Solid State Physics XRD Crystallography Miller Indices Reciprocal Space and Real Space Science Worksheet

Use the book attached in the following link and any other relevant source to answer thefollowing questions. https://drive.google.com/file/d/1kN9QMjGsyeQXoBXHLt8kP9IwC1oswz-/view?usp=sharing
Make your answers short and cite the source (and page number or link) of each of your
answer. Kindly, I should be able to reproduce you answer from the page you cite.
2. 2.5. In general, reciprocal lattice vectors are parallel to lattice directions having the same
indices only in a cubic crystal. Nevertheless, this equivalence is found for some zones of high
symmetry in other noncubic Bravais lattices. Give two examples.
4. 2.24. Cu3Au has a cubic structure. When fully ordered, the atoms are located in the following
sites: Au at (0,0,0) and Cu at (1/2, 1/2, 0); ( 1/2, 0, 1/2 ); (0, 1/2, 1/2 ). Under certain
conditions the structure becomes disordered and occupancy of these sites becomes random
(see image and caption below). Assuming the occupancy factor is given by the atomic
concentration of the compound:
(a) Describe the structure factor for the ordered and disordered states.
(b) For which reflections will the structure factor remain unchanged when going through the
order–disorder transition?
5. Examine figure 2.57 in Brandon & Kaplan which shows a TEM diffraction pattern from a
single grain of α-Fe (BCC Fe). Examine the apparent symmetry of the pattern (note spot diam
is used to indicate the intensity of the diffracted beams) – what is the zone axis for this
pattern? (hint:the diffraction pattern MUST reflect AT LEAST the symmetry of the real crystal
along that direction but could appear to be higher symmetry. Index the pattern, remembering
the
following rules:
a. The zone axis (vector that the e-beam is travelling along) points straight out of the page and
is centered in the center of the pattern
b. The in-plane vectors to the diffraction spots are the momentum transfer vectors and
reciprocal lattice vectors ∆k=G. Since the zone axis is normal to the page and all the G vectors
are in plane, the dot products of G*zone axis vector must be zero.
c. The radius from pattern center to each spot is the magnitude of the G vector times a scale
factor. So you can take ratios of these radii and invert them to get the ratios of d-spacings in
real space. The d-spacing as function of hkl for a cubic crystal is equation 2.38 in B&K book.
6. Briefly explain the origins of Kikuchi lines (see chapter 2 near end for discussion in context
of a TEM diffraction pattern. Also read sections 5.6.4-5.6.6. What changes in the discussion for
the OIM / EBSD context as opposed to TEM context?
Use the book attached in the following link and any other relevant
source to answer the following questions. https://drive.google.com/file/
d/1-kN9QMjGsyeQXoBXHLt8kP9IwC1oswz-/view?usp=sharing
Make your answers short and cite the source (and page number or
link) of each of your answer. Kindly, I should be able to reproduce you
answer from the page you cite.
2. 2.5. In general, reciprocal lattice vectors are parallel to lattice directions
having the same indices only in a cubic crystal. Nevertheless, this equivalence
is found for some zones of high symmetry in other noncubic Bravais lattices.
Give two examples.
4. 2.24. Cu3Au has a cubic structure. When fully ordered, the atoms are
located in the following sites: Au at (0,0,0) and Cu at (1/2, 1/2, 0); ( 1/2, 0,
1/2 ); (0, 1/2, 1/2 ). Under certain conditions the structure becomes disordered
and occupancy of these sites becomes random (see image and caption below).
Assuming the occupancy factor is given by the atomic concentration of the
compound:
(a) Describe the structure factor for the ordered and disordered states.
(b) For which reflections will the structure factor remain unchanged when
going through the order–disorder transition?
5. Examine figure 2.57 in Brandon & Kaplan which shows a TEM diffraction
pattern from a single grain of α-Fe (BCC Fe). Examine the apparent symmetry -Fe (BCC Fe). Examine the apparent symmetry Fe (BCC Fe). Examine the apparent symmetry
of the pattern (note spot diam is used to indicate the intensity of the diffracted
beams) – what is the zone axis for this pattern? (hint:the diffraction pattern
MUST reflect AT LEAST the symmetry of the real crystal along that direction
but could appear to be higher symmetry. Index the pattern, remembering the
following rules:
a. The zone axis (vector that the e-Fe (BCC Fe). Examine the apparent symmetry beam is travelling along) points straight out
of the page and is centered in the center of the pattern
b. The in-Fe (BCC Fe). Examine the apparent symmetry plane vectors to the diffraction spots are the momentum transfer
vectors and reciprocal lattice vectors ∆k=G. Since the zone axis is normal to
the page and all the G vectors are in plane, the dot products of G*zone axis
vector must be zero.
c. The radius from pattern center to each spot is the magnitude of the G
vector times a scale factor. So you can take ratios of these radii and invert
them to get the ratios of d-Fe (BCC Fe). Examine the apparent symmetry spacings in real space. The d-Fe (BCC Fe). Examine the apparent symmetry spacing as function of
hkl for a cubic crystal is equation 2.38 in B&K book.
6. Briefly explain the origins of Kikuchi lines (see chapter 2 near end for
discussion in context of a TEM diffraction pattern. Also read sections 5.6.4-Fe (BCC Fe). Examine the apparent symmetry
5.6.6. What changes in the discussion for the OIM / EBSD context as opposed
to TEM context?
Crystal Structure factor calculations
(A) X-Ray Scattering by an Atom
❑ The conventional UC has lattice points as the vertices
❑ There may or may not be atoms located at the lattice points
❑ The shape of the UC is a parallelepiped in 3D
❑ There may be additional atoms in the UC due to two reasons:
➢ The chosen UC is non-primitive
➢ The additional atoms may be part of the motif
Scattering by the Unit cell (uc)
▪ Coherent Scattering
▪ Unit Cell (UC) is representative of the crystal structure
▪ Scattered waves from various atoms in the UC interfere to create the diffraction pattern
The wave scattered from the middle plane is out of phase with the ones
scattered from top and bottom planes
AC = d h 00 =
a
h
MCN :: AC :: 
RBS :: AB :: x
AB x
x
= =
AC  a
h
 R R = MCN = 2d h 00 Sin( ) = 
1 2
 R R = RBS =
1 3
R R =
1 3
AB
x
=

a
AC
h
2 x
x
 = 2 h
 a
a
h
Extending to 3D
=
2




=
2
x
→ fractional coordinate → x
a
 = 2 ( h x + k y + l z )
 R R = 2 h x
1 3
Independent of the shape of UC
Note: R1 is from corner atoms and R3 is from atoms in additional positions in UC
 = 2 ( h x + k y + l z )
In complex notation
E = Aei = fei[2 ( h x+ k y+l z)]
▪ If atom B is different from atom A → the amplitudes must be weighed by the respective
atomic scattering factors (f)
▪ The resultant amplitude of all the waves scattered by all the atoms in the UC gives the
scattering factor for the unit cell
▪ The unit cell scattering factor is called the Structure Factor (F)
Scattering by an unit cell = f(position of the atoms, atomic scattering factors)
F = Structure Factor =
n
hkl
n
F
=  fj e
j =1
Amplitude of wave scattered by all atoms in uc
Amplitude of wave scattered by an electron
i j
n
=  fj e
i[2 ( h xj + k y j + l z j )]
I  F2
For n atoms in the UC
j =1
Structure factor is independent of the shape and size of the unit cell
If the UC distorts so do the planes in it!!
Structure factor calculations
Simple Cubic
A
e
ni
= (−1)
n
e ( odd n ) i = −1
Atom at (0,0,0) and equivalent positions
e ( even n ) i = +1
e ni = e − ni
e i + e − i
= Cos ( )
2
F = fj e
F= f
F2 = f 2
i j
= fj e
i[2 ( h xj + k y j + l z j )]
ei[2 ( h 0+ k  0+l  0)] = f e0 = f
 F is independent of the scattering plane (h k l)
B
Atom at (0,0,0) & (½, ½, 0) and equivalent positions
F = fj e
i j
= fj e
i[2 ( h xj + k y j + l z j )]
F = f ei[2 ( h 0+ k  0+l  0)] + f e
= f e0 + f e
i[ 2  (
C- centred Orthorhombic
h +k
)]
2
1
1
i[2 ( h  + k  + l  0)]
2
2
= f [1 + ei ( h + k ) ]
Real
F =2f
F = f [1 + e
i ( h + k )
]
F2 = 4 f 2
e.g. (001), (110), (112); (021), (022), (023)
F =0
F2 = 0
e.g. (100), (101), (102); (031), (032), (033)
 F is independent of the ‘l’ index
▪ If the blue planes are scattering in phase then on C- centering the red planes will scatter out
of phase (with the blue planes- as they bisect them) and hence the (210) reflection will
become extinct
▪ This analysis is consistent with the extinction rules: (h + k) odd is absent
▪ In case of the (310) planes no new translationally equivalent planes are added on lattice
centering  this reflection cannot go missing.
▪ This analysis is consistent with the extinction rules: (h + k) even is present
C
Atom at (0,0,0) & (½, ½, ½) and equivalent positions
F = fj e
i j
= fj e
i[2 ( h xj + k y j + l z j )]
F = f ei[2 ( h 0+ k  0+l  0)] + f e
= f e0 + f e
i[ 2  (
Body centred
Orthorhombic
h + k +l
)]
2
1
1
1
i[2 ( h  + k  + l  )]
2
2
2
= f [1 + ei ( h + k +l ) ]
Real
F =2f
F = f [1 + e
i ( h + k + l )
]
F2 = 4 f 2
e.g. (110), (200), (211); (220), (022), (310)
F =0
F2 = 0
e.g. (100), (001), (111); (210), (032), (133)
D
Face Centred Cubic
Atom at (0,0,0) & (½, ½, 0) and equivalent positions
F = fj e
i j
= fj e
(½, ½, 0), (½, 0, ½), (0, ½, ½)
i[2 ( h xj + k y j + l z j )]
 i[ 2 ( 0 )] i[ 2 ( h +2 k )] i[ 2 ( k 2+l )] i[ 2 ( l +2h )] 
F = f e
+e
+e
+e



= f [1 + e i ( h + k ) + e i ( k +l ) + e i ( l + h ) ]
Real
F = f [1 + e i ( h + k ) + e i ( k +l ) + e i ( l + h ) ]
(h, k, l) unmixed
F =4f
F 2 = 16 f 2
e.g. (111), (200), (220), (333), (420)
(h, k, l) mixed
F =0
F2 = 0
e.g. (100), (211); (210), (032), (033)
Two odd and one even (e.g. 112); two even and one odd (e.g. 122)
Mixed indices
Two odd and one even (e.g. 112); two even and one odd (e.g. 122)
Mixed indices
CASE
h
k
l
A
o
o
e
B
o
e
e
CASE A : [1 + ei ( e ) + ei ( o ) + ei ( o ) ] = [1 + 1 − 1 − 1] = 0
CASE B : [1 + ei ( o ) + ei ( e ) + ei ( o ) ] = [1 − 1 + 1 − 1] = 0
(h, k, l) mixed
Unmixed indices
F =0
F 2 = 0 e.g. (100), (211); (210), (032), (033)
All odd (e.g. 111); all even (e.g. 222)
Unmixed indices CASE
h
k
l
A
o
o
o
B
e
e
e
CASE A : [1 + ei ( e ) + ei ( e ) + ei ( e ) ] = [1 + 1 + 1 + 1] = 4
CASE B : [1 + ei ( e ) + ei ( e ) + ei ( e ) ] = [1 + 1 + 1 + 1] = 4
(h, k, l) unmixed
F =4f
F 2 = 16 f 2
e.g. (111), (200), (220), (333), (420)
E
Na+ at (0,0,0) + Face Centering Translations → (½, ½, 0), (½, 0, ½), (0, ½, ½)
Cl− at (½, 0, 0) + FCT → (0, ½, 0), (0, 0, ½), (½, ½, ½)
NaCl:
 i[ 2 ( 0 )]
F = f Na + e
+e

h+k
i [ 2 (
)]
2
+e
k +l
i [ 2 (
)]
2
+e
l +h
i [ 2 (
)]
2

+

Face Centred Cubic
 i[ 2 ( h2 )] i[ 2 ( k2 )] i[ 2 ( 2l )] i[ 2 ( h +2k +l )] 
f Cl −  e
+e
+e
+e



F = f Na + [1 + e i ( h + k ) + e i ( k +l ) + e i ( l + h ) ] +
f Cl − [e i ( h ) + e i ( k ) + e i (l ) + e i ( h + k +l ) ]
F = f Na + [1 + e i ( h + k ) + e i ( k +l ) + e i (l + h ) ] +
f Cl − e i ( h + k +l ) [e i ( − k −l ) + e i ( − l − h ) + e i ( − h − k ) + 1]
F = [ f Na + + f Cl − e i ( h + k +l ) ][1 + e i ( h + k ) + e i ( k +l ) + e i ( l + h ) ]
F = [ f Na + + fCl − ei ( h+k +l ) ][1 + ei ( h+k ) + ei ( k +l ) + ei (l +h) ]
F = [Term − 1][Term − 2]
Mixed indices
Mixed indices
Zero for mixed indices
CASE
h
k
l
A
o
o
e
B
o
e
e
CASE A : Term − 2 = [1 + e i ( e ) + e i ( o ) + e i ( o ) ] = [1 + 1 − 1 − 1] = 0
CASE B : Term − 2 = [1 + e i ( o ) + e i ( e ) + e i ( o ) ] = [1 − 1 + 1 − 1] = 0
(h, k, l) mixed
F =0
F 2 = 0 e.g. (100), (211); (210), (032), (033)
Unmixed indices
Unmixed indices CASE
h
k
l
A
o
o
o
B
e
e
e
CASE A : Term − 2 = [1 + e i ( e ) + e i ( e ) + e i ( e ) ] = [1 + 1 + 1 + 1] = 4
CASE B : Term − 2 = [1 + e i ( e ) + e i ( e ) + e i ( e ) ] = [1 + 1 + 1 + 1] = 4
(h, k, l) unmixed
F = 4[ f Na + + f Cl − e i ( h + k +l ) ]
e.g. (111), (222); (133), (244)
F = 4[ f Na + + f Cl − ]
If (h + k + l) is even
F 2 = 16[ f Na + + f Cl − ]2
e.g. (222),(244)
F = 4[ f Na + − f Cl − ]
If (h + k + l) is odd
F 2 = 16[ f Na + − f Cl − ]2
e.g. (111), (133)
 Presence of additional atoms/ions/molecules in the UC can alter the
intensities of some of the reflections
F
NiAl: Simple Cubic (B2- ordered structure)
Al at (0, 0, 0)
Ni at (½, ½, ½)
Click here to know more about ordered structures
F = f Al ei[2 ( h 0+ k  0+l  0)] + f Ni e
= f Al e0 + f Ni e
i[ 2  (
h + k +l
)]
2
1
1
1
i[2 ( h  + k  + l  )]
2
2
2
= f Al + f Ni ei [ h + k +l ]
Real
F = f Al + f Ni
F = f Al + f Ni e
i [ h + k + l ]
SC
F 2 = ( f Al + f Ni ) 2
e.g. (110), (200), (211), (220), (310)
F = f Al − f Ni
F 2 = ( f Al − f Ni ) 2
e.g. (100), (111), (210), (032), (133)
▪ When the central atom is identical to the corner ones− we have the BCC case.
▪ This implies that (h+k+l) even reflections are only present in BCC.
This term is zero for BCC
Reciprocal lattice/crystal of NiAl
e.g. (110), (200), (211); (220), (310)
I  ( f Al + f Ni ) 2
I  ( f Al − f Ni ) 2
e.g. (100), (111), (210), (032), (133)
Click here to know more about
Click here to know more about ordered structures
G
Al Atom at (0,0,0)
Ni atom at (½, ½, 0) and equivalent positions
Simple Cubic (L12 ordered structure)
(½, ½, 0), (½, 0, ½), (0, ½, ½)
 i[2 ( h +2 k )] i[2 ( k 2+l )] i[2 ( l +2h )] 
 + f Ni e
F = f Al e
+e
+e



= f Al + f Ni [ei ( h + k ) + ei ( k +l ) + ei (l + h ) ] Real
i[2 ( 0)]
Ni
Al
F = f Al + f Ni [ei ( h + k ) + ei ( k +l ) + ei (l + h ) ]
(h, k, l) unmixed
F = f Al + 3 f Ni
h,k,l → all even or all odd
(h, k, l) mixed
F 2 = ( f Al + 3 f Ni ) 2
e.g. (111), (200), (220), (333), (420)
F = f Al − f Ni
Two odd and one even (e.g. 112); two even and one odd (e.g. 122)
F 2 = ( f Al − f Ni ) 2
e.g. (100), (211); (210), (032), (033)
Reciprocal lattice/crystal of Ni3Al
Click here to know more about
e.g. (111), (200), (220), (333), (420)
F 2 = ( f Al + 3 f Ni ) 2
F 2 = ( f Al − f Ni ) 2
e.g. (100), (211); (210), (032), (033)
Selection / Extinction Rules
Bravais Lattice
Reflections which
may be present
Reflections
necessarily absent
Simple
all
None
Body centred
(h + k + l) even
(h + k + l) odd
Face centred
h, k and l unmixed
h, k and l mixed
End centred
h and k unmixed
C centred
h and k mixed
C centred
Bravais Lattice
Allowed Reflections
SC
All
BCC
(h + k + l) even
FCC
h, k and l unmixed
DC
h, k and l are all odd
Or
all are even
& (h + k + l) divisible by 4
h2 + k2 + l2
SC
1
100
2
110
3
111
111
4
200
200
5
210
6
211
FCC
BCC
DC
110
111
200
211
7
8
220
220
9
300, 221
10
310
11
311
311
12
222
222
13
320
14
321
220
220
310
311
222
321
15
16
400
17
410, 322
18
411, 330
19
331
400
400
400
411, 330
331
331
Relative intensity of peaks in powder patterns
❑ We have already noted that absolute value of intensity of a peak (which is the area under a
given peak) has no significance w.r.t structure identification.
❑ The relative value of intensities of the peak gives information about the motif.
❑ One factor which determines the intensity of a hkl reflection is the structure factor.
❑ In powder patterns many other factors come into the picture as in the next slide.
❑ The multiplicity factor relates to the fact that we have 8 {111} planes giving rise to single
peak, while there are only 6 {100} planes (and so forth). Hence, by this very fact the
intensity of the {111} planes should be more than that of the {100} planes.
❑ A brief consideration of some these factors follows. The reader may consult Cullity’s book
for more details.
Relative Intensity of diffraction lines in a powder pattern
Structure Factor (F)
Scattering from UC (has Atomic Scattering Factor included)
Multiplicity factor (p)
Number of equivalent scattering planes
Polarization factor
Effect of wave polarization
(
)
I P = 1 + Cos 2 (2 )
Lorentz factor
Combination of 3 geometric factors
 1 
 1 
Lorentz factor = 
(Cos )

 Sin 2 
 Sin 2 
Absorption factor
Specimen absorption
Temperature factor
Thermal diffuse scattering
Multiplicity factor
Actually only 3 planes !
Lattice
Index
Multiplicity
Planes
Cubic
(with highest
symmetry)
(100)
6
[(100) (010) (001)] ( 2 for negatives)
(110)
12
[(110) (101) (011), (110) (101) (011)] ( 2 for negatives)
(111)
12
[(111) (111) (111) (111)] ( 2 for negatives)
(210)
24*
(210) → 3! Ways, (210) → 3! Ways,
(210) → 3! Ways, (210) → 3! Ways
(211)
24
(211) → 3 ways, (211) → 3! ways,
(211) → 3 ways
(321)
48*
(100)
4
[(100) (010)] ( 2 for negatives)
(110)
4
[(110) (110)] ( 2 for negatives)
(111)
8
[(111) (111) (111) (111)] ( 2 for negatives)
(210)
8*
(210) = 2 Ways, (210) = 2 Ways,
(210) = 2 Ways, (210) = 2 Ways
(211)
16
[Same as for (210) = 8]  2 (as l can be +1 or −1)
(321)
16*
Same as above (as last digit is anyhow not permuted)
Tetragonal
(with highest
symmetry)
* Altered in crystals with lower symmetry
Multiplicity factor
Cubic
Hexagonal
Tetragonal
Orthorhombic
Monoclinic
Triclinic
hkl
48*
hk.l
24*
hkl
16*
hkl
8
hkl
4
hkl
2
hhl
24
hh.l
12*
hhl
8
hk0
4
h0l
2
hk0
24*
h0.l
12*
h0l
8
h0l
4
0k0
2
hh0
12
hk.0
12*
hk0
8*
0kl
4
hhh
8
hh.0
6
hh0
4
h00
2
h00
6
h0.0
6
h00
4
0k0
2
* Altered in crystals with lower symmetry (of the same crystal class)
00.l
2
00l
2
00l
2
Lorentz factor
Polarization factor
(
)
 1 
 1 
Lorentz factor = 
(Cos )

 Sin 2 
 Sin 2 
I P = 1 + Cos 2 (2 )
 1 + Cos 2 (30
2 ) 

Lorentz Polarization factor = 
2
 Sin  Cos 
Click here for details
XRD pattern from Polonium
All peaks present
Look at general trend line!
Example of effect of Polarization factor on
power pattern
Lorentz-Polarization factor
25
20
15
10
5
0
0
20
40
60
Bragg Angle (, degrees)
80
Intensity of powder pattern lines (ignoring Temperature & Absorption factors)
 1 + Cos 2 2 

I = F p
2
 Sin  Cos 
2
▪ Valid for Debye-Scherrer geometry
▪ I → Relative Integrated “Intensity”
▪ F → Structure factor
▪ p → Multiplicity factor
❑ POINTS
➢ As one is interested in relative (integrated) intensities of the lines constant factors are
omitted
• Volume of specimen • me , e • (1/dectector radius)
➢ Random orientation of crystals → in a material with Texture relative intensities are modified
➢ I is really diffracted energy (as Intensity is Energy/area/time)
➢ Ignoring Temperature & Absorption factors  valid for lines close-by in pattern