Chapter 5: Defects in Solids
• Perfect crystals don’t 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)
Point defects:
Interstitials 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!
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 Avogadro’s 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:
 
3  8.4g  100cm 
1m 

3 
 cm  m 
3
 mol  6.022 x1023atoms 
  8.0x1028 atoms  N


mol
 63.5g 

Measuring Activation Energy
-Q v
Nv

= exp 
 kT
N



By measuring NV versus T,
can get value of QV.
Nv
N
Nv
ln
N
slope
- Q v /k
T
defect concentration
1/T
One method of determining NV is to make very precise measurements
of lattice constant and density.
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).
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
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 pair
i.e. move a cation from a lattice site to a nearby interstitial site
• Schottky defect: cation vacancy – anion vacancy pair
i.e. remove both an anion and a nearby cation at the same time
7
Concentration of Frenkel and Schottky defects
 Qf r 
Nf r  N exp  
 2kT 
 Qs 
Ns  N exp  
 2kT 
• 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.
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).
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)
Second phase particle
-- different composition
-- usually a different structure
Definition: alloy = mixture
10
Impurities in ionic compounds
• Electroneutrality must be maintained
• Example: Ca2+ substituting for 2Na+ in NaCl
Na+
Cl-
cation
vacancy
Ca 2+
Na +
Na +
without impurity
Ca 2+ impurity
• Example: O2- substituting for 2ClO2-
without impurity
Cl Cl O2- impurity
Ca 2+
with impurity
anion vacancy
with impurity
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:
conduction
electron
• 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.
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
Specification of alloy compositions
• Three common methods:
1. Molar concentration (moles per volume) or
mass concentration (mass per volume)
2. Mass fraction or weight percent
3. Atom fraction of atom percent
(mole fraction and mole percent for compounds)
• Need to be able to convert between these.
• Can’t remember all of the equations, and they are a pain to derive.
• So it’s better to pick a basis and then rely on the units.
• Best shown by examples.
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/mol
Atomic 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.
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 Pb
Mass 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 % Pb
Weight % of Sn = (11,158.74 g Sn)/(12,401.94 g alloy)(100%) = 90.0 % Sn
Conversion to molar concentration
• Need the density of the alloy.
• If this isn’t 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 doesn’t 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 g·cm−3
Density of white Sn at room T is 7.365 g·cm−3
• Volume of Pb: (10 g Pb)(cm3 Pb /11.34 g Pb) = 0.8818 cm3 Pb
Volume of Sn: (90 g Sn)(cm3 Sn/7.365 g Sn) = 12.22 cm3 Sn
Total alloy volume if alloy forms ideal solution: 13.102 cm3
• Mass concentration of Pb: (10 g)/(13.102 cm3) = 0.763 g Pb/cm3 alloy
Mass concentration of Sn: (90 g)/(13.102 cm3) = 6.869 g Sn/cm3 alloy
Density 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 don’t know the actual density and you don’t want to measure it.