# 8-10 - University of Warwick

```The Muppet’s Guide to:
The Structure and Dynamics of Solids
Phase Diagrams
Binary Phase Diagrams
Phase B
Phase A
Nickel atom
Copper atom
• When we combine two elements...
what equilibrium state do we get?
• In particular, if we specify...
--a composition (e.g., wt.% Cu – wt.% Ni), and
--a temperature (T )
then...
How many phases do we get?
What is the composition of each phase?
How much of each phase do we get?
Phase Diagrams
• Indicate phases as function of T, Co, and P.
• For this course:
-binary systems: just 2 components.
-independent variables: T and Co (P = 1 atm is almost always used).
T(&deg;C)
• 2 phases:
1600
• Phase
Diagram
for Cu-Ni
at P=1 atm.
L (liquid)
a (FCC solid solution)
L (liquid)
1500
1400
1300
a
(FCC solid
solution)
1200
1100
1000
0
20
40
60
Figure adapted from Callister, Materials science and engineering, 7th Ed.
80
• 3 phase fields:
L
L+a
a
100
wt% Ni
Phase Diagrams
• Indicate phases as function of T, Co, and P.
• For this course:
-binary systems: just 2 components.
-independent variables: T and Co (P = 1 atm is almost always used).
T(&deg;C)
1600
• Phase
Diagram
for Cu-Ni
at P=1 atm.
Liquidus:
Separates the liquid
from the mixed L+a
phase
L (liquid)
1500
1400
Solidus:
Separates the
mixed L+a phase
from the solid
solution
1300
a
(FCC solid
solution)
1200
1100
1000
0
20
40
60
Figure adapted from Callister, Materials science and engineering, 7th Ed.
80
100
wt% Ni
Number and types of phases
• Rule 1: If we know T and Co, then we know:
- the number and types of phases present.
• Examples:
T(&deg;C)
1600
A(1100&deg;C, 60):
1 phase: a
B (1250&deg;C,35)
L (liquid)
1500
1400
B(1250&deg;C, 35):
2 phases: L + a
1300
a
(FCC solid
solution)
1200
A(1100&deg;C,60)
1100
1000
Cu-Ni
phase
diagram
0
20
Figure adapted from Callister, Materials science and engineering, 7th Ed.
40
60
80
100
wt% Ni
Composition of phases
• Rule 2: If we know T and Co, then we know:
--the composition of each phase.
• Examples:
T(&deg;C)
Cu-Ni
system
A
TA
Co = 35 wt% Ni
1300 L (liquid)
At T A = 1320&deg;C:
Only Liquid (L)
B
TB
CL = Co ( = 35 wt% Ni)
At T D = 1190&deg;C:
1200
D
Only Solid ( a)
TD
Ca = Co ( = 35 wt% Ni)
20
3032 35
At T B = 1250&deg;C:
CLCo
Both a and L
CL = C liquidus ( = 32 wt% Ni here)
Ca = C solidus ( = 43 wt% Ni here)
Figure adapted from Callister, Materials science and engineering, 7th Ed.
tie line
a
(solid)
4043
50
Ca wt% Ni
Cooling a Cu-Ni Binary - Composition
• Phase diagram:
Cu-Ni system.
• System is:
--binary
i.e., 2 components:
Cu and Ni.
T(&deg;C) L (liquid)
130 0
L: 35 wt% Ni
a: 46 wt% Ni
• Consider
Co = 35 wt%Ni.
Cu-Ni
system
A
35
32
--isomorphous
i.e., complete
solubility of one
component in
another; a phase
field extends from
0 to 100 wt% Ni.
L: 35wt%Ni
B
C
46
43
D
24
L: 32 wt% Ni
36
120 0
a: 43 wt% Ni
E
L: 24 wt% Ni
a: 36 wt% Ni
a
(solid)
110 0
20
Figure adapted from Callister, Materials science and engineering, 7th Ed.
30
35
Co
40
50
wt% Ni
USE LEVER RULE
The Lever Rule – Weight %
• Tie line – connects the phases in equilibrium
with each other - essentially an isotherm
T(&deg;C)
How much of each phase?
Think of it as a lever
tie line
1300
L (liquid)
B
TB
a
(solid)
1200
R
20
30C
S
40 C
a
L Co
R
50
wt% Ni
WL 
Ma
ML
C  C0
ML
S

 a
ML  M a R  S Ca  CL
Figure adapted from Callister, Materials science and engineering, 7th Ed.
S
M a S  M L R
Wa 
C  CL
R
 0
R  S Ca  CL
Weight fractions of phases – ‘lever rule’
• Rule 3: If we know T and Co, then we know:
--the amount of each phase (given in wt%).
• Examples:
Cu-Ni
system
T(&deg;C)
C o = 35 wt% Ni
At T A : Only Liquid (L)
W L = 100 wt%, W a = 0
At T D : Only Solid ( a)
W L = 0, W a = 100 wt%
At T B : Both a and L
43  35
 73 wt %
43  32
WL 
S
R +S

Wa 
R
R +S
= 27 wt%
Figure adapted from Callister, Materials science and engineering, 7th Ed.
A
TA
1300
TB
1200
TD
20
tie line
L (liquid)
B
S
R
D
3032 35
C LC o
a
(solid)
40 4 3
50
C a wt% Ni
Cooling a Cu-Ni Binary – wt. %
• Phase diagram:
Cu-Ni system.
• System is:
--binary
i.e., 2 components:
Cu and Ni.
T(&deg;C) L (liquid)
130 0
L: 92 wt%
a: 8 wt%
i.e., complete
solubility of one
component in
another; a phase
field extends from
0 to 100 wt% Ni.
• Consider
Co = 35 wt%Ni.
Cu-Ni
system
A
32
--isomorphous
L: 35wt%Ni
34
B
C
46
43
D
24
L: 73 wt%
36
120 0
a: 27 wt%
E
L: 8 wt%
a: 92 wt%
a
(solid)
110 0
20
Figure adapted from Callister, Materials science and engineering, 7th Ed.
30
35
Co
40
50
wt% Ni
Equilibrium cooling
• Multiple freezing sites
– Polycrystalline materials
– Not the same as a single crystal
• The compositions that freeze are a
function of the temperature
• At equilibrium, the ‘first to freeze’
composition must adjust on further cooling
by solid state diffusion
Diffusion is not a flow
Concept behind mean free path in
scattering phenomena conductivity
Our models of diffusion are based on a random walk approach and
not a net flow
http://mathworld.wolfram.com/images/eps-gif/RandomWalk2D_1200.gif
Diffusion in 1 Dimension
• Fick’s First Law
J = flux – amount of material per unit area per unit time
C = concentration
dC
J   D T 
dx
Diffusion coefficient which we expect
is a function of the temperature, T
Diffusion cont….
• Requires the solution of the continuity equation:
The change in concentration as a function of time in a
volume is balanced by how much material flows in per time
unit minus how much flows out – the change in flux, J:
C
J

0
t
x
•
BUT
dC
J   D T 
dx
giving Fick’s second law (with D being constant):
C
  C 
2C

D
  D T  2
t
x  x 
x
Solution of Ficks’ Laws
t=0
Co
For a semi-infinite sample
the solution to Ficks’ Law
gives an error function
distribution whose width
increases with time
C
C
t=t
x
Consider slabs of Cu and Ni.
Interface region will be a
mixed alloy (solid solution)
Interface region will grow as a
function of time
Slow Cooling in a Cu-Ni Binary
T(&deg;C) L (liquid)
L: 35wt%Ni
Co = 35 wt%Ni.
Enough time is
allowed at each
temperature
change for atomic
diffusion to occur.
–
Thermodynamic
ground state
Each phase is
homogeneous
130 0
L: 35 wt% Ni
a: 46 wt% Ni
Cu-Ni
system
A
B
C
D
120 0
L: 32 wt% Ni
a: 43 wt% Ni
E
L: 24 wt% Ni
a: 36 wt% Ni
a
(solid)
110 0
20
Figure adapted from Callister, Materials science and engineering, 7th Ed.
30
35
Co
40
50
wt% Ni
L
Non –
equilibrium
cooling
No-longer in the
thermodynamic
ground state
Reduces the
melting
temperature
Figure adapted from Callister, Materials science and engineering, 7th Ed.
α+L
α
Cored vs Equilibrium Phases
• Ca changes as we solidify.
• Cu-Ni case:
First a to solidify has Ca = 46 wt% Ni.
Last a to solidify has Ca = 35 wt% Ni.
• Fast rate of cooling:
Cored structure
• Slow rate of cooling:
Equilibrium structure
First a to solidify:
46 wt% Ni
Last a to solidify:
&lt; 35 wt% Ni
Figure adapted from Callister, Materials science and engineering, 7th Ed.
Uniform C a:
35 wt% Ni
Binary-Eutectic Systems – Cu/Ag
2 components
has a special composition
with a min. melting temperature
a
phase:
b
phase:
Mostly
copper
Solvus line
– the
solubility
limit
Mostly
Silver
• Limited solubility – mixed phases
• 3 phases regions, L, a and b and 6 phase fields - L, a and b, L+a, L+b, ab
Figure adapted from Callister, Materials science and engineering, 7th Ed.
Binary-Eutectic Systems
Cu-Ag system
T(&deg;C)
The Eutectic point
1200
L (liquid)
TE, Eutectic temperature, 779&deg;C
1000
CE, eutectic composition, 71.9wt.%
a
TE
• TE
: No liquid below TE
Min. melting TE
800
E
779&deg;C
CaE=8.0
600
CE=71.9
L +b b
CbE=91.2
ab
400
200
• Eutectic transition
L(CE)
L+a
0
a(CaE) + b(CbE)
Figure adapted from Callister, Materials science and engineering, 7th Ed.
Co
20
40
60 CE 80
wt% Ag in Cu/Ag alloy
100
Any other composition, Liquid
transforms to a mixed L+solid phase
Pb-Sn (Solder) Eutectic System (1)
• For a 40 wt% Sn-60 wt% Pb alloy at 150&deg;C, find...
--the phases present: a + b
--compositions of phases:
CO = 40 wt% Sn
Ca = 11 wt% Sn
Cb = 99 wt% Sn
--the relative amount
of each phase:
Wa =
S
R+S
C -C
= b O
Cb - Ca
99 - 40
59
=
= 67 wt%
99 - 11
88
CO - Ca
Wb = R
=
Cb - Ca
R+S
40 - 11
29
=
= 33 wt%
=
99 - 11
88
Pb-Sn
system
T(&deg;C)
300
200
L (liquid)
a
L+a
18.3
150
L +b b
183&deg;C
61.9
R
97.8
S
100
a+ b
=
0 11 20
Ca
40
Co
60
80
C, wt% Sn
99100
Cb
Figure adapted from Callister, Materials science and engineering, 7th Ed.
Microstructures in Eutectic
Systems: II
L: Co wt% Sn
T(&deg;C)
• 2 wt% Sn &lt; Co &lt; 18.3 wt% Sn
• Result:
400
 Initially liquid → liquid + a
 then a alone
 finally two phases
 a poly-crystal
 fine b-phase inclusions
L
300
L +a
a
200
TE
a: Co wt% Sn
a
b
100
a+ b
0
10
20
Co
Co ,
2
(sol. limit at T room )
18.3
(sol. limit at TE)
Figure adapted from Callister, Materials science and engineering, 7th Ed.
L
a
Pb-Sn
system
30
wt% Sn
Microstructures
in Eutectic Systems: Co=CE
• Result: Eutectic microstructure (lamellar structure)
--alternating layers (lamellae) of a and b crystals.
T(&deg;C)
Micrograph of Pb-Sn
eutectic
microstructure
L: Co wt% Sn
300
Pb-Sn
system
a
200
L+a
L
TE
100
ab
0
Lb b
183&deg;C
20
18.3
40
b: 97.8 wt% Sn
a: 18.3 wt%Sn
60
CE
61.9
80
100
97.8
C, wt% Sn
Figures adapted from Callister, Materials science and engineering, 7th Ed.
160m
Microstructures
in Eutectic Systems: Co=CE
• Result: Eutectic microstructure (lamellar structure)
--alternating layers (lamellae) of a and b crystals.
T(&deg;C)
L: Co wt% Sn
300
Pb-Sn
system
a
200
L+a
Wa 
Lb b
183&deg;C
TE
Pb
rich
L
Wb 
100
ab
0
20
18.3
40
b: 97.8 wt% Sn
a: 18.3 wt%Sn
60
CE
61.9
80
100
97.8
C, wt% Sn
Figures adapted from Callister, Materials science and engineering, 7th Ed.
97.8  61.9
 45.2%
97.8  18.3
61.9  18.3
 54.8%
97.8  18.3
Sn Rich
Lamellar Eutectic Structure
At interface, Pb moves to a-phase
and Sn migrates to b- phase
Lamellar form to minimise diffusion
distance – expect spatial extent to
depend on D and cooling rates.
Figure adapted from Callister, Materials science and engineering, 7th Ed.
Sn
Pb
Microstructures IV
• 18.3 wt% Sn &lt; Co &lt; 61.9 wt% Sn
• Result: a crystals and a eutectic microstructure
L: Co wt% Sn
T(&deg;C)
300
Pb-Sn
system
a
200
L+a
L
a
L
R
a L
L+b b
S
TE
a+ b
100
0
20
18.3
40
60
61.9
80
Co, wt% Sn
Figure adapted from Callister, Materials science and engineering, 7th Ed.
100
• Just above TE :
C a = 18.3 wt% Sn
CL = 61.9 wt% Sn
Wa= S
= 50 wt%
R +S
WL = (1- W a) = 50 wt%
Microstructures IV
• 18.3 wt% Sn &lt; Co &lt; 61.9 wt% Sn
• Result: a crystals and a eutectic microstructure
L: Co wt% Sn
T(&deg;C)
300
Pb-Sn
system
a
200
L+a
TE
R
100
a+ b
L
a
L
R
a L
L+b b
S
• Just below TE :
C a = 18.3 wt% Sn
C b = 97.8 wt% Sn
Wa = S
= 73 wt%
R +S
W b = 27 wt%
S
Primary, a
0
20
18.3
40
60
61.9
80
Co, wt% Sn
Figure adapted from Callister, Materials science and engineering, 7th Ed.
100
97.8
Eutectic, a
Eutectic, b
Intermetallic Compounds
a
phase:
Mostly
Mg
Mg2Pb
b
phase:
Mostly
Note: intermetallic compound forms a line - not an area because stoichiometry (i.e. composition) is exact.
Figure adapted from Callister, Materials science and engineering, 7th Ed.
Eutectoid &amp; Peritectic
Peritectic transition  + L
Cu-Zn Phase diagram
Eutectoid transition 
+
Figure adapted from Callister, Materials science and engineering, 7th Ed.

mixed liquid and solid to
single solid transition
Solid to solid ‘eutectic’ type
transition
Iron-Carbon (Fe-C) Phase Diagram
L
 + Fe3C
-Eutectoid (B):

a + Fe3C
T(&deg;C)
1600

L
1400
1200
 +L

(austenite)
 
 
1000
800
a
600
Result: Pearlite =
alternating layers of
a and Fe3C phases
S
 +Fe3C
727&deg;C = T eutectoid
R
S
1
0.76
C eutectoid
120 m
0
(Fe)
Figure adapted from Callister, Materials science and engineering, 7th Ed.
L+Fe3C
R
B
400
A
1148&deg;C
Fe3C (cementite)
• 2 important
points
-Eutectic (A):
2
3
a+Fe3C
4
5
6
6.7
4.30
Co, wt% C
Fe3C (cementite-hard)
a (ferrite-soft)
Iron-Carbon
http://www.azom.com/work/pAkmxBcSVBfns037Q0LN_files/image003.gif
The Muppet’s Guide to:
The Structure and Dynamics of Solids
The Final Countdown
Characterisation
Over the course so far we have seen how thermodynamics plays an
important role in defining the basic minimum energy structure of a
solid.
Small changes in the structure (such as the perovskites) can produce
changes in the physical properties of materials
Kinetics and diffusion also play a role and give rise to different metastable structures of the same materials – allotropes / polymorphs
Alloys and mixtures undergo multiple phase changes as a function of
temperature and composition
BUT how do we
characterise samples?
Probes
Resolution better than the inter-atomic spacings
• Neutrons
• Electrons
Probes
Treat all probes as if they were waves:
Wave-number, k:
Momentum, p:
k  k  2
p  k;
Photons

p  mv  h
‘Massive’ objects

Xavier the X-ray
Speed of Light
E  hc
Planck’s constant

Wavelength
Ex(keV)=1.2398/(nm)
Elastic scattering as Ex&gt;&gt;kBT
Norbert the Neutron
De Broglie equation:
  h mv
mass
velocity
2
2
1
h
Kinetic Energy: En  2 mv 
2mn  2
En(meV)=0.8178/2(nm)
Strong inelastic scattering as En~kBT
Eric the Electron
• Eric’s rest mass: 9.11 &times; 10−31 kg.
•
electric charge: −1.602 &times; 10−19 C
• No substructure – point particle
De Broglie equation:
  h mv
mass
velocity
Ee depends on accelerating voltage :–
Range of Energies from 0 to MeV
Probes
Resolution better than the interatomic spacings
Absorption low – we want a ‘bulk’ probe
• Electrons - Eric
• quite surface sensitive
• Optical – spectroscopy
• X-rays :
• VUV and soft (spectroscopic and surfaces)
• Hard (bulk like)
• Neutrons - Norbert
Interactions
Englebert
Xavier
Norbert
1. Absorption
2. Refraction/Reflection
3. Scattering
Diffraction
a
Crystals are 2D with planes separated by dhkl. There will only
be constructive interference when a== - i.e. the reflection
condition.
Basic Scattering Theory
The number of
scattered particles per
second is defined
using the standard
expression
Unit solid angle
d
I s  I 0 
d
Differential
cross-section
Defined using Fermi’s Golden Rule
d
 Final INTERACTION POTENTIAL Initial
d
d

d
Spherical
Scattered
Wavefield

exp i k  r  V r   r  dr
Scattering
Potential
2
Incident
Wavefield
Different for X-rays, Neutrons and Electrons
d

d

exp i k  r  V r   r  dr
2
BORN approximation:
• Assumes initial wave is also spherical
• Scattering potential gives weak interactions
d

d
q  k0  k 
 exp i k  r  V r  exp i k  r  dr
2
0
d

d
 V r  exp i q  r  dr
2
Scattered intensity is proportional to the Fourier
Transform of the scattering potential
Scattering from Crystal
As a crystal is a periodic repetition of atoms in 3D we can formulate
the scattering amplitude from a crystal by expanding the scattering
from a single atom in a Fourier series over the entire crystal
f 
 (r ) exp iq  r  dV
V
A(q) 
f
T
Atomic Structure
Factor
j
j
 q exp  i q   T  r j  
Real Lattice Vector:
T=ha+kb+lc
The Structure Factor
Describes the Intensity of the diffracted beams in reciprocal space
 exp i q  r    exp i  hu  kv  lw  2 
j
j
j
hkl are the diffraction planes, uvw
are fractional co-ordinates within
the unit cell
If the basis is the same, and has a
scattering factor, (f=1), the structure
factors for the hkl reflections can be found
Weight  phase
hkl
The Form Factor
Describes the distribution of the diffracted
beams in reciprocal space
The summation is over the entire crystal
which is a parallelepiped of sides:
L  q 
N1a  N2b  N3c
N1
 exp i q  T    exp  n  i q  a 
1
T
n1 1
N3
N2
 exp  n  i q  b  exp  n  i q  c 
2
n2 1
3
n3 1
The Form Factor
Measures the translational symmetry of the lattice
q  d  s
Redefine q:
L  q 
Deviation from reciprocal
lattice point located at d*
Ni
  exp i n s   
i i
i 1,2,3 ni
i 1,2,3
sin  Ni si 
sin  si 


sin  Ni si 
i 1,2,3
sj
5
2.5x10
5
2
2.0x10
[L(s1)]
The Form Factor has low
intensity unless q is a
reciprocal lattice vector
associated with a reciprocal
lattice point
1.5x10
5
1.0x10
5
0.5x10
0
-0.02
N=2,500; FWHM-1.3”
N=500
5
-0.01
0
0.01
0.02
The Form Factor
L  q 

sin  Ni si 
sj
5
100
2.5x10
80
2.0x10
2
5
60
[L(s1)]
[L(s1)]
2
i 1,2,3
40
20
0
5
1.5x10
5
1.0x10
5
0.5x10
-0.6
-0.3
0
0.3
0.6
0
-0.02
-0.01
0
0.01
0.02
N=10
N=500
The square of the Form Factor in one dimension
Scattering in Reciprocal Space

  exp i  q  T 
A  q   f j  q exp i q  r j 


j
T
Peak positions and intensity tell us about the structure:
POSITION
OF PEAK
PERIODICITY
WITHIN
SAMPLE
WIDTH
OF PEAK
EXTENT OF
PERIODICITY
INTENSITY
OF PEAK
POSITION OF
ATOMS IN
BASIS
Powder Diffraction
It is impossible to grow some materials in a single crystal form or
we wish to study materials in a dynamic process.
Powder Techniques
Allows a wider range of materials to be studied under
different sample conditions
1. Inductance Furnace  290 – 1500K
2. Closed Cycle Cryostat  10 – 290K
3. High Pressure  Up-to 5 million Atmospheres
• Phase changes as a function of
Temp and Pressure
• Phase identification
Search and Match
Powder Diffraction often used to identify phases
 Cheap, rapid, non-destructive and only small quantity of sample
 Monochromatic xrays
Intensity
Diffractometer
High Dynamic range
detector
2 Angle
JCPDS Powder Diffraction File lists
materials (&gt;50,000) in order of their dspacings and 6 strongest reflections
OK for mixtures of up-to 4
components and 1% accuracy
Single Crystal Diffraction
  2d hkl sin  hkl
sample needs to moved to the
Bragg condition….
Angular resolution is
the Darwin width of
analyser crystal
(Typically 10-20”)
Detailed Lateral Information obtained
XMaS Beamline - ESRF
Strain
Peak positions defined by the lattice parameters:
L  q 
N1
  exp i  q  n 
i
i  a, b, c n1 1
Strain is an extension or compression of the lattice,

dhkl  dhkl
Results in a systematic shift of all the peaks
Ho Thin Films
Intensity (arb. units)
XRD measured as a function of temperature
10
4
10
2
10
0
10
-2
10
-4
20
T=294K
T=244K
T=194K
T=94K
T=144K
T=42K
T=10K
T=300K
40
60
Scattering Angle ()
80
100
Ho Thin Films
Substrate and Ho film follow have different behaviour
Intensity (arb. units)
1000
T=294K
T=244K
T=194K
T=144K
T=94K
T=42K
T=10K
100
10
1
30
32
Scattering Angle ()
34
36
Whole film refinement
Diffraction peaks can also be broadened in qz by:
1. Grain Size
2. Micro-Strains OR Both
The crystal is made up of particulates which
all act as perfect but small crystals
Number of planes sampled is finite
Recall form factor:
L  q 

i  a,b,c
sin  Ni si 
si
Scherrer Equation
 Size 2  

D cos B 
NixMn3-xO4+ (400 Peak)
As Grown at 200&ordm;C
AFM images (1200 x 1200 nm)
1.0
Annealed at 800&ordm;C
Intensity
0.8
800C
850C
900C
650C
700C
750C
0.6
0.4
0.2
0
-1.0
400
-0.5
0
0.5
2
450nm thick films
1.0
Diffraction peaks can also be broadened in qz by:
1. Grain Size
2. Micro-Strains OR Both
The crystal has a distribution of inter-planar spacings dhkl &plusmn;dhkl.
Diffraction over a range, , of angles
Differentiate Bragg’s Law:
  d
d
Strain 2   2 tan B 
Width in
Strain
Bragg
angle
Diffraction peaks can also be broadened in qz by:
1. Grain Size
2. Micro-Strains OR Both
Total Broadening in 2 is sum of Strain and Size:
Rearrange
Total 2  
Williamson-Hall
plot

D cos B 
 2 tan B 




hkl 2  cos Bhkl  2 sin Bhkl 
y  mx  c

D
Powder Diffraction
0.25
2
0.20
400
0.10
222
311
400
0
220
0
10
20
2
2
30
2
(h +k +l )
200
0.008
100
0
30
40
50
60
Width * cos(B)
Intensity (arb. units)
0.05
300
333
422
0.15
2
sin (B)
Powder of Nickel Manganite
CUBIC Structure
2
y=(1.541 /(4*a ))x
a=8.348 &plusmn; 0.0036
0.007
y=((1.541/d))+(2s)x
Grain
&plusmn; 19.5
Grain
sizeSize=299
= 30&plusmn;2nm
Strain
= 0.005&plusmn;0.001
= 0.005
&plusmn; 0.001
0.006
Detector Angle (&deg;)
0.005
0.05
0.10
0.15
sin(B)
0.20
0.25
Cubic-Tetragonal Distortions
CUBIC
ac
TETRAGONAL
ac
High Temperature Powder XRD
Intensity (counts)
Intensity (counts)
0.4BiSCO3 - 0.6PbTiO3 (K. Datta)
Tetragonal → Cubic phase transition
30000
30000
20000
10000
20000
30.8
30.9
31.0
31.1
31.2
31.3
31.4
31.5
31.6
31.7
2Theta (&deg;)
10000
0
25
30
35
Courtesy, D. Walker and K. Datta University of Warwick
40
45
50
55
60
65
70
2Theta (&deg;)
CsCoPO4
Variable temperature powder X-ray diffraction data show a marked change in the
pattern at 170 &deg;C.
Dr. Mark T. Weller, Department of Chemistry, University of Southampton, www.rsc.org/ej/dt/2000/b003800h/
Sn in a Silica Matrix
1. What form of tin
2. Particle size
3. Strain
4. Melting Temperature
Eutectic’s
Following Structural Changes
• Consider Cu/Ni with 35
wt.% Ni
T(&deg;C) L (liquid)
130 0
L: 35 wt% Ni
a: 46 wt% Ni
32
43
L: 32 wt% Ni
36
120 0
a: 43 wt% Ni
E
L: 24 wt% Ni
a: 36 wt% Ni
a
(solid)
D.
E. Solid
B
C
D
24
C.
Cu-Ni
system
A
A. Liquid
B. Mixed Phase
L: 35wt%Ni
110 0
20
Figure adapted from Callister, Materials science and engineering, 7th Ed.
30
35
Co
40
50
wt% Ni
USE LEVER RULE
Cored
Samples
L
α+L
α
Issues:
Lattice Parameter
Particle Size
Strain Dispersion




hkl 2  cos Bhkl  2 sin Bhkl 

D
NiCr
Structural
Changes?
Fcc: hkl are
either all
odd or all
even.
Bcc: sum
of hkl must
be even.
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