1. 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

2. 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

3. 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]

4. Exercise - What is the Miller index of the plane below?
Find intercepts on a,b,c:
Take reciprocals
Multiply up to integers:

5. 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

6. Using the same set of axes draw the planes with the following Miller indices:
(0 0 1)
(1 1 1)

7. 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?

8. 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

9. d-spacing formula For orthogonal crystal systems (i.e. ===90) :-
For cubic crystals (special case of orthogonal) a=b=c :-
e.g. for (1 0 0) d = a
(2 0 0) d = a/2
(1 1 0) d = a/2 etc.

10. A cubic crystal has a=5. 2 Å (=0. 52nm)
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

11. 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.

12. X-ray Diffraction

13. Diffraction - an optical grating
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

14. 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 Å
so  = Å
X-rays, electrons, neutrons suitable

15. Diffraction from crystals
X-ray Tube
Detector ?

16. Beam 2 lags beam 1 by XYZ = 2d sin 
so 2d sin  = n Bragg’s Law

17. e. g. X-rays with wavelength 1. 54Å are reflected from planes with d=1
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 m, d = 1.2 x 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  = 

18. 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

19. 2dhkl sin  =  2d sin  = n or
Use Bragg’s law and the d-spacing equation to solve a wide variety of problems
2d sin  = n or
2dhkl sin  = 

20. X-rays with wavelength 1.54 Å are “reflected” from the
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.
d = 4.24 Å

21. d = 4.24 Å n = 1 :  = 10.46° n = 2 :  = 21.30° n = 3 :  = 33.01°
= (1 1 0)
= (2 2 0)
= (3 3 0)
= (4 4 0)
= (5 5 0)
2dhkl sin  = 

22. 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