Chapter 3 X-ray Diffraction Bragg’s Law Laue’s .

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Chapter 3 X-ray diffraction Bragg’s law Laue’s condition Equivalence of Bragg’s law and Laue’s condition Ewald construction geometrical structure factor1

Bragg’s lawConsider a crystal as made out of parallel planes of ions, spaced a distance d apart. The conditions for asharp peak in the intensity of the scattered radiation are1.2.That the x-rays should be specularly reflected by the ions in any one plane andThat the reflected rays from successive planes should interfere constructivelyPath difference between two rays reflected fromadjoining planes:2d sin θFor the rays to interfere constructively, this pathdifference must be an integral number of wavelengthλnλ 2d sin θBragg’s condition.2

Bragg angleθis just the half of the total angle2θbywhich the incident beam is deflected.There are different ways of sectioning the crystal into planes, each of which will it self produce further reflection.The same portion of Bravais lattice shown in theprevious page, with a different way of sectioning thecrystal planes. The incident ray is the same. But boththe direction and wavelength (determined by Braggcondition with d replaced by d’) of the reflected ray aredifferent from the previous page.3

Von Laue formulation of X-ray diffraction by a crystal No particular sectioning of crystal planesr Regard the crystal as composed of identical microscopic objects placed at Bravais lattice site R Each of the object at lattice site reradiate the incident radiation in all directions. Diffraction peaks will be observed in directions that the rays scattered from all lattice points interfere constructivelyIncident x-ray:n̂along directionλwavelengthwave vectorA scattered wave:r 2πk nˆdirectionwavelengthwave vectorλnˆ ′λr 2πk′ nˆ ′λrd cos θ d cos θ ′ d (nˆ nˆ ′)Condition for constructive interferenceMultiply 2πr r rd ( k k ′) 2πmrd (nˆ nˆ′) mλfor integer m4

The conditionr r rd (k k ′) 2πmrdr r rR ( k k ′) 2πmsoOr equivalentlyethereforeorholds for all possibler v ri ( k ′ k ) R 1r rk′ kis a reciprocal lattice vectorr r rG k′ kLaue condition:Constructive interference will occur provided that the change in wave vectorlatticer r rG k′ kis a vector of reciprocalAlternative formulation of Laue conditionThis formulation involves only incident wave vectorSincer rk′ krkis a reciprocal lattice vector, so isr r rk k′ G, and does not involver rk k′rk′r r rk′ k Gr rr r′k k k GSquaring both sidesr rG 2k G2r rk k G 2k G2orrrGG k 2G225

rrGG k 2Gis interpreted as:The component of rthe incident wave vectorof the length ofGrkalong the reciprocal lattice vector1 rG2rGmust be half1 rG2rGAn incident wave vector will satisfy the Laue condition if and only if the tip of the vector lies in a plane that is theperpendicular bisector of a line joining the origin of the reciprocal space to a reciprocal lattice point.Such k-space planes are called Bragg planes.6

Equivalence of the Bragg and von Laue formulationsrrSuppose the incident and scattered wave vectors k and k ′ , satisfy the Laue conditionthatr r rG k′ kr r rG k′ kr rk k′Elastic scattering:It follows thatbe a reciprocal lattice vectorrk′andrkmake the same angleθwith therplane perpendicular to G . Therefore the scattering can beviewed as a Bragg reflection with Bragg angleθ, from thefamily of direct lattice planes perpendicular to the reciprocallattice vectorrG.The distance between successive planes in this family mustsatisfy:r2πG0 drrG0 is the shortest wave vector parallel to GrrGG must be an integral multiple of0 , since reciprocal lattice is a Bravais latticewhereFrom the figure,rrG nG0rr2πnG n G0 drG 2k sin θ7

k sin θ Note thatk πndnλ 2d sin θ2πBragg conditionλA Laue diffraction peak corresponding to a change in the wave vector given by the reciprocal lattice vectorcorresponds to a Bragg reflection from the family of direct lattice planes perpendicular torBragg reflection is just the length of GrrG i.e. the set of planes perpendicular to G , isλ 2 sin θλ 2d (h, k , l ) sin θh, k, l are the coordinates of the reciprocal lattice vector associated with the diffractionrrG nG0rGdnd 2π2πd (h, k , l ) r rn G n G0Rewrite Bragg condition:So. The order, n, of thedivided by the length of the shortest reciprocal lattice vector parallel torG is the measurement direction. Only structural information alongbeing measured.Define d spacing:rGrrrrG hb1 kb2 lb3So h, k, l have a common factor n.For example (222) direction peak is actually the the 2nd order (n 2) diffraction peak of the (111) plane.8

r 2πG (hxˆ kyˆ lzˆ )aFor cubic system:r 2πG ah2 k 2 l 22πad (h, k , l ) r Gh2 k 2 l 2Tetragonal system:Hexagonal:Orthorhombic:1d21d21 h2 k 2 l 2 2d2a2c22 4 h hk k l 2 2 3 a2 c222hkl 2 2 2abcExperimental geometries suggested by the Laue conditionGeneral observationsrAn incident wave vector k will lead to a diffraction peak (or “Bragg reflection”) if and only if the tip ofthe wave vector lies on a reciprocal space Bragg plane.Since Bragg planes are a discrete family of planes, a fixed incident wave vector – i.e., for a fixed x-ray wavelength andfixed incident direction relative to the crystal axes – there will be in general no diffraction peaks at allrIf one wishes to search experimentally for Bragg peaks, one must therefore relax the constraint for fixed k , eitherrvarying the magnitude of k (i.e. varying wavelength) or varying its direction (in practice, varying the orientation ofthe crystal with respect to incident direction).9

The Ewald constructionDraw a sphere in reciprocalspace centered on the tip of therincident wave vector kof radius k (so that it passed throughthe origin)There will be some wave vectorsatisfying theLaue condition if and only if some reciprocal lattice point(in addition to the origin lies on the surface of thesphere)In general, a sphere in reciprocal space with the origin on its surface will have no other reciprocal lattice points on its surfaceSo in general, there are no diffraction peaks.The following methods are used to relax the constrains in order to achieve diffraction peaks1. Laue methodFix the orientation of the single crystal. Search for Bragg peaks by using not a monochromatic x-ray beam, but oneλ1containing wavelength forup to λ0 .10

The Ewald sphere will expand into the region, containedr 2πk0 nˆbetween the two spheres determined byλ0r 2πk1 nˆ , and Bragg peaks will be observedandλ1corresponding to any reciprocal lattice vectors layingwithin the region.Laue method is best suited for determining theorientation of a single crystal specimen whose stuctureis known.2. The rotating crystal methodFix the wavelength, allow the angle of incidence to vary in practice, the incident direction is fixed, and theorientation of the crystal varies.The crystal is rotated about the same fixed axis. As the crystal rotates, the reciprocal lattice it determines will rotate bythe same amount about the same axis.11

The Ewald sphere is fixed in space, while the entire reciprocallattice rotates about the the axis of rotation of the crystal.During this rotation, each reciprocal lattice point traversesa circle about the roataion zxis, and a Bragg reflectionoccurs whenever this circle intersects the Ewald sphere.12

3. The powder or Debye-Scherrer methodEquivalent to a rotating crystal experiment in which, in addition, the axis of rotation is varied over all possible orientations.In practice this is achieved by using a polycrystalline sample or powder, grains of which are still enormous on the atomicscale.Bragg reflections are determined by fixing the incidentrkvector, and with it the Ewald sphere, and allowing thereciprocal lattice to rotate through all possible angles about the origin, so that each reciprocal lattice vectorgenerates a sphere of radiusrGabout the origin.13

Diffraction by a lattice with a basis: the geometrical structure factorThe diffraction conditions (either Bragg or Laue) tells only the location of the diffraction peaks, but not the magnitude.The magnitude of the diffraction peaks are determined by the electron density distribution of the basis. It is relatedto the Fourier transformation of the basis.Phase difference between beams scattered from volume elementsrrapart:er r ri ( k k ′ ) rThe amplitude of the wave scattered from a volume element is proportional to local electron concentrationThe total scattering amplitude:Under diffraction conditionrn(r )r i ( kr kr′) rrr i kr rrF dV n(r ) e dV n(r ) er r k G14

r iGr rrr r iGr ( rr Rr )r iGr rrr iGr rrF dV n(r ) e dV n(r R)e dV n(r )e N dV n(r )er r rGR cellR cellr rr iG rFGr N dV n(r ) e NSGrcellFourier transform of wave vectorcellNote thatrGrn(r ) nGr err riG ror magnitude of Fourier termFrG rGVGer riG rer riG rsrr rn( r ) n j ( r r j )j 1rrjContribution of the jth atom in the basis to the electron density atrrposition vector of the jth atomrn j (ρ )The electron density of the jth atom in its own coordinates.r rSGr dV n j (r rj )er r iG rj er r iG r jr dV n j ( ρ )er r iG ρr rr rjrrrrjOjrr rρ r rjrAtomic form factorf j dV n j ( ρ ) eSGr f j ejr r iG ρr r iG r j15

Diffraction by a monatomic lattice with a basisr r iG r jrGjS f eBcc structureRegard as simple cubic with a basis:r ad 2 ( xˆ yˆ zˆ)2rBasis: d1 0r 2πG (n1 xˆ n2 yˆ n3 zˆ )aSGr f (1 e iπ ( n1 n2 n3 ) )SGr 0When n1 n2 n3 odd integerSGr 2 fWhen n1 n2 n3 even integer(n1 , n2 , n3 )are actually the coordinates of reciprocal lattice vector (h,k,l)To have an diffraction peak,h k l evenDiffraction pattern does not contain lines such as (100), (300), (111), or (221), but lines such as (200),(110) and (222) will be present.ad (h, k , l ) The corresponding d spacing can be calculated usingh2 k 2 l 216

Physical interpretation of the fact that (100) reflection vanishes:Decipher the reciprocal lattice from the structural factor analysisSGr 0When n1 n2 n3 odd integer, no diffraction.SGr 2 fWhen n1 n2 n3 even integer, diffraction peak existsReciprocal lattice is fcc17

Fcc structure:Regard as simple cubic cell with a basisBasis:rd1 0r ad 2 ( yˆ zˆ )2r ad 3 ( xˆ zˆ )2r ad 4 ( xˆ yˆ )2r 2πG (n1 xˆ n2 yˆ n3 zˆ )aSGr f (1 e iπ ( n2 n3 ) e iπ ( n1 n3 ) e iπ ( n1 n2 ) )SGr 4 fSGr 0If n1, n2, and n3 are all even or all oddotherwiseTo have diffraction peaks, h,k,l all even or all odd(111), (200), (220), (311), (222), (400), etc. peaks will be observed.d spacings can be calculated byd (h, k , l ) ah2 k 2 l 218

Von Laue formulation of X-ray diffraction by a crystal No particular sectioning of crystal planes Regard the crystal as composed of identical microscopic objects placed at Bravais lattice site Each of the object at lattice site reradiate the incident radiation in all directions.

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