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Atomic Bonding and Crystal Structure

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Page 1: 01 Crystal Structure

Atomic Bonding and

Crystal Structure

Page 2: 01 Crystal Structure

Metallic Bond

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������������ ��������������� !��!�ก����# ����$$

Electron ��-�ก�� ���������#������$�!��.�������ก����� ��# /�� � 0 1 � � �� 1 � $ $ � � Electron Charge Cloud

Page 3: 01 Crystal Structure

.#�� ���8����# ก�$�����# ����# �!��.������ �����-��ก�ก���ก!�01�� ����� ��8���9�� �ก�� :��ก�� �!��.����/����1���;�����ก��-�0ก��ก!�01�� ����� ��8���9�� �ก�� �� ����� ก�;#����ก�!���ก����

�� ��8 � . � � �# � � � �# �!��.������ ��.#.#���� /��/<���0���� ���</���;� ������8�� ����ก�# =Interatomic DistanceC

Ra 20 =

Page 4: 01 Crystal Structure

Bonding Energy = E0 - Emin.# Bonding Energy /���� ��8� ��:�/<����;���������� ������ ��8�

Page 5: 01 Crystal Structure

Crystal Structure

Page 6: 01 Crystal Structure
Page 7: 01 Crystal Structure
Page 8: 01 Crystal Structure

Principle Metallic Crystal Structures

Body Centered Cubic

BCCFace Centered Cubic

FCCHexagonal Closed Pack

HCP

Page 9: 01 Crystal Structure

Body Centered Cubic

Page 10: 01 Crystal Structure

a

a

a

aaa

2

2 222

=

=+

a

a2

( )a

aaaa

3

22 2222

=

+=+

Page 11: 01 Crystal Structure
Page 12: 01 Crystal Structure

Atomic Packing Factor (APF)

APF = total sphere volume

total unit cell volume

=

Page 13: 01 Crystal Structure

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

=

=

=

=

Page 14: 01 Crystal Structure

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.

Page 15: 01 Crystal Structure

Face Centered Cubic

Page 16: 01 Crystal Structure
Page 17: 01 Crystal Structure

( ) ( )

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

=

=

=

=

=

Page 18: 01 Crystal Structure

Exercises

Page 19: 01 Crystal Structure

Hexagonal Close-Packed

Page 20: 01 Crystal Structure

����������� !"#$"�%&�'&(��)�*ก,-- HCP

Page 21: 01 Crystal Structure

caVcell 60sin3 2=

Page 22: 01 Crystal Structure

Exercises

Page 23: 01 Crystal Structure

),,( zyx −

./( comma %�#0 1233(����4�5-

Atomic Position in Cubic Unit Cell

6(��73�-.�(�734%&8#���-3(�0�0(�

Page 24: 01 Crystal Structure

1. 09�:;3<7ก�3=���9�,�0��.0'�$$7�7 x, y ,� z $�4="�0.0��4�5-��$�9�3�- x,y,z

2. �3�ก9��03.�(,��� 3(�0=�� Cubic $"%��$���4!��ก�- 1

3. ��ก�9�,�0��$"%��0(��ก��� 1 .�(4="�0.0&>14?@'��0

Atomic Position in Cubic Unit Cell

�7A"ก�&

Page 25: 01 Crystal Structure

Atomic Position in Cubic Unit Cell

Page 26: 01 Crystal Structure

][ zyx

B$�./( comma %�#0 1233(�� Square Bracket

6(��73�-./(="33(�0-0

Direction in Cubic Unit Cell

Page 27: 01 Crystal Structure

Direction in Cubic Unit Cell

1. 09�:;3<7ก�3 Destination ��C��-3(��:;3<7ก�3=�� Origin

2. 4$8#�B3()�����,�(�!9����4�=.�(41D0:9�0�04�5$ (ก9�:�34?@'��03(��ก�&09����'��0$�%>F)

3. 09�%��!"#B3($�4="�0.0 Square Bracket 4&"����$�9�3�- xyz �3�B$�$" comma

4. ��ก���4�=�73�-.�(4="�04%&8#���$���-3(�0-0=�����4�=

�7A"ก�&

Page 28: 01 Crystal Structure

Direction in Cubic Unit Cell

Page 29: 01 Crystal Structure

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

Page 30: 01 Crystal Structure

Direction in Cubic Unit Cell

]312[

],1,[3

],1,[)0,0,0(),1,(

31

32

31

32

31

32

=

−×=

−=−−

Example

Page 31: 01 Crystal Structure

Direction in Cubic Unit Cell

]123)[

]101)[

]112)[

]110[]100)[

d

c

b

anda

Exercises

Draw the following direction vectors in cubic unit cells

Page 32: 01 Crystal Structure
Page 33: 01 Crystal Structure

[???] [???]

Page 34: 01 Crystal Structure

Crystallographic Plane in Cubic Unit Cells

Page 35: 01 Crystal Structure

1. ��:;3��3=��& 0�-!"#,ก0 xyz

2. ก��-4?@'��0=��:;3��3!�C�'�$3. ก9�:�34?@'��0,� !�04�=.�(�#9�!"#';34. ��ก���4�=�73�-.�(4="�04%&8#���$���-3(�0-0

=�����4�=5. 4="�04&"����$�9�3�-H��.0��4�5- �3�B$�$"

comma %�#0

�7A"ก�&

Crystallographic Plane in Cubic Unit Cells

Page 36: 01 Crystal Structure

Crystallographic Plane in Cubic Unit Cells

)100()0,0,1(),,1( →→∞∞

)110()0,1,1(),1,1( →→∞

)111()1,1,1()1,1,1( →→

Page 37: 01 Crystal Structure

)632()1,,3()1,,(23

32

31 →→

Page 38: 01 Crystal Structure

)110)(

)221)(

)011)(

)101)(

d

c

b

a

Exercises

Draw the following crystallographic planes in cubic unit cells

Crystallographic Plane in Cubic Unit Cells

Page 39: 01 Crystal Structure
Page 40: 01 Crystal Structure

Home Work

Page 41: 01 Crystal Structure

222 lkh

adhkl

++=

Interplanar spacing

Page 42: 01 Crystal Structure

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?

Page 43: 01 Crystal Structure

Crystallographic Planes and Directions in Hexagonal Unit Cells

$" 4 ,ก0 %8� a1, a2, a3 ,� c./(���4�= 4 ���.0ก�&& -; direction ,� plane

Page 44: 01 Crystal Structure

Crystallographic Plane in Hexagonal Unit Cell

Page 45: 01 Crystal Structure

)1110(

)1,1,0,1(

)1,1,,1(

1

1

1

3

2

1

−∞

=

−=

∞=

=

c

a

a

a

Page 46: 01 Crystal Structure

)0022(

)0,0,2,2(

),,,(21

21

3

21

2

21

1

∞∞−

∞=

∞=

−=

=

c

a

a

a

Page 47: 01 Crystal Structure

∞=

−=

=

−=

∞−−

−−

4

3

21

2

1

21

1

1

),1,,1(

)0,1,2,1(

)0121(

a

a

a

a

Page 48: 01 Crystal Structure

Crystallographic Direction in Hexagonal Unit Cell

Page 49: 01 Crystal Structure

Home Work

Determine Miller-Bravais indices of

cratal plane in figure on the left.

Page 50: 01 Crystal Structure

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

Page 51: 01 Crystal Structure

Comparison of FCC and HCP

Page 52: 01 Crystal Structure
Page 53: 01 Crystal Structure

Volume and Planar Density Calculations

Volume Density

3/

/

/

cmg

cellunitvolume

cellunitmassv

=

Page 54: 01 Crystal Structure

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

Page 55: 01 Crystal Structure

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

Page 56: 01 Crystal Structure

Volume and Planar Density Calculations

Planar Density

2/

.

mmatoms

areaselected

atomofcenterofnoequivalentp

=

Page 57: 01 Crystal Structure

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

×=

==

=×+

Page 58: 01 Crystal Structure

Crystal Structure Analysis

Page 59: 01 Crystal Structure
Page 60: 01 Crystal Structure

Off-Phase

In-Phase

Page 61: 01 Crystal Structure

θλ

θλλ

sin2

sin2

hkl

hkl

d

dn

PNMPn

=

=

+=

Page 62: 01 Crystal Structure

X-Ray Diffractometer

Page 63: 01 Crystal Structure
Page 64: 01 Crystal Structure
Page 65: 01 Crystal Structure

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