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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./( comma %�#0 1233(����4�5-

Atomic Position in Cubic Unit Cell

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Atomic Position in Cubic Unit Cell

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Atomic Position in Cubic Unit Cell

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

6(��73�-./(="33(�0-0

Direction in Cubic Unit Cell

Direction in Cubic Unit Cell

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

�7A"ก�&

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

comma %�#0

�7A"ก�&

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

=

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

=

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

×=

==

=×+

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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