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

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നേടുക Crystal Structures ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Crystal Structures MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Crystal Structures MCQ Objective Questions

Top Crystal Structures MCQ Objective Questions

Crystal Structures Question 1:

What is the diameter of the largest sphere in terms of lattice parameter α, which will fit the void at the center of the cube edge of a BCC crystal?

  1. 0.134 α
  2. 0.25 α
  3. 0.433 α
  4. 0.5 α

Answer (Detailed Solution Below)

Option 1 : 0.134 α

Crystal Structures Question 1 Detailed Solution

Explanation:

F1 M.J Madhu 17.04.20 D4

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 

Crystal Structures Question 2:

Which of the following determines the frequency of atomic vibrations in the crystal

  1. Crystal element
  2. Crystal temperature
  3. Stiffness of the bond it makes with the neighbour
  4. Number of atoms per unit in crystal

Answer (Detailed Solution Below)

Option 3 : Stiffness of the bond it makes with the neighbour

Crystal Structures Question 2 Detailed Solution

An atom in a crystal vibrates at a frequency determined by the stiffness of the bonds with neighbour atoms. The crystal frequency is independent of crystal heat current and crystal temperature. Important Points The influence of temperature:
  • 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.

Crystal Structures Question 3:

What is the atomic packing factor for BCC and FCC, respectively?

  1. 0.52, 0.78
  2. 0.74, 0.68
  3. 0.52, 0.74
  4. 0.68, 0.74

Answer (Detailed Solution Below)

Option 4 : 0.68, 0.74

Crystal Structures Question 3 Detailed Solution

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:

Characteristics

BCC

FCC

HCP

a to r relation

\(a = \frac{{4r}}{{√ 3 }}\)

\(a = 2√ 2 r\)

\(a = 2r\)

The average number of atoms

2

4

6

Co-ordination number

8

12

12

APF

0.68

0.74

0.74

Examples

Na, K, V, Mo, Ta, W

Ca, Ni, Cu, Ag, Pt, Au, Pb, Al

Be, Mg, Zn, Cd, Te

 

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

Crystal Structures Question 4:

When a pair of one cation and one anion are absent from ionic crystal, the resulting defect in solid is called:

  1. Substitutional impurity
  2. Interstitial impurity
  3. Frenkel’s defect
  4. Schottky’s defect

Answer (Detailed Solution Below)

Option 4 : Schottky’s defect

Crystal Structures Question 4 Detailed Solution

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:

F2 S.B Madhu 27.07.20 D6

 

Schottky Defect

Frenkel Defect

Forms when oppositely charged ions leave their lattice sites, creating vacancies

Smaller ion (usually the cation) is displaced from its lattice position to an interstitial site

These vacancies are formed in stoichiometric units, to maintain an overall neutral charge in the ion solid

Creates a vacancy defect at its original site and an interstitial defect at its new location

The density of the solid crystal is less than normal

Does not change the density of the solid

Occurs only when there is a small difference in size between cations and anions

Shown in ionic solids with the large size difference between the anion and cation

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. 

Crystal Structures Question 5:

The crystal structure of α-iron is -

  1. Close packed hexagonal
  2. Body centered cubic
  3. Simple cubic
  4. Face centered cubic

Answer (Detailed Solution Below)

Option 2 : Body centered cubic

Crystal Structures Question 5 Detailed Solution

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. 

b

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. 

f

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.

d

Crystal Structures Question 6:

The unit cell of a certain type of crystal is defined by three vectors a, b and c and the angle between a and b is α, b and c is β and a and c is γ. Now if α = β = γ = 90° and a = b ≠ c. The crystal structure is

  1. Triclinic
  2. Tetragonal
  3. Orthorhombic
  4. Monoclinic

Answer (Detailed Solution Below)

Option 2 : Tetragonal

Crystal Structures Question 6 Detailed Solution

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.

F1 Ashik Madhu 01.02.21 D1

There are 7 types of crystal systems:

Crystal System

Angles between Axis

Unit Cell Dimensions

Cubic

α = β = γ = 90°

a = b = c

Tetragonal

α = β = γ = 90°

a = b ≠ c

Orthorhombic

α = β = γ = 90°

a ≠ b ≠ c

Rhombohedral

α = β = γ ≠ 90°

a = b = c

Hexagonal

α = β = 90°, γ = 120°

a = b ≠ c

Monoclinic

α = γ = 90°, β ≠ 90°

a ≠ b ≠ c

Triclinic

α ≠ β ≠ γ

a ≠ b ≠ c

Crystal Structures Question 7:

Molybdenum has a Body-Cantered Cubic (BCC) structure with an atomic radius of 1.36 Ȧ .Then the lattice parameter for BCC molybdenum is

  1. 2.77 Ȧ 
  2. 3.14 Ȧ 
  3. 5.12 Ȧ 
  4. 6.28 Ȧ 

Answer (Detailed Solution Below)

Option 2 : 3.14 Ȧ 

Crystal Structures Question 7 Detailed Solution

Concept:

Body-centered cubic (BCC) structure):

F1 S.B 12.8.20 Pallavi D7

 

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:

Structure

Atomic Radius

Simple cubic

\(r = \frac{a}{2}\)

BCC

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

FCC

\(r = \frac{{\sqrt 2 a}}{4}\)

Diamond cubic

\(r = \frac{{\sqrt 3 \;a}}{8}\)

Crystal Structures Question 8:

At room temperature, sodium crystallizes in a body-centered cubic lattice with edge length a = 4.24 Å. The theoretical density of sodium (At. wt. of Na = 23) is

  1. 1.002 g cm–3 
  2. 2.002 g cm–3 
  3. 3.002 g cm–3 
  4. 4.002 g cm–3 

Answer (Detailed Solution Below)

Option 1 : 1.002 g cm–3 

Crystal Structures Question 8 Detailed Solution

Formula Used:

Diagram

GATE ME 2009 Images-Q15

GATE ME 2009 Images-Q15.1

GATE ME 2009 Images-Q15.2

Effective no. of

lattice points

(n)

\(\Rightarrow \frac{1}{8} × 8 = 1\)

\(\frac{1}{8} × 8 + 1 = 2\)

\(\frac{1}{8} × 8 + \frac{1}{2} × 6 = 4\)

 

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 

Crystal Structures Question 9:

A unit cell of Face centred cubical crystal structure has _______ atoms per unit cell.

  1. 4
  2. 2
  3. 8
  4. 12

Answer (Detailed Solution Below)

Option 1 : 4

Crystal Structures Question 9 Detailed Solution

Explanation:

An FCC structure contains atom at:

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

Characteristics

BCC

FCC

HCP

a to r relation

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

\(a = 2\sqrt 2 r\)

\(a = 2r\)

The average number of atoms

2

4

6

Co-ordination number

8

12

12

APF

0.68

0.74

0.74

Examples

Na, K, V, Mo, Ta, W

Ca, Ni, Cu, Ag, Pt, Au, Pb, Al

Be, Mg, Zn, Cd, Te

Crystal Structures Question 10:

In a crystal lattice, the vacancies created in the absence of certain atoms are known as _______. 

  1. Hertz defects
  2. Schottky defects
  3. Pauli defects
  4. Crystal defects

Answer (Detailed Solution Below)

Option 2 : Schottky defects

Crystal Structures Question 10 Detailed Solution

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
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