crystallography 10
TRANSCRIPT
Diffraction: Intensity(From Chapter 4 of Textbook 2 and Chapter 9 of Textbook 1)
Electron atoms group of atoms or structure Crystal (poly or single)
Scattering by an electron:
r
P
= /2
2
202
22
420
0 sinsin4 r
KIrm
eII
0: 410-7 mkgC-2
a single electron charge e (C), mass m (kg), distance r (meters)
by J.J. Thomson
x
y
z
O
P
Randompolarized
2r
222zy EEE
00022
21
21 IIIEEE zyzy 2
y component
z component
= yOP = /2 20 rKII yPy
2cos220 r
KII zPz = zOP = /2 -2
22cos12cos
2
202
2020
rKI
rKI
rKIIII zyPzPyP
Polarization factor
0 20 40 60 80 100 120 140 160 1800.4
0.6
0.8
1.0
Po
lariz
atio
n Fa
ctor
2 (Degrees)
Pass through a monochromator first (Bragg angle M) the polarization factor is ?
M
M
2cos12cos2cos1
2
22
(Homework)
x
y
z
O
P
polarization isnot complete
randomanymore
2
r
x
y
z
O
PRandompolarized
2M
r P
22cos12cos
2
202
2020M
MzyzPyPP rKI
rKI
rKIIII
0 20 40 60 80 100 120 140 160 1800.4
0.5
0.6
0.7
0.8
0.9
1.0
Po
lariz
atio
n Fa
ctor
2 (Degrees)
Si (111) as monochromatorCu K; = 28.44o
M
Atomic scattering (or form) factor
a single free electron atoms
electron singlea by scattered amplitudeatomby scattered amplitudefactor scattering atomic
Differential atomic scattering factor (df) :
dVerE
df i
e
)]()[/2( 0)(1 ssr Ee: the magnitude of the wave from a bound electron
Os0
R
r
x1dV
s
x2
2
path different (O and dV):R-(x1 + x2).
srsr Rxx 201 ;
Electron density Phase difference
Spherical integration dV = dr(rd) (rsind)
http://pleasemakeanote.blogspot.tw/2010/02/9-derivation-of-continuity-equation-in.html
r: 0 - : 0 - : 0 - 2
rsind
rsin(+d)dr
dr
d
d
rr
ddrrdrrddr
0
2
0
2
0
sin2)sin)((
3
34 r
= 23
3r
r
r
drdrddrr
0
2
0
2 sin2sin2
S0
S S-S0
2
Evaluate (S - S0)r = | S - S0||r|cos |(S - S0)|/2 = sin.
cossin2)( 0 r rSS
dVerE
dff i
e
)]()[/2( 0)(1)( ssr
r
r
ri
e
drdrerE
dff0 0
2/cossin4 sin2)(1)(
Let /sin4kcosd
r
r
kri
e
dedrrrE
f0 0
cos2 cos2)(1)(
krkr
ikree
ikre ikrikrikr sin2
cos
0cos
cos
r
re
drkr
krrrE
f0
2 sin)(41)(
For = 0, only k = 0 sinkr/kr = 1.
For n electrons in an atom
electrons n0
2 sin)(41)(r
re
drkr
krrrE
f
ZdrrrE
fr
re
electrons n
0
2 )(41)0( Number of electronsin the atom
equal to 1 boundelectrons
Tabulated
Anomalous Scattering:Previous derivation: free electrons!Electrons around an atom: free?
kxFdt
xdm 2
2 free electronharmonic oscillator
m
k
tiex 0Assumekmkeeim titi 2
02
000)( 2/1
0 )/( mk
)(2
2
tFkxdt
xdm Forced oscillator
tieFtF 0)( Assume
tiCex Assumetititi eFkCeeiCm 0
2)( 02 FkCCm
20mk
020
2 FCmCm )(
)( 220
00
220
m
FCFmC
Resonance frequency
tiem
Fx
)( 220
0
tiCex
Same frequency as F(t), amplitude(, 0) = 0 C is ; in reality friction term exist no
tieFtFkxdtdxc
dtxdm
02
2
)(
Oscillator with damping (friction v)
assume c = m
Fxmdtdxm
dtxdm 2
02
2
mFxmxixi /)( 002000
2
Assume x = x0eit
)( 220
00 im
Fx
Real part and imaginary part
f +f + if
real imaginary
2
220 )(
1 i
0 E
if 0
Resonance: X-ray frequency; 0: bounded electrons around atoms 0 electron escape # of electrons around an atom f (f correction term)
imaginary part correction: f (linear absorption coefficient)
Examples: Si, 400 diffraction peak, with Cu K (0.1542 nm)
nm 13577.04/54309.0400 d6.341542.0sin2 400 d
1
368.0sin
sin 0.3 0.48.22 7.20
526.7368.03.04.03.0
22.820.722.8
ff
Anomalous Scattering correction 4.0 ;2.0 ff
Atomic scattering factor in this case: 7.526-0.2+0.4i = 7.326+0.4i
f and f: International Table for X-ray Crystallography V.III
Structure factor
atoms unit cell
plane(h00)
A
B
C
N M
S R
1 13 3
2 2
apath difference:11 and 22 (NCM) sin2 001122 hdNCM
How is an atom located in a unit cell affect the h00 diffraction peak?
why:? Meaningful!
AChadh /00
How is the diffraction peaks (hkl) of a structure named? Unit cell
Miller indices (h00):
x̂path difference: 11 and 33 (SBR)
ahx
hax
ACABSBR /1133
phase difference (11 and 33) ahx
ahx
22
1133
position of atom B: fractional coordinate of a: u x/a.
huahx 22
1133
the same argument B: x, y, z x/a, y/b, z/c u, v, wDiffraction from (hkl) plane
)(2 lwkvhu
electron singlea by scattered amplitudecellunit a of atoms allby scattered amplitude F
F: amplitude of the resultant wave in terms of the amplitude of the wave scattered by a single electron.
)(2)(22
)(21
222111 NNN lwkvhuiN
lwkvhuilwkvhui efefefF
N
n
lwkvhuinhkl
NNNefF1
)(2
F (in general) a complex number.
N atoms in a unit cell; fn: atomic form factor of atom n
How to choose the groups of atoms to represent a unit cellof a structure? 1. number of atoms in the unit cell2. choose the representative atoms for a cell properly (ranks of equipoints).
Example 1: Simple cubic 1 atoms/unit cell; 000 and 100, 010, 001, 110, 101, 011, 111: equipoints of rank 1; Choose any one will have the same result!
ffeF lkhihkl )000(2
22 fFhkl for all hkl
Example 2: Body centered cubic 2 atoms/unit cell; 000 and 100, 010, 001, 110, 101, 011, 111: equipoints of rank 1; ½ ½ ½: equipoints of rank 1; Two points to choose: 000 and ½ ½ ½.
)1( )()21
21
21(2)000(2 lkhilkhilkhi
hkl effefeF
fFhkl 2 when h+k+l is even 22 4 fFhkl
0hklF when h+k+l is odd 02 hklF
Example 3: Face centered cubic 4 atoms/unit cell; 000 and 100, 010, 001, 110, 101, 011, 111: equipoints of rank 1; ½ ½ 0, ½ 0 ½, 0 ½ ½, ½ ½ 1, ½ 1 ½, 1 ½ ½: : equipoints of rank 3; Four atoms chosen: 000, ½ ½ 0, ½ 0 ½, 0 ½ ½.
]1[ )()()(
)21
210(2)
210
21(2)0
21
21(2)000(2
lhilkikhi
lkhilkhilkhilkhihkl
eeef
fefefefeF
fFhkl 4 when h, k, l is unmixed (all evens or all odds)
22 16 fFhkl
0hklF when h, k, l is mixed 02 hklF
Example 4: Diamond Cubic 8 atoms/unit cell; 000 and 100, 010, 001, 110, 101, 011, 111: equipoints of rank 1;
½ ½ 0, ½ 0 ½, 0 ½ ½, ½ ½ 1, ½ 1 ½, 1 ½ ½: equipoints of rank 3;¼ ¼ ¼, ¾ ¾ ¼, ¾ ¼ ¾, ¼ ¾ ¾: equipoints of rank 4;Eight atoms chosen: 000, ½ ½ 0, ½ 0 ½, 0 ½ ½ (the same as FCC),¼ ¼ ¼, ¾ ¾ ¼, ¾ ¼ ¾, ¼ ¾ ¾!
)43
43
41(2)
43
41
43(2)
41
43
43(2)
41
41
41(2
)21
210(2)
210
21(2)0
21
21(2)000(2
lkhilkhilkhilkhi
lkhilkhilkhilkhihkl
fefefefe
fefefefeF
)41
41
41(2
)21
210(2)
210
21(2)0
21
21(2
1
1
lkhi
lkhilkhilkhi
hkl
e
eeefF
FCC structure factor
)1(4 ifFhkl when h, k, l are all odd 22 32 fFhkl fFhkl 8 when h, k, l are all even and h + k + l = 4n
22 64 fFhkl 0)11(4 fFhkl
when h, k, l are all even and
02 hklFh + k + l 4n
0hklF when h, k, l are mixed 02 hklF
Example 5: HCP 2 atoms/unit cell 8 corner atoms: equipoints of rank 1; 1/3 2/3 ½: equipoints of rank 1; Choose 000, 1/3 2/3 1/2.
(000)
(001)
(100)(010)
(110)
( 1/3 2/3 1/2)
equipoints
)21
32
31(2)000(2 lkhilkhi
hkl fefeF
Set [h + 2k]/3+ l/2 = g
)1( 2 ighkl efF
gfgfeefF igighkl 2222222 cos4)2cos22()1)(1(
)
232(cos4cos4 22222 lkhfgfFhkl
h + 2k l
3m3m
3m13m1
evenoddevenodd
)23
2(cos2 lkh 2hklF
10
0.250.75
4f 2
0f 2
3f 2
Multiplicity FactorEqual d-spacings equal B
E.g.: Cubic (100), (010), (001), (-100), (0-10), (00-1): Equivalent Multiplicity Factor = 6 (110), (-110), (1-10), (-1-10), (101), (-101), (10-1),(-10-1), (011), (0-11), (01-1), (0-1-1): Equivalent Multiplicity Factor = 12
lower symmetry systems multiplicities .
E.g.: tetragonal (100) equivalent: (010), (-100), and (0-10) not with the (001) and the (00-1). {100} Multiplicity Factor = 4 {001} Multiplicity Factor = 2
Multiplicity p is the one counted in the point groupstereogram.In cubic (h k l)
}{hkl p = 48 3x2x23 = 48}{hhl p = 24 3x23 = 24}0{ kl p = 24 3x23 = 24
}0{ kk p = 12 3x22 = 12
}{hhh p = 8 23 = 8}00{h p = 6 3x2 = 6
Lorentz factor:
dependence of the integrated peak intensities
1. finite spreading of the intensity peak
2. fraction of crystal contributing to a diffraction peak
3. intensity spreading in a cone
2sin1
cos
2sin1
2B
Inte
nsity
Diffraction Angle 2
Imax
Imax/2
2
IntegratedIntensity
B
2 1 2
A Ba
2
C D
Na
1N
B2
1
2 B1
B2
1
path difference for 11-22= AD – CB = acos2 - acos1
= a[cos(B-) - cos (B+)]= 2asin()sinB ~ 2a sinB.
2Na sinB = completelycancellation (1- N/2, 2- (N/2+1) …)
1
BNa sin2
Maximum angular range of the peak
Imax 1/sinB, Half maximum B 1/cosB (will be shown later) integrated intensity ImaxB (1/sinB)(1/cosB) 1/sin2B.
2
number of crystals orientated at or near the Bragg angle
crystal plane
rrN B )90sin(2
2cos
4)90sin(2
2
B
B
rrr
NN
Fraction of crystal: r/2-
)90sin( Br
diffracted energy:equally distributed (2Rsin2B) the relative intensity per unit length 1/sin2B. 2B
3
cossin41
2sincos
2sin1cos
2sin1factor Lorentz
2
2
BBB
B
Lorentz–polarization factor:(omitting constant)
cossin2cos1factor onpolarizati-Lorentz
2
2
Lorentz factor:
0 20 40 60 80 100 120 140 160 180 2000
20
40
60
80
100
Lo
rent
z-Po
lariz
atio
n Fa
ctor
2 (Degrees)
Absorption factor:
X-ray absorbed during its in and out of the sample.Hull/Debye-Scherrer Camera: A(); A() as .
ldx
I0
2
A
B
CdID
x
Incident beam: I0; 1cm2
incident angle . Beam incident on the plate: )(
0ABeI
: linear absorption coefficient
a: volume fraction of the specimen that are at the right angle for diffractionb: diffracted intensity/unit volume
1cmDiffractometer:
volume = l dx 1cm = ldx.actual diffracted volume = aldxDiffracted intensity:Diffracted beam escaping from the sample:
dxeablI AB)(0
dxeeablI BCAB )()(0
sin;
sin;
sin1 xBCxABl
dxeabIdIx
D
sin
1sin
10
sin
If = = dxeabIdI xD
sin/20
sin
2sin2
20
0sin2
00
abIxdeabIdIIx
x
xx
x DD
Infinite thickness ~ dID(x = 0)/dID(x = t) = 1000 and = = ).
Temperature factor (Debye Waller factor):Atoms in lattice vibrate (Debye model)
d
u
d
u
high Blow B
Lattice vibration is more significant at high B
(u/d) as B
Temperature (1) lattice constants 2 ; (2) Intensity of diffracted lines ; (3) Intensity of the background scattering .
Meff 0Formally, the factor is included in f asBecause F = |f 2| factor e-2M shows upWhat is M?
2222
2
22 sinsin222
BB Bu
duM
u0u
02 u
: Mean square displacement
Debye:
2u2
2
2 sin4
)(6
Bxx
mkThM
h: Plank’s constant;T: absolute temperature;m: mass of vibrating atom;: Debye temperature of the substance; x = /T;(x): tabulated function
e-2M
sin /
12
4
2
2 1015.16
A
Tmk
Thm atomic weight (A):
Temperature (Thermal) diffuse scattering (TDS) as I as
peak width B slightly as T
TDSI
2 or sin/
0
SummaryIntensities of diffraction peaks from polycrystalline samples:
MeApFNI 22
222 )(
cossin2cos1
Diffractometer:
MepFNI 22
222
cossin2cos1
Perturbation: preferred orientation; Extinction (large crystal)
Other diffraction methods:
Match calculation? Exactly: difficult; qualitatively matched.
ExampleDebye-Scherrer powder pattern of Cu made withCu radiation
1 2 3 4 5 6 7 8line hkl h2+k2+l2 sin2 sin (o) sin/(Å-1) fCu
12345678
111200220311222400331420
3481112161920
0.13650.18200.3640.5000.5460.7280.8650.910
0.3690.4270.6030.7070.7390.8530.9300.954
21.725.337.145.047.658.568.472.6
0.240.270.390.460.480.550.600.62
22.120.916.814.814.212.511.511.1
Cu: Fm-3m, a = 3.615 Å
1 9 10 11 12 13 14
line |F|2 P Relative integrated intensityCalc.(x105) Calc. Obs.
12345678
78106990452035003230250021201970
86
122486
2424
12.038.503.702.832.743.184.816.15
7.52 3.562.012.380.710.482.452.91
10.04.72.73.20.90.63.33.9
VsSssmwss
cossin2cos1
2
2
If h, k, l are unmixedIf h, k, l are mixed
CufF 4
0F
111200220311222400331420
Structure Factor
2
222
2
1a
lkhdhkl
3111ad
542.1sin3
615.32542.1sin2 111111111 d
24.0542.13694.0sin 111
o111111 68.213694.0sin
0 0.1 0.2 0.3 0.429 27.19 23.63 19.90 16.48
sin
1.2263.2390.19
2.03.090.19
24.03.0 111111
Cu
Cu
ff
7814)4( 21112111 CufF
}111{ p = 8 (23 = 8)
05.12cossin2cos1
1111112
1112
MeApFNI 22
222 )(
cossin2cos1
753370I
Dynamic Theory for Single crystal Kinematical theoryDynamical theory
K0 K0 K1
S0 S
K1
K2 K2
K1 K2(hkl)
RefractionPRIMARY EXTINCTIONK0 & K1 : /2; K1 & K2 : /2 K0 & K2 : ; destructive interference
2|2cos|1
2sin||
38 2
2
2
FNmceI I |F| not |F|2!
Negligible absorption
e: electron charge; m: electron mass; N: # of unit cell/unit volume.
2|2cos|1
2sin||2
2
2
FNmces
FWHM for Darwincurve = 2.12s
5 arcs < < 20 arcs
Width of the diffraction peak (~ 2s)