1. IPCMS-GEMME, BP 43, 23 rue du Loess, 67034 Strasbourg Cedex 2
Characterization of thin films and bulk materials using x-ray and electron scattering
V. Pierron-Bohnes
IPCMS-GEMME, BP 43, 23 rue du Loess, Strasbourg Cedex 2
Plan : lattices
x-ray and electron - matter interaction
real lattice and reciprocal lattice in 3D and 2D samples
experimental set-ups
studies on single crystals
synchrotron radiation
strains measurements using x-ray scattering and TEM
powder scattering measurement
texture analysis
reflectometry
chemical analysis
short and long range order measurements
This part will contain:
crystal lattice, periodicity, symmetries, Bravais lattices, basis
Bragg relation, planes and points in reciprocal space,
periodicity, symmetries, and intensities of peaks (extinctions, superstructures…)
Fourier transform, 3D and 2D cases

2. 2D lattices with single atom basis
square lattice
hexagonal lattice
elementary cell
highest symmetry
cell
When placing points regularly on a plane, we form a lattice. By definition a lattice is formed from an unique unitary cell, formed on the elementary translation vectors, a and b, and all the points of the plane can be attained by translating the cell by a combination of these vectors. There are several choices for these vectors.
Exercise: give other choices for the 2 higher lattices.
We can choose some high symmetry lattices like the square lattice or the hexagonal lattice, or less symmetric like the rectangular lattice. The elementary cell contains the minimum number of atoms (1 if possible) whereas sometimes it is more successful to consider a larger cell with higher symmetry. To describe a structure it is often useful to consider the highest symmetry cell. In this example, the rectangular cell has some obvious symmetries that would be omitted when considered as a parallelepiped cell.
Exercise: what symmetries has the rectangular cell that has not the parallelepiped cell ?
rectangular lattice
centered rectangular lattice

3. 2D lattices with complex basis
square lattice
hexagonal lattice
Most crystals are not so simple: we need several atoms in each cell, they form the basis that will be reproduced by translations to produce the whole crystal. If the basis has a lower symmetry than the lattice, the crystal will have the lowest symmetry.
Exercise: give the symmetries of each lattices shown here.
rectangular lattice
centered rectangular lattice

4. 2D and 3D symmetries Translation symmetries
Rotational symmetries of a crystal: 2-fold, 3-fold, 4-fold, and 6-fold
Mirror symmetries
Centrosymmetry
All the possible symmetries are the translation symmetries, the rotational symmetries of a crystal: 2-fold, 3-fold, 4-fold, and 6-fold, the mirror symmetries, and the centro-symmetry. Do you know all these symmetries?
Exercise: Name the symmetries of the cube shown here.
Exercise: Find all the symmetries of the cube.

5. Naming the planes in 3D
In 3D, we consider the referential based on the elementary vectors
a, b, and c (with any angle between them a priori) which allow to
describe the crystal.
To name a plane family,
we consider one plane.
It cuts the 3 axes directed
by a, b and c at three points
with abcissae A, B and C.
The Miller indices of
the plane family are
the smaller integers hkl
verifying hA=kB=lC (or
proportional to 1/A, 1/B, 1/C). One notes plane (hkl) and family {hkl}. 2 2 3
Exercise: what is the name of the plane on the top figure?
Exercise: draw a 112 plane in the cubic lattice

6. Perpendicular to a planes in 3D
In an orthonormal referential, a vector perpendicular to a (hkl) plane
is the vector with components hkl on the vectors a, b, c.
The distance between
planes is given by: 2 3 2 2 3 3
Exercise: verify the property of the normal vector and the distance in the different cases of the bottom panel of the figure and for the 112 planes.
In a cubic it is given by:

7. Diffraction of a wave 1 A1 A2
Let us consider a plane wave arriving on two atoms A1 and A2 with a wave vector k0. It is elastically diffused with a final wave vector k. The difference of optical path between the wave passing through A1 and that passing through A2 is delta1+delta2. The phase difference is phi if the waves have the wavelength lambda. We define the diffusion vector Q as k-k0. phi is then simply expressed as minus the scalar product of the two vectors Q and A1A2.

8. Diffraction of a wave 2 A1 A2
Because the diffusion is elastic, the norms of k0 and k are equal. The vector Q is bisecting the angle formed by the incident and the diffused beams.
The waves emitted by the point A1 and A2 are in phase if phi is a multiple or 2pi.

9. Diffraction of a wave 3 d characteristic of planes perpendicular to Q
All the points on a plane perpendicular to Q are in phase because phi equals 0. The amplitudes of the wave diffused by all these points are added. If there are N equivalent atoms in these points, with a diffusion amplitude F, the amplitude for the plane will be NA and the intensity N2A2.
The phase difference with A2 will be the same for any atom in the plane passing through A1 and perpendicular to Q. The phase difference with A1 will be the same for any atom in the plane passing through A2 and perpendicular to Q. phi is thus a characteristic of the two planes.
If this phase is a multiple of 2pi, the resulting amplitudes for the two planes will be 2NA (if there are also N atoms on the plane passing across A2); if the phase is an odd multiple of pi, the resulting amplitudes for the two planes will be 0. The coherence condition is dQ=2n pi with n an integer.
There will be measurable intensity in a direction k if there are planes perpendicular to Q=k-k0 with a distance d such that dQ=2n pi. This a a necessary condition, not a sufficient condition, because the intensity will be 0 if there is a supplementary plane at d/2 with the same number and same nature of atoms, because there will be some destructive diffusion interferences.
When the intensity at point Q is not 0, one says that there is diffraction (constructive diffusion interferences) of the beam and that there is a Bragg peak at Q. One defines the reciprocal space as a space where are placed the Bragg peaks. The point at Q from the origin is a point of the reciprocal space of the structure.
characteristic of planes perpendicular to Q
additive interferences: d Q = 2np or exp(idQ)=1

10. square lattice real space reciprocal space lattice parameter a O ?
Let us consider the square lattice in a 2D crystal to make it more easy, it is the (001) plane of the 3Dsimple cubic lattice. What is the reciprocal space ? ?

11. square lattice real space reciprocal space lattice parameter a O a O*
From the horizontal periodicity, we get the two points of the reciprocal space at 2pi/a horizontally. O*
2p/a

12. periodicity parameter 2p/a
square lattice
real space
lattice parameter a a
reciprocal space
If Qd=2pi the relation is also true for 2Q 3Q… we have thus a row of reciprocal space points.
periodicity parameter 2p/a O*
2p/a

13. periodicity parameter 2p/a
square lattice
real space b a
lattice parameter a a
reciprocal space
Considering the horizontal planes, we get the reciprocal space points along the vertical.
2p/a b*
periodicity parameter 2p/a a* O*

14. periodicity parameter 2p/a
square lattice
real space b
lattice parameter a a
The sum of two vectors of the reciprocal space is a vector of the reciprocal space. The reciprocal space is thus a square lattice in 2D and a cubic lattice in 3D.
reciprocal space
= square lattice b* a* O*
periodicity parameter 2p/a

15. simple cubic lattice real space
lattice parameter a
simple cubic reciprocal space
periodicity parameter 2p/a
othorhombic lattice real space
The reciprocal space of a simple cubic lattice is a simple cubic lattice. The reciprocal space of a simple orthogonal lattice is a simple orthogonal lattice.
lattice parameters a ≠ b ≠ c
orthorhombic reciprocal space
periodicity parameters a*=2p/a, b*=2p/b, c*=2p/c

16. reciprocal space, general case
Real space with vectors: with a basis
Reciprocal space with vectors:
such that : with some extinctions
due to the basis
These are the formulae for the general case. ai* is perpendicular to ai per construction. The current point of the lattice in the real space is R=ma1+na2+pa3, the current point of the reciprocal space is Q=ha1*+ka2*+la3* and they verify : R.Q=2 pi . j with j=hm+ln+lp and thus exp(iR.Q)=1.

17. reciprocal space, intensity of peaks
The intensities of the peaks can be calculated by this general formula. We get the product of 2 factors: a factor characteristic of the lattice, with non 0 values at the points of the reciprocal lattice (a comb function with Dirac peaks at these points) multiplied by a factor characteristic of what is on the cell: which atoms and at which positions. This term will give the intensity of the different peaks and the extinctions. It is calculated from the electron density for the x-ray diffraction and from the electrostatic potential due to electrons and nuclei for the electron diffraction.
reciprocal
lattice peaks:
where are the peaks ?
form factor or
structure factor:
what is the peak intensity ?
for x-rays

18. centered cubic lattice real space
lattice parameter a
face centered cubic reciprocal space
periodicity parameter 4p/a
extinction of 1 peak over 2
due to the intermediate planes
extinction of 001 peak
The reciprocal space of any cubic lattice is a simple cubic, but the extinctions make it more clear to consider the lattice with double parameters:
The reciprocal space of a centered cubic lattice can be considered as a face centered cubic lattice with double parameters. As a matter of facts, the center of the cube is neither in the initial (001) planes, nor in the (111) planes. It is in the (110) planes, the 2 pi/a point is thus conserved.
Exercise: find the extinctions from the form factor formula
no extinction of 110 peak
extinction of 111 peak

19. face centered cubic lattice real space
lattice parameter a
centered cubic reciprocal space
periodicity parameter 4p/a
extinction of 1 peak over 2
due to the intermediate planes
extinction of 001 peak
The reciprocal space of a face centered cubic lattice can be considered as a centered cubic lattice with double parameters. As a matter of facts, the center of the (001) face is neither in the initial (001) planes, nor in the (101) planes. It is in the (111) planes, the 2 pi/a point is thus conserved.
Exercise: find the extinctions from the form factor formula
extinction of 110 peak
no extinction of 111 peak

20. Ordered phases: L12 structure in real space
lattice parameter a
simple cubic reciprocal space
periodicity parameter 2p/a
different intensity of 1 peak over 2
due to the intermediate planes
001: superstructure peak
For the ordered structures formed starting with a face centered cubic phase, several sub-lattices have to be defined. To determine if a peak is present or not, the form factor has to be calculated, or the occurrence or planes with different occupancy. For example, in the L12 structure, 4 simple cubic sub-lattices are considered; 1 of them are occupied by large red atoms A (initial simple cubic lattice) and 3 of them is occupied by the small blue atoms B. Between two (110) planes of the initial simple cubic structure, a new plane containing only B atoms is present, the form factor will be non 0 (contrary to the fcc phase) but will be different from that of the other peaks as 000. These peaks are called superstructure peaks (those that were extinguished in the fcc and have a non 0 intensity due to a new occupation). Those that were not extinguished in the fcc structure are called the fundamental peaks. It is the case of the 111 peak, because all the (111) planes are mixed with the compound concentration AB3.
The intensity of fundamental peaks varies as the average of the atomic form factors of the 2 atoms. The intensity of superstructure peaks varies as the difference of the atomic form factors of the atoms.
Exercise: draw the different (001) planes and equivalents. Idem for (110) planes. Occupation ?
Exercise: write the form factor of the L12 structure and calculate its values on the different peaks cited in the text and 000.
fundamental
110:
superstructure peak
111: fundamental peak
superstucture

21. Ordered phases: L10 structure in real space
lattice parameter a
simple cubic reciprocal space
periodicity parameter 2p/a
different intensity of 1 peak over 2
due to the intermediate planes
100: extinguished
001: superstructure peak z x
The ordered L10 structure is also formed starting with a face centered cubic phase. This structure is tetragonal due to the symmetry break between x and z axes. We can define two sublattices. Each of them is a base centered tetragonal. One of them is occupied by A atoms, the other one by B atoms. Between two (110) planes of the initial simple cubic structure, a new plane containing only small blue atoms is present, the form factor will be non 0. These peaks are superstructure peaks. On contrary, the (101) planes have all the AB concentration. The 101 peak is thus extinguished. The 111 peak is a fundamental peak, because all the (111) planes are mixed with the compound concentration AB and because this peak was present in the fcc structure (no supplementary plane in comparison to the simple cubic structure).
Here also the intensity of fundamental peaks varies as the average of the atomic form factors of the 2 atoms. The intensity of superstructure peaks varies as the difference of the atomic form factors of the atoms.
Exercise: draw the different (001) planes and equivalents. Idem for (110) planes. Occupation ?
Exercise: write the form factor of the L12 structure and calculate its values on the different peaks cited in the text and 000.
fundamental
101: extinguished
110: superstructure peak
111: fundamental peak
superstucture

22. periodicity parameters 2p/a, 2p/c
simple hexagonal lattice real space
lattice parameters a, c
simple hexagonal reciprocal space
periodicity parameters 2p/a, 2p/c c c* a3 a1
The hexagonal lattice in a plane can be generated from the elementary cell formed from the vectors a1 and a2 with equal length and with a 120° angle between them. As this diamond cell does not present the 6-fold symmetry of the lattice, it is usually considered that a third axis can be defined a3 at 120° of both a1 and a2. That is why you can find 4 indices in the papers treating of the 3D-hexagonal structures that are completed by the orthogonal vector. This lattice is very complicated and I will not detail it more here. The reciprocal lattice vectors are perpendicular to those in the real space.
a2* a1 a2
a1* a2

23. compact hexagonal lattice real space
lattice parameters a,
c = a = √8/3 a c c a3 a3 a1
The hexagonal compact lattice can be constructed by placing balls on a plane as compactly as possible. They adopt a hexagonal configuration in the plane. The second shell is placed in the hollows above the center of the triangle formed by 3 balls. There are 2 possible ways to place the first ball of the second shell. The others are placed to form the same hexagonal lattice in the second shell. When beginning the third shell, there is again the choice between two places. If the balls of the third shell are vertically aligned on those of the first shell, the compact hexagonal lattice is formed. If the other position is chosen, it is the face centered cubic phase (the horizontal planes are the (111) planes of the fcc). In a crystal, the balls are the atoms. The real size is more like this, but it is very difficult to see the symmetries of the lattice, that is why, the atoms are represented with a much smaller relative dimension than theirdistances. a1 a2 a2

24. periodicity parameters 2p/a, 4p/c
compact hexagonal lattice real space
lattice parameters a,
c = a = √8/3 a
simple hexagonal reciprocal
space with extinctions
periodicity parameters 2p/a, 4p/c c c* a3
The reciprocal space of the hexagonal compact phase is an hexagonal lattice. Considering the horizontal planes, we get the extinction of one peak over two along the vertical. There is no extinction in the basal plane because the supplementary atoms are not at half distance of the initial lattice planes.
Exercise: Apply the general formulae to the hexagonal case. Explain the drawing of the reciprocal space. Find the intensities of the different peaks in plane and out of plane. Find the extinctions. a1 a1
a2* a2
a1* a2

25. Fourier transform
Mathematical operator between functions: Fourier transform
If we come back to the formulae of slide 17, we can rewrite the final expression taking into account the expression of Fk, and if we sum all the contributions to rau to get the total electron density (for x-rays or the total electrostatic potential for electrons), we get a simple expression relating rau_total and the diffracted amplitude. The diffracted amplitude is the Fourier transform of the total electron density. This operator can be inverted: the total density of state can be theoretically deduced from the amplitude, if it can be deduced from the measured intensity (the problem is that the phase of the amplitude is lost most generally).
Inverse Fourier transform
phase lost (→ holography)

26. periodicity parameter 2p/a with rods
2D square lattice
real space b
lattice parameter a
with atoms a
For a surface, the reciprocal space is made of rods: the limitation of the real space to a plane implies that the reciprocal space is infinite in that direction as there is no vertical coordinate.
reciprocal space
= square lattice b*
periodicity parameter 2p/a with rods a*