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
2
2
2
For orthogonal crystal systems
(i.e. ===90) :-
1 h
k
l
 2 2 2
2
d
a
b
c
For cubic crystals (special case
of orthogonal) a=b=c :-
1 h2  k 2  l 2

2
d
a2
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 h 2  k 2 l2

 2
2
2
d
a
c
[a  b]
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
Path difference XY
between diffracted
beams 1 and 2:

2
a
sin = XY/a
1
X
 XY = a sin 
Y

Coherent incident light
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
Detector
“Reflected” radiation
Incident radiation
1
2


X
d
?
Z
Y
Transmitted radiation
“Reflected” radiation
Incident radiation
1
2


X
d
Z
Y
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,
2d sin   n
 n 
  sin  
 2d 
1
d = 1.2 x 10-10 m, =?
n=1 :  = 39.9°
n=2 : X
(n/2d)>1
2d sin  = n
We normally set n=1 and adjust Miller indices, to give
2dhkl 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, =8.86o
n=2, =17.93o
n=3, =27.52o
n=4, =38.02o
n=5, =50.35o
n=6, =67.52o
no reflection for n7
(2 0 0) reflection, d=2.5 Å
n=1,
=17.93o
n=2,
=38.02o
n=3, =67.52o
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
2dhkl sin  = 
2
2
2
1 h k l
2
2
2
2
d
a
b
c
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 h k l

2
d
a2
2
2
d 18 
2
2
11 0

 0.056
2
6
d = 4.24 Å
d = 4.24 Å
 n 
  sin  
 2d 
1
n=1:
 = 10.46°
= (1 1 0)
n=2:
 = 21.30°
= (2 2 0)
n=3:
 = 33.01°
= (3 3 0)
n=4:
 = 46.59°
= (4 4 0)
n=5:
 = 65.23°
= (5 5 0)
2dhkl 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