# Objectives By the end of this section you should:

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

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

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: 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 Å so  = Å X-rays, electrons, neutrons suitable

Diffraction from crystals
X-ray Tube Detector ?

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
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  = 

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

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  = 

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

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  = 

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