pertemuan 3 close packing
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Pertemuan 3 close packingTRANSCRIPT
Universitas Negeri Jakarta
Pertemuan 3
CLOSE PACKING STRUCTURE
Iwan Sugihartono, M.SiJurusan Fisika, FMIPAUniversitas Negeri Jakarta
1
Outline
Crystals
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Crystal structure basics
unit cells
symmetry
lattices
Some important crystal structures and properties
close packed structures
octahedral and tetrahedral holes
basic structuresferroelectricity
Diffraction
how and why - derivation
Objectives
By the end of this section you should:
• understand the concept of close packing
• know the difference between hexagonal and cubic close packing
• know the different types of interstitial sites in a close packed structure
• recognise and demonstrate that cubic close packing is equivalent to a face centred cubic unit cell
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Packing
Can pack with irregular shapes
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Two main stacking sequences:
If we start with one cp layer, two possible ways of adding a second layer (can have one or other, but not a mixture) :
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Two main stacking sequences:
If we start with one cp layer, two possible ways of adding a second layer (can have one or other, but not a mixture) :
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Let’s assume the second layer is B (red). What about the third layer?
Two possibilities:
(1) Can have A position again (blue). This leads to the regular sequence …ABABABA…..
Hexagonal close packing (hcp)
(2) Can have layer in C position, followed by the same repeat, to give …ABCABCABC…
Cubic close packing (ccp)
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Hexagonal close packed Cubic close packed
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No matter what type of packing, the coordination number of each equal size sphere is always 12
We will see that other coordination numbers are possible for non-equal size spheres
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The reasons why a particular metal prefers a particular structure are still not well understood
Metals usually have one of three structure types:ccp (=fcc, see next slide), hcp or bcc (body centred cubic)
Sc Ti V Cr Mn Fe Co Ni Cu Zn
Y Zr Nb Mo Tc Ru Rh Pd Ag Cd
La Hf Ta W Re Os Ir Pt Au Hg
hcp
ccp = fcc
bcc
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ccp = fcc ?
Build up ccp layers (ABC… packing)
Add construction lines -can see fcc unit cell
c.p layers are oriented perpendicular to the body diagonal of the cube
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Hexagonal close packed structures (hcp)
hcp bcc
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Recurring themes...
Foot and mouth virus
Body centred cubic
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Packing Fraction
We (briefly) mentioned energy considerations in relation to close packing (low energy configuration)
Rough estimate - C, N, O occupy 20Å3
Can use this value to estimate unit cell contents
Useful to examine the efficiency of packing - take c.c.p. (f.c.c.) as example
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Calculate unit cell side in terms of r:
2a2 = (4r)2
a = 2r 2
Volume = (16 2) r3
Face centred cubic - so number of atoms per unit cell =corners + face centres = (8 1/8) + (6 1/2) = 4
So the face of the unit cell looks like:
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Packing fraction
The fraction of space which is occupied by atoms is called the “packing fraction”, , for the structure
spaceavailable
atomsby occupied space =
4
4
3
16 2 3 2074
3
3
r
r.
For cubic close packing:
The spheres have been packed together as closely as possible, resulting in a packing fraction of 0.74
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Group exercise:
Calculate the packing fraction for a primitive unit cell
A = 2 r
52,01
3
3
34
a
r
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Mencari Fraksi Packing
Jumlah atom efektif dalam unit cell = 12(1/6)+2(1/2)+3=6
%7474,060sin
3/420
3
caa
r
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Primitive
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Close packing
Cubic close packing = f.c.c. has =0.74
Calculation (not done here) shows h.c.p. also has =0.74 -equally efficient close packing
Primitive is much lower: Lots of space left over!
A calculation (try for next time) shows that body centred cubic is in between the two values.
THINK ABOUT THIS! Look at the pictures - the above values should make some physical sense!
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Hitunglah efisiensi packing dan kerapatan dari NaCl bila diberikan data sebagai
berikut:
Jari-jari ion Na = 0,98 A
Jari-jari ion Cl = 1,81 A
Massa atom Na = 22,99 amu
Massa atom Cl = 35,45 amu
?Solusinya:
Parameter kisi, a = 2 (Jari-jari ion (Na + Cl)) = 5.58 A
Fraksi Packing:
= Volume ion yang ada dalam sebuah unit cell
Volume unit cellnya
= 4 (4/3) phi (r3Na + r3
Cl) / a3 = 66,3 %
Density:
= Massa unit cell / Volumenya
= 2234 kg m-3
1 amu = 1,66 x 10-27 kg
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Summary By understanding the basic geometry of a
cube and use of Pythagoras’ theorem, we
can calculate the bond lengths in a fcc
structure
As a consequence, we can calculate the
radius of the interstitial sites
we can calculate the packing efficiency for
different packed structures
h.c.p and c.c.p are equally efficient packing
schemes
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THANK YOU
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