the muppet’s guide to:
DESCRIPTION
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... - PowerPoint PPT PresentationTRANSCRIPT
The Muppet’s Guide to:The Structure and Dynamics of Solids
Phase Diagrams
• 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?
Binary Phase DiagramsPhase BPhase A
Nickel atomCopper atom
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).
• PhaseDiagramfor Cu-Niat P=1 atm.
• 2 phases: L (liquid)
a (FCC solid solution)
• 3 phase fields: LL + aa
wt% Ni20 40 60 80 10001000
1100
1200
1300
1400
1500
1600T(°C)
L (liquid)
a (FCC solid solution)
L + aliquidus
solid
us
Figure adapted from Callister, Materials science and engineering, 7 th Ed.
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).
• PhaseDiagramfor Cu-Niat P=1 atm.
wt% Ni20 40 60 80 10001000
1100
1200
1300
1400
1500
1600T(°C)
L (liquid)
a (FCC solid solution)
L + aliquidus
solid
us
Figure adapted from Callister, Materials science and engineering, 7 th Ed.
Liquidus:Separates the liquid from the mixed L+ aphase
Solidus:Separates the mixed L+ a phase from the solid solution
wt% Ni20 40 60 80 10001000
1100
1200
1300
1400
1500
1600T(°C)
L (liquid)
a (FCC solid solution)
L + a
liquidus
solid
us
Cu-Niphase
diagram
Number and types of phases• Rule 1: If we know T and Co, then we know: - the number and types of phases present.
• Examples:
A(1100°C, 60): 1 phase: a
B(1250°C, 35): 2 phases: L + a
B (
1250
°C,3
5) A(1100°C,60)
Figure adapted from Callister, Materials science and engineering, 7 th Ed.
wt% Ni20
1200
1300
T(°C)
L (liquid)
a(solid)L + a
liquidus
solidus
30 40 50
L + a
Cu-Ni system
Composition of phases• Rule 2: If we know T and Co, then we know: --the composition of each phase.
• Examples:TA
A
35Co
32CL
At TA = 1320°C:
Only Liquid (L) CL = Co ( = 35 wt% Ni)
At TB = 1250°C:
Both a and L CL = C liquidus ( = 32 wt% Ni here)
Ca = C solidus ( = 43 wt% Ni here)
At TD = 1190°C:
Only Solid ( a) Ca = Co ( = 35 wt% Ni)
Co = 35 wt% Ni
BTB
DTD
tie line
4Ca3
Figure adapted from Callister, Materials science and engineering, 7 th Ed.
wt% Ni20
1200
1300
30 40 50110 0
L (liquid)
a (solid)
L + a
L + a
T(°C)
A
35Co
L: 35wt%Ni
Cu-Nisystem
• Phase diagram: Cu-Ni system.
• System is: --binary i.e., 2 components: Cu and Ni. --isomorphous i.e., complete solubility of one component in another; a phase field extends from 0 to 100 wt% Ni.
• Consider Co = 35 wt%Ni.
Cooling a Cu-Ni Binary - Composition
4635
4332
a: 43 wt% Ni
L: 32 wt% Ni
L: 24 wt% Ni
a: 36 wt% Ni
Ba: 46 wt% NiL: 35 wt% Ni
C
D
E
24 36
Figure adapted from Callister, Materials science and engineering, 7 th Ed. USE LEVER RULE
• Tie line – connects the phases in equilibrium with each other - essentially an isotherm
The Lever Rule – Weight %
How much of each phase? Think of it as a lever
ML M
R S
RMSM L
L
L
LL
LL CC
CC
SR
RW
CC
CC
SR
S
MM
MW
00
wt% Ni
20
1200
1300
T(°C)
L (liquid)
a(solid)L + a
liquidus
solidus
30 40 50
L + aB
T B
tie line
CoC L Ca
SR
Figure adapted from Callister, Materials science and engineering, 7 th Ed.
• Rule 3: If we know T and Co, then we know: --the amount of each phase (given in wt%).• Examples:
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%
C o = 35 wt% Ni
Weight fractions of phases – ‘lever rule’
wt% Ni20
1200
1300
T(°C)
L (liquid)
a(solid)L + a
liquidus
solidus
30 40 50
L + a
Cu-Ni system
TA A
35C o
32C L
BT B
DT D
tie line
4Ca3
R S
= 27 wt%
43 3573 %
43 32wt
At T B : Both a and L
WL= S
R + S
Wa= R
R + S
Figure adapted from Callister, Materials science and engineering, 7 th Ed.
wt% Ni20
120 0
130 0
30 40 50110 0
L (liquid)
a (solid)
L + a
L + a
T(°C)
A
35C o
L: 35wt%Ni
Cu-Nisystem
• Phase diagram: Cu-Ni system.• System is: --binary i.e., 2 components: Cu and Ni. --isomorphous i.e., complete solubility of one component in another; a phase field extends from 0 to 100 wt% Ni.• Consider Co = 35 wt%Ni.
Cooling a Cu-Ni Binary – wt. %
46344332
a: 27 wt%
L: 73 wt%
L: 8 wt%
a: 92 wt%
Ba: 8 wt% L: 92 wt%
C
D
E
24 36
Figure adapted from Callister, Materials science and engineering, 7 th Ed.
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
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
Concept behind mean free path in scattering phenomena - conductivity
Diffusion in 1 Dimension
• Fick’s First Law
dCJ D T
dx
J = flux – amount of material per unit area per unit timeC = concentration
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:
• giving Fick’s second law (with D being constant):
2
2
C C CD D T
t x x x
0C Jt x
dC
J D Tdx
BUT
Solution of Ficks’ Laws
C
x
CCo
t = 0
t = t
For a semi-infinite sample the solution to Ficks’ Law gives an error function distribution whose width increases with time
Consider slabs of Cu and Ni.
Interface region will be a mixed alloy (solid solution)
Interface region will grow as a function of time
wt% Ni20
120 0
130 0
30 40 50110 0
L (liquid)
a (solid)
L + a
L + a
T(°C)
A
35C o
L: 35wt%Ni
Cu-Nisystem
Co = 35 wt%Ni.
Slow Cooling in a Cu-Ni Binary
a: 43 wt% Ni
L: 32 wt% Ni
L: 24 wt% Ni
a: 36 wt% Ni
Ba: 46 wt% NiL: 35 wt% Ni
C
D
E
Figure adapted from Callister, Materials science and engineering, 7 th Ed.
Enough time is allowed at each temperature change for atomic diffusion to occur. – Thermodynamic ground state
Each phase is homogeneous
Non – equilibrium
cooling α
L
α + L
Figure adapted from Callister, Materials science and engineering, 7 th Ed.
Reduces the melting
temperature
No-longer in the thermodynamic
ground state
• Ca changes as we solidify.• Cu-Ni case:
• Fast rate of cooling: Cored structure
• Slow rate of cooling: Equilibrium structure
First a to solidify has Ca = 46 wt% Ni.
Last a to solidify has Ca = 35 wt% Ni.
Cored vs Equilibrium Phases
First a to solidify: 46 wt% Ni
Uniform C a:
35 wt% Ni
Last a to solidify: < 35 wt% Ni
Figure adapted from Callister, Materials science and engineering, 7 th Ed.
2 componentshas a special compositionwith a min. melting temperature
Binary-Eutectic Systems – Cu/Ag
• 3 phases regions, L, a and b and 6 phase fields - L, a and , b L+ , a L+ , +b a b
• Limited solubility – mixed phases
Figure adapted from Callister, Materials science and engineering, 7 th Ed.
a phase:
Mostly copper
b phase:
Mostly Silver
Solvus line – the solubility limit
Min. melting TE
Binary-Eutectic Systems
• Eutectic transitionL(CE) (CE) + (CE)
• TE : No liquid below TE
TE, Eutectic temperature, 779°CCE, eutectic composition, 71.9wt.%
The Eutectic point
Cu-Ag system
L (liquid)
a L + a L +b b
a + b
Co wt% Ag in Cu/Ag alloy20 40 60 80 1000
200
1200T(°C)
400
600
800
1000
CE
TE CaE=8.0 CE=71.9 CbE=91.2
779°C
Figure adapted from Callister, Materials science and engineering, 7 th Ed.
Any other composition, Liquid transforms to a mixed L+solid phase
E
L + aL +b
a + b
200
T(°C)
18.3
C, wt% Sn20 60 80 1000
300
100
L (liquid)
a 183°C 61.9 97.8
b
• For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find... --the phases present: Pb-Sn
system
Pb-Sn (Solder) Eutectic System (1)
a + b--compositions of phases:
CO = 40 wt% Sn
--the relative amount of each phase:
150
40
Co
11
C
99
C
SR
Ca = 11 wt% SnCb = 99 wt% Sn
Wa =C - CO
C - C
= 99 - 4099 - 11
= 5988 = 67 wt%
SR+S =
W =CO - C
C - C=R
R+S
= 2988
= 33 wt%= 40 - 1199 - 11
Figure adapted from Callister, Materials science and engineering, 7 th Ed.
• 2 wt% Sn < Co < 18.3 wt% Sn• Result: Initially liquid → liquid + then alone finally two phases
a poly-crystal fine -phase inclusions
Microstructures in Eutectic Systems: II
Pb-Snsystem
L + a
200
T(°C)
Co , wt% Sn10
18.3
200Co
300
100
L
a
30
a + b
400
(sol. limit at TE)
TE
2(sol. limit at T room)
La
L: Co wt% Sn
ab
a: Co wt% Sn
Figure adapted from Callister, Materials science and engineering, 7 th Ed.
• Result: Eutectic microstructure (lamellar structure) --alternating layers (lamellae) of a and b crystals.
Microstructures in Eutectic Systems: Co=CE
160 m
Micrograph of Pb-Sn eutectic microstructurePb-Sn
systemL
a
200
T(°C)
C, wt% Sn
20 60 80 1000
300
100
L
a b
L+ a
183°C
40
TE
18.3
: 18.3 wt%Sn
97.8
: 97.8 wt% Sn
CE61.9
L: Co wt% Sn
Figures adapted from Callister, Materials science and engineering, 7 th Ed.
• Result: Eutectic microstructure (lamellar structure) --alternating layers (lamellae) of a and b crystals.
Microstructures in Eutectic Systems: Co=CE
Pb-Snsystem
L
a
200
T(°C)
C, wt% Sn
20 60 80 1000
300
100
L
a b
L+ a
183°C
40
TE
18.3
: 18.3 wt%Sn
97.8
: 97.8 wt% Sn
CE61.9
L: Co wt% Sn
Figures adapted from Callister, Materials science and engineering, 7 th Ed.
97.8 61.945.2%
97.8 18.3W
61.9 18.354.8%
97.8 18.3W
Pb rich
Sn Rich
Lamellar Eutectic Structure
Figure adapted from Callister, Materials science and engineering, 7 th Ed.
PbSn
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.
• 18.3 wt% Sn < Co < 61.9 wt% Sn• Result: a crystals and a eutectic microstructure
Microstructures IV
18.3 61.9
SR WL = (1- Wa) = 50 wt%
Ca = 18.3 wt% Sn
CL = 61.9 wt% SnS
R + SWa = = 50 wt%
• Just above TE :
Pb-Snsystem
L+b200
T(°C)
Co, wt% Sn
20 60 80 1000
300
100
L
a b
L +a
40
a +b
TE
L: Co wt% Sn LL
Figure adapted from Callister, Materials science and engineering, 7 th Ed.
• 18.3 wt% Sn < Co < 61.9 wt% Sn• Result: a crystals and a eutectic microstructure
Microstructures IV
18.3 61.9
SR
97.8
SR
Primary, a
Eutectic, a Eutectic, b
• Just below TE :Ca = 18.3 wt% SnCb = 97.8 wt% Sn
SR + S
Wa = = 73 wt%
Wb = 27 wt%
Pb-Snsystem
L+b200
T(°C)
Co, wt% Sn
20 60 80 1000
300
100
L
a b
L +a
40
a +b
TE
L: Co wt% Sn LL
Figure adapted from Callister, Materials science and engineering, 7 th Ed.
Intermetallic Compounds
Mg2Pb
Note: intermetallic compound forms a line - not an area - because stoichiometry (i.e. composition) is exact.
Figure adapted from Callister, Materials science and engineering, 7 th Ed.
a phase:
Mostly Mg
b phase:
Mostly Lead
Eutectoid & Peritectic
Cu-Zn Phase diagram
Eutectoid transition +
Peritectic transition + L
Figure adapted from Callister, Materials science and engineering, 7 th Ed.
mixed liquid and solid to single solid transition
Solid to solid ‘eutectic’ type transition
Iron-Carbon (Fe-C) Phase Diagram• 2 important points
-Eutectoid (B): g a + Fe3C
-Eutectic (A): L g + Fe3C
Fe3C
(cem
entit
e)
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
g(austenite)
g+L
g +Fe3C
a +Fe3C
a+g
L+Fe3C
d
(Fe) Co, wt% C
1148°C
T(°C)
a 727°C = T eutectoid
ASR
4.30Result: Pearlite = alternating layers of
a and Fe3C phases
120 mm
g ggg
R S
0.76
Ceu
tect
oid
B
Fe3C (cementite-hard)a (ferrite-soft)
Figure adapted from Callister, Materials science and engineering, 7 th Ed.
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
CharacterisationOver 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 meta-stable 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
• Electromagnetic Radiation
• Neutrons
• Electrons
Probes
Treat all probes as if they were waves:
;
hp k p mv
Wave-number, k:2
k k
Momentum, p:
Photons ‘Massive’ objects
Xavier the X-ray
hcE
Ex(keV)=1.2398/l(nm)
Speed of Light
Planck’s constant Wavelength
Elastic scattering as Ex>>kBT
Norbert the Neutron
hmv
222
12 2n
n
hE mvm
En(meV)=0.8178/l2(nm)
De Broglie equation:
mass velocity
Kinetic Energy:
Strong inelastic scattering as En~kBT
Eric the Electron
• Eric’s rest mass: 9.11 × 10−31 kg.• electric charge: −1.602 × 10−19 C• No substructure – point particle
hmvDe Broglie equation:
mass velocity
Ee depends on accelerating voltage :– Range of Energies from 0 to MeV
ProbesResolution better than the interatomic spacings
Absorption low – we want a ‘bulk’ probe
• Electrons - Eric
• quite surface sensitive
• Electromagnetic Radiation - Xavier
• Optical – spectroscopy
• X-rays :
• VUV and soft (spectroscopic and surfaces)
• Hard (bulk like)
• Neutrons - Norbert
Interactions
1. Absorption
2. Refraction/Reflection
3. Scattering Diffraction
EnglebertXavier
Norbert
Crystals are 2D with planes separated by dhkl. There will only be constructive interference when == - i.e. the reflection
condition.
a
Basic Scattering Theory
The number of scattered particles per
second is defined using the standard
expression
I Id
ds 0
Unit solid angle Differentialcross-section
Defined using Fermi’s Golden Rule
INTERACTION POTENTId
dFi Ana L i ll Init a
Spherical Scattered Wavefield
ScatteringPotential
Incident Wavefield
Different for X-rays, Neutrons and Electrons
2
exp k r r r r
d
dd
i V
BORN approximation:• Assumes initial wave is also spherical• Scattering potential gives weak interactions
0
2rexp kp rk r rex V i
ddi
d
2
exp q rr r id
dd
V
Scattered intensity is proportional to the Fourier Transform of the scattering potential
q k k0
2
exp k r r r rd
i V dd
Scattering from CrystalAs 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
( ) exp q r V
f r i dV
(q) q exp q T rj jT j
A fi
Atomic Structure Factor
Real Lattice Vector: T=ha+kb+lc
The Structure FactorDescribes the Intensity of the diffracted beams in reciprocal space
exp q r exp u v w 2jj j
i i h k l
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 hkl
Weight phase
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:
1
1
32
2 3
1T 1
2 31 1
q exp q T exp q a
exp q b exp q c
N
n
NN
n n
L i n i
n i n i
1 2 3N a N b N c
The Form FactorMeasures the translational symmetry of the lattice
The Form Factor has low intensity unless q is a
reciprocal lattice vector associated with a reciprocal
lattice point
1,2,3 1,2,3 1,2,3
sin s sin sq exp s
sin s
i
i
Ni i i i
i ijini i i
N NL i n
s
0
0.5x105
1.0x105
1.5x105
2.0x105
2.5x105
-0.02 -0.01 0 0.01 0.02
Deviation parameter, s1 (radians)
[L(s
1)]
2
N=2,500; FWHM-1.3”
N=500
q d s Deviation from reciprocal lattice point located at d*
Redefine q:
The Form Factor
0
20
40
60
80
100
-0.6 -0.3 0 0.3 0.6
Deviation parameter, s1 (radians)
[L(s
1)]
2
0
0.5x105
1.0x105
1.5x105
2.0x105
2.5x105
-0.02 -0.01 0 0.01 0.02
Deviation parameter, s1 (radians)
[L(s
1)]
2
The square of the Form Factor in one dimension
N=10 N=500
1,2,3
sin sq i i
ji
NL
s
Scattering in Reciprocal Space
T
q q exp q r exp q Tj jj
A f i i 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 DiffractionIt 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 MatchPowder Diffraction often used to identify phases
Cheap, rapid, non-destructive and only small quantity of sample
Inte
nsi
ty
2 A ngle
JCPDS Powder Diffraction File lists materials (>50,000) in order of their d-
spacings and 6 strongest reflectionsOK for mixtures of up-to 4
components and 1% accuracy
Monochromatic x-rays
Diffractometer
High Dynamic range detector
Single Crystal Diffraction
2dhkl hklsin Monochromatic radiation so 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
StrainPeak positions defined by the lattice parameters:
1
1 1, ,
q exp qN
ini a b c
L i n
Strain is an extension or compression of the lattice,
hkl hkld d
Results in a systematic shift of all the peaks
Ho Thin FilmsXRD measured as a function of temperature
10-4
10-2
100
102
104
20 40 60 80 100
T=294KT=244KT=194KT=94KT=144KT=42KT=10KT=300K
Scattering Angle ()
Inte
nsity
(ar
b. u
nits
)
Ho Thin FilmsSubstrate and Ho film follow have different behaviour
1
10
100
1000
30 32 34 36
T=294KT=244KT=194KT=144KT=94KT=42KT=10K
Scattering Angle ()
Inte
nsi
ty (
arb
. u
nits
)
Whole film refinement
Peak BroadeningDiffraction 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
, ,
sin sq i i
ii a b c
NL
s
Number of planes sampled is finite
Recall form factor: Scherrer Equation
2
cosSizeBD
NixMn3-xO4+ (400 Peak)As Grown at 200ºC AFM images (1200 x 1200 nm)
400
0
0.2
0.4
0.6
0.8
1.0
-1.0 -0.5 0 0.5 1.0
900C850C
800C
750C700C
650C
2
Inte
nsi
ty
D
450nm thick films
Annealed at 800ºC
Peak BroadeningDiffraction 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 ±Ddhkl.
Diffraction over a range, ,Dq of angles
Differentiate Bragg’s Law: 2 2 tanStrain B
Width in radians
Strain Bragg angle
dd
Peak BroadeningDiffraction peaks can also be broadened in qz by:
1. Grain Size 2. Micro-Strains OR Both
Total Broadening in 2q is sum of Strain and Size:
2 2 tancosTotal B
BD
2 cos 2 sinhkl hkl hklB B D
Rearrange
Williamson-Hall plot
y mx c
Powder Diffraction
0
100
200
300
400
30 40 50 60
Detector Angle (°)
Inte
nsity
(a
rb.
units
)
Powder of Nickel ManganiteCUBIC Structure
0
0.05
0.10
0.15
0.20
0.25
0 10 20 30
333422
400
222311
220
y=(1.5412/(4*a2))xa=8.348 ± 0.0036
(h2+k2+l2)
sin2
(B)
0.005
0.006
0.007
0.008
0.05 0.10 0.15 0.20 0.25
y=((1.541/d))+(2s)xGrain Size=299 ± 19.5a/a = 0.005 ± 0.001
sin(B)
Wid
th *
cos
(B) Grain size = 30±2nm
Strain Dispersion = 0.005±0.001
Cubic-Tetragonal Distortions
CUBICTETRAGONAL
a c a c
High Temperature Powder XRD
25 30 35 40 45 50 55 60 65 702Theta (°)
0
10000
20000
30000
Inte
ns
ity
(c
ou
nts
)
30.8 30.9 31.0 31.1 31.2 31.3 31.4 31.5 31.6 31.72Theta (°)
10000
20000
30000
Inte
ns
ity
(c
ou
nts
)
0.4BiSCO3 - 0.6PbTiO3 (K. Datta)
Tetragonal → Cubic phase transition
Courtesy, D. Walker and K. Datta University of Warwick
CsCoPO4
Dr. Mark T. Weller, Department of Chemistry, University of Southampton, www.rsc.org/ej/dt/2000/b003800h/
Variable temperature powder X-ray diffraction data show a marked change in the pattern at 170 °C.
Sn in a Silica Matrix
1. What form of tin
2. Particle size
3. Strain
4. Melting Temperature
Eutectic’s
wt% Ni20
120 0
130 0
30 40 50110 0
L (liquid)
a (solid)
L + a
L + a
T(°C)
A
35C o
L: 35wt%Ni
Cu-Nisystem
• Consider Cu/Ni with 35 wt.% Ni
Following Structural Changes
4332
a: 43 wt% Ni
L: 32 wt% Ni
L: 24 wt% Ni
a: 36 wt% Ni
Ba: 46 wt% NiL: 35 wt% Ni
C
D
E
24 36
Figure adapted from Callister, Materials science and engineering, 7 th Ed. USE LEVER RULE
A. Liquid
B. Mixed Phase
C.
D.
E. Solid
Cored Samples
α
L
α + L
Issues:
Lattice Parameter
Particle Size
Strain Dispersion
2 cos 2 sinhkl hkl hklB B D
NiCrStructural Changes?
Fcc: hkl are either all odd or all
even.
Bcc: sum of hkl must be even.