MENA 3100 32 10 OBK Xray diffraction XRD

MENA 3100, 3/2 -10, OBK X-ray diffraction XRD Røntgendiffraksjon Single crystal Powder

Diffraction

Crystal Atomic arrangement Lattice

Lattice + base The 14 Bravais lattices 1848

Bragg’s law 2 d sinq = nl

Bragg’s law 2 d sinq = nl n=2 2 nd. order But XRD is typically done with monochromatic radiation

Bragg’s law 2 d sinq = nl 2 nd. order from 001 1 th. order from 002

The 14 Bravais lattices Face centered cubic lattice g-iron Na. Cl g-brass

Extinction Utslukning Example: Face centered cubic Special example: Cubic close packed structure

Extinction Utslukning Example: Face centered cubic Special example: Cubic close packed structure

Don’t confuse with. . .

X-ray diffraction Need X-rays Wavelenght Cu. Ka 1 = 1. 54056 Å Mo. Ka 1 = 0. 70930 Å Characteristic X-rays Brehmsstrahlung = white radiation Cu. Ka 1 = 1. 54056 Å Cu. Ka 2 = 1. 54433 Å I(Cu. Ka 1) = 2·I(Cu. Ka 2) Cu. Ka = 1. 5418 Å

Need sample Single crystal Powder

Need experimental setup Bragg-Brentano geometry (Primary monochromator) Diffractometer (Secondary monochromator)

Want monochromatic radiation Characteristic X-rays Brehmsstrahlung = white radiation Monochromators: Many possibilities

Euclid, The Elements (300 B. C. ) Proposition 21, Book III: The angles in the same segment of a circle are equal to another Johansson monochromator

Bragg-Brentano geometry The source S can be:

Need detector … Need some slits … Fixed slits: Smaler area at high angle Smaller area at high angles Variable slits: Same area at all angles S Sample

The experiment Got some crystals Want a powder Gets lot of small crystals

Want powder Random orientation

The result The diffractogram Quartz (Don’t call it spectrum)

Why?

Identification PDF = JCPDS

Indexing

Lattice dimensions

Strain Particle size

Fluorescens

Cu. Ka 1 = 1. 54056 Å Cu. Ka 2 = 1. 54433 Å Primary and secondary monochromators

Rietveld refinement