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SOLIDS
Diah Susanti, Ph.D
Materials and Metallurgical Engineering Department
Institut Teknologi Sepuluh Nopember (ITS)
Surabaya - INDONESIA
solidification
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• Solid
A solid is a state of matter with a defined shape and volume. Atoms,
ions, and molecules in a solid pack tightly together and may form
crystals. Examples of solids include rocks, ice, diamond, and wood.
• Liquid
A liquid is a state of matter with a defined volume, but no defined
shape. In other words, liquids take the shape of their container.
Particles in a liquid have more energy than in a solid, so they are
further apart and less organized (more random). Examples of liquids
include water, juice, and vegetable oil.
• Gas
A gas is a state of matter lacking either a defined volume or defined
shape. Like a liquid, a gas takes the shape of a container. Unlike a
liquid, a gas easily expands or contracts to fill the entire volume of the
container. Particles in a gas have more energy than in solids or liquids.
They tend to be further apart and more random than in a liquid.
Examples of gases include air, water vapor, and helium.
• Plasma
Plasma is a state of matter similar to a gas, except all of the particles
carry an electrical charge. Also, plasma tends to exist at very low
pressure, so the particles are even further apart than in a gas. Plasma
can consist of ions, electrons, or protons. Examples of plasma include
lightning, the aurora, the Sun, and the inside of a neon sign.
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SOLIDS
• Solids differ from gases and liquids in many ways: (1)
they retain their shape and volume when transferred
from one container to another, (2) they are virtually
incompressible (3) they exhibit extremely slow rates of
diffusion.
• There are two types of solids, amorphous and
crystalline. If the temperature at which the solid is
formed is approached slowly so as to allow the array of
particles to become well ordered, a crystalline solid
results.
• If, on the other hand, the temperature is lowered very
rapidly, there is a chance that the particles will be
‘frozen’ in a chaotic state. In this case the particles are
arranged in a random fashion and the resulting solid is
said to be amorphous (without form). Examples: glass,
rubber and most plastics.
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Crystalline Solid
• The flat planes in crystalline solid are called
faces and the characteristics angles are called
interfacial angles.
• These characteristics are present no matter how
crystal was formed and are completely
independent of the size of the crystal.
• This situation may be contrasted with what
happens when an amorphous solid is broken.
Crystalline Solid
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Crystalline Solid
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Amorphous Material
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X-Ray Diffraction
• When a crystal is bathed in X-rays, each atom of
crystal within the path of an X-ray absorbs some
of its energy and then reemits it in all direction.
Thus each atom is a source of secondary
wavelets, and the X-rays are said to be scattered
by the atoms.
• These secondary wavelets from the different
sources interfere with each other, either by
reinforcing (it is said in phase) or by cancelling
each other (it is said out of phase)
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William Bragg and Lawrence Bragg proposed Bragg’s
equation:
This equation implies that the extra distance travelled by
the more penetrating beam to be some integral multiple of
the X-rays wavelength so that the waves have to be in
phase and have intensity at the detector
n = an integer; 1,2,3, … dhkl = interplanar spacing
2T = angle between the X-rays enter and leave the
particular set of layer
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• https://www.youtube.com/watch?v=4e8EB1jzH
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• https://www.youtube.com/watch?v=xBA09PXP
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• https://www.youtube.com/watch?v=QHMzFUo
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Example 7.1
• X-rays of wavelength 154 pm strike a crystal and
are observed to be reflected at an angle of 22.5 o.
Assuming that n = 1 calculate the spacing
between the planes of atoms that are responsible
for this reflection.
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1. Calculate the angles at which X-rays of
wavelength 229 pm will be observed to be
reflected from crystal planes spaced (a) 1 nm
apart, (b) 200 pm apart. Assume that n=1.
2. Calculate the interplanar spacings that
correspond to reflection at T = 20o, 27.4o, and
35.8o by X-rays of wavelength 141 pm. Assume
that n=1.
3. From the following list of angles, determine
the angles at which X-rays of wavelength 141
pm, diffracted from planes of atoms 200 pm
apart, are in phase: T = 17.3o ,20.5o, 44.4o, and
55.3o
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Lattice
Square Lattice point
lattice Unit cell: the smallest repeating unit
generating the entire lattice
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Bravais Lattice
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Type of lattice point
Type of lattice point Contribution to One Unit Cell
Corner 1/8
Edge ¼
Face center ½
Body Center 1
How many lattice points do SC, BCC, FCC and end-centered orthorhombic
possess?
The number of lattice points refers to the number of atom.
SC =
BCC =
FCC =
End centered orthorhombic =
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Avogadro’s number
Metallic copper is found to crystallize with a FCC lattice in which a copper
atom is located at each lattice point. The length of the edge of the unit
cell is found to be 361.5 pm (from XRD). Copper has an atomic weight of
63.54 g mol-1 and a density of 8.936 gcm-3. What is the value of
Avogadro’s number?
Answer : a=b=c= 361.5 pm = 3.615x10-8 cm
AW = 63.54 g mol-1 U = 8.936 gcm-3
the volume of 1 mol Copper: 7.111 cm3
The volume of a unit cell: (361.5 x 10-10)3 = 4.724x10-23 cm3. Thus in 1 mol
Cu there are 1.505x1023 unit cells. In a unit cell there are 4 Cu atoms.
The number of atoms of Cu per 1 mol Cu = 4x1.505x1023 = 6.02x1023
atoms = Avogadro’s number.
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1. Metallic sodium crystallizes with a BCC lattice.
The element has a density of 0.97 gcm-3. What is
the length of the edge of the unit cell in Na (AW =
23 g mol-1)?
2. CsCl crystallizes with a cubic unit cell of edge
length 412.3 pm. The density of CsCl is 3.99 gcm-3.
Show that the unit cell cannot be face-centered or
body-centered (CsCl MW = 168.5 g mol-1)
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ATOMIC AND IONIC RADII
B AB = 4r = (AC2 + BC2)1/2
r 4r = (2AC2)1/2
4r = AC (2)1/2 = a(2)1/2
A C r = 0.25a(2)1/2
For Cu, a = 361.5 pm
a
So the radius = 127.8 pm
FCC Volume = a3 = (4r/21/2)3
Au, a= 407.86 pm, r = 144.2
pm, volume = 6.785x107 pm3
Besides Cu, other familiar metals such as Al, Pb, Ag, and
Au also crystallize in FCC structures.
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ATOMIC AND IONIC RADII
B AB = 4r = (AC2 + BC2)1/2
r a 4r = ((a(2)1/2)2+a2)1/2
4r = (2a2+a2)1/2 = a(3)1/2
A C r = 0.25a(3)1/2
a(2)1/2
Example: Cr, Na
BCC
Example simple cubic : CsCl
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1. Chromium, used to protect and beautify other metals, crystallizes
in a BCC structure in which the Cr atoms are in contact along the
body diagonal of the unit cell. The edge of the unit cell is 288.4 pm.
Calculate the atomic radius of Cr.
2. The density of Cr is 7.19 gcm-3 and the unit cell edge is 288.4 pm.
Use these data to compute the Avogadro’s number. (AW Cr:52
gram/mol)
3. Al crystallizes in a FCC structure (cubic closest packed). If the Al
atom has an atomic radius of 143 pm, what is the length of the unit
cell edge in Al.
4. Silver has an atomic radius of 144 pm. What would be the density
of Ag if it were to crystallize in the following structures: (a) simple
cubic, (b) bcc, and (c) fcc. The actual density of Ag is 10.6 gcm-3.
Which of these corresponds to the correct structure for Ag? (AW
Ag: 108 gram/mol)
5. Calculate the amount of vacant (unoccupied) space in a primitive
cubic, bcc, and fcc of identical spheres of diameter 100 pm.
6. Gold crystallizes in a FCC structure. The length of the unit cell edge
in Au is 407.86 pm. What is the atomic radius of Au?
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The Face-Centered Cubic Lattice
Na+ cation (positive ion) surrounded by 6 anions
Cl- anion (negative ion) octahedrally
surrounded by cation.
Na+ atom = 1/8 * 8 + 1/2*6 = 4 atoms
Cl- atom = 1/4 * 12 + 1 = 4 atoms
The ratio of Na+ and Cl- is 4:4 = 1:1
Other substances possessing the rock salt
structure include:
1. All alkali halide except CsCl (below 450o C),
The structure of NaCl is termed CsBr, and CsI.
rock salt structure 2. AgF, AgBr, AgI.
3. Oxides and sulfides of the alkaline earth
metals.
4. NiO.
5. Some alloys.
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The Face-Centered Cubic Lattice
Zn+ cation (positive ion)
S- anion (negative ion) tetrahedrally
surrounded by 4 cations
Zn+ atom = 1/8 * 8 + 1/2*6 = 4 atoms
S- atom = 4 atoms
The ratio of Zn+ and S- is 4:4 = 1:1
Other substances possessing the zinc blende
structure include: ZnS, ZnO, CuCl, CuBr, and
BeO
The structure of ZnS is termed
zinc blende structure
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The Face-Centered Cubic Lattice
Ca+ cation (positive ion)
F- anion (negative ion) tetrahedrally
surrounded by 4 cations
Ca+ atom = 1/8 * 8 + 1/2*6 = 4 atoms
F- atom = 8 atoms
The ratio of Zn+ and S- is 4:8 = 1:2
Other substances possessing the fluorite
structure include: BaF2, SrCl2 and ThO2
The structure of CaF2 is termed
fluorite structure
Another structure called antifluorite structure is the same as the fluorite
structure except that the positions of positive and negative ions are reversed
to give cation-to-anion ratio of 2:1. Examples: M2S, M2Se, and M2Te, where M
= Na+ or K+
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Closest-Packed Structures
• When all atoms in a substance are the same, its
crystal structure is often determined by the
most efficient packing of sphere, that is, the
arrangement of atoms that gives rise to the
smallest amount of unoccupied space.
• The most efficiently packing arrangements are
called closest-packed structures.
• There are two structures: cubic closed packed
(ccp) and hexagonal closed packed (hcp).
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A B
• A first layer of spheres is packed together as tightly as possible. A second layer
of spheres is placed on the first with each sphere in the second layer resting in
the depression, or hole, between spheres in the first layer. There are now two
ways of arranging the spheres in third layer.
1. Placing the third layer directly above the first layer, the fourth above the
second layer, the fifth above the third layer and so on, so that we have
ABAB…pattern as shown in Fig. A. This is a hexagonal closed packed
(hcp) structure.
2. Placing the third layer above holes in the first layer. If we continue building
the structure by laying the fouth layer directly above the first, the fifth
above the second, and so on, we arrive at an ABCABC…pattern as shown in
Fig B, a cubic closed packed (ccp) structure, having a face centered
cubic (fcc) structure.
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• In both the ccp and the hcp structures each atom is in contact
with 12 other “nearest neighbours”. We can compare this with a
body-centered cubic structure where each atom would in
contact with its eight nearest neighbours and with a simple
cubic structure where each atom would have only six nearest
neighbours.
• There is a relatively large number of substances that crystallize
with these closest packed structures (for example: most metals).
• In the ccp structure there are two types of vacant sites:
tetrahedral and octahedral. A tetrahedral site is an empty
space surrounded by six spheres and an octahedral site is an
empty space surrounded by six spheres.
• There are twice as many tetrahedral sites as there are
octahedral sites.
• These vacant “holes” may be used to accommodate other atoms
and, when filled in this way, give rise to a variety of different
structures.
• Sodium chloride, for example, may be considered as a ccp
arrangement of Cl- ions with Na+ ions located in octahedral sites
(recall that in NaCl each cation is surrounded octahedrally by
anions.)
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