01 crystal structure
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Atomic Bonding and
Crystal Structure
Metallic Bond
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Ra 20 =
Bonding Energy = E0 - Emin.# Bonding Energy /���� ��8� ��:�/<����;���������� ������ ��8�
Crystal Structure
Principle Metallic Crystal Structures
Body Centered Cubic
BCCFace Centered Cubic
FCCHexagonal Closed Pack
HCP
Body Centered Cubic
a
a
a
aaa
2
2 222
=
=+
a
a2
( )a
aaaa
3
22 2222
=
+=+
Atomic Packing Factor (APF)
APF = total sphere volume
total unit cell volume
=
Ra 43 =
( ) ( )
3
33
373.8
3
48
8
1
3
41
R
RRVatom
=
+
= ππ
3aVunit =
68.032.12
373.83
3
===R
R
V
VAPF
unit
atom
3
3
3
3
33
32.12
33
64
3
4
3
4
R
R
Ra
Ra
=
=
=
=
Exercises
Lithium at 20 °C is BCC and has a lattice constant of 0.35092 nm. Calculate a
value for the atomic radius of lithium in nanometers.
Face Centered Cubic
( ) ( )
3
33
3
16
3
4
8
18
3
4
2
16
R
RRVatom
π
ππ
=
+
=
3aVunit =
74.0216
3
16
3
3
===R
R
V
VAPF
unit
atom
π
Ra 42 =
3
3
3
3
3
33
216
2
2
2
32
22
64
2
4
2
4
R
R
R
Ra
Ra
=
=
=
=
=
Exercises
Hexagonal Close-Packed
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caVcell 60sin3 2=
Exercises
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Atomic Position in Cubic Unit Cell
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3. ��ก�9�,�0��$"%��0(��ก��� 1 .�(4="�0.0&>14?@'��0
Atomic Position in Cubic Unit Cell
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Atomic Position in Cubic Unit Cell
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Direction in Cubic Unit Cell
Direction in Cubic Unit Cell
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Direction in Cubic Unit Cell
Direction in Cubic Unit Cell
]121[
],1,[2
],1,[
]0,01,0[)0,0,0(),1,(
21
21
21
21
21
21
21
21
=
×=
=
−−−=−
Example
Direction in Cubic Unit Cell
]312[
],1,[3
],1,[)0,0,0(),1,(
31
32
31
32
31
32
=
−×=
−=−−
Example
Direction in Cubic Unit Cell
]123)[
]101)[
]112)[
]110[]100)[
d
c
b
anda
Exercises
Draw the following direction vectors in cubic unit cells
[???] [???]
Crystallographic Plane in Cubic Unit Cells
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comma %�#0
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Crystallographic Plane in Cubic Unit Cells
Crystallographic Plane in Cubic Unit Cells
)100()0,0,1(),,1( →→∞∞
)110()0,1,1(),1,1( →→∞
)111()1,1,1()1,1,1( →→
)632()1,,3()1,,(23
32
31 →→
)110)(
)221)(
)011)(
)101)(
d
c
b
a
Exercises
Draw the following crystallographic planes in cubic unit cells
Crystallographic Plane in Cubic Unit Cells
Home Work
222 lkh
adhkl
++=
Interplanar spacing
nm
lkh
adhkl
128.0
)0()2()2(
361.0
222
222
=
++=
++=
Interplanar spacing
ExampleCopper has an FCC crystal structure and a unit cell with a lattice constant of 0.361 nm. What is its interplanar spacing d220?
Crystallographic Planes and Directions in Hexagonal Unit Cells
$" 4 ,ก0 %8� a1, a2, a3 ,� c./(���4�= 4 ���.0ก�&& -; direction ,� plane
Crystallographic Plane in Hexagonal Unit Cell
)1110(
)1,1,0,1(
)1,1,,1(
1
1
1
3
2
1
−
−∞
=
−=
∞=
=
c
a
a
a
)0022(
)0,0,2,2(
),,,(21
21
3
21
2
21
1
−
∞∞−
∞=
∞=
−=
=
c
a
a
a
∞=
−=
=
−=
∞−−
−−
4
3
21
2
1
21
1
1
),1,,1(
)0,1,2,1(
)0121(
a
a
a
a
Crystallographic Direction in Hexagonal Unit Cell
Home Work
Determine Miller-Bravais indices of
cratal plane in figure on the left.
Family of Direction and Plane
]100[],010[],001[],001[],010[],100[
Some planes and directions are “Crystallographic equivalent”
><100
)001(),010(),100( }100{
Directions
Planes
Comparison of FCC and HCP
Volume and Planar Density Calculations
Volume Density
3/
/
/
cmg
cellunitvolume
cellunitmassv
=
=ρ
Volume and Planar Density Calculations
Cu has FCC crystal structure and an atomic radius of 0.1278 nm. Calculate a theoretical density of copper in g/cm3. The atomic mass of copper is 63.54 g/mol
gmolatoms
molgatomm
v
m
nmR
a
Ra
v
22
231022.4
/1002.6
)/54.63)(4(
361.02
)1278.0)(4(
2
4
42
−×=×
=
=
===
=
ρ
3323
22
323
32939
3
98.81070.4
1022.4
1070.4
1070.4)10361.0(
cm
g
cm
g
v
m
cm
mm
av
v =××
==
×=
×=×=
=
−
−
−
−−
ρ
Volume and Planar Density Calculations
Planar Density
2/
.
mmatoms
areaselected
atomofcenterofnoequivalentp
=
=ρ
Volume and Planar Density Calculations
Calculate the planar atomic density on the (110) plane of α iron in atoms/mm2. The lattice constant of α iron is 0.287 nm.
( )( )
2
12
22
2
41
102.17
2.17)287.0(2
2
2
2
)(2
41
mm
atoms
nm
atoms
aaap
×=
==
=×+
=ρ
Crystal Structure Analysis
Off-Phase
In-Phase
θλ
θλλ
sin2
sin2
hkl
hkl
d
dn
PNMPn
=
=
+=
X-Ray Diffractometer
Example
A sample of BCC iron was placed in an x-ray diffractrometer using incoming x-rays with a wavelength 0.1541 nm. Diffraction from the {110} planes was
obtained at 2θ = 44.704°. Calculate a value for the lattice constant a of BCC iron. (Assume first order diffraction with n = 1)
nmnm
lkhda
nmnm
d
d
hkl
hkl
hkl
287.0)414.1)(2026.0(
0112026.0
2026.0)3803.0(2
1541.0
sin2
sin2
35.22
704.442
222
222
==
++=
++=
==
=
=
=
=
θλ
θλθ
θo
o
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