Chapter 2 Elastic Scattering of Waves
1. Experimental setting of X-ray diffraction
Detector
 
r r'
Sample
(Scattering
centers)

r'
2
 
s 's

s

r
O
2
X-ray source

s'
 Incident
X-ray is a plane wave
 

 

r
r

 ( , t) = A exp i( s - t), where s is the propagation

vector.  s = 2/
 X-ray interacts with electrons.
 The interaction is regarded as elastic scattering of



photons, so that  s =  s ' = 2/. s ' is the wave
vector of the diffracted beam.
    
 s // r ' , s ' // r  r '
 Define Scattering Angle = 2
  
  s = s'  s

 s = 2s sin
= 2 sin   2/
= 4 sin /
1
2. Scattered wave from a small element of electron cloud
Scattering
center
Scattered
spherical
wave

s

 At t1, the X-ray reaches a scattering center at r ' , such
 

that the E-field is  ( r ' , t1) = A exp i( s  r ' -  t1)

 The X-ray is scattered from r ' as a spherical wave,
such that the E-field reaching
the detector at t1+t is :
 
  

A exp i( s  r 't1 )
 scatt( r , t) =
exp i [ s '  ( r  r ' )  t]
 
r  r'
 
  
A' ' exp is  r '
=
exp i [ s '  ( r  r ' )  t].
 
r  r'
 
 For a small sample, the variation of r  r ' is small

 
when r ' changes. r  r ' in the amplitude is regarded as
unchanged.
  
 The variation of the phase s '  ( r  r ' )  r'/  r'/0.1
nm and changes very fast when r' changes. Hence the
 
r
factor  r ' in the phase cannot be regarded as a
constant.
 
  

 scatt( r , t) = A' exp (i s  r ' ) exp i [ s '  ( r  r ' )  t]
 
  
r

= A' exp i( s '  t) exp i[( s  s ' )  r ' ]
 
 
= A' exp i( s '  r  t) exp i( s  r ' )
2
3. Scattering by one atom :

scatt,1 atom( r ,t)
 
 
 
= A' exp i( s '  r t)  exp(i s  r ' ) n( r ' )d r '3 ,

where n( r ' ) is the electron density.
 
n( r ' )d r '3 is the number of electrons inside the

volume d r '3 .
4. Atomic form factor
 Let the center of the jth atom to be rj .



 Decompose r ' as rj + r '' , where r '' is the position of
the electron cloud element w.r.t. the position of the
atom.
 
 scatt,1 atom= A' exp i( s '  r t)




 expi s  ( rj + r '' ) nj( r '' )d r '' 3
 

=A' exp i( s '  r t) exp (i s  rj )
 

 3

exp(i
)
n
(
)d
r
'
'
r
'
'
s
r
'' ,
j

 


Define fj =  exp(i s  r '' ) nj( r '' )d r '' 3 as the Atomic
Form Factor of the jth atom.
 

 scatt,1 atom= A' fj exp i( s '  r t) exp (i s  rj )

 fj is determined by the electron density and  s .

r ''

r'
r'

rj
3
5. Diffraction by all atoms
 

 scatt( r ,t) = A' exp i( s '  r t)



 
     fm expi s  (n1 a +n2 b +n3 c + pm ),
n1 n n3 m
2
where n1, n2 and n3 scan over all integers (lattice
points),
over all atoms in the basis.
 and m scans
 
 
scatt( r ,t) = A' exp i( s '  r t) [  fm expi s  pm ]
m
N
N
3
2
 
 
 
[  exp(in1 s  a )][  exp(in2 s  b )][  exp(in3 s  c )]
N
1
n11
n3 1
n2 1
 
 Define Structure Factor F =  fm expi s  pm ,
m
which is related to the structure of the basis.
The intensity of the scattering wave is:
I  scatt scatt* = scatt2
N
N
 

1
1

= A2 F2 [  exp(in1 s  a )][  exp(in1 s  a )]*
n11
n11
N
  N2
 
2
 [  exp(in2 s  b )][  exp(in2 s  b )]*
n2 1
n2 1
N
  N3
 
3
[  exp(in3 s  c )][  exp(in3 s  c )]*
n3 1
n3 1
N1a
N3c
N2b
c
a
4
6. Bragg Condition
N
 Consider the sum  exp(inx) = e-ix [1 exp(iNx)]
1 exp(ix)
n1

[1 exp(iNx)] [1 exp(iNx)]*
1 exp(ix) 1 exp(ix)
iNx  e- iNx ) 1
1

exp(

i
Nx
)
1

exp(
i
Nx
)
1

(
e
][
]=
= [
1 exp(ix) 1 exp(ix)
1 (eix  e- ix ) 1
2 Nx / 2)
2 ( Nx / 2)
1

(
1

2
sin
sin
2

2
cos
Nx
=
=
=
2
2  2cos x
1 (1 2sin x / 2)
sin 2 ( x / 2)
2 ( Nx / 2)
sin

has maximum value when x/2 = m,
sin 2 ( x / 2)
where m = 0, 1, 2,.. (integer).
 
 We may replace x by  s  a , so that

2 ( N s  a / 2) sin 2 ( N s  b / 2) sin 2 ( N s  c / 2)
sin
3
1
2 
I =A2F2
 



sin 2 (s  c / 2)
sin 2 (s  a / 2)
sin 2 (s  b / 2)
 I is maximum when
 
 s  a /2 = nh,

 s  b /2 = nk,
 
 s  c /2 = nl,
where n, h, k and l are integers.






 Consider  s = nh A +nk B +nl C = n G , where G is a
reciprocal lattice vector,
  

 
 s  a /2 = n(h A +k B +l C )  a /2 = nh, and

 s  b /2 = nk,
 
 s  c /2 = nl,

Laue condition :  s in this form gives maximum
intensity of the scattered wave.
5


 From Laue condition :  s = n G .

Sec. B.1  s = 4 sin /
Sec. A.9   G  = 2/d
4 sin / = 2n/d, gives
Bragg condition
2d sin  = n
When n = 1, h, k and l do not have common factor, 2d
sin = 

s'

d
2

2
 
 s =G

s

lattice
point


G (=  s ) is  to the (hkl) planes, i.e. (hkl) planes are
involved in the diffraction.
 For a set of lattice planes with a fixed spacing d and
fixed , n = 1 corresponds to the lowest . n > 1
corresponds to larger , namely, higher order
diffraction.
 2(d/n) sin  =  describes the diffraction from planes
with separation smaller than d. These planes would
contain lattice points only when a non-primitive lattice
is used for discussion.
6
7. Real Lattice Space and Reciprocal Lattice Space
Reciprocal
Real space
 space

 

Fundamental A = 2 b  c /,
a =2 B  C /Vc,



lattice vectors B = 2 c  a /,
b =2 C  A /Vc,




C = 2 a  b /.
c
 B /V
c.


  =2 A


Lattice vector G =h A + k B + l C R = n1 a + n2 b + n3 c
Note: Vc = volume of a primitive
  unit cell of the -3
reciprocal lattice = A  B C . The unit is [L] , the


reciprocal of that of a  b  c .
Exercise : The fundamental lattice vectors of B.C.C. :

a =a( x̂ + ŷ  ẑ )/2, b =a( x̂ + ŷ + ẑ )/2, and

c = a( x̂  ŷ + ẑ )/2.
Show that the fundamental reciprocal lattice
vectors
are :


A =2( x̂ + ŷ )/a, B =2( ŷ + ẑ )/a, and

C =2( x̂ + ẑ )/a.
Exercise : Fundamental
lattice vectors of s.c. lattice are:



a =a x̂ , b =a ŷ , and c = a ẑ . Show that the
fundamental
 vectors are:

 reciprocal lattice
A =2 x̂ /a, B =2 ŷ /a, and C =2 ẑ /a.
Exercise : Find the reciprocal lattice vectors of :
(a) A linear atom chain with atomic
separation a.


(b) F.C.C. with a =a( x̂ + ŷ )/2, b =a( ŷ + ẑ )/2

and c =a( x̂ + ẑ )/2.
7
8. Structure factor effect:
 I  2 F2 […]2 […]2 […]2
 Laue condition ensures […]2 are maximized
 F2 also affects I
Example : Find F of FCC
 
Assuming that the condition of s = G is satisfied.
 
F =  fm exp (i s  pm )
m
 
=  fm exp (i G  pm )
m




G =h A + k B + l C




pm = um a + vm b + wm c , where um, vm and wm  1
 
G  pm = 2 (h um + k um + l wm)
Then F =  fm exp i2 (h um + k um + l wm)
m
=f {1+expi(h+k) + expi(k+l)+ expi(h+l)}
[Consider the four atoms at 000, 12 12 0, 0 12 12 , 12 0 12 . fm = f.]
(i)
(ii)
(iii)
(iv)
h,k,l all even, F = 4f  maximum
h,k,l all odd, F = 4f  maximum
2 even, 1 odd, F =f (1+1-1-1) = 0
2 odd, 1 even, F = 0.
Note : contributions of the (100)
and (200) planes cancel each other.
x
8
z
y
Example : Structure factor of CsCl.
A basis contains atoms
at:


  
p = 0, p = ( a + b + c )/2.
Cs
Cl
F =  fm exp i2 (h um + k um + l wm)
m
= fCs exp (0) + fCl exp i2 (h/2 + k/2 + l/2)
= fCs + fCl exp i (h + k + l)
(i) h + k + l is odd, F = fCs  fCl  weak line (not
completely cancelled)
(ii) h + k + l is even, F = fCs + fCl  strong line
z
z
(100)
(100)
d
d
(200)
y
y
x
x
For s.c. cubic, (100) planes
give 2d sin  = 
 constructive interference
For CsCl, contribution
from the (100) and (200)
planes partially cancelled
with each others 
weakening of I
Exercise : For B.C.C., show that
F = f [1+expi(h+k+l)]
9
9. Table of diffraction lines of cubic systems
h2+k2+l2
1
2
3
4
5
6
8
9
10
11
12
13
14
16
S.C.
100
110
111
200
210
211
220
300,221
310
311
222
320
321
400
Cubic
hkl
F.C.C. B.C.C. Diamond
--------110
--111
--111
200
200
----------211
--220
220
220
--------310
--311
--311
222
222
----------321
--400
400
400
10
10. Ewald Construction
Step 1 : Draw the reciprocal lattice space

Step 2 : Draw the incident wave vector s , with the
end of the vector located at one reciprocal
lattice point

Step 3 : Draw a circle with the other end of the s

vector as the center and a radius of  s .
Step 4 : If a reciprocal lattice points is cut by the

circle, the vector s ' connecting the point
and the center of the circle would be the
wavevector of the constructive scattered Xray
 beam.
 The reciprocal lattice vector
G =h A +k B +l C connecting the two
reciprocal points as shown is perpendicular
to the (hkl) planes given the strong
diffraction peak.


    (hkl) plane

s
'
     
  s    
     
11
11. Experimental techniques
 X-ray tube : hot e bombard a metal target
 Bremstrahlung radiation : When electrons are
decelerated, a continuous spectrum of X-ray radiation
is generated.
 If the electrons knock out some electrons from the
inner electron shells of the atoms, the electrons of the
outer electrons shells would drop to fill up the
vacancies. This gives K and K lines of radiation.
Example
Radiation
Cu K1
Wavelength (nm) 0.1540562
Intensity
very strong
Cu K2
Cu K
0.154439 0.1392218
strong
weak
Structure of an X-ray tube
glass sheild
metal sheild
water cooling
filament
leads
high voltage
target
Be window
K (Characteristic radiation)
Counts
K
25 kV
20 kV
15kV
Wavelength
12
5kV
0.3 nm

Monochromators
Absorbing filters
Target
Filter
Mo
Zr
Cu
Ni
Co
Fe
Fe
Mn
Cr
V
Transmittance
0.29
0.42
0.45
0.48
0.49
No filter
Ni filter
K
Counts
K
K
K
0.12
0.14
0.16
 (nm)
0.18
Single crystal filters
13
0.12
0.14
0.16
0.18
 Collimator
Pinhole collimator
S

Detectors : films, proportional counter, GeigerMuller counter, scintillation counter,
semiconductor detector.
Laue method :
 Single crystal sample
 White light (Bremstrahlung radiation)
 The shortest and longest wavelengths min and max


give  smax = 2/min and  smin = 2/max
 The Ewald construction shows that all the reciprocal
lattice points between the two spheres give strong
peaks.
    
    
D F
    
    





14
2
r
Powder method
 Powder sample (random orientation) and single 
(monochromatic)
 A film is rolled in a cylindrical shape with the sample
placed at the center
 The diffraction pattern consists of concentric rings. 2
= U /2
R

s'
Normal
film
2

s
R
U
X-ray diffractometer
 2 scan : when the sample is moved by , the
detector is moved by 2
Detector
X-ray tube

15