Transcript Energy in Hydrogen Atom

Body Centered Cubic (BCC)
Crystal Structure
•
•
Each atom has 8 nearest neighbors.
Therefore, coordination number is 8.
•
Examples : Chromium (a=0.289 nm)
 Iron (a=0.287 nm)
 Sodium (a=0.429 nm)
Figure 3.4 a&b
BCC Crystal Structure (Cont..)
Each unit cell has
(8x1/8) + 1 = 2 atoms
•
Atoms contact each
other at cube diagonal
Therefore, lattice
constant a =
4R
3
Figure 3.5
Atomic Packing Factor of BCC
Structure
Atomic Packing Factor =
Vatoms =
 4 R 3 

2.


 3 
V unit cell = a3 =
Therefore APF =
Volume of atoms in unit cell
Volume of unit cell
= 8.373R3
 4R 




 3
3
= 12.32 R3
8.723 R3
12.32 R3 = 0.68
Face Centered Cubic (FCC)
Crystal Structure
•
Coordination number for FCC structure is 12
•
•
Atomic Packing Factor is 0.74
Examples : Aluminum (a = 0.405)
 Gold (a = 0.408)
Figure 3.6 a&b
FCC Crystal Structure (Cont..)
Each unit cell has
(8 x 1/8)+ (6 x ½) = 4 atoms
Therefore, lattice
constant a
=
4R
2
Figure 3.7
Structure of Diamond
• Four sp3 orbitals -- symmetrical - regular tetrahedron.
• This structure gives high hardness, high bonding strength
(711 KJ/mol) and high melting temperature (3550oC).
Carbon Atom
Tetrahedral arrangement in diamond
Hydrogen Bonds
•
Hydrogen bonds are Dipole-Dipole interaction
• Between polar bonds containing hydrogen atom.
• In water, dipole  asymmetrical arrangement of
hydrogen atoms.
• Attraction - positive oxygen pole & negative
hydrogen pole.
H
105 0
O
Figure 2.28
H
Hydrogen
Bond
Permanent Dipoles
•
Dipoles that do not fluctuate with time are called Permanent
dipoles.
 Examples:-
BP = -128 oC
CH4
BP = -14 oC
CH3Cl
Symmetrical
Arrangement
Of 4 C-H bonds
Asymmetrical
Tetrahedral
arrangement
No Dipole
moment
Creates
Dipole
Procedure to Find Direction Indices
Produce the direction vector till it
emerges from surface of cubic cell
z
(1,1/2,1)
Determine the coordinates of point
of emergence and origin
(1,1/2,1) - (0,0,0)
= (1,1/2,1)
y
(0,0,0)
Subtract coordinates of point of
Emergence by that of origin
Are all
integers?
NO
YES
Are any of the direction
vectors negative?
YES
Represent the indices in a square
bracket without comas with a
over negative index (Eg: [121])
x
2 x (1,1/2,1)
= (2,1,2)
The direction indices are [212]
Convert them to
smallest possible
integer by multiplying
by an integer.
NO
Represent the indices in a square
bracket without comas (Eg: [212] )
Miller Indices - Procedure
Choose a plane that does not pass
through origin
Determine the x,y and z intercepts
of the plane
Find the reciprocals of the intercepts
Fractions?
Place a ‘bar’ over the
Negative indices
Clear fractions by
multiplying by an integer
to determine smallest set
of whole numbers
Enclose in parenthesis (hkl) where h,k,l
are miller indices of cubic crystal plane
for x, y and z axes. Eg: (111)
A simple may to solve difficult Miller Indices problem
z
(2/3,1,1)
(1,2/3,1/3)
y
(1,1,1/3)
x
Conventional Method
(complicated)
z
(2/3,1,1)
(1,2/3,1)
y
(1,1,1/3)
x
A Simple Method
Use this equation
h x1 + k y1 +l z1 = h x2 + k y2 +l z2 = h x3 + k y3 +l z3
h + 2/3 k + l = 2/3 h + k + l = h + k +1/3 l
3 h + 2 k +3 l = 2 h + 3 k + 3 l = 3 h + 3 k + l
3 h + 2 k +3 l = 2 h + 3 k + 3 l
h =k
h = 2l

h = k = 2l
If l = 1  h=2, k=2, l=1  (221)
X-Ray Diffraction
For rays reflected from different planes to be in
phase, the extra distance traveled by a ray should be
a integral multiple of wave length λ .
nλ = MP + PN
(n = 1,2…)
n is order of diffraction
If dhkl is interplanar distance,
Then MP = PN = dhkl.Sinθ
Therefore,
Figure 3.28
λ = 2 dhkl.Sinθ
Crystal Structure of Unknown Metal
Unknown
metal
Crystallographic
Analysis
Sin2 A
 0.75
Sin  B
2
FCC
Crystal
Structure
Sin2 A
Sin2 B
 0.5
BCC
Crystal
Structure
Interstitial Solid Solution
•
•
Solute atoms fit in  voids (interstices) of solvent atoms.
Interstitial solid solution of carbon in γ iron (FCC).
Iron atoms r=0.129nm
Carbon atoms r=0.075nm
Figure 4.15a
Stress and Strain in Metals
6-10
•
Metals undergo deformation under uniaxial tensile
force.
•
Elastic deformation: Metal
returns to its original
dimension after tensile
force is removed.
•
Plastic deformation: The
metal is deformed to
such an extent such
that it cannot return
to its original dimension
Elastic
deformation
Plastic
deformation
Engineering Stress and Strain
F (Average uniaxial tensile force)
A0 (Original cross-sectional area)
Engineering stress σ =
Units of Stress are PSI or N/M2 (Pascals)
Δl
A0

0
Engineering strain = ε =
0
A
1 PSI = 6.89 x 103 Pa
Change in length
Original length

  0
0
Units of strain are in/in or m/m.
Figure 5.15
6-11



Poisons Ratio
 (lateral)
Poisons ratio =
.
y
 

 (longitudinal)
z
w  w0 
0
w0
w
 

   0 
w0
0
Usually poisons ratio ranges from
0.25 to 0.4.
Example: Stainless steel
Copper
6-12
0.28
0.33
Shear Stress and Shear Strain
S (Shear force)
A (Area of shear force application)
τ = Shear stress =
Figure 5.17
Shear strain γ =
Amount of shear displacement
Distance ‘h’ over which shear acts.
Elastic Modulus G = τ / γ
6-13
Yield Strength
•
Yield strength is strength at which
metal or alloy show significant
amount of plastic deformation.
•
0.2% offset yield strength is that
strength at which 0.2% plastic
deformation takes place.
•
Construction line, starting at 0.2%
strain and parallel to elastic region
is drawn to find 0.2% offset yield
strength.
6-18
Ultimate tensile strength
•
•
Ultimate tensile strength (UTS) is the maximum strength reached by
the engineering stress strain curve.
Necking starts after UTS is reached.
Al 2024-Tempered
Figure 5.24
•
More ductile the metal is, more
is the necking before failure.
•
Stress increases till failure. Drop
in stress strain curve is due to stress
calculation based on original area.
S
T
R
E
S
S
Mpa
Necking Point
Al 2024-Annealed
Strain
Stress strain curves of
Al 2024 With two different
heat treatments. Ductile
annealed sample necks more
6-19
True Stress – True Strain
•
True stress and true strain are based upon instantaneous
cross-sectional area and length.
F
• True Stress = σt =
• True Strain = εt =
Ai (instantaneous area)
i
li
A0
d
   ln l0  ln Ai
0
• True stress is always greater than engineering stress.
6-22
Gibbs Phase Rule
•
P+F = C+2
P = number of phases that coexist in a system
C = Number of components
F = Degrees of freedom
• For pure water, at triple point, 3 phases coexist.
• There is one component (water) in the system.
• Therefore 3 + F = 1 + 2
F = 0.
• Degrees of freedom indicate number of variables
that can be changed without changing number of
phases.
8-4
Cooling Curves
• Used to determine phase transition
temperature.
• Temperature and time data of cooling molten
metal is recorded and plotted.
Pure Metal
Solid
Nucleation
Iron
• Thermal arrest : heat lost = heat supplied by
solidifying metal
• Alloys solidify over a range of temperature (no
thermal arrest)
Phase Diagram from Cooling
Curves
•
Series of cooling curves at different metal
composition are first constructed.
• Points of change of slope of cooling curves (thermal
arrests) are noted and phase diagram is constructed.
• More the number of cooling curves, more accurate
is the phase diagram.
Figure 8.4
8-6
The Lever Rule
•
The Lever rule gives the weight % of phases in any
two phase regions.
Xs  Xl 1
wo  X s ws  X l wl
Wt fraction of solid phase
 w0  wl  LO
 
X s  
 ws  wl  LS
Wt fraction of liquid phase
Figure 8.5
8-7
 ws  wo  OS
 
X l  
 ws  wl  LS
Non Equilibrium Solidification of
Alloys
•
Fast rate of cooling (non-equilibrium) gives rise to
cored structure.
• Rapid cooling delays solidification.
• Homogenization: Cast ingots heated
to elevated temperature to eliminate
cored structure.
• Temperature of homogenization
must be lower than lowest melting
point of any of the alloy components.
Figure 8.7
Figure 8.8
•
Binary Eutectic Alloy System
In some binary alloy systems, components have limited solid
solubility.
Example : Pb-Sn alloy.
• Eutectic composition
freezes at lower
temperature than all
other compositions.
• This lowest
temperature is called
eutectic temperature.
Figure 8.11
Eutectic temperature
Liquid
Cooling
α solid solution + β solid solution
Binary Eutectic Alloy System
12%
15%
8-9
48%