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X- Generation X-ray • X-ray tube (sealed) • Pure metal target (Cu) • remove inner-shell electrons Mineral identification from target. Mode analysis • Other electrons “fall” Structure Studies into hole.

X-ray from Tube X-ray Generation

• The incoming must have enough energy to remove inner 1s electrons from the copper . • This energy corresponds to the Cu absorption edge • The 2s and 2p electrons fall back into the 1s shell and emit the Kα1 Kα2 lines.

Energy Calculations Energy Calculations • What is the minimum potential in KV that is required to excite Cu K-series • Planck’s constant (h) = 6.6 * 10-34 joule-sec from a Cu-target X-ray tube? • Absorption edge of Cu = 1.380Å • 1 electron-volt = 1.6016 * 10-19 λ joule • E = hc/ = (6.60 10-34)(3*108)/(1.380*10-10) -15 • Speed of (c) = 3.0 * 108 m/s • E = 1.435*10 joule -15 -19 • ν = c/λ • E = 1.435*10 /1.6016*10 = 8958 ev • E = hν = hc/λ • The potential on the tube must exceed 8.958 KV

1 Diffraction X-Ray Diffraction

• Diffraction is the coherent of from a periodic array of scatterers. • The of light is about half a micron • Atoms separated by distance d will scatter • Light is diffracted by the tracks in a in when the path length difference is CD. an integral number of . • The wavelengths of X-rays is about • Path length difference B-C-D = nλ the same as the interatomic distances • nλ = 2d sin θ in crystals.

X-ray Diffraction Experiment

• We use the ‘monochromatic’ Kα1-2 lines for our diffraction experiment. • The wavelength is 1.5405Å • We use a diffracted beam monochro- mator to clean up the X-rays entering the detector. • We use a powdered sample so that all orientations are present in the sample. • We move the detector through angle 2θ.

2 Miller Indices

• The real use of Miller indices is to describe diffraction planes. • For a lattice plane with Miller indices h k l in an orthorhombic lattice a b c, • d = 1 / [(h/a)2+(k/b)2+(l/c)2]1/2 • For cubic: • d = a/[h2+k2+l2]1/2

Diffraction Calculations XPOW • XPOW uses the unit cell and position data to • For forsterite a = 4.75; b = 10.20; c = calculate the diffraction pattern. α=β=γ 5.98Å orthorhombic = 90º • Intensities can be calculated knowing the position • Calculate 2θ for the (201) lattice and scattering characteristics of each atom. spacing for Cukα (λ = 1.5405Å) • Fhkl = square root of integrated intensity. • d = 1 / [(h/a)2+(k/b)2+(l/c)2]1/2 • fj = scattering of atom j at angle 2θ • d = 1/ [(2/4.75)2+(1/5.98)2]1/2 •Atomj located at fractional coordinates xj, yj, zj. • d = 1/0.4530 = 2.207Å • 2θ = 2 sin-1 λ/2d = 2* sin-1 (1.5405/4.414) • 2θ = 2 * 20.43 = 40.86º

Uses of X-ray

• Mineral identification • Determination of Unit Cell Parameters • Modal (phase percentage) Analysis • Crystal Structure Determination

3 X-ray

Electron Microprobe X-ray Fluorescence

• Chemical analysis • Major and minor element • Uses Ag kα to excite secondary X-rays from sample. • Powdered or -fused glass sample.

Electron Microprobe Electron Microprobe

• Quantitative Chemical analysis • Major and minor element • Uses electrons to excite secondary X- rays from sample. • Electrons can be focussed onto a 10μm spot • Sample is polished thin section

4 Mineral

(Mössbauer Spectroscopy) • X-ray Fluorescence • Raman spectroscopy • Optical absorbance spectroscopy • Spectroscopy • NMR (Nuclear Magnetic Resonance) Visible Light: 7700 - 3900Å

Mössbauer Spectroscopy Mössbauer spectroscopy • Resonant Gamma Ray spectroscopy Looks at γ-ray interaction at • Uses 57Fe gamma decay at 14.4 KeV nucleus of 57Fe (2%) • Source is 57Co (291 days) • Source is accelerated mechanically to produce ultra-fine doppler energy shifts • Absorption as a function of source velocity • Looks at effects at nucleus due to d-orbital occupancy and perturbations from local coordination effects

Mössbauer spectroscopy Mössbauer spectroscopy

5 FTIR Spectrometer

• Near IR 5000 - 13000cm-1 – orbital transitions • Mid-IR 2500 - 5000cm-1 – N-H and O-H bond vibrations • Far IR 500 - 2500 cm-1 – Cation-Oxygen bond vibrations – Structural .

Mid IR spectroscopy Mid IR spectroscopy

Raman Spectroscopy Raman Spectroscopy

• Looks at wavelength shifts in scattered light. • Shifts are in atomic vibrational part of spectrum • 0 - 5000cm-1. (same as mid to far IR) • Excitation is usually by a monochromatic source in the visible region (commonly a ).

6 Electromagnetic Spectrum

Visible Light: 7700 - 3900Å

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