X-ray diffraction

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X-ray diffraction Meet in the LGRT lab Again, will hand in worksheet, not a formal lab report Revision exercise – hand in by April 17 th class.

description

X-ray diffraction. Meet in the LGRT lab Again, will hand in worksheet, not a formal lab report Revision exercise – hand in by April 17 th class. Diffraction summarized. The 6 lattice parameters ( a,b,c, a,b,g ) of a crystal determine the position of x-ray diffraction peaks. - PowerPoint PPT Presentation

Transcript of X-ray diffraction

Page 1: X-ray diffraction

X-ray diffraction

• Meet in the LGRT lab• Again, will hand in worksheet, not a formal lab report

• Revision exercise – hand in by April 17th class.

Page 2: X-ray diffraction

Diffraction summarized

• The 6 lattice parameters (a,b,c,) of a crystal determine the position of x-ray diffraction peaks.

• The contents of the cell (atom types and positions) determine the relative intensity of the diffraction peaks.

• If a diffraction peak can be identified with a Miller Index, the unit cell on the phase can usually be determined.

Page 3: X-ray diffraction

Miller indices

• Choose origin• Pick 1st plane away from origin

in +a, +b direction• Find intercepts in fractional

coordinates (“none = ∞”).

• h = 1/(a-intercept)• k = 1/(b-intercept)• l = 1/(c-intercept)

• If no intercept, index = 0

• Plane equation:ha + kb + lc = 1

Miller index, hkl

(2-10)

(110) (100)

(150)

• Same set of planes will be described if all 3 Miller indices are inverted (2-10) ≡ (-210)

• Minus sign → bar 10)2(0)1(2

Page 4: X-ray diffraction

Crystals facets correspond to Miller indices

• Haüy, 1784

• Crystals (like calcite) are made of miniscule identical subunits

ba

(-120)

Woolfson

• Facets can be described by low-order Miller indices

(-120)

(-110)(-100)

(-1-10)

(-2-10)

(3-40)

(340)

Page 5: X-ray diffraction

Miller indices

• For a lattice of known dimensions, the Miller indices can be used to calculate the d-spacing between hkl planes (a,b,c = lattice parameters).

• This d-spacing will determine where powder diffraction peaks are observed

Miller index, hkl

(2-10)

(-110) (100)

(150)2

2

2

2

2

2

2

1cl

bk

ah

d

Origin at orange dot

Page 6: X-ray diffraction

Distance between planes (Miller indices)

Page 7: X-ray diffraction

Bragg’s law derivation(angle of incidence = angle of diffraction)

• x = d sin • extra distance = 2d sin = n• n = 2d sin

d

x

d

x x

a b

= 2d sin (for x-ray diffraction)

Page 8: X-ray diffraction

Powder diffractometer

Page 9: X-ray diffraction

KI (CsCl-type) x-ray pattern

Page 10: X-ray diffraction

Crystals, diffraction, and Miller indices

(100)

(010)

(001)

(0-10)

Page 11: X-ray diffraction

Coherent scattering from a row of atoms

Will only happen when emission from all atoms is simultaneously stimulated.(Solid lines represent spatial regions where phase = 0 deg).

Page 12: X-ray diffraction

a

Laue condition – vector description• Extra distance = a cos - a cos • Extra distance = -(a·S0) + (a·S)• Extra distance = (a·s)

a

S

S0

a

• Will have diffraction when: a cos - a cos = h (h = integer)• Will have diffraction when: a·s = h (h = integer)

s = (S – S0)/

v1

v2

cos = v1·v2

Page 13: X-ray diffraction

a

Laue condition – 2D• Must have coherent scattering from ALL ATOMS in the lattice, no just from one row.• If color indicates phase of radiation scattered from each lattice point when observed at a distant site P, we see

that scattering from rows is in phase while columns is out of phase, making net scattering from all 35 points incoherent and therefore NOT observable

S

S0

b

• Will have diffraction when: a·s = h (h = integer)• Will have diffraction when: b·s = k (k = integer)• Will have diffraction when: c·s = l (l = integer)

Every lattice point related by translational symmetry will scattering in phase when conditions are met

Page 14: X-ray diffraction

Single crystal diffraction

• Used to solve molecular structure– Co(MIMT)2(NO3)2 example

• Data in simple format – hkl labels + intensity + error

0 0 1 0.00 0.10 0 0 2 42.60 1.40 0 0 3 1.10 0.30 0 0 4 100.30 2.50 0 0 5 -0.30 0.50 0 0 6 822.30 16.70 0 0 7 -0.40 0.50 0 0 8 656.40 13.90 0 0 9 1.00 0.80 0 0 10 73.40 3.00 0 0 11 0.00 1.40 0 0 12 4.70 1.60 0 0 13 1.00 1.70 0 2 1 611.40 14.40 0 -2 -1 613.90 12.10 0 2 2 443.90 8.90 0 -2 -2 443.00 8.70 0 2 3 59.90 1.50 0 -2 -3 56.90 1.50 0 2 4 55.20 1.60 0 -2 -4 51.80 1.50

Page 15: X-ray diffraction

h=2

h=1

h=0

h=-1

Page 16: X-ray diffraction

l=6l=7l=8

h=2

l=9l=10

Page 17: X-ray diffraction

3D view of reciprocal lattice

Massa

(Lines connecting dots are unecessary)

Page 18: X-ray diffraction

Relation between direct and reciprocal cells

Stout and Jensen

Page 19: X-ray diffraction

Direct vs. reciprocal lattice

direct or real

reciprocal

Lattice vectors a, b, c

Vector to a lattice point: d = ua + vb + wc

Lattice planes (hkl)

Lattice vectors a*, b*, c*

Vector to a reciprocal lattice point: d* = ha* + kb* + lc*

Each such vector is normal to the real space plane (hkl)

Length of each vector d* = 1/d-spacing (distance between hkl planes)

Page 20: X-ray diffraction

Hexagonal lattice

changes from 120 to 60 deg

ab

b*a*

direct or real reciprocal

Page 21: X-ray diffraction

Distances between planes (d*)

direct or real

reciprocal

Vector to a reciprocal lattice point: d* = ha* + kb* + lc*

Length of each vector d* = 1/d-spacing (distance between hkl planes)

|d*| = (d*·d*)1/2

=(ha* + kb* + lc*) (ha* + kb* + lc*)

=(ha*)2 + (kb*)2 + (lc*)2 + 2(ha*) (kb*) + 2(ha*) (lc*) + 2(kb*) (lc*)

= h2a2* + k2b*2 + l2c*2 + 2klb*c*cos* + 2lhc*a*cos* + 2hka*cos*

Page 22: X-ray diffraction

• Density = mass / volume

• Mass of unit cell = (# Ca)(mCa) + (# Ru)(mRu) + (# O)(mO)

• Volume of unit cell = abc = (3.7950 Å)3

• mamuNA = mgrams

Density of CaRuO3

VNnMV

NnM

AA

/

• M = molar mass (g/mol)

• n = # f.u. per cell• NA = Avagadro’s #

(1/mol)• V = cell volume