Question 1

Which of the following defects is observed in steel?

Accepted Answer

Steel is an alloy composed primarily of iron and a small amount of carbon. In the crystal lattice of iron, the much smaller carbon atoms occupy the interstitial voids between the iron atoms. When foreign atoms occupy the interstitial sites of a host lattice, the defect is known as an interstitial impurity defect. Therefore, steel exhibits an interstitial impurity defect. Answer: Interstitial impurity defect

Question 2

A metal has a fcc structure if the edge length of unit cell is 200 pm, calculate the volume occupied by particles in a unit cell.

Accepted Answer

Edge length of the unit cell, a = 200 pm = 2 10^ -8 cm Volume of the unit cell, V = a^3 = (2 10^ -8 )^3 = 8 10^ -24 cm ^3 For a face-centered cubic (fcc) structure, the packing efficiency is 74\%. Volume occupied by particles in the unit cell = 0.74 V = 0.74 8 10^ -24 cm ^3 = 5.92 10^ -24 cm ^3 Answer: 5.92 10^ -24 \ cm ^3

Question 3

A metal crystallizes in a bcc unit cell having unit cell volume 2.5 10^ -23 \ cm ^3. Find the number of unit cells in 18 g metal if the density of metal is 7.2\ g cm ^ -3 .

Accepted Answer

Total volume of the metal is given by the ratio of mass to density. V_ total = Mass Density = 18 7.2 = 2.5\ cm ^3 The number of unit cells is the total volume divided by the volume of one unit cell. Number of unit cells = V_ total V_ unit cell = 2.5 2.5 10^ -23 = 1.0 10^ 23 Answer: 1.0 10^ 23

Question 4

Calculate the density of an element having molar mass 225\ g mol ^ -1 forming bcc structure [a^3 N_A = 75\ cm ^3 mol ^ -1 ]

Accepted Answer

For a body-centered cubic (bcc) structure, the number of atoms per unit cell is Z = 2. The formula for the density of a unit cell is given by: d = Z M a^3 N_A Substituting the given values Z = 2, M = 225\ g mol ^ -1 , and a^3 N_A = 75\ cm ^3 mol ^ -1 : d = 2 225 75 d = 450 75 = 6.0\ g cm ^ -3 Answer: 6.0 g cm^ -3

Question 5

What is the coordination number of particle in simple cubic structure ?

Accepted Answer

In a simple cubic structure, each particle is located at the corners of the cubic unit cell. The nearest neighbours of any particle are the particles situated at the adjacent corners along the edges of the unit cells. For any given particle, there are 2 nearest neighbours along the x-axis, 2 along the y-axis, and 2 along the z-axis. Total number of nearest neighbours = 2 + 2 + 2 = 6. Therefore, the coordination number of a particle in a simple cubic structure is 6. Answer: 6

Question 6

Calculate the radius of atom of metal forming fcc unit cell having edge length 360 pm

Accepted Answer

For a face-centered cubic (fcc) unit cell, the atoms touch each other along the face diagonal. The relationship between the edge length a and the atomic radius r is given by: 2 a = 4r r = a 2 2 Given a = 360 pm: r = 360 2 2 = 90 2 Using 2 = 1.414: r = 90 1.414 = 127.26 pm Answer: 127.26 pm

Question 7

In ionic solid, anions are arranged in hcp array and cations occupy 1 2 tetrahedral voids. What is the formula of ionic compound ? [Consider A = Cation ; B = anion]

Accepted Answer

Let the number of anions (B) forming the hcp array be N. The total number of tetrahedral voids generated in the lattice is 2N. Since cations (A) occupy 1 2 of the tetrahedral voids, the number of cations (A) is 1 2 2N = N. The ratio of the number of cations (A) to the number of anions (B) is N : N = 1 : 1. Therefore, the formula of the ionic compound is AB . Answer: AB

Question 8

Which of the following statements is true about schottky defect ?

Accepted Answer

Schottky defect is a type of point defect in a crystal lattice where an equal number of cations and anions are missing from their normal lattice sites. Since ions are missing from the crystal, the overall mass decreases while the volume remains unchanged, leading to a decrease in the density of the substance. Electrical neutrality is maintained because the number of missing positive and negative charges is equal. This defect is typically found in ionic compounds where the sizes of the cation and

Question 9

Calculate the total volume of simple cubic unit cell in cm ^3 if void volume is 2.0 10^ -23 \, cm ^3

Accepted Answer

For a simple cubic unit cell, the relation between edge length a and radius of the atom r is a = 2r. The packing fraction is given by Volume of atoms Total volume = 4 3 r^3 a^3 = 4 3 r^3 8r^3 = 6 0.524. The fraction of void volume in the unit cell is 1 - 0.524 = 0.476. Given the void volume is 2.0 10^ -23 \, cm ^3. Total volume of the unit cell V = Void volume Fraction of void volume = 2.0 10^ -23 0.476 . V 4.2 10^ -23 \, cm ^3. Answer: 4.2 10^ -23

Question 10

What is the number of tetrahedral voids present in 0.5 mole of compound forming hcp structure?

Accepted Answer

Number of atoms in 0.5 mole of the compound = 0.5 N_A = 0.5 6.022 10^ 23 = 3.011 10^ 23 . In a close-packed structure, the number of tetrahedral voids is twice the number of atoms forming the lattice. Number of tetrahedral voids = 2 3.011 10^ 23 = 6.022 10^ 23 . Answer: 6.022 10^ 23

Question 11

Gold crystallizes as fcc unit cell, the edge length of unit cell is 408 pm. What is the radius of gold atom?

Accepted Answer

For a face-centered cubic (fcc) unit cell, the relationship between the edge length a and the radius of the atom r is given by: 4r = 2 a r = a 2 2 Given a = 408 pm: r = 408 2 2 = 204 2 = 102 2 pm Substituting 2 = 1.414: r = 102 1.414 = 144.228 pm 144.2 pm Answer: 144.2 pm

Question 12

A metal forms fcc structure. Calculate the volume of fcc unit cell in cm ^3 if void volume is 1.66 10^ -23 \ cm ^3.

Accepted Answer

For an fcc (face-centered cubic) structure, the packing efficiency is 74\%. The percentage of void volume in the fcc unit cell is 100\% - 74\% = 26\%. Let the total volume of the unit cell be V. Void volume = 0.26 V 0.26 V = 1.66 10^ -23 \ cm ^3 V = 1.66 10^ -23 0.26 \ cm ^3 V = 6.3846 10^ -23 \ cm ^3 6.385 10^ -23 \ cm ^3 Answer: 6.385 10^ -23 \ cm ^3

Question 13

What is the total number of adjacent unit cells shared by each face particle of fcc unit cell ?

Accepted Answer

In a face-centered cubic (fcc) unit cell, particles are located at the corners and at the center of each face. A particle present at the center of a face lies exactly on the boundary between two adjacent unit cells. Therefore, each face-centered particle is shared equally by exactly 2 unit cells. Answer: 2

Question 14

Calculate the number of unit cell in 0.5 g metal if the product of volume and density of unit cell is 6.65 10^ -23 g

Accepted Answer

Mass of one unit cell is given by the product of its volume and density. Mass of one unit cell = 6.65 10^ -23 g Total mass of the metal = 0.5 g Number of unit cells = Total mass Mass of one unit cell Number of unit cells = 0.5 6.65 10^ -23 Number of unit cells = 7.518 10^ 21 Answer: 7.518 10^ 21

Question 15

What is the total number of Bravais lattices in orthorhombic crystal system?

Accepted Answer

The orthorhombic crystal system has 4 Bravais lattices. These are: 1. Primitive (P) 2. Body-centered (I) 3. Face-centered (F) 4. Base-centered or End-centered (C) Therefore, the total number of Bravais lattices in the orthorhombic crystal system is 4. Answer: 4

Question 16

Calculate the void volume in fcc unit cell if total volume of a unit cell is 6.4 10^ -23 \ cm ^3 ?

Accepted Answer

The packing efficiency of a face-centered cubic (fcc) unit cell is 74\%. The percentage of void space in the fcc unit cell is 100\% - 74\% = 26\%. Given the total volume of the unit cell is V = 6.4 10^ -23 \ cm ^3. Void volume = 0.26 V Void volume = 0.26 6.4 10^ -23 \ cm ^3 = 1.664 10^ -23 \ cm ^3 Answer: 1.664 10^ -23 \ cm ^3

Question 17

Calculate the number of atoms present in 1\ g metal that forms simple unit cell structure if product of density and volume of unit cell is 66.5 10^ -24 \ g

Accepted Answer

For a simple cubic unit cell, the number of atoms per unit cell is Z = 1. The mass of one unit cell is equal to the product of its density and volume. Mass of one unit cell = V = 66.5 10^ -24 g . Since there is 1 atom per unit cell, the mass of one atom is 66.5 10^ -24 g . The number of atoms in 1 g of the metal is given by: Number of atoms = Total mass Mass of one atom Number of atoms = 1 66.5 10^ -24 Number of atoms = 0.015037 10^ 24 Number of atoms 1.504 10^ 22

Question 18

Calculate the void volume of bcc unit cell if volume of unit cell is 8 0 10^ -23 cm ^3

Accepted Answer

The packing efficiency of a body-centered cubic (bcc) unit cell is 68\%. The percentage of void space in the unit cell is 100\% - 68\% = 32\%. Given the volume of the unit cell is 8 0 10^ -23 cm ^3. Void volume = 32 100 8 0 10^ -23 cm ^3 Void volume = 2 56 10^ -23 cm ^3 Answer: 2 56 10^ -23 cm ^3

Question 19

Which of the following dopants is used in germanium to form n-type semiconductor ?

Accepted Answer

Germanium (Ge) is a Group 14 element. To form an n-type semiconductor, it must be doped with a pentavalent impurity, which is an element from Group 15 of the periodic table. Among the given options, Boron (B), Indium (In), and Gallium (Ga) are Group 13 elements (trivalent impurities) and will form p-type semiconductors when doped with Germanium. Phosphorus (P) is a Group 15 element (pentavalent impurity). When Germanium is doped with Phosphorus, it provides an extra electron, resulting in an n-t

Question 20

Calculate the number of atoms in 1 g metal that forms bcc crystal structure [ a^3 = 6 6 10^ -22 g ]

Accepted Answer

The mass of one unit cell is given by the product of its density and volume: Mass of unit cell = a^3 = 6 6 10^ -22 g For a body-centered cubic (bcc) crystal structure, the number of atoms per unit cell is Z = 2. Mass of one atom = Mass of unit cell Z = 6 6 10^ -22 2 = 3 3 10^ -22 g The number of atoms in 1 g of the metal is: Number of atoms = 1 g Mass of one atom = 1 3 3 10^ -22 = 3 03 10^ 21 Answer: 3 03 10^ 21

Solid State — Previous Year Questions & Solutions