Transcript 投影片 1
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:
P
r
0
I I0
4
2
e4
2 2
m r
= /2
K
2
2
sin I 0 2 sin
r
0: 410-7 mkgC-2
by J.J. Thomson
a single electron charge e (C), mass m (kg),
distance r (meters)
z
P
E
2
Ey Ez
2
2
r
2
x
y component
z component
O
Random
polarized
y
1
2
E E E I0 y I0z
2
2
y
= yOP = /2
= zOP = /2 -2
I P I Py I Pz I 0 y
K
r
2
I0z
2
z
I Py I 0 y
I Pz I 0 z
1
2
I0
K
r
2
K
r
2
cos 2
2
2
K
1
cos
2
2
cos 2 I 0 2
2
r
r
2
K
Polarization factor
Polarization Factor
1.0
0.8
0.6
0.4
0
20
40
60
80
100 120 140 160 180
2 (Degrees)
Pass through a monochromator first (Bragg angle M)
the polarization factor is ?
z
z
P
2M
Random
polarized
r
P
2
y
O
P
r
O
polarization is
not complete
random
anymore
y
x
x
I P I P y I P z I 0 y
K
r
2
I0z
1 cos 2 cos 2 M
2
K
r
2
cos 2 M
2
2
K 1 cos 2 M
I 0 2
r
2
2
1 cos 2 M
2
(Homework)
1.0
Si (111) as monochromator
o
Cu K; M= 28.44
Polarization Factor
0.9
0.8
0.7
0.6
0.5
0.4
0
20
40
60
80
100 120 140 160 180
2 (Degrees)
Atomic scattering (or form) factor
a single free electron atoms
atomic scattering
factor
amplitude
amplitude
scattered
scattered
by atom
by a single electron
x2
path different (O and dV):
R-(x1 + x2).
x1 r s 0 ;
x2 R r s
x1 dV
O r
s0
2
s
R
Differential atomic scattering factor (df) :
df
1
( r )e
( 2 i / )[ r ( s s 0 )]
Ee
Electron density
dV
Ee: the magnitude of the wave
from a bound electron
Phase difference
Spherical integration dV = dr(rd) (rsind)
rsind
r: 0 -
:0-
: 0 - 2
dr
rsin(+d)d
d
r
d
http://pleasemakeanote.blogspot.tw/2010/02/9-derivation-ofcontinuity-equation-in.html
2
dr ( rd )( r sin d ) 2
r 0 0
4
r
3
r
3
sin d
0
=2
3
r dr
2
r
2
r
3
2
r dr
sin d
0
r
0
2 r sin d dr
2
Evaluate (S - S0)r = | S - S0||r|cos
|(S - S0)|/2 = sin.
S
S0
S-S0
( S S 0 ) r 2 r sin cos
2
1
( 2 i / )[ r ( s s 0 )]
f ( ) df
( r )e
dV
Ee
1 r
4 ri sin cos /
2
f ( ) df
(
r
)
e
2
r
sin d dr
r
0
0
Ee
d cos
Let k 4 sin /
1 r
2
kri cos
f ( )
(
r
)
2
r
dr
e
d cos
0
E e r0
e
ikr cos
cos
ikr
cos 0
e
ikr
e
ikr
ikr
2 sin kr
kr
1
f ( )
Ee
r
r0
4 r ( r )
2
sin kr
dr
kr
For n electrons in an atom
1
f ( )
Ee
n electrons
r
r0
4 r ( r )
2
sin kr
dr
Tabulated
kr
For = 0, only k = 0 sinkr/kr = 1.
f (0)
1
Ee
n electrons
r
r0
4 r ( r ) dr Z
2
equal to 1 bound
electrons
Number of electrons
in the atom
Anomalous Scattering:
Previous derivation: free electrons!
Electrons around an atom: free?
m
d x
dt
2
k
free electron
harmonic oscillator
2
F kx
m
Assume x e i t
0
m (i 0 ) e
2
i 0 t
ke
i 0 t
m k
2
0
0 (k / m )
Resonance frequency
2
m
d x
dt
2
1/ 2
kx F ( t )
Forced oscillator
Assume F ( t ) F0 e i t
Assume x Ce i t
Cm ( i ) e
2
i t
kCe
i t
k m0
2
F0 e
i t
Cm
2
kC F0
Cm m C F0 mC ( ) F0 C
2
2
0
2
0
2
F0
m ( 0 )
2
2
x Ce
i t
F0
x
m ( 0 )
2
2
e
i t
Same frequency as F(t), amplitude(, 0)
= 0 C is ; in reality friction term exist no
Oscillator with damping (friction v)
2
m
d x
dt
2
c
dx
dt
2
m
d x
dt
2
m
kx F ( t ) F0 e
dx
dt
assume c = m
i t
Assume x = x0eit
m0 x F
2
( i ) x 0 i x 0 m 0 x 0 F0 / m
2
x0
2
F0
m ( 0 i )
2
2
Real part and imaginary part
if 0
2
1
( 0 i )
2
2
E
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)
f +f + if
real
imaginary
Examples:
Si, 400 diffraction peak, with Cu K (0.1542 nm)
d 400 0 . 54309 / 4 0 . 13577
nm
2 d 400 sin 0 . 1542 34 . 6
sin
sin
1
0 . 368
8 . 22 7 . 20
8 . 22 f
0 .3 0 .4
0 . 3 0 . 368
0.3
0.4
8.22 7.20
f 7 . 526
Anomalous Scattering correction
f 0 . 2 ; f 0 . 4
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
How is the diffraction peaks (hkl) of a structure named? Unit cell
How is an atom located in a unit cell affect the h00 diffraction peak?
Miller indices (h00):
d h 00 a / h AC
3
1
A
2
1
S B R
2
C
plane a
(h00)
path difference:11 and 22 (NCM)
2211 NCM 2d h 00 sin
why:? Meaningful!
path difference: 11 and 33 (SBR)
AB
x
hx
3311 SBR
AC
a/h
a
N
3
xˆ
M
phase difference (11 and 33)
3311
2 hx
2hx
a
a
position of atom B: fractional coordinate of a: u x/a.
2hx
3311
2hu
a
the same argument B: x, y, z x/a, y/b, z/c u, v, w
Diffraction from (hkl) plane
2 (hu kv lw)
amplitude scattered by all atoms of a unit cell
F
amplitude scattered by a single electron
F: amplitude of the resultant wave in terms of the amplitude
of the wave scattered by a single electron.
N atoms in a unit cell; fn: atomic form factor of atom n
F f1e 2i ( hu1 kv1 lw1 ) f 2 e 2i ( hu2 kv2 lw2 ) f N e 2i ( hu N kvN lwN )
N
Fhkl
fne
2 i ( hu N kv N lw N )
n 1
F (in general) a complex number.
How to choose the groups of atoms to represent a unit cell
of a structure?
1. number of atoms in the unit cell
2. 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!
Fhkl fe 2i ( h 0 k 0l 0 ) f
Fhkl f
2
2
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 1 1
2i ( h k l )
2 2 2
Fhkl fe 2i ( h 0 k 0l 0 ) fe
Fhkl 2 f when h+k+l is even
Fhkl 0
when h+k+l is odd
f (1 ei ( h k l ) )
Fhkl 4 f
2
Fhkl 0
2
2
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.
Fhkl fe 2i ( h 0 k 0l 0 ) fe
1 1
2i ( h k l 0 )
2 2
fe
1
1
2i ( h k 0 l )
2
2
fe
1 1
2i ( h 0 k l )
2 2
f [1 ei ( h k ) ei ( k l ) ei ( h l ) ]
Fhkl 4 f
when h, k, l is unmixed (all evens or all odds)
Fhkl 16 f
2
Fhkl 0
2
when h, k, l is mixed
Fhkl 0
2
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!
Fhkl fe 2i ( h 0 k 0l 0 ) fe
fe
Fhkl
1 1 1
2i ( h k l )
4 4 4
1 1
2i ( h k l 0 )
2 2
fe
fe
3 3 1
2i ( h k l )
4 4 4
1
1
2i ( h k 0 l )
2
2
fe
fe
3 1 3
2i ( h k l )
4 4 4
1 1
2i ( h 0 k l )
2 2
fe
1 3 3
2i ( h k l )
4 4 4
1 1
1
1
1 1
2i ( h k l 0 )
2i ( h k 0 l )
2i ( h 0 k l )
2
2 2
f 1 e 2 2 e 2
e
1 1 1
2i ( h k l )
4 4 4
1
e
FCC structure factor
Fhkl 4 f (1 i ) when h, k, l are all odd
Fhkl 8 f
Fhkl 32 f
when h, k, l are all even and h + k + l = 4n
Fhkl 64 f
2
2
Fhkl 4 f (1 1) 0 when h, k, l are all even and
2
h + k + l 4n
Fhkl 0
2
Fhkl 0 when h, k, l are mixed
Fhkl 0
2
2
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.
Fhkl fe 2i ( h 0 k 0l 0 ) fe
1 2 1
2i ( h k l )
3 3 2
Set [h + 2k]/3+ l/2 = g
(001) ( 1/3 2/3 1/2)
(000)
(010)
(100) (110)
equipoints
Fhkl f (1 e 2ig )
Fhkl f 2 (1 e 2ig )(1 e 2ig ) f 2 (2 2 cos 2g ) 4 f 2 cos 2 g
2
Fhkl
2
h 2k l
4 f cos g 4 f cos (
)
3
2
2
2
2
2
h + 2k
l
h 2k l
cos (
)
3
2
3m
3m
3m1
3m1
even
odd
even
odd
1
0
0.25
0.75
2
Fhkl
4f 2
0
f2
3f 2
2
Multiplicity Factor
Equal 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 group
stereogram.
In cubic (h k l)
{hkl }
{hhl }
{ 0 kl }
p = 48
3x2x23 = 48
p = 24
3x23 = 24
p = 24
3x23 = 24
{ 0 kk }
p = 12
3x22 = 12
{hhh }
{h 00 }
p=8
p=6
23 = 8
3x2 = 6
Lorentz factor:
dependence of the integrated peak intensities
1. finite spreading of the intensity peak
1
sin 2
2. fraction of crystal contributing to a diffraction peak cos
sin 2
3. intensity spreading in a cone
1
2
1
Imax
B
1
2
2
Intensity
1 B
2 B
path difference for 11-22
= AD – CB = acos2 - acos1
= a[cos(B-) - cos (B+)]
= 2asin()sinB ~ 2a sinB.
2Na sinB = completely
cancellation (1- N/2, 2- (N/2+1) …)
1
Imax/2
B
Integrated
Intensity
2B
Diffraction Angle 2
1
2
2
D
C
2
a
A
1
B
N
Na
Maximum angular range of the peak
2 Na sin B
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
N 2r sin(90 B ) r
r sin( 90 B )
/2-
Fraction of crystal:
N 2r sin(90 B ) r
2
N
4r
cos B
2
crystal plane
r
3
diffracted energy:
equally distributed (2Rsin2B)
the relative intensity per unit length 1/sin2B.
2B
Lorentz factor:
1
1
cos
cos B
Lorentz factor
2
sin
2
sin
2
sin
2 B
B
B
1
4 sin 2 cos
Lorentz–polarization factor:
(omitting constant)
1 cos2 2
Lorentz - polarization factor 2
sin cos
Lorentz-Polarization Factor
100
80
60
40
20
0
0
20
40
60
80 100 120 140 160 180 200
2 (Degrees)
Absorption factor:
X-ray absorbed during its in and out of the sample.
Hull/Debye-Scherrer Camera: A(); A() as .
Diffractometer:
Incident beam: I0; 1cm2
incident angle .
Beam incident on the plate: I 0 e ( AB )
a: volume fraction of the specimen that
are at the right angle for diffraction
b: diffracted intensity/unit volume
I0 1cm dID
C
x A
B
dx
l
2
: linear absorption
coefficient
volume = l dx 1cm = ldx.
actual diffracted volume = aldx
Diffracted intensity: ablI 0 e ( AB) dx
Diffracted beam escaping from the sample: ablI 0 e ( AB ) e ( BC ) dx
1
x
x
l
; AB
; BC
sin
sin
sin
I 0 ab
dI D
e
sin
If = =
ID
x
x 0
1
1
x
sin sin
dx
I 0 ab 2 x / sin
dI D
e
dx
sin
2 x
I 0 ab x sin 2 x I 0 ab
dI D
e
d
x
0
2
sin 2
Infinite thickness ~ dID(x = 0)/dID(x = t) = 1000 and = = ).
Temperature factor (Debye Waller factor):
Atoms in lattice vibrate (Debye model)
(1) lattice constants 2 ;
Temperature (2) Intensity of diffracted lines ;
(3) Intensity of the background scattering .
u
u
d
low B
d
high B
Lattice vibration is more significant at high B
(u/d) as B
Formally, the factor is included in f as f f 0 e M
Because F = |f 2| factor e-2M shows up
What is M?
2
2
2
u
2 sin B
sin B
M 2 2 2 2 2 u 2
B
d
u 2 : Mean square displacement
Debye: M
2
6h T
mk 2
x sin B
(
x
)
4
2
h: Plank’s constant;
T: absolute temperature;
m: mass of vibrating atom;
: Debye temperature of the substance; x = /T;
(x): tabulated function
u 0
u
u2 0
e-2M
6h 2T 1.15 10 4 T
m atomic weight (A):
2
mk
A 2
1
0
sin /
I
TDS
2 or sin/
Temperature (Thermal) diffuse scattering (TDS) as
I as
peak width B slightly as T
Summary
Intensities of diffraction peaks from polycrystalline samples:
Diffractometer:
I N
2
2
F
1 cos 2 2
p
2
sin
cos
2M
e
Other diffraction methods:
I N
2
F
2
1 cos 2 2
p
2
sin
cos
A ( ) e 2 M
Match calculation? Exactly: difficult; qualitatively matched.
Perturbation: preferred orientation; Extinction (large crystal)
Example
Debye-Scherrer powder pattern of Cu made with
Cu radiation
Cu: Fm-3m, a = 3.615 Å
1
line
1
2
3
4
5
6
7
8
2
3
hkl h2+k2+l2
111
3
200
4
220
8
311
11
222
12
400
16
331
19
420
20
4
sin2
0.1365
0.1820
0.364
0.500
0.546
0.728
0.865
0.910
5
sin
0.369
0.427
0.603
0.707
0.739
0.853
0.930
0.954
6
7
(o) sin/(Å-1)
21.7
0.24
25.3
0.27
37.1
0.39
45.0
0.46
47.6
0.48
58.5
0.55
68.4
0.60
72.6
0.62
8
fCu
22.1
20.9
16.8
14.8
14.2
12.5
11.5
11.1
Structure Factor
F 4 f Cu
F 0
1
9
If h, k, l are unmixed
If h, k, l are mixed
10
11
1 cos 2
2
111
200
220
311
222
400
331
420
line
|F|2
P
1
2
3
4
5
6
7
8
7810
6990
4520
3500
3230
2500
2120
1970
8
6
12
24
8
6
24
24
sin cos
2
12.03
8.50
3.70
2.83
2.74
3.18
4.81
6.15
12
13
14
Relative integrated intensity
Calc.(x105) Calc.
Obs.
7.52
10.0
Vs
3.56
4.7
S
2.01
2.7
s
2.38
3.2
s
0.71
0.9
m
0.48
0.6
w
2.45
3.3
s
2.91
3.9
s
d
h k l
2
1
2
hkl
2
a
2
2
d 111
2 d 111 sin 111 1 . 542 2
sin 111 0 . 3694 111
sin 111
sin
0 . 3694
a
3 . 615
3
sin 111 1 . 542
3
o
21 . 68
0 . 24
1 . 542
0
0.1
0.2
0.3
0.4
29 27.19 23.63 19.90 16.48
0 . 3 0 . 24
19 . 90 f
111
Cu
0 .3 0 .2
19 . 90 23 . 63
f Cu 22 . 1
111
F111
2
( 4 f Cu ) 7814
111
p=8
{111 }
1 cos 2 111
2
(23 = 8)
2
sin 111 cos 111
2
I N
2
F
2
I 753370
12 . 05
1 cos 2 2
p
2
sin
cos
2 M
A ( ) e
Dynamic Theory for Single crystal
Kinematical theory
Dynamical theory
S0
Refraction
PRIMARY EXTINCTION
K0
S
K0
K1
K1
K2
K2
K0 & K1 : /2; K1 & K2 : /2
K0 & K2 : ; destructive interference
K1 K2
(hkl)
Negligible absorption
8
I
3
e 2 N2 | F | 1 | cos 2 |
2
2
mc sin 2
I |F| not |F|2!
e: electron charge; m: electron mass; N: # of unit cell/unit volume.
Width of the diffraction peak (~ 2s)
e 2 N2 | F | 1 | cos 2 |
s 2
2
mc sin 2
FWHM for Darwin
curve = 2.12s
5 arcs < < 20 arcs