XY

Z

d

Incident radiation “Reflected” radiation

Transmitted radiation

1

2

We normally set n=1 and adjust Miller indices, to give

2dhkl sin =

2d sin = n

e.g. X-rays with wavelength 1.54Å are reflected from planes with d=1.2Å. Calculate the Bragg angle, , for constructive interference.

= 1.54 x 10-10 m, d = 1.2 x 10-10 m, =?

d2

nsin

nsind2

1

n=1 : = 39.9°

n=2 : X (n/2d)>1

Example of equivalence of the two forms of Bragg’s law:

Calculate for =1.54 Å, cubic crystal, a=5Å

2d sin = n

(1 0 0) reflection, d=5 Å

n=1, =8.86o

n=2, =17.93o

n=3, =27.52o

n=4, =38.02o

n=5, =50.35o

n=6, =67.52o

no reflection for n7

(2 0 0) reflection, d=2.5 Å

n=1, =17.93o

n=2, =38.02o

n=3, =67.52o

no reflection for n4

1d

ha

kb

lc2

2

2

2

2

2

2

Use Bragg’s law and the d-spacing equation to solve a wide variety of problems

2d sin = n

or

2dhkl sin =

X-rays with wavelength 1.54 Å are “reflected” from the (1 1 0) planes of a cubic crystal with unit cell a = 6 Å.

Calculate the Bragg angle, , for all orders of reflection, n.

Combining Bragg and d-spacing equation

056.06

0112

2

222

2 a

lkh

d

1

18d2 d = 4.24 Å

d = 4.24 Å

d2

nsin 1

n = 1 : = 10.46°

n = 2 : = 21.30°

n = 3 : = 33.01°

n = 4 : = 46.59°

n = 5 : = 65.23°

= (1 1 0)

= (2 2 0)

= (3 3 0)

= (4 4 0)

= (5 5 0)

2dhkl sin =

SummarySummary We can imagine planes within a crystal

Each set of planes is uniquely identified by its Miller index (h k l)

We can calculate the separation, d, for each set of planes (h k l)

Crystals diffract radiation of a similar order of wavelength to the interatomic spacings

We model this diffraction by considering the “reflection” of radiation from planes - Bragg’s Law