September 16, 2010
Class 3
Hexagonal Close-Packed Structure (hcp)
Image from http://www.hull.ac.uk/php/chsajb/general/unitcell.html
The unit cell for an hcp is a right prism rhombohedral base (see figure
21 to see the relationship between the unit cell and the hexagonal cell).
The basis has 2 identical atoms, 1 at a corner and the other at 2/3,1/3,1/2.
Its position coincides with the center of one of the two triangles in the
rhombus.
This is the closest way to packed up spheres, the packing fraction is
0.74, the same as the fcc. In fact depending on how you pile up spheres
you get hcp or fcc
Assuming you packed spheres such that each is in contact to other 6 spheres you would form a
basal plane for an hexagonal structure or the one of the planes in the fcc lattice. Now add a
second layer of sphere with in the center of three of the spheres at the base. Now you can add a
third layer in at least two way, one such that each sphere on the third layer is right above one in
the first layer, then you have and
hcp structure (ABAB…) or add the
third layer such that each ball is
above the center of three spheres in
the first layer not occupied by
spheres on the second layer
(ABCABC…) this would be an fcc
lattice. The table in page 16 show
some crystals with this structure
Diamond Structure
The diamond structure is a fcc with a two atom’s basis, one at 0,0,0 and the other at ¼, ¼, ¼.
Another way to view this lattice, is to consider two fcc structures displaced ¼ in each direction
(1/4 of the body diagonal).
One example of this structure is the “diamond” where the name of
the lattice comes from. Diamond has a basis of 2 C atoms.
Another example is Zinc Sulfide (ZnS), in this case, one atom is Zn
and the other is S.
The table in page 18 shows other crystals with this structure.
Image from
http://www.uncp.edu/home/mcclurem/lattice/diamond.htm
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Index for crystal planes
When dealing with crystals, not only the 3-D structure is important but also the planes associated
with it. For every crystal structure a number of planes can be defined. This is very important for
many applications, for example, when using x-ray diffraction to characterize the crystal, the
diffraction pattern will change as we rotate the crystal depending on what planes are the x-rays
diffracted from.
There are many way to name these planes, one of the way is known as Miller index and they are
obtained as follow
1) find the intersect of the plane with each of the three translation vectors (they may be
primitive vectors or not)
2) Take the reciprocal and reduce them to three smallest integers having the same ratio and
enclose the results between parenthesis (hkl)
When a plane is parallel to an axis the corresponding index is 0 (1/)
For instance, in figure 13, the intersects to the three vectors are respectively (3,2,2).
The reciprocal are 1/3, 1/2, 1/2.
The miller indexes for this plane are then (233). These indexes usually correspond to a family of
parallel planes. For instance, all the planes perpendicular to a1 in the cubic lattice that intersects
the axis on an integer value (1,0,0), (2,0,0), etc have miller index (100).
When the plane intersects the axis on a negative coordinate, then the corresponding miller index
is negative which is indicated by placing a dash above the index, for instance ( hk l ) would be the
Miller indexes of a plane that intersects a1 and a2 on the positive side, while it intersects a3 on the
negative side of the axis.
Question: what would the (200) plane be?
Planes that are equivalent by symmetry can be denoted by curly brackets, thus the set of cube
faces can be denoted as {100}
Finally, a direction in a crystal can be denoted by index defined similarly to the Miller indexes
but enclosing them within square brackets. To define directions in a crystal, the smallest integers
that have the ratio of the components of the vector in the desired direction are used.
For a cubic lattice, the (100) is the plane that is perpendicular to a1 while [100] is the direction
parallel to a1. For cubic crystals [hkl] is a vector that is perpendicular to the (hkl) plane. That is
not true for every crystal structure.
Non-Ideal Crystal Structures
It has not been proven that an ideal crystal is actually the structure that minimizes the energy of a
crystal even at absolute zero; at finite temperature most probably no structure is ideal. This is
true even if no defects (like vacancy, substitutions, etc) are considered
Two examples:
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Random stacking: ABABAB… defines a hcp structure while ABCABC defines an fcc
structure, however some structures are know where the staking is random, i.e. in every
subsequent plane one of A, B or C exist but in a random sequence, this is known as random
stacking.
Polytypism: This is similar to the random stacking but there is a periodicity which is however
very long. For instance, ABACBABACBABACB… notice that this is periodic but the repeat
unit consist on 5 layers ABACB. ZnS is on example where more than 150 polytypes has being
identifies. The longest periodicity is 360 layers.
Notice table 3 and 4, these tables should be used for permanent reference during the quarter.
Group Homework 1: Due Tuesday September 21th.
Diffraction of Waves by Crystals
Double slit experiment
Waves have the property that they can interfere with each other, when two waves coincide on a
point, their amplitudes are added. Assume that two light waves, with the same wave length,
reach a particular point on a screen “in phase” such that the peaks and valleys of both waves
arrive at the same time (constructive interference), in this case a bright spot will be formed, the
intensity of the light at that spot will be twice that of each individual wave. If on the contrary
they arrive totally “out of phase” (destructive interference), meaning the peaks of one coincide
with the valleys of the other, a dark point will be formed. As the shift in phase between the two
waves increase, the intensity of the light in the screen continuously alternate between bright and
dark.
The double slit experiment illustrates this phenomenon, when monochromatic light
(composed with waves of a single frequency) goes through such arrangement, each slit actually
becomes a source in phase with one another.
If the path difference is equal to nλ where n is an integer, a bright image results from constructive
interference. If t is the separation of the slits and  the angle of propagation with respect to the
perpendicular of the plane of the slits, the condition for a bright spot is:
nλ = t sin
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The above phenomenon is called diffraction. In fact the most common technique to sample
crystal structure is x-ray diffraction. Since particles (such as electrons and neutrons) are known
to also behave as waves, electron or neutron diffraction is also used for this study.
It should be noticed that the above condition for diffraction is a general one; a bright spot will be
produce every time the path difference between two waves is equal to a multiple of the wave
length. Notice however that the wave length cannot be larger than t and thus the radiation wave
lengths able to produce diffraction are limited by the characteristics of the diffraction system.
Diffraction in Crystals
Particle Wave Duality of Matter
A way to sample crystal structures is by diffraction. Characteristic lengths in crystal structures
are of the order of 1-10Å and thus suitable radiation should have wave length on that range what
leads to an energy (in the case of photons) of problem 1 activity 2
E
hc
 1.23x10 3 eV  1.23x10 2 eV

h=6.62x10-34J.s and c=3x108m/s
Radiation with energy in the order of keV are known as x-rays.
In 1924, DeBroglie postulated that matter also has a wave-nature depending on their linear
momentum. This was a new idea since Einstein had only hypothesized the particle-wave duality
for EM radiation!!!
DeBroglie – Particle-Wave nature applies to matter having momentum, p!!!
The DeBroglie relation is :  
h
.
p
Matter wave associated with 1kg baseball traveling at 10m/sec: Problem 2 activity 2
p=mv = 1kg(10m/sec) = 10 kg m/sec
h 6.626 x10 34 [kgm2 / s 2 ][ s]
 
 6.626 x10 35 m
p
10kgm / s
TOO SHORT TO MEASURE!!!
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However the wave length associate with a 100eV electron Problem 3 activity 2
100eV(1.6x10-19J J/eV) = 1.602e-17 J
6.626 x10 34 [kgm2 / s 2 ][ s]
6.626 x10 34 [kgm2 / s 2 ][ s]
h
h



 1.2 x10 10 m
 24
31
17
2
2
p
5.4 x10 [kgm / s]
2mK
2  9.1x10 [kg]  1.602 x10 [kgm / s ]
We can measure this wavelength using solids having interplanar spacing near 1-2 Angstroms

In fact, any matter having a mass which is in the range of h, will have a measurable matter wave
associated with it!!! That means neutrons, atoms, ions, and small molecules may (in theory)
ALL be used for diffraction (of course, depending on the wavelength of the matter wave).
Typically electrons, neutrons and x-rays are used for diffraction from crystallographic planes.
Wavelength vs. Particle Energy
100
Neutron (0.01 eV)
Electron (100 eV)
Wavelength (Angstroms)
X-ray Photon (1000 eV)
10
X-ray Photon
Neutron
1
1
Electron
10
100
0.1
Energy (eV)
Reproduction of Figure 1 in Chapter 2 of Kittel.
The reproduced figure above was made in MS Excel using the energy-wavelength relationships
for photons, neutrons, and electrons. Problems 4, 5 and 6 activity 2
Image from http://imagers.gsfc.nasa.gov/ems/visible.html
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There are two ways to understand diffraction by crystals. One is due to W.L. Bragg and it is
based on atomic positions in the “direct lattice”, the other is due to Von Laue and take advantage
of the concept of reciprocal lattice which we will define soon
Bragg Law
This is based on the same concepts as the double slit experiment; the difference in path length of
the light rays must be a multiple of the wave length for constructive interference to occur.


 
dhkl

dhkl sin 

Path Difference of incoming wave is d sin 
Path Difference of outgoing wave is d sin 
Total Path Difference is
2d sin 
Thanks Dr. Dobbins
Hypothesis for Bragg’s Law
A specularly reflected beam (in = out) will occur at each successive plane within the solid.
 Angle between incoming beam and hkl plane:  (normally used in crystallography rather
than the angle with the normal used in optics).
 Total path distance traveled to plane 2= path distance to plane 1 plus 2d sin( )
 Total path distance traveled to plane 3= path distance to plane 1 plus 4d sin( )
 Total path distance traveled to plane 4= path distance to plane 1 plus 6d sin( )
Bragg’s Diffraction Condition
Constructive interference will occur if the neutrons, x-rays, or electrons used to probe the
structure have an integral number of wavelengths equal to the path length difference, Hence, n
= 2d sin .
Bragg’s Law is a consequence of the periodicity of the lattices. Diffraction, in fact, will
occur for any periodic structure (engineered, like quantum dots, or natural, like atomic planes).
The Diffraction condition is not a consequence of the type of basis set of atoms at the
lattice points but of the periodicity of the lattice—however, the diffracted intensity is a function
of the basis. We can, therefore, use diffraction to perform chemical analysis of a crystalline
structure.
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The reason for which the diffracted intensity is a function of the lattice plane chemistry is that xrays and electrons are reflected by the electron density around the atoms while neutrons are
reflected by the heavy core of the atoms (neutrons are much to heavy to be reflected by mere
electrons around the atoms) thus the nature of the atoms is important to determine the scattering
conditions.
Scattered Wave Amplitude
Let’s consider scattering (i.e. specular reflection) from the arrangement of atoms below
(alternating large and small atoms).
Any atomic property (electron density, charge concentration, magnetic moment, etc) when
mapped into a crystal will show the same periodicity as the atomic arrangement. For instance
such an arrangement would result in the periodic electron density given below.
a
a
a
Mathematically, this electron density must satisfy the expression
n(x + T) = n(x) or in 3-D n(r + T) = n(r).
Translations into Fourier Space (or Reciprocal Space)
Problems with such periodic systems are much easier to treat in Fourier space. In 1D, the
Fourier transform for the electron density is
n( x )  n o

 C
p 0
p

 2px 
 2px 
 2ipx 
cos
n p exp
  S p sin 
 or n( x) 


 a 
 a 
 a 
 p  

where p are integers, Cp and Sp are the Fourier Coefficients (Real values). np can be complex
numbers with the condition n* p  n p what ensures that n(x) is real.
2p
For each point x in real space, there is a reciprocal lattice point in Fourier space defined as
.
2 4 6
The reciprocal lattice points are
,
,
,…
a a a
a
So, each of the reciprocal lattice points are like “modes” in the Fourier space, thus a natural way
to see the Fourier transform is that a linear combination of each of these “modes” with a given
amplitude gives out the electron density. Diffraction is thus a technique that, not only allows
finding the crystal structure but also the electron density and thus the chemical composition of
the crystal.
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