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CHXII01:THE SOLID STATE

318684 The edge length of a cubic unit cell of a metal having atomic weight 75 u is 5 $A^{\circ}$ . If the density is $2 g {cm}^{- 3}$ , then the radius of the metal atom is ____ pm. $[N_{A} = 6.0 \times 10^{23} {mol}^{- 1}, \sqrt{3} = 1.7]$

1 602

2 450

3 212.5

4 425

Explanation:

$d = \frac{Z \times M}{N_{A} \times a^{3}}$
$∴ Z = \frac{d \times N_{A} \times a^{3}}{M}$
$= \frac{2 \times 6.0 \times 10^{23} \times {(5 \times 10^{- 8})}^{3}}{75}$
$= 20 \times 10^{23} \times 10^{- 24}$
$= 2$
$∴$ Unit cell is bcc lattice.
For bcc lattice, $r = \frac{\sqrt{3}}{4} \times a$
$∴ r = \frac{\sqrt{3}}{4} \times 5$
$= \frac{1.7}{4} \times 5 = 2.125$ $A^{\circ}$
$= 212 .5 pm$

CHXII01:THE SOLID STATE

318685 Statement A :
Density of the unit cell is same as the density of the substance.
Statement B :
Density of the unit cell has no relation with the volume of unit cell.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Density of the unit cell is related with its volume as
Volume of the unit cell $= \frac{mass of unit cell}{density of unit cell}$
Further the density of unit cell also represent the density of the substance.

CHXII01:THE SOLID STATE

318686 Copper crystallises in face centred cubic lattice with a unit cell length of $361 pm$ . What is the radius of copper atom in pm?

1 157

2 181

3 108

4 128

Explanation:

For fcc, radius of atom
$r = \frac{a}{2 \sqrt{2}}$
$a = 361 pm (given)$
$r = \frac{361}{2 \sqrt{2}} = 127.6 pm \approx 128 pm$

CHXII01:THE SOLID STATE

318687 Number of unit cells in 16 g of Y (atomic mass $= 40 u)$ which crystallises in bcc structure is $x \times N_{A}$ . The value of $x$ is ____ .
$[N_{A} =$ Avogadro number $]$

1 0.02

2 0.5

3 0.2

4 2

Explanation:

No. of atoms in 16 g of $Y = \frac{16}{40} \times N_{A}$ $= 0.4 \times N_{A}$
In bcc unit cell, no. of atoms present in one unit cell is 2 .
$∴$ No. of unit cells $= \frac{0.4 \times N_{A}}{2} = 0.2 \times N_{A}$
$∴$ Value of $x = 0.2$

CHXII01:THE SOLID STATE

318684 The edge length of a cubic unit cell of a metal having atomic weight 75 u is 5 $A^{\circ}$ . If the density is $2 g {cm}^{- 3}$ , then the radius of the metal atom is ____ pm. $[N_{A} = 6.0 \times 10^{23} {mol}^{- 1}, \sqrt{3} = 1.7]$

1 602

2 450

3 212.5

4 425

Explanation:

$d = \frac{Z \times M}{N_{A} \times a^{3}}$
$∴ Z = \frac{d \times N_{A} \times a^{3}}{M}$
$= \frac{2 \times 6.0 \times 10^{23} \times {(5 \times 10^{- 8})}^{3}}{75}$
$= 20 \times 10^{23} \times 10^{- 24}$
$= 2$
$∴$ Unit cell is bcc lattice.
For bcc lattice, $r = \frac{\sqrt{3}}{4} \times a$
$∴ r = \frac{\sqrt{3}}{4} \times 5$
$= \frac{1.7}{4} \times 5 = 2.125$ $A^{\circ}$
$= 212 .5 pm$

CHXII01:THE SOLID STATE

318685 Statement A :
Density of the unit cell is same as the density of the substance.
Statement B :
Density of the unit cell has no relation with the volume of unit cell.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Density of the unit cell is related with its volume as
Volume of the unit cell $= \frac{mass of unit cell}{density of unit cell}$
Further the density of unit cell also represent the density of the substance.

CHXII01:THE SOLID STATE

318686 Copper crystallises in face centred cubic lattice with a unit cell length of $361 pm$ . What is the radius of copper atom in pm?

1 157

2 181

3 108

4 128

Explanation:

For fcc, radius of atom
$r = \frac{a}{2 \sqrt{2}}$
$a = 361 pm (given)$
$r = \frac{361}{2 \sqrt{2}} = 127.6 pm \approx 128 pm$

CHXII01:THE SOLID STATE

318687 Number of unit cells in 16 g of Y (atomic mass $= 40 u)$ which crystallises in bcc structure is $x \times N_{A}$ . The value of $x$ is ____ .
$[N_{A} =$ Avogadro number $]$

1 0.02

2 0.5

3 0.2

4 2

Explanation:

No. of atoms in 16 g of $Y = \frac{16}{40} \times N_{A}$ $= 0.4 \times N_{A}$
In bcc unit cell, no. of atoms present in one unit cell is 2 .
$∴$ No. of unit cells $= \frac{0.4 \times N_{A}}{2} = 0.2 \times N_{A}$
$∴$ Value of $x = 0.2$

CHXII01:THE SOLID STATE

318684 The edge length of a cubic unit cell of a metal having atomic weight 75 u is 5 $A^{\circ}$ . If the density is $2 g {cm}^{- 3}$ , then the radius of the metal atom is ____ pm. $[N_{A} = 6.0 \times 10^{23} {mol}^{- 1}, \sqrt{3} = 1.7]$

1 602

2 450

3 212.5

4 425

Explanation:

$d = \frac{Z \times M}{N_{A} \times a^{3}}$
$∴ Z = \frac{d \times N_{A} \times a^{3}}{M}$
$= \frac{2 \times 6.0 \times 10^{23} \times {(5 \times 10^{- 8})}^{3}}{75}$
$= 20 \times 10^{23} \times 10^{- 24}$
$= 2$
$∴$ Unit cell is bcc lattice.
For bcc lattice, $r = \frac{\sqrt{3}}{4} \times a$
$∴ r = \frac{\sqrt{3}}{4} \times 5$
$= \frac{1.7}{4} \times 5 = 2.125$ $A^{\circ}$
$= 212 .5 pm$

CHXII01:THE SOLID STATE

318685 Statement A :
Density of the unit cell is same as the density of the substance.
Statement B :
Density of the unit cell has no relation with the volume of unit cell.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Density of the unit cell is related with its volume as
Volume of the unit cell $= \frac{mass of unit cell}{density of unit cell}$
Further the density of unit cell also represent the density of the substance.

CHXII01:THE SOLID STATE

318686 Copper crystallises in face centred cubic lattice with a unit cell length of $361 pm$ . What is the radius of copper atom in pm?

1 157

2 181

3 108

4 128

Explanation:

For fcc, radius of atom
$r = \frac{a}{2 \sqrt{2}}$
$a = 361 pm (given)$
$r = \frac{361}{2 \sqrt{2}} = 127.6 pm \approx 128 pm$

CHXII01:THE SOLID STATE

318687 Number of unit cells in 16 g of Y (atomic mass $= 40 u)$ which crystallises in bcc structure is $x \times N_{A}$ . The value of $x$ is ____ .
$[N_{A} =$ Avogadro number $]$

1 0.02

2 0.5

3 0.2

4 2

Explanation:

No. of atoms in 16 g of $Y = \frac{16}{40} \times N_{A}$ $= 0.4 \times N_{A}$
In bcc unit cell, no. of atoms present in one unit cell is 2 .
$∴$ No. of unit cells $= \frac{0.4 \times N_{A}}{2} = 0.2 \times N_{A}$
$∴$ Value of $x = 0.2$

CHXII01:THE SOLID STATE

318684 The edge length of a cubic unit cell of a metal having atomic weight 75 u is 5 $A^{\circ}$ . If the density is $2 g {cm}^{- 3}$ , then the radius of the metal atom is ____ pm. $[N_{A} = 6.0 \times 10^{23} {mol}^{- 1}, \sqrt{3} = 1.7]$

1 602

2 450

3 212.5

4 425

Explanation:

$d = \frac{Z \times M}{N_{A} \times a^{3}}$
$∴ Z = \frac{d \times N_{A} \times a^{3}}{M}$
$= \frac{2 \times 6.0 \times 10^{23} \times {(5 \times 10^{- 8})}^{3}}{75}$
$= 20 \times 10^{23} \times 10^{- 24}$
$= 2$
$∴$ Unit cell is bcc lattice.
For bcc lattice, $r = \frac{\sqrt{3}}{4} \times a$
$∴ r = \frac{\sqrt{3}}{4} \times 5$
$= \frac{1.7}{4} \times 5 = 2.125$ $A^{\circ}$
$= 212 .5 pm$

CHXII01:THE SOLID STATE

318685 Statement A :
Density of the unit cell is same as the density of the substance.
Statement B :
Density of the unit cell has no relation with the volume of unit cell.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Density of the unit cell is related with its volume as
Volume of the unit cell $= \frac{mass of unit cell}{density of unit cell}$
Further the density of unit cell also represent the density of the substance.

CHXII01:THE SOLID STATE

318686 Copper crystallises in face centred cubic lattice with a unit cell length of $361 pm$ . What is the radius of copper atom in pm?

1 157

2 181

3 108

4 128

Explanation:

For fcc, radius of atom
$r = \frac{a}{2 \sqrt{2}}$
$a = 361 pm (given)$
$r = \frac{361}{2 \sqrt{2}} = 127.6 pm \approx 128 pm$

CHXII01:THE SOLID STATE

318687 Number of unit cells in 16 g of Y (atomic mass $= 40 u)$ which crystallises in bcc structure is $x \times N_{A}$ . The value of $x$ is ____ .
$[N_{A} =$ Avogadro number $]$

1 0.02

2 0.5

3 0.2

4 2

Explanation:

No. of atoms in 16 g of $Y = \frac{16}{40} \times N_{A}$ $= 0.4 \times N_{A}$
In bcc unit cell, no. of atoms present in one unit cell is 2 .
$∴$ No. of unit cells $= \frac{0.4 \times N_{A}}{2} = 0.2 \times N_{A}$
$∴$ Value of $x = 0.2$