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

yxa ˆ 2

1ˆ

2

31 aa ; yxa ˆ

2

1ˆ

2

32 aa , za ˆ 3 c ,

where x̂ , ŷ , and ẑ are unit vector in the Cartesian coordinate, and a and c are lattice constants.

(a) Show that the volume of the primitive cell is ca22

3.

(b) Show that the primitive translations of the reciprocal lattice are

yxb ˆ2

ˆ3

21

aa

, yxb ˆ

2ˆ

3

22

aa

, zb ˆ

23

c

,

so that the lattice is its own reciprocal, but with a rotation of axes.

(c) Describe and sketch the first Brillouin zone of the hexagonal space lattice. ((Solution)) Hexagonal, space lattice

yxa ˆ 2

1ˆ

2

31 aa , yxa ˆ

2

1ˆ

2

32 aa , za ˆ 3 c

(a) The volume of the primitive cell is

ca

c

aa

aa

Vc2

321 2

3

00

022

3

022

3

)( aaa

(b)

)ˆˆ3

1(

2

00

022

3

ˆˆˆ

2

3

2

)(2

2

321

yx

zyx

aab

a

c

aa

ca

Vc

)ˆˆ3

1(

2

022

300

ˆˆˆ

2

3

2

)(2

2

132

yx

zyx

aab

a

aa

c

ca

Vc

z

zyx

aab

ˆ2

022

3

022

3

ˆˆˆ

2

3

2

)(2

2

213

c

aa

aa

ca

Vc

____________________________________________________________________________

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

j

jC ifS )exp( RG .

where Cf 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

nvvv 4321 , where all indices are even ( n is any integer), and satisfy

14321 nvvv , where all indices are odd. (Notice that h, k, l may be written for 1v ,

2v , 3v and this is often done.).

Hint: The reciprocal lattice for the conventional cubic cell is given by

332211 bbbG vvv

with xa

eb2

1 , ya

eb2

2 , and za

eb2

3 , a is the lattice constant of the conventional

cubic cell, and xe , ye , and ze are the unit vectors in the Cartesian co-ordinate.

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

)0,0,0(0 R , )0,21

,2

1(1 R , )

2

1,

2

1,0(2 R , )2

1,0,

2

1(3 R ,

)4

1,

4

1,

4

1(4 R , 145 )

4

1,

4

3,

4

3( RRR , 246 )4

3,

4

3,

4

1( RRR

347 )4

3,

4

1,

4

3( RRR

((Solution))

The structure factor S is given by

j

jC ifS )exp( RG ,

where Cf is the atomic form factor of C atom. G is the reciprocal lattice vector for the

conventional cubic unit cell with side a,

R0

R1

R2

R3

R4

R5

R6

R7

a

),,(2

321332211 vvva

vvv

bbbG ,

with xa

eb2

1 , ya

eb2

2 , and za

eb2

3 . In each cubic unit cell, there are 8 atoms

)0,0,0(0 R , )0,21

,2

1(1 R , )

2

1,

2

1,0(2 R , )2

1,0,

2

1(3 R ,

044 )4

1,

4

1,

4

1( RRR , 145 )4

1,

4

3,

4

3( RRR ,

246 )4

3,

4

3,

4

1( RRR , 347 )4

3,

4

1,

4

3( RRR

in the units of a. Then we get the expression of S as

)]exp()exp()exp())][exp(exp(1[

)exp()]exp()exp()exp()[exp(

)]exp()exp()exp()[exp(

32104

43210

3210

RGRGRGRGRG

RGRGRGRGRG

RGRGRGRG

iiiiif

iiiiif

iiiiS

C

C

The second term is the fcc structure factor. S can be rewritten as

)()(

)]}(exp[

)](exp[)](exp[1)]}{(2

exp[1{

13

3221321

fccSbasisSf

vvi

vvivvivvvifS

C

C

where

)](2

exp[1)( 321 vvvibasisS

,

)](exp[)](exp[)](exp[1)( 133221 vvivvivvifccS .

(b)

(i) Evaluation of S(basis)

2)( basisS for nvvv 4321 .

ibasisS 1)( for 14321 nvvv .

0)( basisS for 24321 nvvv .

ibasisS 1)( for 14321 nvvv .

(ii) Product of )( fccS and )(basisS

If all indices are even, 4)( fccS

nvvv 4321 2)( basisS

24321 nvvv 0)( basisS 5

If all indices are odd, 4)( fccS

,14321 nvvv ibasisS 1)(

,14321 nvvv ibasisS 1)(

If one of indices is odd and the other two indices are even 5

0)( fccS

If one of indices is even and other two indices are odd 5

0)( fccS

The selection rule for diamond are compared with the data of neutron power diffraction

(111) )1(421 iSS

(220) 821 SS

(311) )1(421 iSS

(400) 821 SS

(331) )1(421 iSS

(422) 821 SS

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

aaa 21 and ca 3 .

Fig. Crystal structure of GaN.

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.

(b) Write down the explicit expressions of a set of primitive reciprocal lattice vectors

1b , 2b , and 3b in Cartesian co-ordinates (in terms of a and c).

(c) Draw the first Brillouin zone.

(d) Find the volume of the first Brillouin zone in terms of a and c.

(e) How many atoms of each kind are in the basis of this crystal structure?

((Solution))

(a)

a1

a2

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 xa ea 1 , yx aa eea 23

2

12

Since 22211 baba ,and 01221 baba

211 ba 2)30cos(11 ba

leading to

aab

3

4

)30cos(

2

1

1

222 ba 2)30cos(22 ba

leading to

a1

a2

b1

b2

aab

3

4

)30cos(

2

2

2

Thus we have

)0,2

1,

2

3(

3

4)0,30sin,30(cos

3

41

aa

b

)0,1,0(3

4)0,90sin,90(cos

3

42

aa

b

Since 233 ba , we have c

b2

3 , or

)2

,0,0(3c

b

(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 ),( yx kk plane is illustrated by the shaded area in Fig. Sweep this shaded

area along zk axis from c

kz

to c

kz

, we get the whole first Brillion zone.

\

Fig. First Brillouin zone in the (kx, ky) plane.

(c) The volume of the first Brillouin zone

3

32)2(2

2

3

3

4)60sin(

2

32

321caca

bbb

or

b1

b2

b1 2

b2 2

3

32)2(

2

3)

3

4(

2

200

03

40

02

1

3

4)

2

3(

3

4

)(2

32

321caac

c

a

aa

bbb

(d) 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 1b , 2b , and 3b in Cartesian co-

ordinates (in terms of a and c) for GaN structure.

(b) Suppose that we have a powdered x-ray diffraction of GaN. The reciprocal lattice

vector G is defined by

321 bbbG lkh (h, k, l: integers).

Calculate the magnitude |G| in terms of h, k, l, a, and c.

(c) In the powder x-ray diffraction, the Bragg reflection occurs when the condition

GQ

sin4

,

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))

Fig. The relation between the reciprocal lattice vectors and lattice vectors.

a = 3.186 Å, c = 5.186 Å, 54184.1 Å (CuK line) (a)

211 ba 2)30cos(11 ba leading to

aab

3

4

)30cos(

2

1

1

222 ba 2)30cos(22 ba leading to

aab

3

4

)30cos(

2

2

2

a1

a2

b1

b2

Thus we have

)2

1,

2

3(

3

4)30sin,30(cos

3

41

aa

b

)1,0(3

4)90sin,90(cos

3

42

aa

b

(b)

The reciprocal lattice vector:

321 bbbG lkh .

where h, k, and l are integers. We note that

031 bb , 032 bb , c

b2

3

2

1

2

121 2

1)60cos( bb bb

Then we have

22222

2

322

122

23

222

221

21

2

321321

2

)2

()3

4)((

)(

2

)()(

cl

akhkh

blbkhkh

blbkhkbh

lkhlkh

bb

bbbbbbG

or

22222 )2

()3

4)((

cl

akhkh

G

(c) The Bragg condition:

22222 )2

()3

4)((sin

4

cl

akhkh

GQ

The scattering angle (2) is obtained as

))2

()3

4)((

4(arcsin 22 22222

cl

akhkh

Using Mathematica we calculate 2 (degrees) as a function of the index (h, k, l).

(h, k, l) 2

(1,0,0) 32.4493° (0,0,2) 34.592°

(1,0,1) 36.9013° (1,0,2) 48.1577° (1,1,0) 57.8864° (1,0,3) 63.5061° (2,0,0) 67.9469° (1,1,2) 69.2179° (2,0,1) 70.6541° (0,0,4) 72.9706°

(2,0,2) 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, 54184.1 Å (CuK line)

5. x-ray powder diffraction (Debye-Scherrer)

(a) What is the wavelength of the FeK? (b) 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 ? (c) 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.

(d) 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.

((Solution)) (a), (b) (c)

= 2��� � ���3 = 1.933��

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

2 2 2 3, 4,8,11,12,16h k l (fcc structure)

2

0.06074a

leading to

3.92a Å (fcc lattice) There are four atoms in conventional unit cell for the fcc structure. The density is

3

4m

a

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 �−+$���

,��

#-.

or

%��� = 1 � exp�−+��� � exp�−+2��� � ⋯ . � exp[−+�1 − 1���1 or

%��� = ��234 ��5,67���234 ��567�

%��� =exp 8−+ 1��2 9 [exp 8+

1��2 9 − exp 8−+

1��2 9

exp 8− +��2 9 [exp 8+��

2 9 − exp 8−+��

2 9]

or

%��� =exp 8−+ 1��2 9 sin �

1��2 �

exp 8− +��2 9 sin ���2 �

The intensity is

;��� = ⌊%���⌋� = >?@A �BCDA �

>?@A �CDA � (4)

Suppose that

� = E �6 � �� (4) or

�� = 2E� �� leading to

;��� = ⌊%���⌋� = >?@A �BC�A �

>?@A �C�A � (4)

(a) When ��0, ;��� tends to N2, where � = E �6 (Bragg point). (3) (b) ;��� becomes zero at F = 1�� = 2$ or � = �,6 $

x Ma Q2

an

y I x I x 0

4 2 0 2 4

Fig. � = �6 E � �. F = 1�� = 1��� −�6 E�. The normalized intensity shows a peak at x = 0

(Bragg point; � = �6 E). Note that N = M in this Fig. (5)

The Bragg peak appears at � = �6 E. The width of the Bragg peak is

1��� = 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) Determine the pr imit ive latt ice vectors appropriate to this latt ice and calculate

the area of the unit cell. (b) Determine the reciprocal latt ice vectors appropriate to this latt ice and draw

the first Brillouin zone. (c) Derive the equat ion for the phonon modes using the first nearest-neighbor

approximation.

(d) 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)

The primitive lattice vectors:

)0,0,1()0),0sin(),0(cos(1 aa a ,

)0,2

3,

2

1()0),60sin(),60(cos(2 aa a ,

)1,0,0(3 a (this primitive vector is used only for the calculation of b1 and b2).

Area A is given by

2

360sin 202121 aA aaaa =

a1

a2

b1

b2

60

30

30

The reciprocal lattice vectors; b1, b2 (b) We note that

22211 baba , 01221 baba Then we have

230cos1111 baba a

b3

41

,

230cos2222 baba a

b3

42

.

The reciprocal lattice vector b2 is perpendicular to a1 and b1 is perpendicular to a2. Mathematically,

)0,2

1,

2

3(

3

4

],,[2

321

321

a

aaa

aab ,

)0,1,0(3

4

],,[2

321

132

a

aaa

aab .

or more conveniently, we use

)2

1,

2

3(

3

41

a

b , )1,0(

3

42

a

b

for the 2D reciprocal lattice vectors. (c)

Consider the following two-dimensional close-packed lattice and its unit cell:

Fig. 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 two-non-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.

)],(6)1,1()1,(),1(

)1,1()1,(),1([),(

mlmlmlml

mlmlmlCmlM

uuuu

uuuu

ɺɺ (1)

where M is the mass of atom, C is the spring constant, and ml ,1u is the displacement vector for

the atom at the position (la1, ma2). We assume the solution of the form as

])(exp[)0(),( 21 tmliml aakuu .

with

mliml ,1)exp(),1( uaku ,

Hl,mL

Hl,m+1LHl-1,m+1L

Hl-1,mLHl-1,mL

Hl,m-1LHl+1,m-1L

Hl+1,mLa1

a2

mliml ,2 )exp()1,( uaku .

Substituting these into Eq.(1), we get

),(]6)(exp[)exp()exp(

)(exp[)exp()[exp(),(

2121

21212

mliii

iiiCmlM

uakakakak

akakakaku

Then we have the dispersion relation of �vs k�as

)]cos()cos()cos(3[2

2121

2

akakakak C

M

The Brillouin zone for the triangular lattice is shown below.

Fig. 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.

Hl,mL

Hl,m+1LHl-1,m+1L

Hl-1,mLHl-1,mL

Hl,m-1LHl+1,m-1L

Hl+1,mLa1

a2

b1

b2

(d) At the middle of a Brillouin zone (zone boundary)

At the point, k = 0.

0 . At the M point

221 bbk

.

111 21

baak , 222 21

baak .

Then we get

4)cos()cos()cos(32

2

C

M

or

M

C82

At the K point

3

2 21 bbk

3

2

3

1111

baak ,

3

4

3

2222

baak

Then we get

2

9)

3

2cos()

3

4cos()

3

2cos(3

2

2

C

M

or

M

C92

Fig. 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°.

-6 -4 -2 0 2 4 6

-6

-4

-2

0

2

4

6

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 222 lkh = 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.

2222sin4

lkha

.

((Solution)) We use the Bragg condition given by

2222sin4

lkha

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

222sin2

lkha

Suppose that Al has a fcc structure. In this case, it is expected that

222 lkh = 3, 4, 8, 11, 12, 16, 19, 20, 24, 27, 32,.....

We make a plot of the date of

sin2

as a function of 222 lkh .

= 19.48°, 3222 lkh

= 22.64°, 4222 lkh

= 33.00°, 8222 lkh

= 39.68°, 11222 lkh

= 41.83 12222 lkh

= 50.35°, 16222 lkh

= 57.05°, 19222 lkh

= 59.42°. 20222 lkh

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

107.2

27

M

VA

where M = 27 g/mol and = 2.7 g/cm3. There are NA (Al atoms) in this volume VA. Then we have

4

3a

N

V

A

A

The Avogadro number is evaluated as

3

4

a

VN AA =6.229 x 10

23

Note that the correct value is 6.023 x 1023.

2

lsinHqL

h2 + k2 + l2

0.2 0.4 0.6 0.8 1.0 1.2

1

2

3

4