Crystal Structures MCQ Quiz in मल्याळम - Objective Question with Answer for Crystal Structures - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

The void formed at the center is octahedral in shape. The largest radius that can be fit in it-

\(R = \frac{\alpha }{2} - r\)

For BCC crystal structure \(\alpha = \frac{{4\;r}}{{\sqrt 3 }}\)

r is the radius of the corner atoms, α is the cube edge

\(R = \frac{\alpha }{2} - \frac{{\sqrt 3 }}{4}\alpha = 0.067\)

∴ D = 2 × R = 2 × 0.067

∴ D = 0.134 α 

Important Points

• Temperature can influence the magnetic characteristics of materials.
• The rise in the temperature of solid increases the magnitude of the thermal vibrations of atoms.
• The atomic magnetic moments are free to rotate; hence, with rising temperature, the increased thermal motion of the atoms tends to randomize the directions of any moments that may be aligned.

Explanation:

The atomic packing factor is defined as the ratio of the volume occupied by the average number of atoms in a unit cell to the volume of the unit cell.

Mathematically, Atomic Packing Factor (APF):

APF \( = \frac{{{N_{atoms}} ~\times ~{V_{atoms}}}}{{{V_{unit\;cell}}}}\) ...(1)

Characteristics of various types of structures are shown in the table below:

For Cubic Unit Cell

Nav = \(N_c\over 8\) + \(N_f\over 2\) + \(N_i \over 1\)

Nav = Average no of atoms in unit cell, Nc = No of corner atoms, Ni = No of interior atoms, Nf = No of face centre atoms

Calculation:

No of atoms in f.c.c unit cell = 4

\(APF = \frac{{{N_{atoms}}{V_{atom}}}}{{{V_{crystal}}}} = \frac{{4\left( {\frac{4}{3}} \right)\pi {r^3}}}{{{{\left( {a } \right)}^3}}}= \frac{{4\left( {\frac{4}{3}} \right)\pi {r^3}}}{{{{\left( {{{2√2r}}{}} \right)}^3}}}\)

for FCC a = 2√2 r where a is side of the cube and r is atomic radius.

APF = 0.74

For BCC:

Nav = \(8\over 8\) + 0 + \(1\over 1\) = 2

√3a = 4r

Put all values in equation 1

(APF)BCC = 0.68

Explanation:

Imperfections in crystal structure involving either a single atom or a few numbers of atoms are called as point imperfections/defects. 

Types of point defects or imperfections are:

• Vacancy defects
• Interstitial Defects
• Substitutional defects
• Schottky defects
• Frenkel defects

The difference between Schottky and Frenkel is explained with the help of the following diagram:

Additional Information

Interstitial impurity:

• When some constituent particles (atoms or molecules) occupy an interstitial site, the crystal is said to have an interstitial defect
• This defect increases the density of the substance.
• Ionic solids must always maintain electrical neutrality. 

Concept:

Crystal structure of Material is classified as follows:

BCC: BCC stands for Body-Centered Cubic. In one unit cell, there is one atom at center, 1 atom at each corner. The crystal structure is used for Brittle materials only. 

E.g. V, Mo, Ta, W, Ferrite or α-iron, δ-ferrite or δ-iron

FCC: FCC stands for Face Centered Cubic. In one unit cell, there is one atom at center, 1 atom at each corner, and 1 atom on each face. The crystal structure is used for Ductile materials only. 

E.g. Ni, Cu, Ag, Pt, Au, Pb, Al, Austenite or γ-iron

The iron-carbon diagram shows the different crystal structures for different phases.

Explanation:

If the atoms or atom groups in the solid are represented by points and the points are connected, the resulting lattice will consist of an orderly stacking of blocks or unit cells.

• The orthorhombic unit cell is distinguished by three lines called axes of twofold symmetry about which the cell can be rotated by 180° without changing its appearance.
• This characteristic requires that the angles between any two edges of the unit cell be right angles but the edges may be any length.

There are 7 types of crystal systems:

Concept:

Body-centered cubic (BCC) structure):

In ΔABC:

(BC)² = (AB)² + (AC)²

\({\left( {4r} \right)^2} = {\left( {\sqrt 2 a} \right)^2} + {\left( a \right)^2}\)

(4r)² = 3a²

\(4r = \sqrt 3 \;a\)

\(\therefore r = \frac{{\sqrt 3 }}{4}a\)

Where, lattice parameters of BCC are:

r = atomic radius

a = edge length of the unit cell

Calculation:

Given that, r = 1.36 A°

\(\because r = \frac{{\sqrt 3 }}{4}a\)

\(\therefore a = \frac{{4r}}{{\sqrt 3 }} = \frac{{4 \times 1.36}}{{1.732}} = 3.14\;A^\circ \)

Important points:

Formula Used:

Density = \(\frac{n× W}{Av.No × a^3}\)

Here, W is the atomic weight

Av. No is avogadro's number = 6.023 × 10²³

n is Effective no. of lattice points

Application:

We have,

n = 2 (For BCC)

W = 23 (for Na)

a = 4.24 Å = 4.24 × 10^-8 cm

Hence,

Density = \(\frac{2\times 23}{6.023\times 10^{-23}\times (4.24\times 10^{-8} )^3}\) = 1.002 g cm–3 

Explanation:

An FCC structure contains atom at:

• All the eight-corner position
• All the centre position of six sides.

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Schottky defect:

• This kind of vacancy defect is found in Ionic Solids. But in ionic compounds, we need to balance the electrical neutrality of the compound so an equal number of anions and cations will be missing from the compound.
• It reduces the density of the substance.
• In this, the size of cations and anions are almost the same.

Frenkel defect:

• In ionic solids generally, the smaller ion (cation) moves out of its place and occupies an intermolecular space.
• In this case, a vacancy defect is created in its original position and the interstitial defect is experienced at its new position.

• The density of a substance remains unchanged.
• It happens when there is a huge difference in the size of anions and cations