Problems and solutions
x-ray diffraction
Masatsugu Sei Suzuki
Department of Physics, SUNY at Binghamton
(Date: January 28, 2019)
1.
Reciprocal lattice and Brillouin zone
The primitive translation vectors of the hexagonal space lattice may be taken as
a1 
3
1
a xˆ  a yˆ ;
2
2
a2  
3
1
a xˆ  a yˆ ,
2
2
a3  c zˆ ,
where x̂ , ŷ , and ẑ are unit vector in the Cartesian coordinate, and a and c are lattice constants.
(a)
(b)
3 2
a c.
2
Show that the primitive translations of the reciprocal lattice are
Show that the volume of the primitive cell is
b1 
(c)
2
2
xˆ 
yˆ ,
a
3a
b2  
2
2
xˆ 
yˆ ,
a
3a
b3 
2
zˆ ,
c
so that the lattice is its own reciprocal, but with a rotation of axes.
Describe and sketch the first Brillouin zone of the hexagonal space lattice.
((Solution))
Hexagonal, space lattice
a1 
3
1
3
1
a xˆ  a yˆ , a 2  
a xˆ  a yˆ ,
2
2
2
2
(a) The volume of the primitive cell is
3a
2
3a
Vc  a1  (a 2  a3 )  
2
0
(b)
a
2
a
2
0
0
0
c
3 2
a c
2
a3  c zˆ
b1 
2
(a 2  a 3 )
Vc
xˆ
2
3a


2
3 2
a c 0
2
2 1

(
xˆ  yˆ )
a
3
b2 
zˆ
0
c
2
(a3  a1 )
Vc
xˆ
2

0
3 2
a c 3a
2
2
2
1

(
xˆ 
a
3
b3 
yˆ
a
2
0
yˆ
zˆ
0
a
2
c
0
yˆ )
2
(a1  a 2 )
Vc
xˆ
3a
2

2
3 2
a c
3a
2

2
2

zˆ
c
yˆ
a
2
a
2
zˆ
0
0
____________________________________________________________________________
2.
x-ray diffraction (diamond)
The crystal structure of diamond is described in Fig.1. The basis consists of eight atoms
if the cell is taken as the conventional cube. (a) Find the structure factor S of this basis,
where the structure factor S is given by
S 

f C exp(  i G  R j ) .
j
where f C is the atomic form factor of C atom, and G is the reciprocal lattice vector.
(b) Find the zeros of S and show that the allowed reflections of the diamond structure satisfy
v1  v2  v3  4n , where all indices are even ( n is any integer), and satisfy
v1  v2  v3  4n  1 , where all indices are odd. (Notice that h, k, l may be written for v1 ,
v2 , v3 and this is often done.).
Hint: The reciprocal lattice for the conventional cubic cell is given by
G  v1b1  v2b2  v3b3
2
2
2
e x , b2 
e y , and b3 
e z , a is the lattice constant of the conventional
a
a
a
cubic cell, and e x , e y , and e z are the unit vectors in the Cartesian co-ordinate.
with b1 
R6
R7
R2
R5
R3
R4
R1
R0
a
Fig.
Crystal structure of diamond in the conventional cubic unit cell with the lattice
constant a. There are 8 C atoms in this cell. The positions are given by
1 1
1 1
1 1
R0  (0,0,0) , R1  ( , ,0) , R2  (0, , ) , R3  ( ,0, ) ,
2 2
2 2
2 2
1 1 1
3 3 1
1 3 3
R4  ( , , ) ,
R5  ( , , )  R4  R1 ,
R6  ( , , )  R4  R2
4 4 4
4 4 4
4 4 4
3 1 3
R7  ( , , )  R4  R3
4 4 4
((Solution))
The structure factor S is given by
S  f C  exp(iG  R j ) ,
j
where f C is the atomic form factor of C atom. G is the reciprocal lattice vector for the
conventional cubic unit cell with side a,
G  v1b1  v2b2  v3b3 
with b1 
2
(v1 , v2 , v3 ) ,
a
2
2
2
e x , b2 
e y , and b3 
e z . In each cubic unit cell, there are 8 atoms
a
a
a
1 1
1 1
1 1
R0  (0,0,0) , R1  ( , ,0) , R2  (0, , ) , R3  ( ,0, ) ,
2 2
2 2
2 2
1 1 1
3 3 1
R4  ( , , )  R4  R0 ,
R5  ( , , )  R4  R1 ,
4 4 4
4 4 4
1 3 3
3 1 3
R6  ( , , )  R4  R2 ,
R7  ( , , )  R4  R3
4 4 4
4 4 4
in the units of a. Then we get the expression of S as
S  [exp(iG  R0 )  exp(iG  R1 )  exp(iG  R2 )  exp(iG  R3 )]
 fC [exp(iG  R0 )  exp(iG  R1 )  exp(iG  R2 )  exp(iG  R3 )] exp(iG  R4 )
 f C [1  exp(iG  R4 )][exp(iG  R0 )  exp(iG  R1 )  exp(iG  R2 )  exp(iG  R3 )]
The second term is the fcc structure factor. S can be rewritten as

(v1  v2  v3 )]}{1  exp[i (v1  v2 )]  exp[i (v2  v3 )]
2
 exp[i (v3  v1 )]}
S  fC {1  exp[i
 fC S (basis )S ( fcc )
where
S (basis )  1  exp[i

2
(v1  v2  v3 )] ,
S ( fcc )  1  exp[i (v1  v2 )]  exp[i (v2  v3 )]  exp[i (v3  v1 )] .
(b)
(i)
Evaluation of S(basis)
S (basis)  2
for
v1  v2  v3  4n .
S (basis)  1  i
(ii)
for
v1  v2  v3  4n  1 .
S (basis)  0
for v1  v2  v3  4n  2 .
S (basis)  1  i
for
v1  v2  v3  4n  1 .
Product of S ( fcc) and S (basis)
If all indices are even,
S ( fcc)  4
v1  v2  v3  4n
S (basis)  2
v1  v2  v3  4n  2
S (basis)  0
If all indices are odd,
5
S ( fcc)  4
v1  v2  v3  4n  1,
S (basis)  1  i
v1  v2  v3  4n  1,
S (basis)  1  i
If one of indices is odd and the other two indices are even
S ( fcc)  0
5
If one of indices is even and other two indices are odd
S ( fcc)  0
5
The selection rule for diamond are compared with the data of neutron power diffraction
(111)
S1S 2  4(1  i )
(220)
S1S 2  8
(311)
S1S 2  4(1  i )
(400)
S1S 2  8
(331)
S1S2  4(1  i)
(422)
S1S 2  8
Fig.
Neutron powder diffraction patterns recorded from powered graphite and diamond
(reproduced from Wollan and Shull, 1948)
3.
Structure of GaN
A model of GaN crystal is given in the figure below. It has a hexagonal lattice with
a1  a2  a and a3  c .
Fig.
Crystal structure of GaN.
a2
a1
Fig.
Hexagonal in-plane structure with the primitive lattice vectors a1 and a2.
(a)
Write down the explicit expressions of a set of the basis vectors a1, a2, a3 in
Cartesian co-ordinates, which form the unit cell of this crystal.
Write down the explicit expressions of a set of primitive reciprocal lattice vectors
b1 , b2 , and b3 in Cartesian co-ordinates (in terms of a and c).
(b)
(c)
(d)
(e)
Draw the first Brillouin zone.
Find the volume of the first Brillouin zone in terms of a and c.
How many atoms of each kind are in the basis of this crystal structure?
((Solution))
(a)
b2
a2
b1
a1
Fig.
In-plane (hexagonal) structure of GaN. b1 and b2 are the reciprocal lattice vector. a1 and
a2 are the primitive lattice vector.
(a)
In the above figure, we have a1  a e x ,
Since a1  b1  a2  b2  2 ,and a1  b2  a2  b1  0
a1  b1  2
a1b1 cos(30)  2
leading to
b1 
2
4

a1 cos(30)
3a
a2  b2  2
leading to
a2b2 cos(30)  2
1
3
a2   a e x 
a ey
2
2
b2 
2
4

a2 cos(30)
3a
Thus we have
b1 
4
4
3 1
(cos 30, sin 30,0) 
( , ,0)
3a
3a 2 2
b2 
4
4
(cos 90, sin 90,0) 
(0,1,0)
3a
3a
Since a3  b3  2 , we have b3 
b3  (0,0,
2
, or
c
2
)
c
(b)
The first Brillouin zone
The first Brillion zone of the hexagonal lattice is also a hexagonal structure. The cross section
of the Brillouin zone in (k x , k y ) plane is illustrated by the shaded area in Fig. Sweep this shaded
area along kz axis from k z  

c
to k z 

c
, we get the whole first Brillion zone.
b2
b2 2
b1
b1 2
\
Fig.
First Brillouin zone in the (kx, ky) plane.
(c)
The volume of the first Brillouin zone
2
 4  3 2 (2 )3 2 3
  ( b1 b2 sin 60) b3  
 2

ac 3
 3a  2 c
or
4
3
( )
3a 2
  (b1  b2 )  b3 
0
0
(d)
4 1
3a 2
4
3a
0
0
2 4 2 3 (2 )3 2 3
0 
(
)
 2
c
ac 3
3a 2
2
c
As shown in the figure below there are two 2 atoms of each kind per unit cell (denoted by
black thick lines).
________________________________________________________________________
4.
X-ray powder diffraction of GaN)
In order to solve this problem, first you need to solve the problem-2 (a).
(a)
Calculate the primitive reciprocal lattice vectors b1 , b2 , and b3 in Cartesian co(b)
ordinates (in terms of a and c) for GaN structure.
Suppose that we have a powdered x-ray diffraction of GaN. The reciprocal lattice
vector G is defined by
G  hb1  kb2  lb3
(c)
(h, k, l: integers).
Calculate the magnitude |G| in terms of h, k, l, a, and c.
In the powder x-ray diffraction, the Bragg reflection occurs when the condition
Q 
4

sin   G ,
is satisfied, where Q is the scattering vector,  is the wave-length of x-ray (CuK
line, 1.54184Å). In GaN, we have a = 3.186 Å and c = 5.186 Å. Calculate the
scattering angle (2) for typical indices (h, k, l) and compare your result with the
experimental result (powder x-ray diffraction of GaN with CuK) shown below.
Fig.
Power x-ray diffraction of GaN. The x-ray intensity vs the scattering angle 2.
((Solution))
b2
a2
b1
a1
Fig.
The relation between the reciprocal lattice vectors and lattice vectors.
a = 3.186 Å, c = 5.186 Å,
  1.54184 Å (CuK line)
(a)
a1  b1  2
a1b1 cos(30)  2
leading to
b1 
2
4

a1 cos(30)
3a
a2  b2  2
a2b2 cos(30)  2
leading to
b2 
2
4

a2 cos(30)
3a
Thus we have
b1 
4
4
3 1
(cos 30, sin 30) 
( , )
3a
3a 2 2
b2 
4
4
(cos 90, sin 90) 
(0,1)
3a
3a
(b)
The reciprocal lattice vector:
G  hb1  kb2  lb3 .
where h, k, and l are integers. We note that
b1  b3  0 ,
b3 
b2  b3  0 ,
2
c
1 2
2
b1  b2  b1 cos(60)  b1
2
Then we have
2
G  (hb1  kb2  lb3 )  (hb1  kb2  lb3 )
2
2
 h 2b1  2hkb1  b2  k 2b2  l 2b3
2
 (h 2  hk  k 2 )b1  l 2b3
2
2
 (h 2  hk  k 2 )(
4 2 2 2 2
) l ( )
c
3a
G  (h 2  hk  k 2 )(
4 2 2 2 2
) l ( )
c
3a
or
(c)
The Bragg condition:
Q
4

sin   G  (h 2  hk  k 2 )(
4 2 2 2 2
) l ( )
c
3a
The scattering angle (2) is obtained as
2  2 arcsin (
4 2 2 2 2

(h 2  hk  k 2 )(
) l ( ) )
4
c
3a
Using Mathematica we calculate 2 (degrees) as a function of the index (h, k, l).
(h, k, l)
2
(1,0,0)
(0,0,2)
(1,0,1)
(1,0,2)
(1,1,0)
(1,0,3)
(2,0,0)
(1,1,2)
(2,0,1)
(0,0,4)
(2,0,2)
32.4493°
34.592°
36.9013°
48.1577°
57.8864°
63.5061°
67.9469°
69.2179°
70.6541°
72.9706°
78.5401°
The experimental values of the scattering angle 2 for each (h, k, l) for GaN can be well explained
by the above calculations.
Fig.
Power x-ray diffraction of GaN. The x-ray intensity vs the scattering angle 2. a = 3.186
Å,
c = 5.186 Å,   1.54184 Å (CuK line)
5.
x-ray powder diffraction (Debye-Scherrer)
(a)
(b)
What is the wavelength of the FeK?
What is the Bragg condition for the conventional cubic cell? Show that the values of
ℎ
is equal to 3, 4, 8, 10, 12,… for fcc ?
Determine the structure of the system. What is the lattice constant a for the conventional
cubic cell? Assume that there is only one kind of atom.
Suppose that the mass of each atom is m= 26.9815 amu, what is the density of the
system? 1 atomic mass unit (amu) = 1.660538782 x 10-24 g.
(c)
(d)
((Solution))
(a), (b) (c)
2
=

3
= 1.933Å
In a Debye-Schrerrer experiment, we have the Bragg condition,
= 2 sin  =
4

sin  =
=
2
ℎ
or
sin
=

4
ℎ
Experimentally we have
sin
= 0.1843 = 0.0607 × 3
= 0.2459 = 0.0607 × 4
= 0.4887 = 0.0607 × 8
= 0.6707 = 0.0607 × 11
= 0.7314 = 0.0607 × 12
= 0.9739 = 0.0607 × 16
Thus
h2  k 2  l 2  3, 4,8,11,12,16
2
4a
(fcc structure)
 0.0607
leading to
a  3.92 Å
(fcc lattice)
There are four atoms in conventional unit cell for the fcc structure. The density is

4m
a3
where
m= 26.9815 amu, with 1 atomic mass unit (amu) = 1.660538782 x 10-24 g.
 = 2.976 g/cm3
6.
x-ray scattering intensity
Calculate the scattered intensity from a linear chain of atoms with ordered domains of N
atoms. Assume that there is no phase correlation between atoms in different domains.
Hint: Consider a domain where there are N atoms. Hint: discuss the structure factor.
((Solution))
In a linear crystal there are identical point scattering centers at every lattice point # = $ ,
where m is an integer. The total amplitude is proportional to F;
,
,
%
= & exp −+ #
= & exp −+$
%
=1
exp −+2
%
=
%
1
1
1
exp 8−+ 2 9 [exp 8+ 2 9 − exp 8−+ 2 9
=
+
+
+
exp 8− 2 9 [exp 8 2 9 − exp 8− 2 9]
#-.
#-.
or
exp −+
⋯.
exp[−+ 1 − 1
or
234
234
5,67
567
or
%
The intensity is
1
1
exp 8−+ 2 9 sin 2
=
+
exp 8− 2 9 sin 2
1
= ⌊%
;
⌋ =
BCD
A
CD
>?@A
A
>?@A
(4)
Suppose that
=E

(4)
6
or
= 2E
leading to
;
= ⌊%
(a)
When
(b)
;
0, ;
⌋ =
BC
A
C
>?@A
A
>?@A
(4)
tends to N2, where
becomes zero at F = 1
=E

6
(Bragg point).
(3)

= 2$ or = ,6 $
y I x I x 0
x Ma Q
2
n
a
4
2
0
2
4
=
Fig.

6
E
. F=1
=
(Bragg point;
The Bragg peak appears at
1

6
=1
−

6
E . The normalized intensity shows a peak at x = 0
E). Note that N = M in this Fig.
=

6
(5)
E. The width of the Bragg peak is
= 2
or
=G =

,6
(5)
When N becomes large, the width of the Bragg peak becomes sharper.
7.
Triangular lattice
The figure below shows a triangular lattice in two dimensions, where a is the distance
between the atoms.
(a)
(b)
(c)
(d)
Determine the pr imit ive lattice vectors appropriate to this lattice and calculate
the area of the unit cell.
Determine the reciprocal lattice vectors appropriate to this lattice and draw
the first Brillouin zone.
Derive the equation for the phonon modes using the first nearest-neighbor
approximation.
Calculate the phonon frequency at , M, and K points in the Brillouin zone
(BZ) ( is the center of the BZ, M is the center of any edge of the BZ, and K is
any corner of the BZ).
____________________________________________________________________________
((Solution))
(a)
b2
a2
30
60
a1
30
b1
The primitive lattice vectors:
a1  a (cos(0), sin(0),0)  a (1,0,0) ,
1 3
a2  a(cos(60), sin(60),0)  a ( ,
,0) ,
2 2
a3  (0,0,1)
(this primitive vector is used only for the calculation of b1 and b2).
Area A is given by
A  a1  a 2  a1 a2 sin 600  a 2
3
=
2
The reciprocal lattice vectors; b1, b2
(b)
We note that
a1  b1  a2  b2  2 ,
a1  b2  a2  b1  0
Then we have
a1  b1  a1b1 cos 30  2
b1 
4
,
3a
a2  b2  a2b2 cos 30  2
b2 
4
.
3a
The reciprocal lattice vector b2 is perpendicular to a1 and b1 is perpendicular to a2.
Mathematically,
b1  2
a 2  a3
4
3 1

( ,  , 0) ,
[a1 , a2 , a3 ]
2
3a 2
b2  2
a3  a1
4

(0,1,0) .
[a1 , a2 , a3 ]
3a
or more conveniently, we use
b1 
4
3 1
( , ) ,
3a 2 2
b2 
4
(0,1)
3a
for the 2D reciprocal lattice vectors.
(c)
Consider the following two-dimensional close-packed lattice and its unit cell:
Hl,m+1L
Hl-1,m +1L
a2
Hl,mL
a1
Hl-1,mL
Hl,m- 1L
Fig.
Hl+1,mL
Hl+1,m- 1L
Real space for the 2D triangular lattice.
Primitive cell vectors a1 and a2 are shown. Determine the first Brillouin zone appropriate to this
lattice. Arrange that the reciprocal lattice vectors b1 and b2 are correctly oriented. Then, using the
nearest-neighbor approximation for a vibrating net of point masses, determine the dispersion
equation [let the central mass points have coordinates (la1, ma2)]. Calculate the frequency at twonon-equivalent symmetry points on the zone boundary.
We set up the equation of the motion for the atom at the center. There are nearest neighbor
interactions from atoms surrounding the atom at the center.
ɺɺ(l , m)  C[u(l  1, m)  u(l , m  1)  u(l  1, m  1)
Mu
 u(l  1, m)  u(l , m  1)  u(l  1, m  1)  6u(l , m)]
(1)
where M is the mass of atom, C is the spring constant, and ul 1, m is the displacement vector for
the atom at the position (la1, ma2). We assume the solution of the form as
u(l , m)  u(0) exp[ik  (la1  ma2 )  t ] .
with
u(l  1, m)  exp( ik  a1 )ul , m ,
u(l , m  1)  exp( ik  a2 )ul , m .
Substituting these into Eq.(1), we get
 M 2 u(l , m)  C[exp(ik  a1 )  exp(ik  a2 )  exp[i(k  a1  k  a2 )
 exp(ik  a1 )  exp(ik  a 2 )  exp[i(k  a1  k  a2 )  6]u(l , m)
Then we have the dispersion relation of  vs k as
M 2
 [3  cos(k  a1 )  cos(k  a2 )  cos(k  a1  k  a 2 )]
2C
The Brillouin zone for the triangular lattice is shown below.
b2
Hl,m+ 1L
Hl- 1,m +1L
a2
Hl,mL
Hl-1,m L
Hl,m- 1L
Fig.
a1
Hl+1,m-1L
Hl+ 1,mL
b1
Relation between the real space (lattice vectors a1 and a2) and the reciprocal lattice
(b1 and b2). The magnitude of a1 and b1 are chosen appropriately.
(d)
At the middle of a Brillouin zone (zone boundary)
At the  point,
k = 0.
  0.
At the M point
k
b1  b2
.
2
1
k  a1  a1  b1   ,
2
1
k  a2  a 2  b2   .
2
Then we get
M 2
 3  cos( )  cos( )  cos(   )  4
2C
or
2 
8C
M
At the K point
k
b1  2b2
3
1
2
k  a1  a1  b1 
,
3
3
2
4
k  a2  a2  b2 
3
3
Then we get
M 2
2
4
2
9
 3  cos( )  cos( )  cos( ) 
2C
3
3
3
2
or
2 
9C
M
6
4
2
0
-2
-4
-6
-6
Fig.
-4
-2
0
2
4
6
Contour plot of constant energy in the reciprocal lattice plane. The first Brillouin
zone is shown by the blue solid line. The energy of the point K (the corner of the
first Brillouin zone) is higher than that of the point M (the middle of the zone
boundary).
9.
x-ray powder diffraction (Aluminium)
In a Debye-Scherrer diffractogram, we obtain a measure of the Bragg angles . In a particular
experiment with Al (Aluminium) powder, the following  data were obtained when CuK
radiation (the wavelength  = 1.54184 Å) was used:
19.48°, 22.64°, 33.00°, 39.68°, 41.83°, 50.35°, 57.05°, 59.42°.
Aluminium has atomic weight 27 g/mol and density 2.7 g/cm3. (c) Show that Al has a fcc (face
centered cubic) structure, where h 2  k 2  l 2 = 3, 4, 8, 11, 12, 16, 19, 20, 24, 27, 32, 35, 40, 36,
43, 44, 48. (d) What is the lattice parameter of Al of the conventional cubic unit cell? (e) Calculate
the Avogadro’s number.
Hint: The basis consists of four lattice points if the cell is taken as the conventional cube for
fcc.
4

sin  
2
a
h2  k 2  l 2 .
((Solution))
We use the Bragg condition given by
4

sin  
2
a
h2  k 2  l 2
where h, k, l are the integers, a is the lattice constant of the conventional cubic lattice.  = 1.54184
Å for CuK. This formula can be rewritten as
2
a sin   h 2  k 2  l 2

Suppose that Al has a fcc structure. In this case, it is expected that
h 2  k 2  l 2 = 3, 4, 8, 11, 12, 16, 19, 20, 24, 27, 32,.....
We make a plot of the date of
2

sin  as a function of
h2  k 2  l 2 .
 = 19.48°,
h2  k 2  l 2  3
 = 22.64°,
h2  k 2  l 2  4
 = 33.00°,
h2  k 2  l 2  8
 = 39.68°,
h2  k 2  l 2  11
 = 41.83
h 2  k 2  l 2  12
 = 50.35°,
h 2  k 2  l 2  16
 = 57.05°,
h 2  k 2  l 2  19
 = 59.42°.
4
2
l
h2  k 2  l 2  20
sinHqL
3
2
1
h2 + k 2 + l2
0.2
0.4
0.6
0.8
1.0
1.2
The least-squares fit of the data yield the lattice constant a as a = 4.0045 Å.
The Avogadro number can be evaluates as follows, based on the lattice constant a. In Al (fcc),
there are 4 Al toms in the conventional cubic lattice.
The volume for Al 1mol is
VA 
M


27
 10
2.7
where M = 27 g/mol and  = 2.7 g/cm3. There are NA (Al atoms) in this volume VA. Then we have
VA a 3

NA 4
The Avogadro number is evaluated as
NA 
4VA
=6.229 x 1023
3
a
Note that the correct value is 6.023 x 1023.