L05A: Chapter 5: Defects in Solids https://en.wikipedia.org/wiki/Crystallographic_defect

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L05A Chapter 5Defects in Solidshttps//en.wiki
pedia.org/wiki/Crystallographic_defect

• Perfect crystals dont exist.
• Many of the important properties of materials are
  strongly influenced by defects, or even entirely
  determined by their presence.
• In some cases properties are improved, and in
  others degraded. Depends on material, defect
  type, and the particular property.
• Defect types and amounts depend on composition,
  temperature, pressure, and processing history.
• Classification of defects by dimensionality
• Point defects (zero dimensional)
• Line defects (one dimensional)
• Surface defects (two dimensional)
• Volume defects (three dimensional)

Last revised on February 11, 2014 by W.R. Wilcox,
Clarkson University
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Point defectsInterstitials and vacancies in a
pure metal

• A vacancy is an empty lattice site the atom
  supposed to be there is missing.
• An interstitial is an atom located between
  lattice sites.

• The concentration of both types of defects
  increases with temperature, although the number
  of interstitials in metals is generally
  negligible. Why?
• Because the atoms are close together and crowding
  another of the same size between them severely
  strains the lattice. Just not enough space!

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Vacancy concentration

• The number of vacancies NV in a given amount of
  metal is given by NV N exp(-QV/kT), where
  N is the total number of lattice sites in that
  amount of metal QV is the activation energy
  ? energy required to form a single vacancy T
  is the absolute temperature k R/NA
  1.38x10-23J/atom.K 8.62x10-5eV/atom.K is
  the Boltzmann constant R is the gas constant
  (8.314 J/mol.K) NA is Avogadros number
  (6.02x1023atom/mol)
• An electron volt, eV, is the energy acquired by
  an electron in moving through a potential
  difference of 1 V. (1eV 1.602x10-19 J)
• Near the melting point of metals at 1 atm, NV/N ?
  10-4.
• Example Problem 5.1 How many atoms in 1 m3 of
  Cu? ACu 63.5 g/mol, ?Cu 8.4 g/cm3. From
  the units

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Measuring Activation Energy
By measuring NV versus T, can get value of QV.
One method of determining NV is to make very
precise measurements of lattice constant and
density.
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One manifestation of changing the vacancy
concentration by changing T
Low energy electron microscope image of a (110)
surface of NiAl.

• Increasing the temperature causes the surface
  islands to grow.
• Why? When the equilibrium vacancy concentration
  is increased, expelled atoms diffuse to the
  surface and are incorporated at the edges of
  steps.

Reprinted with permission from Nature (K.F.
McCarty, J.A. Nobel, and N.C. Bartelt, "Vacancies
in Solids and the Stability of Surface
Morphology", Nature, Vol. 412, pp. 622-625
(2001).
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Point Defects in Ceramics
Vacancies -- vacancies exist in ceramics
for both cations and anions
Interstitials -- interstitials exist for
cations -- interstitials are not normally
observed for anions because anions are
large relative to the interstitial sites
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Paired defects in compounds with ionic bonds

• Crystal must maintain charge neutrality.
• This means that a point defect cannot occur by
  itself.
• Frenkel defect cation vacancy cation
  interstitial pairi.e. move a cation from a
  lattice site to a nearby interstitial site
• Schottky defect cation vacancy anion vacancy
  pairi.e. remove both an anion and a nearby
  cation at the same time

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Concentration of Frenkel and Schottky defects

• Note that N is the number of lattice sites, not
  the number of atoms.
• For example, in NaCl there is one Na and one Cl-
  associated with each lattice site.
• The factor of 2 comes about because there are two
  ions associated with each defect.

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Ions with more than one valence state

• Charge neutrality can also be maintained via
  different valence states when these exist.
• For example, iron can exist as Fe2 and Fe3.
• This means that in FeO, the two positive charges
  lost in forming a Fe2 vacancy can be compensated
  by substituting two Fe3 for two Fe2 in nearby
  sites (which leads to departure from
  stoichiometry).

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Impurities in Metals
Two common outcomes when impurity (B) added to
host (A)
1. Solid solution of B in A (usually randomly
distributed). Point defects.
OR
Substitutional solid solution (e.g., Cu in Ni)
Interstitial solid solution (e.g., C in Fe)
2. Solid solution of B in A plus particles of a
new phase (volume defects)
Definition alloy mixture
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Impurities in ionic compounds
Electroneutrality must be maintained
Example Ca2 substituting for 2Na in NaCl
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Impurities in compounds with covalent bonds

• When a substitutional impurity has an extra
  valence electron the extra electron can move in
  an electric field. Since the current is carried
  by negative charges, the semiconductor is called
  "n-type." For example, P in Si

• When the "dopant" has one fewer electrons, the
  missing electron in a covalent bond is called a
  "hole." An electron from an adjacent covalent
  bond can move into this hole in an electric
  field. Since the electric current is carried by
  what appears to be positive charges, the
  semiconductor is called "p-type." For example, B
  in Si

More in section 12.11 in the text.
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Variation in solubility

• Solid solubility depends strongly on temperature
• At high concentrations can get very complex
  behavior Chapter 10
• When solubility is higher at elevated
  temperature, a solid solution can survive to room
  temperature if the cooling rate is sufficient
• The solubility can depend strongly on the
  presence of other components -- which is normal
  in engineering alloys

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Specification of alloy compositions

• Three common methods
• Molar concentration (moles per volume) or mass
  concentration (mass per volume)
• Mass fraction or weight percent
• Atom fraction of atom percent (mole fraction and
  mole percent for compounds)
• Need to be able to convert between these.
• Cant remember all of the equations, and they are
  a pain to derive.
• So its better to pick a basis and then rely on
  the units.
• Best shown by examples.

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Conversion of weight percent to atom percent

• Example Problem 5.5 without equations
• Given Alloy of 97 wt Al, 3 wt Cu
• Atomic weight of Cu 63.55 g/molAtomic weight of
  Al 26.98 g/mol
• Basis 100 g
• Contains 97 g of Al and 3 g of Cu
• Moles of Al (97g Al)(mol Al / 26.98 g Al)
  3.595 mol Al
• Moles of Cu (3 g Cu)(mol Cu / 63.55 g Cu)
  0.0472 mol Cu
• Total moles 3.642 moles
• Atom of Cu (0.0472 mol Cu / 3.642 total
  mol)X100 1.30 at Cu
• Atom of Al (3.595 mol Al / 3.642 total
  mol)X100 98.7 at Al
• Notice that the number of significant figures in
  the final result of a calculation must not be
  excessive you invite ridicule if you copy
  everything down from the computer or calculator.
• If you have more than two components, this type
  of calculation is best done using Excel.

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Conversion of mole percent to weight percent

• Problem 5.16 An alloy consists of 6 atom Pb and
  94 atom Sn.What is its composition in weight ?
• Need atomic weights 207.2 for Pb and 118.710 for
  Sn.
• Basis 100 mol
• In 100 mol of alloy we have 6 mol of Pb and 94
  mol of Sn.
• Mass of Pb (6 mol Pb)(207.2 g Pb / mol Pb)
  1,243.2 g PbMass of Sn (94 mol Sn)(118.710 g
  Sn / mol Sn) 11,158.74 g Sn
• Total mass of 100 mol of alloy 12,401.94 g
• Weight of Pb (1,243.2 g Pb)/(12,401.94 g
  alloy)(100) 10.0 PbWeight of Sn
  (11,158.74 g Sn)/(12,401.94 g alloy)(100) 90.0
  Sn

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Conversion to molar concentration

• Need the density of the alloy.
• If this isnt known, estimate from the densities
  of the pure components by assuming that the alloy
  is an ideal solution, i.e. when the components
  are mixed to form the alloy the total volume
  doesnt change.
• Consider the previous example, 10 weight Pb and
  90 weight Sn.
• Basis 100 g, so have 10 g of Pb and 90 g of Sn.
• Density of Pb at room T is 11.34 gcm-3Density
  of white Sn at room T is 7.365 gcm-3
• Volume of Pb (10 g Pb)(cm3 Pb /11.34 g Pb)
  0.8818 cm3 PbVolume of Sn (90 g Sn)(cm3
  Sn/7.365 g Sn) 12.22 cm3 SnTotal alloy volume
  if alloy forms ideal solution 13.102 cm3
• Mass concentration of Pb (10 g)/(13.102 cm3)
  0.763 g Pb/cm3 alloyMass concentration of Sn
  (90 g)/(13.102 cm3) 6.869 g Sn/cm3
  alloyDensity of alloy if ideal solution 7.63
  g/cm3
• 7.55 g/cm3 given for this alloy at
  http//alasir.com/reference/solder_alloys/
• That is, when you mix lead and tin the volume
  expands slightly!
• Few liquid or solid solutions are exactly ideal,
  so use this method only when you dont know the
  actual density and you dont want to measure it.