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بلورشناسی مواد
حسن علم خواه: گردآوریاستادیار گروه مواد دانشگاه بوعلی سینا
95زمستان Dr. Elmkhah (Bu-Ali Sina Univ.)
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سرفصل مطالب
انواع پیوندهای اتمیساختار بلوری فلزاتاندازه گیری صفحات و جهت بلوریساختار بلوری سرامیک هانحوه اندازه گیری فاصله صفحات بلوری
Dr. Elmkhah (Bu-Ali Sina Univ.)
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© 2
00
3 B
rook
s/Cole P
ub
lishin
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hom
son
Learn
ing
™
© 2
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3 B
rook
s/Cole P
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ing
™
Metals
Ceramic
combination
with a metal
Associated
with polymers
Semiconductor
materials
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مقایسه استحکام مواد مختلف©
20
03
Bro
ok
s/Cole P
ub
lishin
g / T
hom
son
Learn
ing
™
Dr. Elmkhah (Bu-Ali Sina Univ.)
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انواع پیوندهای اتمی
Primary bonding: Metallic bond, Covalent bond, Ionic bond.
Secondary bonding: Van der Waals interactions (London forces, Debye interaction)
Intermetallic compound is a compound such as Al3V formed by two or more metallic atoms
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پیوند فلزی
Dr. Elmkhah (Bu-Ali Sina Univ.)
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پیوند کوواالنت©
20
03
Bro
ok
s/Cole P
ub
lishin
g / T
hom
son
Learn
ing
™
Dr. Elmkhah (Bu-Ali Sina Univ.)
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پیوند یونی
© 2003 Brooks/Cole Publishing / Thomson Learning™
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مقایسه انواع پیوندهای اتمی
Dr. Elmkhah (Bu-Ali Sina Univ.)
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مقدمه
Dr. Elmkhah (Bu-Ali Sina Univ.)
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انواع نظم اتمی
(c) 2003 Brooks/Cole Publishing / Thomson
Learning™ Dr. Elmkhah (Bu-Ali Sina Univ.)
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• Non dense, random packing
• Dense, ordered packing
Dense, ordered packed structures tend to have
lower energies & thus are more stable.
ارتباط انرژی و فشردگیEnergy
r
typical neighborbond length
typical neighborbond energy
Energy
r
typical neighborbond length
typical neighborbond energy
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تغییر آلوتروپی قلع
Dr. Elmkhah (Bu-Ali Sina Univ.)
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سلول واحد
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سیستم های بلوری
7 crystal systems of varying
symmetry are known
These systems are built by
changing the lattice
parameters:
a, b, and c are the edge
lengths
, , and are interaxial
angles
Fig. 3.4, Callister 7e.
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سیستم های بلوری
Crystal structures are divided
into groups according to unit
cell geometry (symmetry).
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ساختار بلوری برخی فلزات
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مفاهیم پایه(Lattice Parameter)پارامتر شبکه یا ثابت •
(در سیستم معکبی)تعداد اتم ها در هر سلول واحد •
(.Coordination No)عدد همسایگی•
( Atomic Packing Factor)فاکتور تراکم اتمی •
(Theoretical Density)چگالی تئوریکی •
( Linear & Planar Density)چکالی خطی و سطحی •
(Crystallographic Planes)صفحات و جهات بلوری •
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• Rare due to low packing density (only Po – Polonium --
has this structure)
• Coordination No. = 6
سادهمکعبیساختارSimple Cubic Structure (SC)
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سایر ساختارهای بلوری متداول
Body Centered Cubic (BCC)
Face Centered Cubic (FCC)
Hexagonal Close-Packed (HCP)
Dr. Elmkhah (Bu-Ali Sina Univ.)
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• Coordination # = 8
Adapted from Fig. 3.2,
Callister 7e.
(Courtesy P.M. Anderson)
(BCC)پر مرکزساختاری مکعبی
ex: Cr, W, Fe (), Tantalum, Molybdenum
2 atoms/unit cell: (1 center) + (8 corners x 1/8)
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• Coordination # = 12
Adapted from Fig. 3.1, Callister 7e.
(Courtesy P.M. Anderson)
(FCC)ساختار مکعبی سطوح پر
ex: Al, Cu, Au, Pb, Ni, Pt, Ag
4 atoms/unit cell: (6 face x ½) + (8 corners x 1/8)
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(1)کار در کالس .را بدست آوریدBCCو FCCرابطه شعاع اتمی و پارامتر شبکه در سیستم -
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• Coordination # = 12
• 3D Projection • 2D Projection
Adapted from Fig. 3.3(a),
Callister 7e.
(HCP)ساختار هگزاگونال
6 atoms/unit cell
ex: Cd, Mg, Ti, Zn
• c/a = 1.633 (ideal)
c
a
A sites
B sites
A sitesBottom layer
Middle layer
Top layer
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(2)کار در کالس : HCPدر سیستم -
چیست؟ cو aرابطه حجم هگزاگونال بر حسب ( 1رابطه شعاع اتمی و پارامتر شبکه چگونه است؟ ( 2تعداد اتمها در سلول واحد چندتا است؟( 3
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تعداد اتم های ثابت شبکهعدد همسایگیموجود درسول واحد
نوع ساختار
62R1Sc
82Bcc
124Fcc
122R6hcp
مشخصات ساختارهای بلوری متداول
؟
؟
؟
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(3)کار در کالس میلیمتر مربع، چند 670و مساحت 0/08mmدر یک ورق نازک طال به ضخامت : تمرین
(آنگستروم است1/44شعاع اتمی آن )سلول واحد شبکه وجود دارد؟
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مفاهیم پایه(Lattice Parameter)پارامتر شبکه یا ثابت شبکه •
(در سیستم بلوری)تعداد اتم ها در هر سلول واحد •
(.Coordination No)عدد همسایگی•
( Atomic Packing Factor)فاکتور تراکم اتمی •
(Theoretical Density)چگالی تئوریکی •
( Linear & Planar Density)چکالی خطی و سطحی •
(Crystallographic Planes)صفحات و جهات بلوری •
Dr. Elmkhah (Bu-Ali Sina Univ.)
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فاکتور تراکم اتمیAtomic Packing Factor (APF)
APF = Volume of atoms in unit cell
Volume of unit cell
کدام ساختار متراکم تر است؟
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APF =
a3
4
3p (0.5a) 31
atoms
unit cellatom
volume
unit cell
volume
دهفاکتور تراکم اتمی برای ساختار معکبی سا
Adapted from Fig. 3.23,
Callister 7e.
close-packed directions
a
R=0.5a
contains (8 x 1/8) = 1 atom/unit cell
Here: a = Rat*2
Where Rat is the ‘handbook’
atomic radius
• APF for a simple cubic structure = 0.52Dr. Elmkhah (Bu-Ali Sina Univ.)
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(4)کار در کالس و وزن 8/57g/cm3نانومتر و چگالی 0/143دارای شعاع اتمی ( Nb)نیوبیوم : تمرینمی FCCیا BCCمشخص کنید ساختمان کریستالی آن . است 92/91g/molاتمی
باشد؟
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BCCفاکتور تراکم اتمی برای ساختار
a
APF =
4
3p ( 3a/4)32
atoms
unit cell atom
volume
a3
unit cell
volume
length = 4R =
Close-packed directions:
3 a
• APF for a body-centered cubic structure = 0.68
aR
Adapted from
Fig. 3.2(a), Callister 7e.
a2
a3
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• APF for a face-centered cubic structure = 0.74
FCCفاکتور تراکم اتمی برای ساختار
The maximum achievable APF!
APF =
4
3p ( 2a/4)34
atoms
unit cell atom
volume
a3
unit cell
volume
Close-packed directions:
length = 4R = 2 a
Unit cell contains:6 x 1/2 + 8 x 1/8
= 4 atoms/unit cella
2 a
Adapted from
Fig. 3.1(a),
Callister 7e.
(a = 22*R)
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APFمقایسه دو ساختار بلوری با بیشترین
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(5)کار در کالس به cو a ،bاورانیوم دارای شبکه بلوری ارتورومبیک بوده و پارامتر شبکه آن : تمرین
اگر چگالی، وزن اتمی و شعاع . نانومتر است0/495و 0/587، 0/286ترتیب برابر باشد، درصد 0/1385nmو 19/05g/cm3 ،236g/molاتمی آن به ترتیب برابر
.فاکتور تراکم اتمی را محاسبه کنید
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rتئوریکیچگالی
where n = number of atoms/unit cell
A = atomic weight
VC = Volume of unit cell = a3 for cubic
NA = Avogadro’s number
= 6.023 x 1023 atoms/mol
Density = r =
VCNA
n Ar =
CellUnitofVolumeTotal
CellUnitinAtomsofMass
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• Ex: Cr (BCC)
A = 52.00 g/mol
R = 0.125 nm
n = 2
a = 4R/3 = 0.2887 nma
R
r = a3
52.002
atoms
unit cellmol
g
unit cell
volume atoms
mol
6.023 x 1023
rtheoretical
ractual
= 7.18 g/cm3
= 7.19 g/cm3
rتئوریکیچگالی
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(6)کار در کالس و وزن اتمی آن 10/5g/cm3چگالی نقره . )شعاع اتمی نقره را محاسبه کنید: تمرین
107/87g/mol)
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(7)کار در کالس ، چگالی HCPساختار بلوری منیزیم . )را محاسبه کنید( Mg)شعاع اتمی منیزیم : تمرین(استc/a=1/62و نسبت 24/3g/molو وزن اتمی آن 1/74g/cm3نقره
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(8)کار در کالس 5با ثابت شبکه FCCسانتیگراد دارای ساختار 850عنصری در دمای باالتر از : تمرین
4با ثابت شبکه BCCآنگستروم می باشد و پایین تر از دمای مذکور دارای ساختار .درصد تغییر حجم حین سرد شدن را محاسبه کنید. آنگستروم است
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ex: linear density of Al in [110]
direction
a = 0.405 nm
)Linear Density( خطیچگالی
Linear Density of Atoms LD =
a
[110]
Unit length of direction vector
Number of atoms
# atoms
length
13.5 nma2
2LD
-==
# atoms CENTERED on the direction of interest!
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(9)کار در کالس را بدست BCCو FCCچگالی اتمی خطی در جهت قطر اصلی معکب در شبکه : تمرین.آورید
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Solution for plane (100)
===
=
536.33536.02
125.04
2
4
125.0
0 nmnmr
FCCa
nmrNi
a0
For (100):
7854.0
2/4
2_
/10600.110536.3
2_
2
2
215
28
==
=
=-
r
rfractionpacking
cmatomscm
atomsdensityplanar
p
FCC
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Solution for plane (110)
5554.0
2/42
2_
/10131.110536.32
2_
2
2
215
28
==
=
=-
r
rfractionpacking
cmatomscm
atomsdensityplanar
p
For (110):
a0
02a
It is important to visualize how the plane is cutting
across the unit cell – as shown in the diagram!
FCC
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Solution for plane (111)
02a
02a
For (111):
Again try to visualize the plane, count the number of
atoms in the plane:
02a
2
000 866.02
32
2
1
2
1_ aaabhareaplane =
==
215
28/10847.1
10536.3866.0
2_ cmatoms
cm
atomsdensityplanar =
=
-
9069.0
866.0
4/22_
2
0
2
0 ==a
afractionpacking
p
Therefore, plane (111) is close-packed!
FCC
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Solution for plane (100)
===
=
510.33510.03
152.04
3
4
152.0
0 nmnmr
BCCa
nmrLi
First, find atomic radius for Nickel from Appendix 2, page 797 of
Textbook (6th Ed. Shackleford) to calculate the lattice parameter:
For (100):
5890.0
4/3_
/10115.810510.3
1_
2
0
2
0
214
28
==
=
=-
a
afractionpacking
cmatomscm
atomdensityplanar
p
BCC
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Solution for plane (110)For (110):
It is important to visualize how the plane is cutting
across the unit cell – as shown in the diagram!
8330.0
2
4/32_
/10148.110510.32
2_
2
0
2
0
215
28
==
=
=-
a
afractionpacking
cmatomscm
atomsdensityplanar
p
BCC
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Solution for plane (111)
02a
For (111):
Note: Since the (111) does NOT pass through the
center of the atom in the middle of the BCC unit cell,
we do not count it!
2
000 866.02
33
2
1
2
1_ aaabhareaplane =
==
214
28/10686.4
10510.3866.0
2/1_ cmatoms
cm
atomdensityplanar =
=
-
3401.0
866.0
4/32/1_
2
0
2
0 ==a
afractionpacking
p
Therefore, there is no close-pack plane in BCC!
There are only (3)(1/6)=1/2 atoms in the plane.
02a
BCC
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FCC Stacking
Highlighting the faces
Highlighting
the stacking
A
B
C
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A sites
B B
B
BB
B B
C sites
C C
CA
B
B sites
• ABCABC... Stacking Sequence
• 2D Projection
• FCC Unit Cell
FCC Stacking Sequence
B B
B
BB
B B
B sitesC C
CA
C C
CA
AB
C
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A
B
HCP Stacking
Highlighting the cell
Figure 3.3
Highlighting
the stacking
A
Layer A
Layer A
Layer B
Self-Assessment: How many atoms/cell?
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FCC HCP
Looking down (111) plane!
Looking down (0001) plane
Comparing the FCC and HCP Planes Stacking
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مفاهیم پایه(Lattice Parameter)پارامتر شبکه یا ثابت شبکه •
(در سیستم بلوری)تعداد اتم ها در هر سلول واحد •
(.Coordination No)عدد همسایگی•
( Atomic Packing Factor)فاکتور تراکم اتمی •
(Theoretical Density)چگالی تئوریکی •
( Linear & Planar Density)چکالی خطی و سطحی •
(Crystallographic Planes)صفحات و جهات بلوری •
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بلوری در سیستم مکعبیتعیین جهات
ex: 1, 0, ½ => 2, 0, 1 => [ 201 ]
-1, 1, 1
z
x
[ 111 ]=>
y
نظروصل کردن مرکز سیستم به نقطه مورد
خواندن محل تقاطع پیکان با محورهایa وb وc
تبدیل اعداد بدست آمده به عدد صحیح
قرار دادن اعداد نهایی در داخل براکت[UVW ]<جهات هم خانواده>
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محل تقاطعمحل تقاطع
تبدیل به عدد صحیحقراردادن در داخل براکت
x y z
a/2 b 0c
1/2 1 0
1 2 0
[120]
مثال دیگر از تعیین جهات بلوری
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[ 1120 ]ex: ½, ½, -1, 0 =>
Adapted from Fig. 3.8(a), Callister 7e.
a1
a2
a3
-a3
2
a2
2
a1-
a3
a1
a2
z
HCPتعیین جهات بلوری در سیستم
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اندیس و بالعکس3اندیس به 4تبدیل جهت بلوری از
=
=
=
'ww
t
v
u
)vu( +-
)'u'v2(3
1-
)'v'u2(3
1-=
]uvtw[]'w'v'u[
Fig. 3.8(a), Callister 7e.
-a3
a1
a2
z
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مثال
-a3
a1
a2
z
1 1 12 ' ' 2 1 1 13 3 3
1 1 12 ' ' 2 1 1 13 3 3
1 1 2 23 3 3
' 0
M-B Indices: [1120]
u u v
v v u
t u v
w w
= - = - =
= - = - =
= - = - = - -
= =
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نحوه تعیین صفحات بلوری
خواندن محل تقاطع صفحه با محورهایa وb وc
وارونه نمودن اعداد فوق
تبدیل اعداد بدست آمده به عدد صحیح
اندیس میلر = قرار دادن اعداد نهایی در داخل پرانتز(hkl ){صفحات هم خانواده}
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صفحات بلوری هم خانواده
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صفحات بلوریz
x
ya b
c
اندیس میلر .4 (110)
مثال a b cz
x
ya b
c
اندیس میلر .4 (100)
محل تقاطع .1 1 1
وارونه نمودن .2 1/1 1/1 1/
1 1 0صحیح کردن .3 1 1 0
محل تقاطع .1 1/2
وارونه نمودن .2 1/½ 1/ 1/
2 0 0صحیح کردن .3 2 0 0
مثال a b c
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صفحات بلوریz
x
ya b
c
4. Miller Indices (634)
example1. Intercepts 1/2 1 3/4
a b c
2. Reciprocals 1/½ 1/1 1/¾
2 1 4/3
3. Reduction 6 3 4
(001)(010),
Family of Planes {hkl}
(100), (010),(001),Ex: {100} = (100),
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x y z
Intercepts
Intercept in terms of lattice parameters
Reciprocals
Reductions
Enclosure
a -b c/2
-1 1/20 -1 2
N/A
(012)
مثال دیگر
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HCPصفحه بلوری در سیستم : مثال دیگر
In hexagonal unit cells the same idea is used
example a1 a2 a3 c
4. Miller-Bravais Indices (1011)
1. Intercepts 1 -1 12. Reciprocals 1 1/
1 0
-1
-1
1
1
3. Reduction 1 0 -1 1
a2
a3
a1
z
Adapted from Fig. 3.8(a), Callister 7e.
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Virtual Materials Science & Engineering (VMSE)
• VMSE is a tool to visualize materials science topics such as
crystallography and polymer structures in three dimensions
• Available in Student Companion Site at www.wiley.com/college/callister
and in WileyPLUSDr. Elmkhah (Bu-Ali Sina Univ.)
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سلول واحد در سرامیک ها
در فلزات فقط یک نوع اتم با شعاع یکسان داریم.
در سرامیکها دو یا چند نوع اتم با شعاع های متفاوت داریم.
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• Coordination Number increases with
عدد همسایگی و شعاع یونی
Adapted from Table 3.3,
Callister & Rethwisch 4e.
2
rcationranion
Coord.
Number
< 0.155
0.155 - 0.225
0.225 - 0.414
0.414 - 0.732
0.732 - 1.0
3
4
6
8
linear
triangular
tetrahedral
octahedral
cubic
Adapted from Fig. 3.5,
Callister & Rethwisch 4e.
Adapted from Fig. 3.6,
Callister & Rethwisch 4e.
Adapted from Fig. 3.7,
Callister & Rethwisch 4e.
ZnS
(zinc blende)
NaCl(sodium
chloride)
CsCl(cesium chloride)
rcationranion
To form a stable structure, how many anions can
surround around a cation?
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محاسبه حداقل نسبت کاتیون به آنیون Determine minimum rcation/ranion for an octahedral site
(C.N. = 6)
a = 2ranion
2ranion + 2rcation = 2 2ranion
ranion + rcation = 2ranion
rcation = ( 2 -1)ranion
arr 222 cationanion =+
414.012anion
cation =-=r
r
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• On the basis of ionic radii, what crystal structure
would you predict for FeO?
• Answer:
5500
1400
0770
anion
cation
.
.
.
r
r
=
=
based on this ratio,
-- coord # = 6 because
0.414 < 0.550 < 0.732
-- crystal structure is NaCl
Data from Table 3.4,
Callister & Rethwisch 4e.
Example Problem: Predicting the Crystal
Structure of FeO
Ionic radius (nm)
0.053
0.077
0.069
0.100
0.140
0.181
0.133
Cation
Anion
Al3+
Fe2+
Fe3+
Ca2+
O2-
Cl-
F-Dr. Elmkhah (Bu-Ali Sina Univ.)
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Rock Salt StructureSame concepts can be applied to ionic solids in general.
Example: NaCl (rock salt) structure
rNa = 0.102 nm
rNa/rCl = 0.564
cations (Na+) prefer octahedral sites
Adapted from Fig. 3.5,
Callister & Rethwisch 4e.
rCl = 0.181 nm
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MgO and FeO
O2- rO = 0.140 nm
Mg2+ rMg = 0.072 nm
rMg/rO = 0.514
cations prefer octahedral sites
So each Mg2+ (or Fe2+) has 6 neighbor oxygen atoms
Adapted from Fig. 3.5,
Callister & Rethwisch 4e.
MgO and FeO also have the NaCl structure
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AX Crystal Structures
939.0181.0
170.0
Cl
Cs ==-
+
r
r
Adapted from Fig. 3.6,
Callister & Rethwisch 4e.
Cesium Chloride structure:
Since 0.732 < 0.939 < 1.0,
cubic sites preferred
So each Cs+ has 8 neighbor Cl-
AX–Type Crystal Structures include NaCl, CsCl, and zinc blende
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ZnS بلوریساختار
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خالصه برخی ساختارهای سرامیکی معروف
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فضاهای خالی در ساختارهای بلوری
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مورفیسمپلی Two or more distinct crystal structures for the
same material (allotropy/polymorphism)
titanium
(HCP), (BCC)-Ti
carbon:
diamond, graphite
BCC
FCC
BCC
1538ºC
1394ºC
912ºC
-Fe
-Fe
-Fe
liquid
iron system:
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SiO4
4-
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• Some engineering applications require single crystals:
• Properties of crystalline materials
often related to crystal structure.
(Courtesy P.M. Anderson)
--Ex: Quartz fractures more easily
along some crystal planes than
others.
--diamond single
crystals for abrasives--turbine blades
Fig. 8.33(c), Callister 7e.
(Fig. 8.33(c) courtesy
of Pratt and Whitney).
(Courtesy Martin Deakins,
GE Superabrasives,
Worthington, OH. Used with
permission.)
Crystals as Building Blocks
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• Most engineering materials are polycrystals.
• Nb-Hf-W plate with an electron beam weld.
• Each "grain" is a single crystal.
• If grains are randomly oriented,overall component properties are not directional.
• Grain sizes typ. range from 1 nm to 2 cm
(i.e., from a few to millions of atomic layers).
Adapted from Fig. K,
color inset pages of
Callister 5e.
(Fig. K is courtesy of
Paul E. Danielson,
Teledyne Wah Chang
Albany)
1 mm
Polycrystals
Isotropic
Anisotropic
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• Single Crystals
-Properties vary with
direction: anisotropic.
-Example: the modulus
of elasticity (E) in BCC iron:
• Polycrystals
-Properties may/may not
vary with direction.
-If grains are randomly
oriented: isotropic.
(Epoly iron = 210 GPa)
-If grains are textured,
anisotropic.
200 mm
Source of data is R.W.
Hertzberg, Deformation
and Fracture Mechanics
of Engineering
Materials, 3rd ed., John
Wiley and Sons, 1989.
courtesy of L.C. Smith
and C. Brady, the
National Bureau of
Standards, Washington,
DC [now the National
Institute of Standards
and Technology,
Gaithersburg, MD].
Single vs PolycrystalsE (diagonal) = 273 GPa
E (edge) = 125 GPa
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Effects of Anisotropy:
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X-Ray Diffraction
Diffraction gratings must have spacings comparable to the wavelength of diffracted radiation.
Can’t resolve spacings
Spacing is the distance between parallel planes of atoms.
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Figure 3.32 Relationship of the Bragg angle (θ) and the
experimentally measured diffraction angle (2θ).
X-ray intensity (from detector)
q
qc
d =n
2 sinqc
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X-Rays to Determine Crystal Structure
X-ray intensity (from detector)
q
qc
d =n
2 sin qc
Measurement of
critical angle, qc,
allows computation of
planar spacing, d.
• Incoming X-rays diffract from crystal planes.
Adapted from Fig. 3.19,
Callister 7e.
reflections must be in phase for a detectable signal!
spacing between planes
d
q
q
extra distance traveled by wave “2”
2 2 2hkl
ad
h k l=
For Cubic Crystals:
h, k, l are Miller IndicesDr. Elmkhah (Bu-Ali Sina Univ.)
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Figure 3.34 (a) An x-ray diffractometer. (Courtesy of
Scintag, Inc.) (b) A schematic of the experiment.
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X-Ray Diffraction Pattern
Adapted from Fig. 3.20, Callister 5e.
(110)
(200)
(211)
z
x
ya b
c
Diffraction angle 2q
Diffraction pattern for polycrystalline -iron (BCC)
Inte
nsity (
rela
tive)
z
x
ya b
c
z
x
ya b
c
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Diffraction in Cubic Crystals:
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• Atoms may assemble into crystalline or
amorphous structures.
• We can predict the density of a material, provided we
know the atomic weight, atomic radius, and crystal
geometry (e.g., FCC, BCC, HCP).
SUMMARY
• Common metallic crystal structures are FCC, BCC, and
HCP. Coordination number and atomic packing factor
are the same for both FCC and HCP crystal structures.
• Crystallographic points, directions and planes are
specified in terms of indexing schemes.
Crystallographic directions and planes are related
to atomic linear densities and planar densities.
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• Some materials can have more than one crystal
structure. This is referred to as polymorphism (or
allotropy).
SUMMARY• Materials can be single crystals or polycrystalline.
Material properties generally vary with single crystal
orientation (i.e., they are anisotropic), but are generally
non-directional (i.e., they are isotropic) in polycrystals
with randomly oriented grains.
• X-ray diffraction is used for crystal structure and
interplanar spacing determinations.
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