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Chapter 2. XChapter 2. X--ray Diffraction ray Diffraction and Reciprocal Latticeand Reciprocal Lattice

Diffraction of waves by crystalsReciprocal LatticeDiffraction of X-raysPowder diffractionSingle 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.

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Diffraction of Waves by Crystals

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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 2πtimes 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 just one unit cell which can be multiplied by translation along all the coordinate axes in the same fashion as a direct lattice.

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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 2 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 2θB.

OG is the g130 vector and thus has magnitude 1/d130, and since |k| = 1/λ:

0G = 2 × (1/λ)sinθB = 1/(d), rearranging this gives Bragg's equation with n = 1:

nλ = 2dsinθ

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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 translationaloperator 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

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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 aBrillouin 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 nextBrillouin zone is entered. In the figure above, the yellow square comprisesthe first zone, the purple comprises the second zone, and so on.

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Diffraction of X-rays

• Powder diffraction

• Single crystal X-ray diffraction

X-ray Tube

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Debye-Scherrer Powder Camera

X- ray Powder Diffractometer

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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

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Powder X-ray diffraction data-sheet

Systematic Absences

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Systematic Absences

• FCC: h,k,l all odd or even

Body Centered Cubic LatticeBCC: h+k+l = even

Face Centered Cube FCC:h,k,l

all odd or even

The Structure Factor

[ ] [ ]∑∑ ++π=κ∆ρ=

j,basisinatomsall

iiij

j,basisinatomsall

jjhkl )lwkvhu(i2expf)(iexpfF

We have shown that the geometrical condition for Bragg diffraction to occur is

hklG=κ∆

where Ghkl is a reciprocal lattice vector, with

∧∧

+∧∧

+∧∧

π=cb.a

balcb.a

ackcb.a

cbh2Ghkl

where uj, vj, and wj are the coordinates of the atoms in the bases expressed as fractions of the real space lattice cell vectors a, b and c. So

Now, the basis vector ρj is cwbvau jjjj ++=ρ

( )

∧∧

+∧∧

+∧∧

++π=ρ=κ∆ρcb.a

balcb.a

ackcb.a

cbh.cwbvau2G.. jjjhkljj

We generally express the scattered amplitude associated with a Bragg reflection in terms of the Structure Factor, Fhkl, ie the scattering from a single unit cell summed over all atoms in the basis, ie

( ).lwkvhu2. jjjj ++π=κ∆ρie

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Calculating the structure factor[ ]∑ ++π=

jiiijhkl )lwkvhu(i2expfF

This structure factor is central to all crystal structure determination. (fj are known as “scattering lengths”). Let us calculate the structure factor for a real crystal

Remember: )nsin(i)ncos(]inexp[ π+π=π therefore if n is odd exp[inπ] = -1 + 0 = -1

if n is even exp[inπ] = 1 + 0 = +1So, for a bcc crystal, we have two types of reflections, one with n=h+k+l oddand one with n=h+k+l even:

for n = h+k+l odd Fhkl = f(1-1) = 0for n = h+k+l even Fhkl = f(1+1) = 2f

For a bcc monatomic crystal (eg elemental iron) we have f1 = f at (u1,v1,w1)= (0,0,0) f2 = f at (u2,v2,w2)= (1/2,1/2,1/2)and

)]lkh(i2exp[f)]000(i2exp[fF 21

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21

hkl ++π+++π=So)]lkh(iexp[1(fFhkl ++π+= for a bcc crystal

Structure factors for a BCC lattice

[ ]∑ ++π=j

iiijhkl )lwkvhu(i2expfFSo, if we evaluate the structure factor

for a bcc crystal we find that it has finite amplitude, and therefore finite intensity,Ihkl ∝ |Fhkl

2|, only if h+k+l is an even integereven if the geometric conditions associated with Bragg’s law are met!!

The absent Bragg reflections associated with h+k+l = an odd integer are called systematic absences.

and Bragg reflections of finite intensity will be observed only from those planes of a bcc crystal for which Nhkl =0, 2, 4, 6, 8, 10, 12……etc

Writing Bragg’s Law as θ=θ++

=θ=λ sinNa2sin

lkha2sind2

hkl

o222

ohkl

h,k,l 0,0,0 1,0,0 1,1,0 2,0,0 2,1,0 2,1,1 2,2,0 etcFhkl 2f 0 2f 2f 0 2f 2f

θλ

= 22

2o

hkl sina4N

Note that h+k+l=0 is counted as even

Using the structure factorSo, the structure factor

determines whether a Bragg reflection from the (hkl) plane of a give crystal structure will have any intensity,

[ ]∑ ++π=j

iiijhkl )lwkvhu(i2expfF

Also, if we measure the intensity of a Bragg refelection, Ihkl ∝ |Fhkl

2|

we can determine what the arrangements of atoms on a given crystal plane might be.

Values of for observed adjacent reflections should be related by specific integer ratios Nhkl and from these inetgers the crystal lattice type can be established

22sin λθ

…..an example….

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BCC and CsCl structuresThe CsCl crystal has

f1 at (u1,v1,w1)= (0,0,0)

f2 at (u2,v2,w2)= (1/2,1/2,1/2)

This is like the bcc crystal, but now f1 and f2 are now different

[ ]∑ ++π=j

iiijhkl )lwkvhu(i2expfF

Evaluating the structure factor

)]lkh(i2exp[f)]000(i2exp[fF 21

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21hkl ++π+++π=we have

)]lkh(iexp[ffF 21hkl ++π+=ie

So for n = h+k+l odd Fhkl = f1- f2for n = h+k+l even Fhkl = f1+ f2

There will now be reflections of finite intensity ∝ (f1-f2)2 even if h+k+l is odd.

These reduce to the bcc case if f1= f2

The face centred cubic structureAn fcc crystal (eg copper) has

f1 =f at (u1,v1,w1) = (0,0,0)f2 = f at (u2,v2,w2) = (1/2,1/2,0)f3 = f at (u3,v3,w3) = (1/2,0,1/2)f4= f at (u4,v4,w4) = (0,1/2,1/2)

[ ]∑ ++π=j

iiijhkl )lwkvhu(i2expfF

So, evaluating the structure factor

)])lk(iexp[)]lh(iexp[)]kh(iexp[1(fFhkl +π++π++π+=we have

By inspection, for Fhkl 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) (2,0,0) (2,2,0) (3,1,1) (2,2,2) (4,0,0) etcNhkl 3 4 8 11 12 16 etc

The diamond structureAn diamond crystal (eg carbon) has the fcc lattice with a basis of (0,0,0) and (1/4,1/4,1/4)

f1 =f at (0,0,0)f2 = f at (1/2,1/2,0)f3 = f at (1/2,0,1/2)f4= f at (0,1/2,1/2)

f5 =f at (1/4,1/4,1/4)f6 = f at (3/4,3/4,1/4)f7 = f at (3/4,1/4,3/4)f8= f at (1/4,3/4,3/4)

∑ ++π=j

)lwkvhu(i2jhkl

iiiefFSo, evaluating the structure factor

)eeeeeee1(fF )l3k3h(i)l3kh3(i)lk3h3(i)lkh(i)lk(i)lh(i)kh(ihkl

2222 +++++++++π+π+π ππππ

+++++++=we have

( )

+++++++= +π+π+π+++π+π+π π

∴ )lk(i)lh(i)kh(i)lkh(i)lk(i)lh(i)kh(ihkl eee1eeee1fF 2

{ }{ }

++++= +π+π+π++π

∴ )lk(i)lh(i)kh(i)lkh(ihkl eee1e1fF 2

latticehkl

basishklhkl FFF ×= This is quite general

structure factor for basis

structure factor for fcc lattice

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The diamond structureThe structure factor for the hkl reflection from the diamond structure is therefore

{ }{ }

++++= +π+π+π++π )lk(i)lh(i)kh(i)lkh(i

hkl eee1e1fF 2

Additional conditions

h, k, l all odd or all even

The fcc part tells us that h, k, and l must be all odd or all even

The basis part says that additionally, for a finite intensity, must be non zero

)lkh(i2e1 ++π

+

(a) If h, k and l are all even and h+k+l =2n such that with n oddin)lkh(i ee2 π++ =π

0F,0F 2hklhkl ==

(b) If h, k and l are all even and h+k+l =4n such that n2i)lkh(i ee2 π++ =π

22hklhkl f64F,f8F ==

(c) If h, k and l are all odd, with n odd)e1(f4F inhkl

2π

+= 2*hklhkl

2hkl f32FFF ==

(h,k,l) (1,1,1) (2,0,0) (2,2,0) (3,1,1) (2,2,2) (4,0,0) etc, as for fcc(c) (a) (b) (c) (a) (b)

Structure factor for ZnSIf the atoms at the 0,0,0 and 1/4, 1/4, 1/4 are different the diamond structure becomes the ZnSstructure. In this case the structure factor for the hkl reflection is

{ }{ }

++++= +π+π+π++π )lk(i)lh(i)kh(i)lkh(i

21hkl eee1effF 2

So, considering the basis contribution

{ })lkh(i21

basishkl

2effF ++π

+=

(a) If h, k and l are all even and h+k+l =2n such that with n oddin)lkh(i ee2 π++ =π

221

2hkl

221

basis

hkl

2ff16F,ffF −=−=

(b) If h, k and l are all even and h+k+l =4n such that n2i)lkh(i ee2 π++ =π

221

2hkl

221

basis

hkl

2ff16F,ffF +=+=

(c) If h, k and l are all odd, and 22

21

2i21

basis

hkl

2ffeffF 2 +=+=

π 22

21

2hkl ff16F +=

X-Ray Powder DiffractometryUses 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 = 2θ.

The computer gives intensity of x-rays as the detector rotates through 2θ. Thus, the angle 2θ 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-planarspacings (d) for all the major peaks and look up a match with ICCD data bank. ICCD = International Center for Crystallographic Data.

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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

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Single Crystal MethodsPut 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 thestructure.

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.