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 Indices

Imagine 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:

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).

Plane perpendicular to y cuts at  , 1,  

(0 1 0) plane

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  ) : For cubic crystals (special case of orthogonal)

a

=

b

=

c

:-

1 d

2 

h a

2 2 

k b

2 2 

l c

2 2

1 d

2 

h

2 

k

2 

l

2

a

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

A cubic crystal has

a

=5.2 Å (=0.52nm). Calculate the d-spacing of the (1 1 0) plane 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 1 d 2  h 2  k 2 a 2  l c 2 2 [ a  b ]

Question 2 in handout: If

a

=

b

=

c

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

 Path difference XY between diffracted beams 1 and 2: sin  = XY/a  XY = a sin 

a

X Y  Coherent incident light 2 1 Diffracted light 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

X-ray Tube

Diffraction from crystals

Incident radiation “Reflected” radiation

Detector

1 2  X Y Z  d ?

Transmitted radiation

Incident radiation “Reflected” radiation 1 2  X Y Z  d Transmitted radiation Beam 2 lags beam 1 by XYZ = 2d sin  so

2d sin



= n



Bragg’s Law

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,  =?

2 d sin   n 

n=1 :



= 39.9

°

  sin  1

n=2 : X (n



/2d)>1

2 d 2d sin  = n  We normally set n=1 and adjust Miller indices, to give

2d hkl sin



=



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,  =8.86

o n=2, n=3, n=4,  =17.93

o  =27.52

o  =38.02

o n=5, n=6,  =50.35

o  =67.52

o no reflection for n  7 (2 0 0) reflection, d=2.5 Å n=1, n=2, n=3,   =17.93

=38.02

 =67.52

o o o no reflection for n  4

Use Bragg’s law and the d-spacing equation to solve a wide variety of problems 2d sin  = n  or 2d hkl sin  =  1 d 2  h 2 a 2  k 2 b 2  l 2 c 2

Combining Bragg and d-spacing equation 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.

1 d

2 

h

2 

k

2 

l

2

a

2  1  1  0 6 2  0 .

056 d 2  18 

d = 4.24 Å

d = 4.24 Å

n = 1 : n = 2 : n = 3 : n = 4 : n = 5 :   sin  1   n 2 d     = 10.46

°  = 21.30

°  = 33.01

°  = 46.59

°  = 65.23

° = (1 1 0) = (2 2 0) = (3 3 0) = (4 4 0) = (5 5 0) 2d hkl sin  = 

Summary

 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