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Vacancy Defects or imperfections can be divided into four categories in crystalline solids: point defect, line defect, surface defect, and volume defect. A point defect is a zero-dimensional (0-D) defect since a point is a unit-less dimensionless quantity by mathematical definition. When one or more crystalline solid atoms leave their original lattice site and/or foreign atoms occupy the crystal's interstitial position/lattice site, a point defect occurs. There are numerous types of point flaws, and one of them is vacancy defects.
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Key Terms: Vacancy Defect, Crystalline Solids, Crystal Lattice, Lattice Site, Crystal Structure, Crystallization, Point Defect, Molecules, Ions
Concept of Crystalline Solids
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Crystalline solids are those in which atoms, molecules, or ions are arranged in a regular pattern that extends in all directions (known as the crystal lattice). A crystal lattice is a three-dimensional structure of atoms, molecules, or ions that is highly organized.
Amorphous solids are referred to those solids in which the organization of atoms, molecules, or ions is very irregular. Polycrystalline solids exist in the middle, where a variety of crystal lattices can be found within the same solid. A crystalline solid is something like a diamond or salt, whereas an amorphous solid is something like wax or glass.
Crystalline and Amorphous Solids
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| Related Articles | ||
|---|---|---|
| Impurity Defects | Interstitial Defect | Frenkel Defect |
| Imperfections in Solids | Schottky Defect | Metal Excess Defect |
Meaning of Vacancy Defect
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When an atom is displaced or removed from its original lattice location, a vacancy is created. As a result, vacancy results in an empty lattice site. Vacancy, like other point defects, is a zero-dimensional defect. A vacancy defect causes stress between nearby atoms. The vacancy defect causes the density of the crystalline solid to decrease due to the drop in the number of atoms. However, the solid's hardness may grow. Vacancy defect happens because of the loss of one atom from its lattice site.
Vacancy Defect
Observation of Vacancy Defects
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Vacancy defects can occur in any and all crystalline solid. It is considered to be inherent. Any substantial part whose temperature is above absolute zero temperature (0K), can contain vacancies. With an increase in temperature, the number of vacancy defects increases exponentially.
Causes of Vacancy Defects
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Vacancy defect which is a point defect might occur due to various reasons, as enlisted below:
- For not permitting directional solidification, usually during casting.
- Due to an upsurge in temperature of the solid for various processes like heat treatment, coating, etc.
- Number of vacancies in a definite number of solid increases exponentially as indicated by the expression provided below.
- Due to irradiation or sputtering effect.
- Due to the occurrence of residual tensile stress within the solid.
Consequences of Vacancies in Solids
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- Although it depends on the material and its crystal structure when considered in general, vacancies can subside the bulk modulus and can increase the young’s modulus.
- Large vacancy concentration can reduce the ductility of any crystalline solid; however, it can also facilitate increasing the hardness.
- It can amend the thermal and electrical resistivity of the solid.
- Common physical properties, like melting point, colour, etc. can also fluctuate due to the presence of vacancies.
Calculation of Vacancies in Solids
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Vacancy defects are temperature-responsive. The number of vacancies present within a particular volume of solid at a particular temperature can be calculated and can be mathematically expressed by the following expression:
\(N_{v}=N \times e^{\left(\frac{-Q_{v}}{R T}\right)}\)
- Nv: The total number of vacancies per unit volume of solid at a particular temperature (vacancies/m3).
- N: Total number of lattice sites per unit volume of solid (lattice sites/m3).
- Qv: Energy required to produce a single vacancy in that solid (J/mole).
- R: Gas constant = 8.314 J/mole-K
- T: Absolute temperature of Solid (K)
Things to Remember
- A particle is missing from its usual location in the crystal lattice during solid crystallization.
- The vacancy in the lattice structure is owed to the missing particle.
- As a result of the missing particles, some of the lattice sites are empty, as indicated in the picture above.
- In such a crystal, the vacancy defect is present.
- When a substance is heated, the vacancy defect might appear.
- The bulk of the substance lowers due to the absence of particles.
- The volume, however, stays unaltered. As a result, the substance's density changes.
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Sample Questions
Ques. Examine the given defective crystal:
X+ Y- X+ Y- X+ Y- X+
X+
Y- O Y- X+ Y- X+ Y-
X+ Y- X+ Y- X+ Y- X+Y- X+ Y- X+ Y- X+ Y-
Answer the following questions:
(a) Is the above defect stoichiometric or non-stoichiometric?
(b) Write down the term used for this type of defect.
(c) Explain why silver halides show this type of defect. (3 Marks)
Ans. (a) It is known to be a stoichiometric defect.
(b) It is known to be a Frenkel defect. In this, an ion is missing from its correct lattice site and occupies the interstitial site so the stoichiometry of the compound is maintained.
(c) Silver halides generally show this type of defect because the size of Ag+ is almost similar to the size of voids, So, the Ag+ ions can move to the interstitial sites of the atom.
Ques. Which point defect in its crystal units leads to an increase in the density of a solid? (2 Marks)
Ans. A solid's density is increased by a metal surplus defect. The presence of additional cations in the interstitial sites is the cause. Foreign atoms or ions occupy interstitial locations as a result of the Interstitial defect.
Ques. Give a reason why Schottky defects lower the density of related solids? (2 Marks)
Ans. Schottky defects are formulated due to missing an identical number of cations and anions from lattice as a consequence of which the density of the lattice solid decreases. That is the main reason why Schottky defects lower the density of the related solids.
Ques. Give a reason for why the conductivity of silicon increases on doping it with phosphorus? (2 Marks)
Ans. The negatively charged additional electrons of doped pentavalent phosphorus boost the conductivity of silicon. As a result, doping silicon with phosphorus boosts its conductivity.
Ques. Define some important points about Schottky's defect. (2 Marks)
Ans. If we consider an ionic crystal of type, supposedly A and B, an equal number of cations and anions are missing from their lattice sites so that the electrical neutrality is maintained, it is called the Schottky defect.
Ques. What is F-centre? (2 Marks)
Ans. The centres which are created by the trapping of electrons in anionic vacancies, and which are responsible for imparting colour to the crystals are called F-centres. (F = Fabre).
Ques. Calculate the density of silver that forms ccp lattice and the length of the edge of its unit cell is 408.6 pm. Here, the atomic mass is 107.9 u. (3 Marks)
Ans. We are given that the lattice is ccp which means that the amount of silver atoms per unit cell = z = 4
Molar mass of silver = 107.9 g mol-1 = 107.9 × 10-3 kg mol-1
Edge length of unit cell = a = 408.6 pm = 408.6 × 10-12 m
Density will come out to be (d) = z.m/a3.NA
= 4 × (107.9 × 10-3 kg mol-1) / (408.6 × 10-12 m)3 (6.022 × 1023 mol-1)
= 10.5 × 103 kg m-3
= 10.5 g cm-3
Ques. The cell edge of a body-centred cubic structure of an element is 288pm and 7.2g/cm3 is the density of the element. Calculate the number of atoms in 208g of the element. (3 Marks)
Ans. Volume of the unit cell = (288pm)3
= (288 × 10-12 m)3
= (288 × 10-10 cm)3
= 2.39 × 10-23 cm3
Volume of 208g of element = mass/density
= 208g/7.2g cm-3
= 28.88 cm3
Number of unit cells in the volume = 28.88 cm3/ (2.39 × 10-23 cm3/unit cell)
= 12.08 × 1023 unit cells
Each bcc cubic unit cell is supposed to contain 2 atoms
The total number of atoms in 208g = 2 (atom/unit cell) × 12.08 × 1023 unit cells
= 24.16 × 1023 atoms
Ques. What is the formula of a compound which has two elements X and Y where the atoms of element X act as cations in order to fill up all the octahedral voids and atoms of element y which act as anions make ccp? (3 Marks)
Ans. The ccp lattice is created by element Y. The number of atoms in element Y is equal to the number of created octahedral voids. The atoms of element X fill all of the octahedral gaps, and their number is comparable to that of element Y. As a result, the equal number of atoms in elements X and Y can be written as a 1:1 ratio, giving the component the formula XY.
Ques. There are two elements in a compound, A and B. Atoms of element A hold 2/3 of tetrahedral voids and the atoms of element B make up the hcp lattice. What is the formula? (3 Marks)
Ans. The number of tetrahedral voids produced comes out to be analogous to the double of the number of atoms of element B. Element A is holding only 2/3rd of the atoms of element A.
Thus, the ratio of the number of atoms of A and B comes out to be = 2 × (2/3): 1 or 4:3
Therefore, the formula of the compound will be = A4B3
Ques. List down which among the following is true if the lattice constant of the semiconductor is reduced: conduction bandwidth, bandgap, and valence bandwidth. (3 Marks)
Ans. Just because the lattice parameter of the semiconductor falls, more and more energy is required to remove one atom when the electrons are closely packed to the atom. This characteristic allows the width of the energy bands in both the conduction bands and valence bands to be reduced. The energy gap between the valence band and conduction bands widens as a result of this. As a result, conduction and valence bandwidths drop while the bandgap increases.
Ques. Describe the nature of a semiconductor at 0 K? (3 Marks)
Ans. It is found to be neutral. The conductivity of a semiconductor is observed to be either increasing or decreasing with the change in temperature. Thus, the semiconductor at 0 K which is absolute zero acts as an insulator as conductivity becomes 0.
Ques. Calculate the number of holes in an N-type semiconductor with 1015 cm-3 donors. (3 Marks)
Ans. Each electron is produced with its particular ionized donor. Thus, n = 1015 cm-3
P = n2i/n = 1020 cm-3/1015 cm-3 = 105 cm-3
With the upsurge in temperature by 60°C. at 1015 cm-3, n remains unvarying while p rises by about a factor of 2300.
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