Analisis Difraksi Sinar- X

Mengapa difraksi sinar-x?Tujuan menggunakan difraksi sinar-xDifraktometer yang mana?Kondisi eksperimenPemrosesan dataAlat analisis yang tepat

Tujuan memakai difraktometerIdentifikasi fasaPenentuan komposisiParameter kisiRegangan-ukuran kristalPenentuan struktur kristal

TeganganDepth profilingPenentuan koefisien ekspansi termalFasa lapisan tipisPreferred orientation kristal

Kondisi EksperimenAmbienSuhu tinggi/rendahTekanan tinggi/rendahMedan magnetTegangan (stress)dll.

Analisis yang tepatApa tujuannya?Apa alat analisisnya?Mudah/sukar aksesnya?PerformaInformasi seputar pemakaian

DS=Divergence SlitSS=Scatter SlitRS= Receiving SlitRSM= Monochromator Receiving SlitLeft-Hand Side (250mm radius) of the Rigaku DiffractometerRSM

Hukum Bragg adalah model sederhana untuk memahami difraksi sinar-XJarak antara bidang-bidang difraksi dari atom menentukan posisi-posisi puncakPuncak-puncak intensitas ditentukan oleh atom-atom dalam bidang difraksi

ChE 440 - 00F

powder diffractometersqw2qX-ray tubeDetector

Kristal tunggal akan menghasilkan hanya satu jenis puncak dalam pola difraksi2qAt 20.6 2q, Braggs law fulfilled for the (100) planes, producing a diffraction peak.The (110) planes would diffract at 29.3 2q; however, they are not properly aligned to produce a diffraction peak (the perpendicular to those planes does not bisect the incident and diffracted beams). Only background is observed.The (200) planes are parallel to the (100) planes. Therefore, they also diffract for this crystal. Since d200 is d100, they appear at 42 2q.

Sampel polikristal mengandung banyak polikristalin. Oleh karena itu semua pola-pola difraksi akan terobservasi2q2q2qFor every set of planes, there will be a small percentage of crystallites that are properly oriented to diffract (the plane perpendicular bisects the incident and diffracted beams). Basic assumptions of powder diffraction are that for every set of planes there is an equal number of crystallites that will diffract and that there is a statistically relevant number of crystallites, not just one or two.

*constructive interference occurs when the path difference is ndestructive interference

*Diffraction from a crystal Constructive interference occurs when the path length the waves travel differs by some multiple of the wavelength:is the spacing between the planes of atomsFor diffraction to occur:hence visible light does not diffract from crystals

In these set of slides we shall consider: scattering from an electron scattering from an atom structure factor calculations (scattering from an unit cell) the relative intensity of reflections in power patternsElements of X-Ray DiffractionB.D. Cullity & S.R. StockPrentice Hall, Upper Saddle River (2001)

Intensity of the Scattered electronsElectronAtomUnit cell (uc)Scattering by a crystalABCPolarization factorAtomic scattering factor (f)Structure factor (F)Click here to jump to structure factor calculations

Structure FactorVector summation of E1 and E2

Structure FactorNotes: 11/9/99pg. 1pg. 2Notes: 11/11/99pg. 1pg. 2pg. 3

If atom B is different from atom A the amplitudes must be weighed by the respective atomic scattering factors (f)The resultant amplitude of all the waves scattered by all the atoms in the UC gives the scattering factor for the unit cellThe unit cell scattering factor is called the Structure Factor (F)Scattering by an unit cell = f(position of the atoms, atomic scattering factors)In complex notationStructure factor is independent of the shape and size of the unit cell!F FhklFor n atoms in the UCIf the UC distorts so do the planes in it!!Note:n is an integer

Structure factor calculationsAAtom at (0,0,0) and equivalent positions F is independent of the scattering plane (h k l)Simple Cubic All reflections are present

BAtom at (0,0,0) & (, , 0) and equivalent positions F is independent of the l indexC- centred OrthorhombicRealBoth even or both oddMixture of odd and evene.g. (001), (110), (112); (021), (022), (023)e.g. (100), (101), (102); (031), (032), (033)(h + k) even(h + k) odd

CAtom at (0,0,0) & (, , ) and equivalent positionsBody centred OrthorhombicReal(h + k + l) even(h + k + l) odde.g. (110), (200), (211); (220), (022), (310)e.g. (100), (001), (111); (210), (032), (133)This implies that (h+k+l) even reflections are only present.The situation is identical in BCC crystals as well.

DAtom at (0,0,0) & (, , 0) and equivalent positionsFace Centred CubicReal(h, k, l) unmixed(h, k, l) mixede.g. (111), (200), (220), (333), (420)e.g. (100), (211); (210), (032), (033)(, , 0), (, 0, ), (0, , )Two odd and one even (e.g. 112); two even and one odd (e.g. 122)h,k,l all even or all odd

(h, k, l) mixede.g. (100), (211); (210), (032), (033)Mixed indicesTwo odd and one even (e.g. 112); two even and one odd (e.g. 122)Unmixed indices(h, k, l) unmixede.g. (111), (200), (220), (333), (420)All odd (e.g. 111); all even (e.g. 222)This implies that in FCConly h,k,l unmixed reflections are present.

Mixed indicesCASEhklAooeBoee

Unmixed indicesCASEhklAoooBeee

ENa+ at (0,0,0) + Face Centering Translations (, , 0), (, 0, ), (0, , ) Cl at (, 0, 0) + FCT (0, , 0), (0, 0, ), (, , )NaCl: Face Centred Cubic

Zero for mixed indices(h, k, l) mixede.g. (100), (211); (210), (032), (033)Mixed indices

Mixed indicesCASEhklAooeBoee

(h, k, l) unmixedIf (h + k + l) is evenIf (h + k + l) is odde.g. (111), (222); (133), (244)e.g. (222),(244)e.g. (111), (133)Unmixed indicesh,k,l all even or all odd

Unmixed indicesCASEhklAoooBeee

FAl at (0, 0, 0) Ni at (, , )NiAl: Simple Cubic (B2- ordered structure)SCReal(h + k + l) even(h + k + l) odde.g. (110), (200), (211), (220), (310)e.g. (100), (111), (210), (032), (133)Click here to know more about ordered structuresWhen the central atom is identical to the corner ones we have the BCC case.This implies that (h+k+l) even reflections are only present in BCC.This term is zero for BCC

Reciprocal lattice/crystal of NiAle.g. (110), (200), (211); (220), (310)e.g. (100), (111), (210), (032), (133)Click here to know more about

GAl Atom at (0,0,0) Ni atom at (, , 0) and equivalent positionsSimple Cubic (L12 ordered structure)Real(h, k, l) unmixed(h, k, l) mixede.g. (111), (200), (220), (333), (420)e.g. (100), (211); (210), (032), (033)(, , 0), (, 0, ), (0, , )Two odd and one even (e.g. 112); two even and one odd (e.g. 122)NiAlh,k,l all even or all oddClick here to know more about ordered structures

Selection / Extinction Rules

Bravais LatticeReflections which may be presentReflections necessarily absentSimpleallNoneBody centred(h + k + l) even(h + k + l) oddFace centredh, k and l unmixedh, k and l mixedEnd centredh and k unmixed C centredh and k mixed C centred

Bravais LatticeAllowed ReflectionsSCAllBCC(h + k + l) evenFCCh, k and l unmixedDCh, k and l are all odd Or all are even & (h + k + l) divisible by 4

h2 + k2 + l2SCFCCBCCDC110021101103111111111420020020052106211211782202202202209300, 2211031031011311311311122222222221332014321321151640040040040017410, 32218411, 330411, 33019331331331

*Identifying structure from a powder pattern (-2) measurementA typical x-ray trace. We see several peaks. In order to identify the structure we have to know the wavelength of radiation and the angles (half the 2 values).Cu= 0.154 nm

*CsCl diffraction: effect of the source on -2 scansCuK=0.154 nmCrK=0.229 nm20120

*Identifying structure from a powder pattern (-2) measurement- A typical XRD trace... X-ray intensity v. 2Cu= 1.54 - In order to identify the structure of the material we need to know the wavelength of radiation and the angles (i.e., half the 2 values).- Several peaks - diagnostic of the material - Each corresponds to an {hkl} for which Braggs Law is satisfied.2dsinq = nl

*Cu= 1.54 2dsinq = nl- assume an order of refraction (n) of 1.- Once you have the angles for each reflection, you can calculate the d values. - Remember were measuring 2q but were using q !d = interplanar spacings for each allowed h, k, lAllowed ???

*- The set of all possible hkl that will actually diffract is limited by the particular Bravais Lattice and the atomic positions.Additional Considerations:The BCC and FCC cubic space lattices have more than one atom per unit cell and there are extra interferences that can occur off of these atoms.- Consider the cubic structures- i.e., not all reflections will be observed in an XRD experiment- This leads to the cancellation of peaks that youd otherwise expect to be there.- Only certain reflections are allowed.

*Allowed Reflections for Cubic Space LatticesRecall

EQ \F(1,d2hkl) = EQ \F((h2+k2+l2),a2)

*Allowable Reflections

*d = interplanar spacings for each allowed {hkl}Back to the story:- Weve calculated the d spacings for each value of q... We want to get a. - Now we need to determine the type of crystal structure (i.e., SC, BCC, or FCC) and formulate a set of Miller Indices to assign to each line in the data set.- Interplanar spacing d is related to Miller Indices h, k, and l and the lattice parameters by:- Were assuming a cubic structure. Recall thatSo

EQ \F(1,d2hkl) = EQ \F((h2+k2+l2),a2)

*2dsinq = nland So for for two different {hkl}s: (e.g., h1k1l1 and h2k2l2)So d2h1k1l1d2h2k2l2=(h22 + k22 + l22)sin2q1sin2q2(h12 + k12 + l12)=So you calculate sin2q values from each value of q obtained from the XRD traceThen the ratio of the sin2q values should equal the ratio of the (h22 + k22 + l22)values (that we determined earlier) for allowable reflections for each structure. (h22 + k22 + l22)sin2q1sin2q2(h12 + k12 + l12)=Once we find a structure, we can assign the correct Miller Indices to each peak in the data. We can then determine the structure (i.e., SC, BCC, or FCC).

EQ \F(1,d2hkl) = EQ \F((h2+k2+l2),a2)

*So from beforeratioPlane (h2 + k2 + l2)SCBCFCC(100)11--(110)222-(111)33-3(200)4444(210)55--(211)666-(220)8888(300)99--(310)101010-(311)1111-11(222)12121212(320)1313--(321)141414-(400)16161616(331)1919-19(h22 + k22 + l22)sin2q1sin2q2(h12 + k12 + l12)= The ratios of sin2q values are consistent with the ratios of the allowed (h22 + k22 + l22) for the FCC structure only0.750.500.730.920.750.840.500.670.750.500.670.750.80

*- We can now assign Miller Indices to each peak in the XRD trace(111)(200)(220)(311)(222)(400)(331)Plane (h2 + k2 + l2)FCC(100)1-(110)2-(111)33(200)44(210)5-(211)6-(220)88(300)9-(310)10-(311)1111(222)1212(320)13-(321)14-(400)1616(331)1919- Once a structure is assigned, we can go back and determine the lattice parameter for this material

*- Once a structure is found, we can go back and determine the lattice parameter for this material:- We can determine the lattice parameter for each line and average to get the most accurate measurement...The FCC structure and the lattice parameter identify this metal as Ni.ora2 = d2hkl(h2 + k2 + l2)aavg = 3.522

EQ \F(1,d2hkl) = EQ \F((h2+k2+l2),a2)

*

*That was the description of perfect crystalsWhat are real crystals and materials like?In real crystals there are errors in the crystal:Point errors missing atoms, extra atoms, impurities, Line errorsdislocations (columns of missing atoms)Planar errorsstacking faults (missing or extra planes of atoms)These types of defects will; give rise to the processes of Solid-state diffusion, phase transformations, strength of materials, the mechanical behavior of materials

**draw the diffraction vector on this slide, or make a second slide explicitly illustrating the diffraction vector**

Where there are N atoms that scatter per unit cell