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Vt liu hcVT 1/2011
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Cu trc chng trnh
1: i cng v tinh th
2: Mt s loi gin pha
3: Tnh cht ca vt liu4: Mt s loi vt liu
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i cng v tinh th
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Mt s khi nim
Mng tinh th l mt tp hp v hn ccnt (nguyn t, phn t hoc ion) sp
xptheo mt trt t nht nh.
Mng c th xem nh c to thnhbng cch spxp lin tip theo cc cnha, b, c
nhng hnh khi ging nhau. Cckhi ny gi l c s v cch spxpcc nt trong c s
l i din chungcho ton mng.
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Cc kiu mng tinh th
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Phng tinh th [u v w]
Cch tm:
T gc trc ta v ng thng song songvi phng cn xc nh
Tm ta nt mng gn gc trc nht trnng thng . Nu ta nt mng l (p, q,r)
th k hiu phng l [pqr]
Nu ta l phn s th qui ng mu s. T
s l u, v, w th k hiu l [uvw] Nu ta c du m th trn u ch s
tng ng ghi du
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Vd
EG c k hiu [0 1]
Cc phng trongcng h c u v w hon
v ch cho nhau [100] [010] [001] thuc
h
O
A
BC
D
E F
G
x
y
z
E'
1
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Mt tinh th (hkl)
L mt phng qua cc tm nguyn t hay ion
Cch tm:
- Tm giao im mt vi 3 trc x, y, z. Ta 3giao im l (p,0,0) (0,q,0)
(0,0,r).
- Ly 1/p, 1/q, 1/r ri qui ng mu s. T s l h,
k, l th k hiu mt l (hkl)- Nu ta c du tr th t du trn uch s tng
ng
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VD
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Miller indices
(hkl) l k hiu ca mt(hkl) v cc mt phngkhc song song vi mt
phng nyH mt: cng cch spxp nt mng:
VD: (100) (010) (001)thuc h {100}
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Gc gia hai phng cho trc
C 2 phng L1 [u1v1w1], L2 [u2v2w2].Tnh gc gia hai phng
i vi h lp phng
vi:
)wwvvuu(NN
1cos 212121'
2
'
1
2
i
2
i
2
i
'
i wvuN
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Gc gia phng v mt tinh th
Tm gc gia phng L [uvw] v mt P (hkl) ivi h lp phng:
Vi
)lwkvhu('N'M
1cos
222
222
wvu'N
lkh'M
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Lp phng tm khi (BCC)
r
r
r
r
a
a
2
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Lp phng tm mt (FCC)
r
r
r
r
a
a
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Bi tp
Xc nh mi quan h gia a (hng smng) v R (bn knh nguyn t),
n (s nguyn t trong 1 ),
s nguyn t gn nht i vi 1 nguyn t,
Mv%(mt th tch)
Ms%
(mt sp xp mt phng) ca bccv fcc?
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Tng kt
Cu trc BCC FCC HCP
Tng s nt 2 4 6
Mi lin hR v a ; ;
R = a/2
Mv% 68% 74% 91%
4
3aR
4
2aR
3
22ac 3
4Ra
2
4Ra
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Mt s khi nim
Mt thng = s nguyn t/chiu di(cm)
lp li = khong cch gia cc nguynt trn phng
Mt phng = s nguyn t/n v dintch(cm)
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Bi tp
Cho rCu = 1,278Ao. Tnh:
Mt thng ca nguyn t theo phng[110] ca Cu (Fcc).
lp li trn phng [211] ca Cu (Fcc).
Mt phng trn (100), (111), caCu (Fcc).
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Khong cch gia cc mt tinh th
l khong cch gn nht gia cc mt tinh thcng ch s Miller (hkl) song
song:
Trong h lp phng a = b = c
2
2
2
2
2
2
2
190 c
l
b
k
a
h
dhkl
o
222 lkh
ad
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Bi tp
So snh d111 v d200 trong Pb (Fcc), chorPb = 4,95 A
0.
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Mt s bi tp
ng (Fcc) c bn knh nguyn t l 1,278Ao. Tnh khi lng ring ca Cu. Cho
MCu=63,5.
St thay i t Bcc sang Fcc 9100C. nhit ny, bn knh nguyn t ca
st
trong hai cu trc l 1,258 A
0
(Bcc) v 1,292A0(Fcc). Tnh % th tch thay i.
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Cu trc cc tinh th v c Dng n cht
VD: Kim cng Graphit
c s: FCC Hnh 6 phng
+ 4 nguyn t trong Cu trc lp
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Cu trc dng hp cht ion AB
NaCl: CsCl c s FCC ca Cl- c s BCC: Cs+ nh, nCS =1
Na+ chim cc l hng Cl- tm khi, nCl- =1
R + r = a/2
nCl- = 4, nNa+ = 42
3arR
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ZnS
c s: Fcc ca S2- v cthm 4 Zn2+ v tr ging 4
nguyn t pha trong ca kimcng.
nS2- = 4; n Zn2+ = 4
4
3arR
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CaF2
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Mt s dng khc
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Cc khuyt tt trong mng tinh th
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Cc k hiu v in tch
Nu M l nguyn t nm ng v tr ca M: MM X l nguyn t nm ng v tr X: XX
Nguyn t M (X) nm v tr xen k: Mi*(Xi)
Khuyt v tr M (X): VM , VX* M (X) nm v tr X (M): MX**(XM)
L (Y) l kim loi (phi kim) ha tr 2 nm v tr M: LM
*
(YX) Dng 1: *, dng 2:**, m 1: , m 2:
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S mt trt t trong vt liu
Mt trt t kiu Schottky l mt trt t kiul trng
Mt trt t kiu Frenken l mt trt t kiuxen k
MX=> VM + VM* Kiu Schottky
MX=> Mi*
+ X
i Kiu Frenken
MX => Mi*+ VM Frenken cation
MX => Xi + VX* Frenken anion
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Nguyn tc:
m bo v kiu mng, v kiu nt cationanion trong tinh th
Trung ha v in
Tun theo quy lut ca phn ng ha hc
VD: NaCl c ph gia CaCl2 tham gia vomng.
Nh vy: Ca2+thay th cho Na+:
CaCl2 => CaNa* +2ClCl + VNa
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Bi tp
Xc nh c ch hnh thnh cc nt trngsau:
Thm LiCl vo CaCl2
Thm CaCl2 vo LiCl
Thm Al2O3 vo SiO2
Thm Al2O3 vo NiO
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Bi tp
Vit c ch mt trt t ca cc lp chtsau:
Mt oxi trng trong MgO
Mt trt t kiu Schottky trong Li3N
Mt ion Ca2+ thay th cho Ba2+ trong BaCl2
Mt ion Na+
thay th cho Ba2+
trong SrCl2
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Cc bi tp
1. Gi s c X3+ thay th Mg2+ trong MgO.
a, Vit phng trnh cu trc cho s thayth
b, Nu t l X3+/Mg2+=0,25. Tnh t lcation/anion
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Gii
Gii: a, 2XMg* + VMg 0.
b, 100Mg2+ c 25X3+ +12,5VMg => tng scation l 125
100Mg2+ => 100O2-
25X3+ => 37,5O2-
Tng anion = 137,5T s cation/anion = 125/137,5 = 0,91
