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Analysis of crystal structure x-rays, neutrons and electrons

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Presentation on theme: "Analysis of crystal structure x-rays, neutrons and electrons"— Presentation transcript:

1 Analysis of crystal structure x-rays, neutrons and electrons
Diffraction Analysis of crystal structure x-rays, neutrons and electrons 2/2-10 MENA3100

2 The reciprocal lattice
g is a vector normal to a set of planes, with length equal to the inverse spacing between them Reciprocal lattice vectors a*,b* and c* These vectors define the reciprocal lattice All crystals have a real space lattice and a reciprocal lattice Diffraction techniques map the reciprocal lattice 2/2-10 MENA3100

3 Radiation: x-rays, neutrons and electrons
Elastic scattering of radiation No energy is lost The wave length of the scattered wave remains unchanged Regular arrays of atoms interact elastically with radiation of sufficient short wavelength CuKα x-ray radiation: λ=0.154 nm Scattered by electrons ~from sub mm regions Neutron radiation λ~0.1nm Scattered by atomic nuclei Several cm thick samples Electron radiation (200kV): λ= nm Scattered by atomic nuclei and electrons Thickness less than ~200 nm 2/2-10 MENA3100

4 Interference of waves Sound, light, ripples in water etc etc
Constructive and destructive interference =2n =(2n+1) 2/2-10 MENA3100

5 Nature of light Newton: particles (corpuscles) Huygens: waves
Thomas Young double slit experiment (1801) Path difference  phase difference Light consists of waves ! Wave-particle duality 2/2-10 MENA3100

6 Discovery of X-rays Wilhelm Röntgen 1895/96 Nobel Prize in 1901
Particles or waves? Not affected by magnetic fields No refraction, reflection or intereference observed If waves, λ10-9 m Røntgen jobbet med en Crookes tube med sort papp rundt. Oppdaget at en fluoriserende skjerm i rommet lyste opp. Fant ut at strålingen gikk gjennom papp og bøker. Så når han holdt et objekt at han kunne se skjelettet sitt. Fotograferte sin kones hånd. 2/2-10 MENA3100

7 Max von Laue The periodicity and interatomic spacing of crystals had been deduced earlier (e.g. Auguste Bravais). von Laue realized that if X-rays were waves with short wavelength, interference phenomena should be observed like in Young’s double slit experiment. Experiment in 1912, Nobel Prize in 1914 2/2-10 MENA3100

8 Laue conditions Scattering from a periodic distribution of scatters along the a axis a ko k The scattered wave will be in phase and constructive interference will occur if the phase difference is 2π. Φ=2πa.(k-ko)=2πa.g= 2πh, similar for b and c 2/2-10 MENA3100

9 The Laue equations The Laue equations give three conditions for incident waves to be diffracted by a crystal lattice Two lattice points separated by a vector r Waves scattered from two lattice points separated by a vector r will have a path difference in a given direction. The scattered waves will be in phase and constructive interference will occur if the phase difference is 2π. The path difference is the difference between the projection of r on k and the projection of r on k0, φ= 2πr.(k-k0) r k k0 k-k0 r*hkl (hkl) Δ=a.(k-ko)=h Δ=b.(k-ko)=k Δ=c.(k-ko)=l If (k-k0) = r*, then φ= 2πn r*= ha*+kb*+lc* Δ=r . (k-k0) 2/2-10 MENA3100

10 Bragg’s law William Henry and William Lawrence Bragg (father and son) found a simple interpretation of von Laue’s experiment Consider a crystal as a periodic arrangement of atoms, this gives crystal planes Assume that each crystal plane reflects radiation as a mirror Analyze this situation for cases of constructive and destructive interference Nobel prize in 1915 2/2-10 MENA3100

11 Derivation of Bragg’s law
θ x dhkl Path difference Δ= 2x => phase shift Constructive interference if Δ=nλ This gives the criterion for constructive interference: Bragg’s law tells you at which angle θB to expect maximum diffracted intensity for a particular family of crystal planes. For large crystals, all other angles give zero intensity. 2/2-10 MENA3100

12 The path difference: x-y
Bragg’s law d θ y x nλ = 2dsinθ Planes of atoms responsible for a diffraction peak behave as a mirror The path difference: x-y Y= x cos2θ and x sinθ=d cos2θ= 1-2 sin2θ 2/2-10 MENA3100

13 von Laue – Bragg equation
θ Vector normal to a plane θ 2/2-10 MENA3100

14 The limiting-sphere construction
Vector representation of Bragg law IkI=Ik0I=1/λ λx-rays>> λe k = ghkl (hkl) k0 k-k0 Diffracted beam Incident beam Reflecting sphere Limiting sphere 2/2-10 MENA3100

15 The Ewald Sphere (’limiting sphere construction’)
Elastic scattering: k k’ The observed diffraction pattern is the part of the reciprocal lattice that is intersected by the Ewald sphere g 2/2-10 MENA3100

16 The Ewald Sphere is flat (almost)
Cu Kalpha X-ray:  = 150 pm => small k Electrons at 200 kV:  = 2.5 pm => large k 2/2-10 MENA3100

17 50 nm 2/2-10 MENA3100

18 Allowed and forbidden reflections
Bravais lattices with centering (F, I, A, B, C) have planes of lattice points that give rise to destructive interference for some orders of reflections. Forbidden reflections y’ y x θ x’ d θ In most crystals the lattice point corresponds to a set of atoms. Different atomic species scatter more or less strongly (different atomic scattering factors, fzθ). From the structure factor of the unit cell one can determine if the hkl reflection it is allowed or forbidden. 2/2-10 MENA3100

19 Structure factors X-ray:
The coordinate of atom j within the crystal unit cell is given rj=uja+vjb+wjc. h, k and l are the miller indices of the Bragg reflection g. N is the number of atoms within the crystal unit cell. fj(n) is the x-ray scattering factor, or x-ray scattering amplitude, for atom j. The structure factors for x-ray, neutron and electron diffraction are similar. For neutrons and electrons we need only to replace by fj(n) or fj(e) . rj uja a b x z c y vjb wjc The intensity of a reflection is proportional to: 2/2-10 MENA3100

20 Example: fcc eiφ = cosφ + isinφ enπi = (-1)n eix + e-ix = 2cosx
Atomic positions in the unit cell: [000], [½ ½ 0], [½ 0 ½ ], [0 ½ ½ ] What is the general condition for reflections for fcc? Fhkl= f (1+ eπi(h+k) + eπi(h+l) + eπi(k+l)) What is the general condition for reflections for bcc? If h, k, l are all odd then: Fhkl= f( )=4f If h, k, l are mixed integers (exs 112) then Fhkl=f( )=0 (forbidden) 2/2-10 MENA3100

21 The structure factor for fcc
The reciprocal lattice of a FCC lattice is BCC What is the general condition for reflections for bcc? 2/2-10 MENA3100

22 The reciprocal lattice of bcc
Body centered cubic lattice One atom per lattice point, [000] relative to the lattice point What is the reciprocal lattice? 2/2-10 MENA3100

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