X-Ray Diffraction HOW IT WORKS

WHAT IT CAN AND WHAT IT CANNOT TELL US

Hanno zur Loye

X-rays are electromagnetic radiation of wavelength about 1 Å (10-10 m), which is about the same size as an atom.

The discovery of X-rays in 1895 enabled scientists to probe crystalline structure ���at the atomic level. X-ray diffraction has been in use in two main areas, for the ���fingerprint characterization of crystalline materials and the determination of their ���structure. Each crystalline solid has its unique characteristic X-ray powder pattern which may be used as a "fingerprint" for its identification. Once the material has ���been identified, X-ray crystallography may be used to determine its structure, i.e. ���how the atoms pack together in the crystalline state and what the interatomic distance ���and angle are etc. X-ray diffraction is one of the most important characterization ���tools used in solid state chemistry and materials science. We can determine the size and the shape of the unit cell for any compound most ���easily using X-ray diffraction.

X-ray Diffraction

­ Structural Analysis ¬ X-ray diffraction provides most definitive

structural information ¬ Interatomic distances and bond angles

­ X-rays ¬ To provide information about structures we

need to probe atomic distances - this requires a probe wavelength of 1 x 10-10 m ~Angstroms

Production of X-rays

­ X-rays are produced by bombarding a metal target (Cu, Mo usually) with a beam of electrons emitted from a hot filament (often tungsten). The incident beam will ionize electrons from the K-shell (1s) of the target atom and X-rays are emitted as the resultant vacancies are filled by electrons dropping down from the L (2P) or M (3p) levels. This gives rise to Ka and Kb lines.

X-rays

Cu, Mo target

20 - 50 kV e -

1s

2p

3p

K

L

M electromagnetic radiation

Cu Kα = 1.5418 Å Mo Kα = 0.7107 Å

X-ray Generation

­ Broad background is called Bremsstrahlung. Electrons are slowed down and loose energy in the form of X-rays

X-ray Production ­ As the atomic number Z of the target element

increases, the energy of the characteristic emission increases and the wavelength decreases.

­ Moseley’s Law (c/l)1/2 ∝ Z ¬ Cu Ka = 1.54178 Å ¬ Mo Ka = 0.71069 Å

­ We can select a monochromatic beam of one wavelength by: ¬ Crystal monochromator Bragg equation ¬ Filter - use element (Z-1) or (Z-2), i.e. Ni for Copper

and Zr for Molybdenum.

X-ray Generation

Brookhaven National Synchrotron Light Source

Single Crystal Diffraction������

A single crystal at random orientations and its corresponding diffraction pattern. Just as the crystal is rotated by a random angle, the diffraction pattern

calculated for this crystal is rotated by the same angle

A 'powder' composed from 4 single crystals in random orientation (left) and the corresponding diffraction pattern (middle). The individual diffraction patterns plotted in the same color as the corresponding crystal start to add up to rings of reflections. With just four reflection its difficult though to recognize the rings. The right image shows a diffraction pattern of 40 single crystal grains (black). The colored spots are the peaks from the 4 grain 'powder' shown in the middle image.

As we have more grains, the ���diffraction pattern looks more ���and more continuous and we ���get the expected powder���pattern shown on the left.

Diffraction from one���single crystal

Diffraction from several randomly���oriented single crystals (powder)

Sample of hundreds of randomly���oriented single crystals (powder)���and film used to collect the data.

Our XRD instruments

Crystal Systems

Axis Axis Angles1 Cubic A = B = C α = β = γ = 902 Tetragonal A = B ≠ C α = β = γ = 903 Orthorhombic A ≠ B ≠ C α = β = γ = 904a Hexagonal A = B ≠ C α = β = 90, γ = 1204b Rhombohedral A = B = C α = β = γ ≠ 90 < 1205 Monoclinic A ≠ B ≠ C α = γ = 90, β > 906 Triclinic A ≠ B ≠ C α ≠ β ≠ γ ≠ 90

The higher the symmetry, the easier it is to index the pattern���and the fewer lines there are in the pattern.

Miller Indices

O

a

c

b

a/2

c/3

The origin is 0, 0, 0

If we drew a third plane, it would pass through the origin.

(1 0 0) (2 0 0) (3 0 0)

The plane cuts the ���x-axis at a/2, the ���b-axis at b, and the ���c-axis at c/3. The the���reciprocals of the ���fractions, (2 1 3), which���are the miller indices.���A plane that is parallel ���to an axis will intersect ���that axis at infinity. One ���over infinity = 0. I.e. the���(100) plane. Parallel to ���b and c and intersects a ���at 1.

The Bragg Equation

Reflection of X-rays from two planes of atoms in a solid. ���x = dsin��

The path difference between two waves: 2 x wavelength = 2dsin(theta) Bragg Equation: nλ = 2dsinθ

d-spacing in different crystal systems

­ Cubic

­ Tetragonal

­ Orthorhombic

­ Hexagonal

­ Monoclinic

­ Triclinic -

�

1d2

= h2 + k 2 + l2

a2

�

1d2

= h2 + k 2

a2+ l2

c 2

�

1d2

= h2

a2+ k 2

b2+ l2

c 2

�

1d2

= 43h2 + hk + k 2

a2⎛

⎝ ⎜

⎞

⎠ ⎟ +

l2

c 2

�

1d2

= 1sin2 β

h2

a2+ k 2 sin2 β

b2+ l2

c 2− 2hlcosβ

ac⎛

⎝ ⎜

⎞

⎠ ⎟

Diffraction pattern

•  Intensity (I) is the total area under a peak

I

2θ (deg)

Indexing Patterns

­ Indexing is the process of determining the unit cell dimensions from the peak positions ¬ Manual indexing (time consuming...but still

useful) ¬ Pattern matching/auto indexing (JADE or other

computer based indexing software)

Cubic pattern

Relationship between diffraction peaks, miller indices and lattice spacings

Simple cubic material a = 5.0 Å

hkl d(Å) 2Θ100 5.00 17.72110 3.54 25.15111 2.89 30.94

Use 1 = h2 + k2 +l2 , and nλ = 2dsinΘ d2 a2 b2 c2

How many lattice planes are possible?���How many d-spacings? The number is large but finite.���nλ = 2dsinθ so if theta = 180, then d = λ/2. For Cu radiation that means that we can only see d-spacings down to 0.77 Å ���for Mo radiation, down to about 0.35 Å

�

1d2

= h2 + k 2 + l2

a2

Bragg Equation: nλ = 2dsinθ

Tetragonal pattern

�

1d2

= h2 + k 2

a2+ l2

c 2

�

sin2θ = λ24

h 2 +k 2

a 2+ l 2

c 2[ ]

Orthorhombic

a ≠ b ≠ c, all angles are 90° Multiplicity is further decreased as the symmetry decreases.

�

1d2

= h2

a2+ k 2

b2+ l2

c 2

�

sin2θ = λ24

h 2

a 2+ k 2

b 2+ l 2

c 2[ ]

Systematic Absences c c' c cc'

a = 2πn

2πn

a = 2πn

Cesium Metal

c c c

c'c'

In order to see diffraction from the (100) plane, the phase���difference must be a multiple of 2p. However, the c and c’���planes are out of phase. Therefore - cancellation of the diffracted ���peak. The (200) plane, however, does not have this problem.

Body Centered Cubic Structure

What Information Do We Get or Can We Get From Powder X-ray Diffraction

­ Lattice parameters ­ Phase identity ­ Phase purity ­ Crystallinity ­ Crystal structure ­ Percent phase composition

What Information Do We NOT Get From Powder X-ray Diffraction

­ Elemental analysis - ¬ How much lithium is in this sample? ¬ Is there iron in this sample ¬ What elements are in this sample

­ Tell me what this sample is ???? ¬ Unless you know something about this sample,���

powder XRD won’t have answers !!!

Powder Preparation

­ It needs to be a powder ­ It needs to be a pure powder ­ Its nice to have about 1/2 g of sample, but

one can work with less ­ The powder needs to be packed tightly in

the sample holder. Lose powders will give poor intensities.

Data Collection

­ The scattering intensity drops as 1/2(1+cos22θ) ­ This means that you don’t get much intensity

past 70 ° 2θ. A good range is 10-70 ° 2θ. ­ How long should you collect (time per step)? ­ Depends on what you want to do!

¬ Routine analysis may only take 30-60 min. ¬ Data for Rietveld analysis may take 12-18 hours to

collect

Rietveld Refinement

• employs a least squares matching algorithm • refinement based on sample parameters and instrumental parameters

Data Analysis

­ If you are trying to confirm that you have made a known material, do a search/match using the JCPDS data base. They have about 100,000 patterns on file.

­ For a new material, you need to index the pattern. Unit cell, lattice parameters and symmetry.

Data Bases

Preferred Orientation

The top image shows 200 random crystallites. The���bottom picture shows 200 oriented crystallites. Despite ���the identical number of reflections,several powder lines ���are completely missing and the intensity of other lines is ���very misleading. Preferred orientation can substantially ���alter the appearance of the powder pattern. It is a serious problem in experimental powder diffraction.

Structure Refinement: ���The Rietveld Method