材料科学基础
Fundamental of Materials
Prof,Tian Min Bo
Tel,62795426, 62772851
E-mail,tmb@mail.tsinghua.edu.cn
Department of Material Science and Engineering
Tsinghua University,Beijing 100084
Lesson four
?Ⅰ.Crystal structure is the real
arrangement of atom in crystals
Crystal structure ＝ Space lattice + Basis
or structure unit
§ 2.3 Crystal Structure and Complex
Lattice
+ =
Fe, Al = 1, 1Fe
Al
The difference between space lattice and crystal structure
2× 3 atoms / cell
1,BCC
? Example,α-Fe,V,Nb,
Ta,Cr,Mo,W,
alkali metals
? n = 2 atoms/cell
? CN=8
? The number of nearest neighbours around
each atom is called —— Coordination Number.
?Ⅱ.Typical crystal structures of metals
? Packing fraction
＝
To determineξ,The atom is looked as a hard
sphere,and the nearest neighbours touch each
other.
∴ For BCC,68.0
)
4
3
(
3
4
2
3
3
?
?
?
a
a?
?
Volume of atoms / cell
Volume of unit cell?
2,FCC
? Example,γ-Fe,Al,Ni,Pb,Cu,
Ag,Au,stainless steal
? n= 8× 1/8+6× 1/2=4 atoms/cell
? CN=12
? 74.0
)
4
2
(
3
4
4
3
3
?
?
?
a
?
?
3,HCP
74.012CN
633.1
3
8
2
3
3
2
)
2
(
2
22
??
???
?
?
?
?
?
?
???
?
a
c
a
c
a
????
? ? ?
????
? ? ?
? ? ?
? Example,Be,Mg,Zn,Cd,Zr,Hf
Ti( low temperature)
? n＝
? CN＝ 12
? ξ＝ 0.74
632211261 ?????
Structure a0 vs,r Atoms per cell Coordination Number Packing factor Examples
SC 1 6 0.52 Polonium (Po),α-Mn
BCC 2 8 0.68
Fe,Ti,W,Mo,
Nb,Ta,K,Na,
V,Zr,Cr
FCC 4 12 0.74 Fe,Cu,Au,Pt,Ag,Pb,Ni
HCP 2 12 0.74 Ti,Mg,Zn,Be,Co,Zr,Cd
ra 20 ?
ra 340 ?
ra 240 ?
00
0
633.1
2
ac
ra
?
?
4,Summary
§ 2.4 Interstices in typical
crystals of metals
?Ⅰ.Two types of Interstitials in
typical crystals
Octahedral interstitial
Tetrahedral interstitial
Definition:
In any of the crystal structures,there are
small holes between the usual atoms into
which smaller atoms may be placed,These
locations are called interstitial sites.
1,Octahedral interstitial
BCC
a23 2
a
2a
FCC
2
a
2a
HCP
2
a
2,Tetrahedral interstitial
BCC
a23
a45
a
FCC
2
a
a43
HCP
?Ⅱ.Determination of the sizes of
interstitials
Definition:
By size of an interstitial we mean diameter of
the maximum hard sphere which can be
accommodated in the interstitial without distorting
the lattice.
di
da
diameter of interstitial atom
diameter of atom in lattice point＝
1,Octahedral interstitial
condition for touching
add ai ??
1??
aa
i
d
a
d
d
For BCC
For FCC
15.01
2
3
2
3
???
?
a
d
d
ad
a
i
a
41.01
2
2
2
2
???
?
a
d
d
ad
a
i
a
2,Tetrahedral interstitial
222 )
2()2()2(
HLdd ia ????
22 HLdd ai ??? 1
22
????
aa
i
d
HL
d
d
H
L
2
ia dd ?
A
D
C
B interstitial
host atom
For BCC
ad a 23?
aL?
2
aH ?
29.0??
a
i
d
d
22.0?
a
i
d
d
aL 22?
ad a 22?
2
aH ?
For FCC
3,Summary
n CN ξ interstices di/da
oct,tete,oct,tete.
BCC 2 8 0.68 6 6/2=3 12 12/2=6 0.15 0.29
FCC 4 12 0.74 4 4/4=1 8 8/4=2 0.41 0.22
HCP 6 12 0.74 6 6/6=1 12 12/6=2 0.41 0.22
Examples and Discussions
1,Both FCC and BCC are close-packed
structures while BCC is more open?
2,The interstitial atoms most likely occupy the
oct,interstitial position in FCC and HCP,while
in BCC two types of interstitial can be occupied
equally.
3,The solid solubility in BCC is much lower than
in FCC.
4,Diffusion of interstitial atoms in BCC diffusion is
much faster than in FCC or HCP at same
temperature.
5,Determine the relationship between the atomic
radius and the lattice parameter in SC,BCC,
and FCC structures when one atom is located
at each lattice point.
6,Determine the density of BCC iron,which has
a lattice parameter of 0.2866nm.
Solution:
For a BCC cell,Atoms/cell = 2
a0 = 0.2866nm = 2.866× 10－ 8cm
Atomic mass = 55.847g/mol
Volume of unit cell = a03 = 23.54× 10 － 24cm3/cell
Density
3
2324-
cm/g882.7
)10) ( 6.0 210( 23,54
)847.55)(2(
)n u m be rsA v og a dr o')(c e l lu n i tofv ol u m e(
)i r onofm a s sa t om i c)(c e l l/a t om sofn u m be r(
?
??
?
??
Exercise
1,Determine the coordinates of centers of both
the octahedral and the tetrahedral interstitials
in HCP referred to a,b and c.
a
b
c
120o
Thank you !
4