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Simple Cubic Lattice Chem 253, UC, Berkeley What we will see in XRD of simple cubic, BCC, FCC? Position Intensity Chem 253, UC, Berkeley 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 Chem 253, UC, Berkeley ' 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 Chem 253, UC, Berkeley For simple cubic: (0,0,0) iK0 Sk e 1 3 Chem 253, UC, Berkeley 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 Chem 253, UC, Berkeley 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 4 Chem 253, UC, Berkeley 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 Chem 253, UC, Berkeley 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 5 Chem 253, UC, Berkeley 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 Chem 253, UC, Berkeley 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 Chem 253, UC, Berkeley 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 Chem 253, UC, Berkeley (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 ) Chem 253, UC, Berkeley Silicon Diffraction pattern 12 Chem 253, UC, Berkeley Chem 253, UC, Berkeley 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 14 Chem 253, UC, Berkeley Nanocrystal X-ray Diffraction Chem 253, UC, Berkeley 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 Chem 253, UC, Berkeley 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) Chem 253, UC, Berkeley 17 Chem 253, UC, Berkeley 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. Chem 253, UC, Berkeley • 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. 18.
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