Chem 253, UC, Berkeley What we will see in XRD of simple cubic, BCC, FCC?
Position Intensity
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Structure Factor: adds up all scattered X-ray from each lattice points in crystal n iKd j Sk e j1 K ha kb lc d j x a y b z c
2 I(hkl) Sk
1 Chem 253, UC, Berkeley X-ray scattered from each primitive cell interfere constructively when:
eiKR 1 2d sin n
For n-atom basis:
sum up the X-ray scattered from the whole basis
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' k d k d di R j ' K k k
Phase difference: K (di d j ) The amplitude of the two rays differ: eiK(di d j )
2 Chem 253, UC, Berkeley
The amplitude of the rays scattered at d1, d2, d3…. are in
the ratios : eiKd j
The net ray scattered by the entire cell:
n iKd j Sk e j1
2 I(hkl) Sk
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For simple cubic: (0,0,0)
iK0 Sk e 1
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For BCC: (0,0,0), (1/2, ½, ½)…. Two point basis
1 2 iK ( x y z ) iKd j iK0 2 Sk e e e j1 1 ei (hk l) 1 (1)hkl
S=2, when h+k+l even S=0, when h+k+l odd, systematical absence
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For BCC: (0,0,0), (1/2, ½, ½)…. Two point basis
S=2, when h+k+l even S=0, when h+k+l odd, systematical absence
(100): destructive
(200): constructive
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Observable diffraction peaks h2 k 2 l 2 Ratio
SC: 1,2,3,4,5,6,8,9,10,11,12.. Simple cubic BCC: 2,4,6,8,10, 12….
FCC: 3,4,8,11,12,16,19, 24…. a dhkl h2 k 2 l 2 2d sin n
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FCC
S=4 when h+k, k+l, h+l all even (h,k, l all even/odd)
S=0, otherwise
i (hk ) i (hl) i (lk ) Sk 1 e e e
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Observable diffraction peaks h2 k 2 l 2 Ratio SC: 1,2,3,4,5,6,8,9,10,11,12..
Simple BCC: 2,4,6,8,10, 12, 14…. cubic (1,2,3,4,5,6,7…)
FCC: 3,4,8,11,12,16,19, 24…. a dhkl h2 k 2 l 2 2d sin n
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Observable diffraction peaks h2 k 2 l 2 Ratio a d hkl 2 2 2 SC: 1,2,3,4,5,6,8,9,10,11,12.. h k l
BCC: 2,4,6,8,10, 12, 14…. (1,2,3,4,5,6,7…) 2d sin n
FCC: 3,4,8,11,12,16, 19, 24…. 2 sin 2 (h2 k 2 l 2 ) 4a2
6 Chem 253, UC, Berkeley What we will see in XRD of simple cubic, BCC, FCC?
2 sin A 2 0.5 (BCC); 0.75 (FCC) sin B
a dhkl h2 k 2 l 2
Chem 253, UC, Berkeley Ex: An element, BCC or FCC, shows diffraction peaks at 2: 40, 58, 73, 86.8,100.4 and 114.7. Determine:(a) Crystal structure?(b) Lattice constant? (c) What is the element? 2theta thetasin 2 h2 k 2 l 2 (hkl) 40 20 0.117 1 (110) 58 29 0.235 2 (200) 73 36.5 0.3538 3 (211) 86.8 43.4 0.4721 4 (220) 100.4 50.2 0.5903 5 (310) 114.7 57.35 0.7090 6 (222)
a =3.18 Å, BCC, W
7 Chem 253, UC, Berkeley Reading : West Chapter 5
Polyatomic Structures
n K ha kb lc S f (k)ei Kd j k j j1 d j x a y b z c
fj: atomic scattering factor No. of electrons n n iKd j 2i(hxkylz) Sk f je f je j1 j1
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Simple Cubic Lattice
Caesium Chloride (CsCl) is primitive cubic Cs (0,0,0) Cl (1/2,1/2,1/2) i (hkl) Sk fCs fCle
What about CsI?
8 Chem 253, UC, Berkeley
(hkl) CsCl CsI (100) (110) (111) (200) (210) (211) (220) (221) (300) (310) (311)
Chem 253, UC, Berkeley FCC Lattices Sodium Chloride (NaCl)
Na: (0,0,0)(0,1/2,1/2)(1/2,0,1/2)(1/2,1/2,0)
Add (1/2,1/2,1/2)
Cl: (1/2,1/2,1/2) (1/2,0,0)(0,1/2,0)(0,0,1/2)
9 Chem 253, UC, Berkeley
i (hk l) i (hk ) i (hl) i (lk ) Sk [ f Na fCle ][1 e e e ]
S=4 (fNa +fCl) when h, k, l, all even;
S=4(fNa-fCl) when h, k, l all odd S=0, otherwise
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(hkl) NaCl KCl (100) (110) (111) (200) (210) (211) (220) (221) (300) (310) (311)
10 Chem 253, UC, Berkeley
Chem 253, UC, Berkeley Diamond Lattice
11 Chem 253, UC, Berkeley i (hk ) i (hl) i (lk ) S fcc 1 e e e
FCC: Diamond Lattice S=4 when h+k, k+l, h+l all even (h,k, l all even/odd) SSi(hkl)=Sfcc(hkl)[1+exp(i/2)(h+k+l)]. S=0, otherwise
SSi(hkl) will be zero if the sum (h+k+l) is equal to 2 times an odd integer, such as (200), (222). [h+k+l=2(2n+1)]
SSi(hkl) will be non-zero if (1)(h,k,l) contains only even numbers and (2) the sum (h+k+l) is equal to 4 times an integer. [h+k+l=4n] Shkl will be non-zero if h,k, l all odd: 4( fsi ifsi )
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Silicon Diffraction pattern
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13 Chem 253, UC, Berkeley
Chem 253, UC, Berkeley Bond charges in covalent solid (1/8,1/8,1/8), (3/8,3/8,1/8),(1/8,3/8,3/8) and (3/8,1/8,3/8). Bond charges form a "crystal" with a fcc lattice with 4 "atoms" per unit cell. origin : (1/8,1/8,1/8). Bond charge position: (0,0,0), (1/4,1/4,0),(0,1/4,1/4) and (1/4,0,1/4).
Sbond-charge(hkl) = Sfcc(hkl)[1+exp(i/2)(h+k)+exp(i/2)(k+l)+exp(i/2)(h+l)].
For h=k=l=2:
Sbond-charge(222)=Sfcc(222)[1+3exp(i/2)4]=4Sfcc(222) non-zero!
n n iKd j 2i(hxkylz) Sk f je f je j1 j1
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Nanocrystal X-ray Diffraction
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Finite Size Effect
15 Chem 253, UC, Berkeley Bragg angle: B in phase, constructive
For 1 B
1 2 Phase lag between two planes: 3 4
j-1 j At j+1 th plane: j+1 Phase lag: j 2j-1 2 2j
2j planes: net diffraction at 1 :0
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Bragg angle: B
in phase, constructive
For 2 B
1 2 Phase lag between two planes: 3 4 5
j-1 At j+1 th plane: j j+1 Phase lag: j 2j-1 2 2j
2j planes: net diffraction at 2 :0
16 Chem 253, UC, Berkeley
How particle size influence the peak width of the diffraction beam.
Full width half maximum (FWHM)
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The width of the diffraction peak is governed by # of crystal planes 2j. i.e. crystal thickness/size
Scherrer Formula: 0.9 t B2 B2 B2 B cosB M S
BM: Measured peak width at half peak intensity (in radians) BS: Corresponding width for standard bulk materials (large grain size >200 nm)
Readily applied for crystal size of 5-50 nm.
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• Suppose =1.5 Å, d=1.0 Å, and =49°. Then for a crystal 1mmindiameter, the breath B, due to the small crystal effect alone, would be about 2x10-7 radian (10-5 degree), or too small to be observable. Such a crystal would contain some 107 parallel lattice planes of the spacing assumed above.
• However, if the crystal were only 500 Å thick,itwould contain only 500 planes, and the diffraction curve would be relatively broad, namely about 4x10-3 radian (0.2°), which is easily measurable.
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