x-rays physics

8/13/2019 X-rays physics

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X-ray Diffraction


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X-ray Generation

X-ray tube (sealed) Pure metal target (Cu)

Electrons removerinner-shell electronsfrom target.

Other electrons fallinto hole.


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X-ray Generation

The incoming electronmust have enough energy

to remove inner 1selectrons from the copperatoms.

This energy correspondsto the Cu absorption edge

The 2sand 2pelectronsfall back into the 1s shelland emit the K1K2

lines.


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X-ray Spectrum from Tube


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Energy Calculations

Plancks constant (h) = 6.6 * 10-34

joule-sec 1 electron-volt = 1.6016 * 10-19joule

Speed of light (c) = 3.0 * 108

m/s Photon frequency= c/ Photon Energy E = h= hc/


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Energy Calculations

What is the minimum potential in KV that isrequired to excite Cu K-series radiation from a Cu-

target X-ray tube? Absorption edge of Cu = 1.380

E = hc/= (6.60 10-34)(3*108)/(1.380*10-10) E = 1.435*10-15joule

E = 1.435*10-15/1.6016*10-19 = 8958 ev The potential on the tube must exceed 8.958 KV


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Diffraction

Diffraction is the coherent scattering of waves

from a periodic array of scatterers. The wavelength of light is about half a micron

Light is diffracted by the tracks in a CD. The wavelengths of X-rays is about the same

as the interatomic distances in crystals. Crystals diffract X-rays.


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X-Ray Diffraction

Atoms separated bydistancedwill scatter inphase when the path length

difference is an integralnumber of wavelengths.

Path length difference B-C-

D = n n= 2dsin


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X-ray Diffraction Experiment

We use the monochromatic K1-2lines forour diffraction experiment.

The wavelength is 1.5405

We use a diffracted beam monochromatorto clean up the X-rays entering the detector.

We use a powdered sample so that allorientations are present in the sample.

We move the detector through angle 2.


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Miller Indices

The real use of Miller indicesis to describe diffractionplanes.

For a lattice plane with Millerindicesh klin anorthorhombic latticea b c,

d = 1 / [(h/a)2+(k/b)2+(l/c)2]1/2

For cubic:

d = a/[h2+k2+l2]1/2


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Diffraction Calculations

For forsteritea= 4.75;b= 10.20; c = 5.98

Calculate 2for the (201) lattice spacing for Cuk(=1.5405) d = 1 / [(h/a)2+(k/b)2+(l/c)2]1/2

d = 1/ [(2/4.75)2+(1/5.98)2]1/2

d= 1/0.4530 = 2.207

2= 2 sin-1/2d =2* sin-1(1.5405/4.414)

2= 2 * 20.43 = 40.86


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XPOW

XPOW uses the unit cell and atom position data tocalculate the diffraction pattern.

Intensities can be calculated knowing the positionand scattering characteristics of each atom.

Fhkl= square root of integrated intensity. fj= scattering of atomjat angle 2

Atomjlocated at fractional coordinatesxj, yj, zj.

=

++=

n

j

lzkyhxi

jhkl

jjjefF1

)(2


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Uses of X-ray Powder Diffraction

Mineral identification Determination of Unit Cell Parameters

Modal (phase percentage) Analysis Crystal Structure Determination

x-rays physics

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