1. X-Ray Diffraction and Reciprocal Lattice
Lecture 4
X-Ray Diffraction and Reciprocal Lattice

2. Diffraction
Diffraction occurs as waves interact with a regular structure whose repeat distance is about the same as the wavelength. The phenomenon is common in the natural world, and occurs across a broad range of scales. For example, light can be diffracted by a grating having scribed lines spaced on the order of a few thousand angstroms, about the wavelength of light.
It happens that X-rays have wavelengths on the order of a few angstroms, the same as typical interatomic distances in crystalline solids. That means X-rays can be diffracted from minerals which, by definition, are crystalline and have regularly repeating atomic structures.

3. X-ray Scattering
When certain geometric requirements are met, X-rays scattered from a crystalline solid can constructively interfere, producing a diffracted beam. In 1912, W. L. Bragg recognized a predictable relationship among several factors.
1. The distance between similar atomic planes in a mineral (the interatomic spacing) which we call the d-spacing and measure in angstroms.
2. The angle of diffraction which we call the theta angle and measure in degrees. For practical reasons the diffractometer measures an angle twice that of the theta angle. We call the measured angle '2-theta'.
3. The wavelength of the incident X-radiation, symbolized by the Greek letter lambda and, in ordinary cases, equal to 1.54 angstroms.

4. Bragg’s Law
Most methods of determining the atomic structure of crystals are based on the idea of scattering of radiation.
X-ray is one of the types of radiation used.
The wavelength of the radiation should have awavelength comparable to a typical interatomic distance which in solids is of a few angstroms (10-8 cm).
The x-ray wavelength, l can be estimated as follows:
Therefore, x-rays of energy 2-10 keV are suitable for studying the crystal structure.
X-rays interact with electronic shells of atoms in a solid. Electrons absorb and re-radiate x-rays which can then be detected. Nuclei are too heavy to respond.
The reflectivity of x-rays is of the order of 10-3 – 10-5, so that the penetration in the solid is deep. Therefore x-rays serve as a bulk probe.,

5. Bragg’s Law
In 1913, Bragg found that crystalline solids have remarkably characteristic patterns of reflected x-ray radiation.
In crystalline materials, for certain wavelength and incident directions, intense peaks of scattered radiation were observed.
Bragg accounted for this by regarding a crystal as made out of parallel planes of atoms spaced d apart.
The conditions for a sharp peak in the intensity of the scattered radiation were:
The x-rays should be specularly (angle of incidence = angle of reflection) reflected in the atomic plane.
The reflected rays from successive planes must interfere constructively.

6. X-ray Diffraction Techniques
The quantities measured:
The scattering angle 2q between the diffracted and incident beams. By substituting sinq into Bragg’s Law, the interplanar spacing and the orientaion of the plane responsible for the diffraction can be determined.
The intensity of the diffracted beam, This quantity determines the cell-structure factor and therefore gives information concerning arrangement of atoms in the unit cell.

7. Laue camera SiGe forward reflexion photo (spots on ellipses)
crystal held
on goniometer
head
Unfiltered X-radiation
through collimator
back reflexion
photo (spots on
hyperbolas)
continuous band of wavelength provided
suitable for determination of crystal orientation & symmetry
each spot corresponds to planes of atoms of spacing d satisfying 2dsinq=l
diffraction pattern exhibits the symmetry of the crystal in the direction of k0
spots corresponding to reciprocal vectors [h,k,l] and [h,k,l] are symmetric
with respect to the centre of the diffraction pattern

8. Rotating crystal r axis of rotation=crystal axis unwrapped film g2
monochromatic
X-ray
beam
trap
film held against
inside wall of
cylinder
Let axis of rotation be the C-axis so that Laue eqn C(cosg-cosg0)=ll
Since g0=p/2, C cosg=ll. For l=0 “cone” is a plane ^ C. For l=1,2
dl=rcosgl. Therefore

9. Powder method powder sample: many crystallites in random orientations
monochromatic
X-ray 2q
wrapped film
cone of x-rays
satisfying the
Laue eqn
powder sample: many crystallites
in random orientations r
unwrapped film
Lack of orientation make crystal structure determination difficult

10. Electron and Neutron diffraction
Electron diffraction
Neutron diffraction
Suitable for study only thin films
and crystals, surfaces because
of strong interaction of the electrons
with atoms and electrons in
solids
LEED –Low Energy Elastic Diffraction:
probes structure of 2D surface of solids.
Surface should be free of absorbed atoms
Wavelength of the electron l(Å)=
=12/ÖEe(eV), l=1Å for Ee=150eV
(l=h/p, Ee=p2/2m)
Weak interaction with
charged particles, but strong
interaction with magnetic nuclei
and electronic magnetisation
Wavelength of the neutron l(Å)=
=0.28/ÖEe(eV), l=1Å for Ee=0.08eV
(l=h/p, Ee=p2/2M) Fe
30o
70o
50o

11. Bragg’s Law
The figure above shows x-rays which are specularly reflected from adjacent planes.
The path difference between the two x-rays is equal to 2dsinq.
For constructive inteference this difference must be an integer number of wavelengths.
This leads to the the Bragg condition:
The number m is known as the order of the corresponding reflection (or order of inteference).

12. Deriving Bragg’s Law
Bragg's Law can easily be derived by considering the conditions necessary to make the phases of the beams coincide when the incident angle is equal to the reflecting angle. The rays of the incident beam are always in phase and parallel up to the point at which the top beam strikes the top layer at atom z (Fig). The second beam continues to the next layer where it is scattered by atom B. The second beam must travel the extra distance AB + BC if the two beams are to continue traveling adjacent and parallel. This extra distance must be an integral (n) multiple of the wavelength (l) for the phases of the two beams to be the same:
nl = AB +BC (2).

13. Deriving Bragg’s Law
Recognizing d as the hypotenuse of the right triangle AbZ, we can use trigonometry to relate d and q to the distance (AB + BC). The distance AB is opposite q so,
AB = d sinq (3).
Because AB = BC and nl = AB +BC becomes,
nl = 2AB (4)
Substituting eq. (3) in eq. (4) we have,
nl = 2 d sinq, (1)
and Bragg's Law has been derived. The location of the surface does not change the derivation of Bragg's Law.

14. Bragg’s Law
There are a number of various setups for studying crystal structure using x-ray diffraction. In most cases, the wavelength is fixed, and the angle is varied to observe diffraction peaks corresponding to reflections from different crystallographic planes.
Using Bragg law, the distance between planes can be determined.
Sharp x-ray diffraction peaks are observed only in the directions and wavelengths for which the x-rays scattered from all lattice points interferes constructively.

15. Diffraction Condition and Reciprocal Lattice
To find condition of constructive interference two scatterers separated by a lattice vector are considered.
X-rays are incident from infinity, along direction with wavelength and wavevector
Assume the scattering is elastic i.e the x-rays are scattered in direction with same wavelength , so that the wavevector .
The path difference between the x-ray scattered from the two atoms should be an integer number wavelengths. Therefore, the condition for constructive interference is
Where m is an integer. Multiplying both sides of the above equation by leads to a condition on the incident and scattered wave vectors

16. Diffraction Condition and Reciprocal Lattice
Defining the scattering wave vector , the diffraction condition can be written as
Where is, a vector where
A set of vectors satisfying this condition form a reciprocal lattice. Vectors are called reciprocal lattice vectors.
A reciprocal lattice is defined with reference to a particular Bravais lattice which is determined by a set of lattice vectors .
The Bravais lattice that determines a particular reciprocal lattice is referred to as the direct lattice, when viewed in relation to its reciprocal. .

17. Algorithm for determining reciprocal of direct lattice
Let and be a set of primitive vectors of the direct lattice. The reciprocal lattice that can be generated from the primitive vectors are
Where is the volume of a unit cell so that .

18. To proof the relation between direct and reciprocal lattice is true
To show that satisfies the equation

19. 1D and 2-D: Direct to Reciprocal Lattice
For 1-D and 2-D cases use equation
For 3-D case use

20. 3-D Direct Lattice to Reciprocal Lattice
From the condition: it can be said that constructive interference occurs provided that the scattering wave vector is a vector of the reciprocal lattice.

21. Bragg’s Law using Reciprocal Lattice Vector
In elastic scattering the photon energy is conserved, so the magnitudes of k and k’ are equal, therefore k2 =k’2. There from
This another way of stating Bragg’s Law.

22. To Show: Reciprocal Lattice Vector is Orthogonal to the plane (hkl)
Definition of Miller Indices
If G is normal to the plane (hkl)
Therefore G is orthogonal to the plane (hkl).

23. To show: The distance between two adjacent parallel planes of direct lattice is d = 2p/G
Note: The nearest plane which is parallel to (hkl) goes through the origin of the Cartesian coordinates.
Therefore the interplanar distance is given by the projection of one of the vectors xa1, ya2, za3 to the direction normal to the (hkl) plane.
The direction is given by unit vector G/G, since G is normal to the plane. Therefore

24. To show: is equivalent to Bragg Law
From :
This angle is opposite to the angle between k and G.
Therefore,
k – same direction as X-ray beam G q
Plane