September 16, 2010
Class 3
Hexagonal Close-Packed Structure (hcp)
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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.
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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 () 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:
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
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
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 : .
Matter wave associated with 1kg baseball traveling at 10m/sec:Problem 2 activity 2
p=mv = 1kg(10m/sec) = 10 kg m/sec
TOO SHORT TO MEASURE!!!
However the wave length associate with a 100eV electronProblem 3 activity 2
100eV(1.6x10-19J J/eV) = 1.602e-17 J
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.
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
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 latticewhich 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.
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
Total path distance traveled to plane 3= path distance to plane 1 plus
Total path distance traveled to plane 4= path distance to plane 1 plus
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.
The reason for which the diffracted intensity is a function of the lattice plane chemistry is that x-rays and electrons are reflected by the electron density around the atoms whileneutrons 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.
ScatteredWaveAmplitude
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.
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
or
where p are integers, Cp and Sp are the Fourier Coefficients (Real values). np can be complex numbers with the condition what ensures that n(x) is real.
For each point x in real space, there is a reciprocal lattice point in Fourier space defined as . The reciprocal lattice points are ,,,…
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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