Chapter 2. X-ray X. Diffraction and Reciprocal Lattice. Scattering from Lattices

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1 Chapter. X-ray X Diffraction and Reciprocal Lattice Diffraction of waves by crystals Reciprocal Lattice Diffraction of X-rays Powder diffraction Single crystal X-ray diffraction Scattering from Lattices Diffraction techniques, which is really a realization of quantum-mechanical scattering on the order of the de- Broglie wavelength, make direct use of the reciprocal lattice. Scattering is VERY useful, as it provides a good method for directly probing the lattice structure of various materials. X-Ray Diffraction is aptly suited for measuring inter-atomic spacings of most physical lattices. Neutron scattering provides other data, as the spin-½ magnetic moment of the neutron interacts with the material. Neutron scattering gives a good way to directly observe phonons, or lattice vibrations, for example.

2 Diffraction of Waves by Crystals

3 Constructing a reciprocal lattice from a direct lattice Pick some point as an origin, then: a) from this origin, lay out the normal to every family of parallel planes in the direct lattice; b) set the length of each normal equal to times the reciprocal of the inter-planar spacing for its particular set of planes; c) place a point at the end of each normal. Relationship between the reciprocal and the direct lattice? The relationship between the period and frequency is similar to that of the reciprocal and the direct lattice. Therefore Fourier transformation is used in the studies of the Real lattice to yield the Reciprocal lattice in the same fashion as with the studies of any other Periodic function, therefore the reciprocal space is also called Fourier space. Advantage of Reciprocal Lattice It looks quite difficult to lay out normals to millions of parallel planes of the real crystal? But fortunately the reciprocal lattice can be described by ust one unit cell which can be multiplied by translation along all the coordinate axes in the same fashion as a direct lattice.

5 Indexing Reciprocal Lattice Ewald Construction n 1913 Peter Ewald published details of a geometrical construction which has been used ever since for interpreting diffraction patterns. When a beam hits a crystal, Ewald's sphere shows which sets of planes are at (or close to) their Bragg angle for diffraction to occur. In D the sphere becomes a circle. The incident wave is represented by a reciprocal vector k. Draw the incident wave vector, k, ending at 0. Construct a circle with radius 1/λ (i.e. k ), which passes through 0. Wherever a reciprocal point touches the circle, Bragg's Law is obeyed and a diffracted beam will occur. CO represents the incident beam and CG is a diffracted beam. The angle between them must be θ B. OG is the g 130 vector and thus has magnitude 1/d 130, and since k = 1/λ: 0G = (1/λ)sinθ B = 1/(d), rearranging this gives Bragg's equation with n = 1: nλ = dsinθ

This indicates, as Arnold Schwarzenegger is quoting, that the wave function, can be represented as the product of a plane wave (eik x) with a periodic function (unk[r]).

6 Bragg Law in Reciprocal Space Bloch Waves Bloch Waves are the most important effect due to the discrete lattice translational symmetry. This arises because the Hamiltonian must commute with the translational operator for any discrete integer lattice translation. This indicates, as Arnold Schwarzenegger is quoting, that the wave function, can be represented as the product of a plane wave (eik x) with a periodic function (unk[r]). The translational invariance of the wave function is of utmost importance. This basically indicates that all the information about the system is stored within a subset of the system, the rest of the information is redundant. Wigner-Seizt Primitive Cell of Reciprocal Space: First Brillouin zones

Brillouin zones Brillouin Zones in 3-D space Brillouin zones The area described on

7 Brillouin zones Brillouin Zones in 3-D space Brillouin zones The area described on the previous slide that contains all the unique wavefunction information is referred to as a Brillouin Zone. There are an infinite number of Brillouin zones, and they are universally referred to according to the rules above. Essentially each time a Bragg Plane is crossed, the next Brillouin zone is entered. In the figure above, the yellow square comprisesthe first zone, the purple comprises the second zone, and so on.

8 Diffraction of X-rays Powder diffraction Single crystal X-ray diffraction X-ray Tube

9 Debye-Scherrer Powder Camera X- ray Powder Diffractometer

$Powder Diffraction Pattern Powder$ and the equation for interplanar

10 Powder Diffraction Pattern Powder Data We know Bragg's Law: nλ =dsinθ and the equation for interplanar spacing, d, for cubic crystals is given by: where a is the lattice parameter this gives: Indexing Powder Data

$diffraction$ data-sheet

11 Powder X-ray diffraction data-sheet Systematic Absences

$Factor We have shown that the geometrical condition for Bragg diffraction to occur is κ = G hkl where G hkl is a reciprocal lattice vector, with b c c a a$ b G hkl = h + k + l a.b c a.

$b c Now, the basis vector ρ is ρ = ua + v b + w c where u, v, and w are the coordinates of the atoms in the bases expressed as fractions of the real space$

12 Systematic Absences FCC: h,k,l all odd or even Face Centered Cube FCC: h,k,l all odd or even Body Centered Cubic Lattice BCC: h+k+l = even The Structure Factor We have shown that the geometrical condition for Bragg diffraction to occur is κ = G hkl where G hkl is a reciprocal lattice vector, with b c c a a b G hkl = h + k + l a.b c a.b c a.b c Now, the basis vector ρ is ρ = ua + v b + w c where u, v, and w are the coordinates of the atoms in the bases expressed as fractions of the real space lattice cell vectors a, b and c. So b c c a a b ρ. κ = ρ.g = ( u a + v b + w c). h + k + l hkl a.b c a.b c a.b c ie ρ. κ = ( hu + kv + lw ). We generally express the scattered amplitude associated with a Bragg reflection in terms of the Structure Factor, F hkl, ie the scattering from a single unit cell summed over all atoms in the basis, ie Fhkl = f exp[ i( ρ κ) ] = f exp[ i(hui + kvi + lwi) ] all atoms all atoms in basis, in basis,

13 Calculating the structure factor = f exp[ i(hui + kvi + lwi ] Fhkl ) This structure factor is central to all crystal structure determination. (f are known as scattering lengths ). Let us calculate the structure factor for a real crystal For a bcc monatomic crystal (eg elemental iron) we have f 1 = f at (u 1,v 1,w 1 )= (0,0,0) and f = f at (u,v,w )= (1/,1/,1/) 1 So Fhkl = f exp[i( )] + f exp[i( 1 h + 1 k + l)] F hkl = f(1+ exp[ i(h + k + l)] for a bcc crystal Remember: exp[ in] = cos(n) + isin(n) therefore if n is odd exp[in] = = -1 if n is even exp[in] = = +1 So, for a bcc crystal, we have two types of reflections, one with n=h+k+l odd and one with n=h+k+l even: for n = h+k+l odd F hkl = f(1-1) = 0 for n = h+k+l even F hkl = f(1+1) = f Structure factors for a BCC lattice So, if we evaluate the structure factor Fhkl = f exp[ i(hui + kvi + lwi) ] for a bcc crystal we find that it has finite amplitude, and therefore finite intensity, I hkl F hkl, only if h+k+l is an even integer even if the geometric conditions associated with Bragg s law are met!! h,k,l 0,0,0 1,0,0 1,1,0,0,0,1,0,1,1,,0 etc F hkl f 0 f f 0 f f The absent Bragg reflections associated with h+k+l = an odd integer are called systematic absences. Note that h+k+l=0 is counted as even ao ao Writing Bragg s Law as λ = dhkl sinθ = sinθ = sinθ + + N h k l hkl ao Nhkl = 4 sin θ λ and Bragg reflections of finite intensity will be observed only from those planes of a bcc crystal for which N hkl =0,, 4, 6, 8, 10, 1 etc Using the structure factor So, the structure factor Fhkl = f exp i(hui + kvi + lwi) [ ] determines whether a Bragg reflection from the (hkl) plane of a give crystal structure will have any intensity, Values of sin θ λ for observed adacent reflections should be related by specific integer ratios N hkl and from these inetgers the crystal lattice type can be established Also, if we measure the intensity of a Bragg refelection, I hkl F hkl we can determine what the arrangements of atoms on a given crystal plane might be...an example.

BCC and CsCl structures The CsCl crystal has f 1 at (u 1,v 1,w 1 )= (0,0,0) f at (u,v,w )= (1/,1/,1/) we have ie This is like the bcc crystal, but now f 1 and f are now different Evaluating the

bcc case if f for n = h+k+l even F hkl = f 1 + f 1 = f There will now be reflections of finite intensity (f 1 -f ) even if h+k+l is odd.

(0,1/,1/) So, evaluating the structure factor = f exp[ i(hui + kvi + lwi ] Fhkl ) we have F hkl = f(1+ exp[ i(h + k)] + exp[ i(h + l)] + exp[ i(k + l)]) By inspection, for F hkl to be non-zero (in

14 BCC and CsCl structures The CsCl crystal has f 1 at (u 1,v 1,w 1 )= (0,0,0) f at (u,v,w )= (1/,1/,1/) we have ie This is like the bcc crystal, but now f 1 and f are now different Evaluating the structure factor Fhkl = f exp[ i(hui + kvi + lwi) ] 1 Fhkl = f1 exp[i( )] + f exp[i( 1 h + 1 k + l)] Fhkl = f1 + f exp[ i(h + k + l)] So for n = h+k+l odd F hkl = f 1 -f These reduce to the bcc case if f for n = h+k+l even F hkl = f 1 + f 1 = f There will now be reflections of finite intensity (f 1 -f ) even if h+k+l is odd. The face centred cubic structure An fcc crystal (eg copper) has f 1 =f at (u 1,v 1,w 1 ) = (0,0,0) f = f at (u,v,w ) = (1/,1/,0) f 3 = f at (u 3,v 3,w 3 ) = (1/,0,1/) f 4 = f at (u 4,v 4,w 4 ) = (0,1/,1/) So, evaluating the structure factor = f exp[ i(hui + kvi + lwi ] Fhkl ) we have F hkl = f(1+ exp[ i(h + k)] + exp[ i(h + l)] + exp[ i(k + l)]) By inspection, for F hkl to be non-zero (in which case it takes the value 4f) h, k and l must be all odd or h, k and l must be all even The only Bragg reflections with finite intensity, from an fcc crystal, are thus (h,k,l) (1,1,1) (,0,0) (,,0) (3,1,1) (,,) (4,0,0) etc N hkl etc The diamond structure An diamond crystal (eg carbon) has the fcc lattice with a basis of (0,0,0) and (1/4,1/4,1/4) f 1 =f at (0,0,0) f = f at (1/,1/,0) f 3 = f at (1/,0,1/) f 4 = f at (0,1/,1/) f 5 =f at (1/4,1/4,1/4) f 6 = f at (3/4,3/4,1/4) f 7 = f at (3/4,1/4,3/4) f 8 = f at (1/4,3/4,3/4) So, evaluating the structure factor i(hui+ kvi + lwi ) Fhkl = fe we have i(h k) i(h l) i(k l) i(h+ k+ l) i(3h+ 3k+ l) i(3h+ k+ 3l) i(h+ 3k+ 3l) Fhkl = f(1+ e + e + e + e + e + e + e ) i(h+ k) i(h+ l) i(k+ l) i(h+ k+ l) i(h+ k) i(h+ l) i(k+ l) F = ( ) hkl f 1 e e e e 1 e e e i(h+ k+ l) i(h+ k) i(h+ l) i(k+ l) F = { + }{ } hkl f 1 e 1 e e e structure factor structure factor for fcc lattice for basis basis lattice F hkl = Fhkl Fhkl This is quite general

The diamond structure The structure factor for the hkl reflection from the diamond structure is therefore i(h+ k+ l) i(h+ k) i(h+ l) i(k+ l) Fhkl = f 1+ e 1+ e + e + e { }{ } Additional h, k, l all

15 The diamond structure The structure factor for the hkl reflection from the diamond structure is therefore i(h+ k+ l) i(h+ k) i(h+ l) i(k+ l) Fhkl = f 1+ e 1+ e + e + e { }{ } Additional h, k, l all odd or all even conditions The fcc part tells us that h, k, and l must be all odd or all even i(h k l) The basis part says that additionally, for a finite intensity, e be non zero (a) If h, k and l are all even and h+k+l =n such that (b) If h, k and l are all even and h+k+l =4n such that e must i(h + k + l) in = e with n odd Fhkl = 0, Fhkl = 0 i(h k l) in e + + = e F hkl = 8f, Fhkl = 64f in * (c) If h, k and l are all odd, Fhkl = 4f(1+ e ) with n odd F hkl = FhklFhkl = 3f (h,k,l) (1,1,1) (,0,0) (,,0) (3,1,1) (,,) (4,0,0) etc, as for fcc (c) (a) (b) (c) (a) (b) Structure factor for ZnS If the atoms at the 0,0,0 and 1/4, 1/4, 1/4 are different the diamond structure becomes the ZnS structure. In this case the structure factor for the hkl reflection is i(h+ k+ l) i(h+ k) i(h+ l) i(k+ l) { f + }{ + + } 1 fe 1 e e e F hkl = + So, considering the basis contribution basis Fhkl = { f } i(h k l) 1 fe i(h k l) in (a) If h, k and l are all even and h+k+l =n such that e + + = e with n odd basis F = f1 f, Fhkl = 16 f1 f hkl i(h k l) in (b) If h, k and l are all even and h+k+l =4n such that e + + = e basis F = f1 + f, Fhkl = 16 f1 + f hkl basis i (c) If h, k and l are all odd, F = f1 + fe = f1 + f and F hkl = 16 f1 + f hkl X-Ray Powder Diffractometry Uses monochromatic x-rays on powder mounted on sample holder attached to a stage (omega) which systematically rotates into the path of the x-ray beam through θ = 0 to 90. The diffracted x-rays are detected electronically (detector) and recorded on a computer. The detector rotates simultaneously with the stage, but rotates through angles = θ. The computer gives intensity of x-rays as the detector rotates through θ. Thus, the angle θ at which diffractions occur and the relative intensities can be read directly from the position and heights of the peaks on graph. Then use the Bragg equation to solve for the inter-planar spacings (d) for all the maor peaks and look up a match with ICCD data bank. ICCD = International Center for Crystallographic Data.

$Single crystal X-ray diffraction Primary application is to determine atomic structure (symmetry, unit cell dimensions, space group, etc.,).$ $The diffracted x-rays exiting the crystal all have different θ and thus produce "spots" on a photographic plate. The diffraction spots show the symmetry of the crystal.$

16 Single crystal X-ray diffraction Primary application is to determine atomic structure (symmetry, unit cell dimensions, space group, etc.,). Older methods (Laue method) used a stationary crystal with "white x- ray" beam (x-rays of variable l) such that Bragg's equation would be satisfied by numerous atomic planes. The diffracted x-rays exiting the crystal all have different θ and thus produce "spots" on a photographic plate. The diffraction spots show the symmetry of the crystal. Modern methods (rotation, Weissenberg, precession, 4-circle) utilize various combination of rotating-crystal and camera setup to overcome limitations of the stationary methods (mainly the # of diffractions observed). These methods use monochromatic x-rays, but vary θ by moving the crystal mounted on a rotating stage. Usually employ diffractometers and computers for data collection and processing. Rotation Single Crystal Camera 4-circle Single Crystal Diffractometer

17 Single Crystal Methods Put a crystal in the beam, observe what reflections come out at what angles for what orientations of the crystal with what intensities. Advantages In principle you can learn everything there is to know about the structure. Disadvantages You may not have a single crystal. It is time-consuming and difficult to orient the crystal. If more than one phase is present, you will not necessarily realize that there is more than one set of reflections.

Keble College - Hilary 2012 Section VI: Condensed matter physics Tutorial 2 - Lattices and scattering

Tomi Johnson Keble College - Hilary 2012 Section VI: Condensed matter physics Tutorial 2 - Lattices and scattering Please leave your work in the Clarendon laboratory s J pigeon hole by 5pm on Monday of