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
K 2
 2 2  sin   I 0 2 sin 
r
m r 
by J.J. Thomson
0: 410-7 mkgC-2
2
a single electron charge e (C), mass m (kg),
distance r (meters)

z
E 2  E y2  Ez2
P
r
2
x
y component
z component
O
Random
polarized
y


1 2
1
2
2
E  E y  Ez  I 0 y  I 0 z  I 0
2
2
 = yOP = /2
 = zOP = /2 -2
K
r2
K
 I 0 z 2 cos2 2
r
I Py  I 0 y
I Pz
2

K
K
K
1

cos
2 
2

I P  I Py  I Pz  I 0 y 2  I 0 z 2 cos 2  I 0 2 
r
r
r 
2

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
P
2M
r
z
Random
polarized
O
P
P
r
2
y
O
polarization is
not complete
random
anymore
y
x
x
I P  I Py  I Pz  I 0 y
2

K
K
K
1

cos
2 M
2
 I 0 z 2 cos 2 M  I 0 2 
2
r
r
r 
2
1  cos2 2 cos2 2 M
1  cos2 2 M
(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 scattered by atom
amplitude scattered by a single electron
x2
path different (O and dV):
R-(x1 + x2).
x1  r  s0 ; x2  R  r  s
x1 dV
O r
s0
2
s
R
Differential atomic scattering factor (df) :
1
df 
 ( r )e( 2i /  )[ r( ss0 )]dV
Ee
Ee: the magnitude of the wave
from a bound electron
Electron density
Phase difference
Spherical integration dV = dr(rd) (rsind)
rsind
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
  
r

2
dr
(
rd

)(
r
sin

d

)

2

r

 dr  sin d
0 0
4 3

r
3
3
r

3

2  r dr  sin d  
2
r
 0
r
 0
r


0
=2
2r sin ddr
2
Evaluate (S - S0)r = | S - S0||r|cos
|(S - S0)|/2 = sin.
S
S0
S-S0
 (S  S0 )  r  2r sin  cos 
2
1
f ( )   df    ( r )e( 2i /  )[ r( ss0 )]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 


r

0


0
Ee
ikr cos cos
e
eikr  e ikr 2 sin kr



ikr cos 0
ikr
kr
1
f ( ) 
Ee

r 
r 0
sin kr
4r  ( r )
dr
kr
2
For n electrons in an atom
1
f ( ) 
Ee
 
n electrons
r 
r 0
sin kr
4r  ( r )
dr
kr
2
Tabulated
For  = 0, only k = 0  sinkr/kr = 1.
1
f ( 0) 
Ee
 
n electrons
r 
r 0
4r 2  ( r )dr  Z
equal to 1 bound
electrons
Number of electrons
in the atom
Anomalous Scattering:
Previous derivation: free electrons!
Electrons around an atom: free?
2
k
free electron
harmonic oscillator
d x
m 2  F  kx
dt
Assume x  ei0t
m
1/ 2


(
k
/
m
)
m(i0 ) e  ke  m  k
0
Resonance frequency
2
d x
m 2  kx  F (t )
Forced oscillator
dt
Assume F (t )  F0 e it
2
k

m

0
Assume x  Ceit
i 0 t
2
Cm(i ) e
2
it
i0t
it
2
0
 kCe  F0 e
it
 Cm 2  kC  F0
F0
 Cm  m C  F0  mC (   )  F0  C 
m(02   2 )
2
2
0
2
0
2
x  Ce
i t
x
F0
i t
e
m(02   2 )
Same frequency as F(t), amplitude(, 0)
 = 0  C is ; in reality friction term exist  no 
Oscillator with damping (friction  v)
d 2x
dx
m 2  c  kx  F (t )  F0eit
assume c = m
dt
dt
d 2x
dx
it
2
Assume
x
=
x
e
0
m 2  m
 m0 x  F
dt
dt
(i)2 x0  ix0  m02 x0  F0 / m
 x0 
F0
m(    i )
2
0
2
Real part and imaginary part
 if   0
1
2
(02   2  i )

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 + if 
real
imaginary
Examples:
Si, 400 diffraction peak, with Cu K (0.1542 nm)
d 400  0.54309 / 4  0.13577 nm
2d 400 sin   0.1542    34.6
sin 
 1

 0.368 
sin 

0.3
0.4
8.22 7.20
8.22  7.20
0.3  0.4

 f  7.526
8.22  f
0.3  0.368
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)
 2211  NCM  2d h 00 sin   
why:?  Meaningful!
path difference: 11 and 33 (SBR)
AB
x
hx
 3311  SBR 

 
AC
a/h
a
N
3
x̂
M
phase difference (11 and 33)
3311
2 hx
2hx


 a
a
position of atom B: fractional coordinate of a: u  x/a.
2hx
 3311 
 2hu
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 2i ( hu1  kv1 lw1 )  f 2 e 2i ( hu2  kv2 lw2 )    f N e 2i ( hu N  kvN lwN )
N
Fhkl   f n e2i ( huN kv N lwN )
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 2i ( h 0 k 0l 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
2i ( h  k  l )
2 2 2
Fhkl  fe 2i ( h 0 k 0l 0 )  fe
Fhkl  2 f when h+k+l is even
Fhkl  0
when h+k+l is odd
 f (1  ei ( 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 2i ( h 0 k 0l 0 )  fe
1 1
2i ( h  k  l 0 )
2 2
 fe
1
1
2i ( h  k 0  l )
2
2
 fe
1 1
2i ( h 0  k  l )
2 2
 f [1  ei ( h  k )  ei ( k l )  ei ( 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 2i ( h 0 k 0l 0 )  fe
 fe
Fhkl
1 1 1
2i ( h  k  l )
4 4 4
1 1
2i ( h  k  l 0 )
2 2
 fe
 fe
3 3 1
2i ( h  k  l )
4 4 4
1
1
2i ( h  k 0  l )
2
2
 fe
 fe
3 1 3
2i ( h  k  l )
4 4 4
1 1
2i ( h 0  k  l )
2 2
 fe
1 3 3
2i ( h  k  l )
4 4 4
1 1
1
1
1 1
2i ( h  k  l 0 )
2i ( h  k 0  l )
2i ( h 0  k  l ) 

2
2 2
 f 1  e 2 2  e 2
e



1 1 1
2i ( 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 2i ( h 0 k 0l 0 )  fe
1 2 1
2i ( 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 2ig )
Fhkl  f 2 (1  e 2ig )(1  e 2ig )  f 2 (2  2 cos 2g )  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
3m1
3m1
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} p = 48 3x2x23 = 48
{hhl} p = 24 3x23 = 24
{0kl} p = 24 3x23 = 24
{0kk} p = 12 3x22 = 12
23 = 8
{hhh} p = 8
3x2 = 6
{h00} p = 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  
 1 
3. intensity spreading in a cone 

 sin 2 
2
1
Imax
B
1
2
2
Intensity

1   B  
 2   B  
path difference for 11-22
= AD – CB = acos2 - acos1
= a[cos(B-) - cos (B+)]
= 2asin()sinB ~ 2a sinB.
2Na sinB =   completely
cancellation (1- N/2, 2- (N/2+1) …)
1
Imax/2
B
Integrated
Intensity
2B
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/sinB,
Half maximum B  1/cosB (will be shown later)
 integrated intensity  ImaxB  (1/sinB)(1/cosB)  1/sin2B.
2
number of crystals orientated at or near the Bragg angle
N  2r sin(90   B )  r
r sin( 90   B )

/2-
Fraction of crystal:
N 2r sin(90   B )  r


2
N
4r
 cos  B

2
crystal plane
r

3
diffracted energy:
equally distributed (2Rsin2B)
 the relative intensity per unit length  1/sin2B.
2B
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:
2

1

cos
2  2 M
2
2
e
I  N F p 2
 sin  cos  
Other diffraction methods:
2

1

cos
2 
2
 A( )e 2 M
I  N F p 2
 sin  cos  
2
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
111
200
220
311
222
400
331
420
If h, k, l are unmixed
If h, k, l are mixed
1
9
10
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
11
12
13
14
1  cos 2 2 Relative integrated intensity
sin 2  cos  Calc.(x105) Calc.
Obs.
12.03
7.52
10.0
Vs
8.50
3.56
4.7
S
3.70
2.01
2.7
s
2.83
2.38
3.2
s
2.74
0.71
0.9
m
3.18
0.48
0.6
w
4.81
2.45
3.3
s
6.15
2.91
3.9
s
1
h2  k 2  l 2

2
2
d hkl
a
a
 d111 
3
3.615
2d111 sin 111  1.542  2
sin 111  1.542
3
sin 111  0.3694  111  21.68o
sin 111
sin 


0
0.3694

 0.24
1.542
0.1
0.2
0.3
0.4
29 27.19 23.63 19.90 16.48
0.3  0.24
0.3  0.2
111


f
Cu  22.1
111
19.90  f Cu
19.90  23.63
111 2
 F111  ( 4 f Cu
)  7814
2
{111}
p=8
(23 = 8)
1  cos 2111
 12.05
2
sin 111 cos 111
2
2

1  cos 2 
2
2
2 M


I  N F p 2
A( )e

 sin  cos  
I  753370
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  N2 | 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  N2 | F |  1 | cos 2 | 
s   2 


2

 mc   sin 2 
FWHM for Darwin
curve = 2.12s
5 arcs <  < 20 arcs