Solid State Test

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Solid State Test - 62

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Question 1
1 / -0

Directions For Questions

In a face centred unit cell with all the positions occupied by A atoms, the body centred octahedral holes in it is occupied by an atom B of an appropriate size.

...view full instructions

Calculate the void fraction per unit volume of unit cell for such crystal.

A
0.236

B
0.67

C
0.45

D
0.82

Solution

When the cation radii are greater or equal to 0.414 R, but less than 0.732 R, where R is the radius of an atom, the cations occupy the octahedral sites. Let us approximate it to 0.5
Thus, the radius of atom B is approximately equal to one half of the radius of atom A.
The packing fraction is $$\displaystyle \frac {\text {volume occupied by the atoms}}{\text {total volume of the unit cell}}=\frac {\displaystyle4 \times \frac {4}{3} \pi r^3+\frac {4}{3} \pi (\frac {r}{2})^3}{(\displaystyle\frac {4r}{\sqrt {2}})^3}=0.764$$
The void fraction is $$1-0.764=0.236$$
Question 2
1 / -0

An element (atomic mass $$100\ g/mol$$) having $$BCC$$ structure has unit cell edge $$400\ pm$$. The density of element is:
(No. of atoms in $$BCC, Z = 2$$)

A
$$2.144\ g/cm^{3}$$

B
$$7.289\ g/cm^{3}$$

C
$$5.188\ g/cm^{3}$$

D
$$10.376\ g/cm^{3}$$

Solution

Mass of two atoms $$= \dfrac {100}{6.02\times 10^{23}} \times 2\ g$$

$$= \dfrac {2}{6.02} \times \dfrac {10^{-21}}{10^{3}} kg$$

volume of cell $$= (4\times 10^{-10})^{3} = 64\times 10^{-30}\ m^3$$

$$Density = \dfrac {mass}{volume} = \dfrac {2\times 10^{24}}{6\times 10^{-30} \times 6.02}$$

$$Density = \dfrac {2}{6.02\times 64} \times 10^{6} kg/ m^{3}$$

$$= \dfrac {2\times 10^{6} \times 10^{3}}{6.02\times 64 \times 10^{6}} gm/ cc$$

$$= \dfrac {2}{6.02\times 64} \times 10^{3} gm/ cc = 5.188/ gm/cc$$
Question 3
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The crystal system of a compound with unit cell dimension $$"a = 0.387, b = 0.387$$ and $$c = 0.504\ nm$$ and $$\alpha = \beta = 90^{\circ}$$ and $$\gamma = 120^{\circ}"$$ is :

A
cubic

B
hexagonal

C
orthohombic

D
rhombohedral

Solution

Thus it is hexagonal crystal system as satisfies the criterion
$$a = b\neq c, \alpha = \beta = 90^{\circ}, \gamma = 120^{\circ}$$
These are the characteristics of a hexagonal system.
Question 4
1 / -0

Directions For Questions

In crystalline solids, atoms or molecules are arranged in a regular and long range order fashion in a three dimensional pattern. These have sharp melting point, flat faces, sharp edges, bounded by well defined planes. A large number of unit cells each of which possess a definite geometry bounded by plane faces give rise to the formation of a crystal. A point at the corner of unit cell contributes for $$\dfrac 18$$ of each such point to unit cell. A point along an edge contribute for $$\dfrac 14$$ of each such point to unit cell. A body centred point contributes for $$1$$ each such points to unit cell. Co-ordination number is the number of nearest neighbours that each ion is surrounded by an oppositely charged ions. Radius of unit cell in sc, fcc and bcc is $$\displaystyle\frac{a}{2},\:\frac{a}{2\sqrt2}$$ and $$\dfrac{\sqrt3a}{4}$$, where $$a$$ is edge length of cell.

...view full instructions

Packing factors for a body centred cubic and simple cubic structure are, respectively :

A
$$0.68$$ and $$0.524$$

B
$$0.524$$ and $$0.68$$

C
$$0.48$$ and $$0.52$$

D
$$0.52$$ and $$0.48$$

Solution

$$\text{Body centred cubic}:$$

Body centred cubic lattice has radius $$=r=\displaystyle\cfrac{\sqrt3a}{4}$$

The unit cell of body centred cubic has $$2$$ atoms on the average.

$$\therefore$$ Volume of atoms $$=\displaystyle2\times \cfrac{4}{3}\pi r^3$$ $$=\displaystyle2\times\cfrac{4}{3}\times\pi\times\begin{bmatrix}\cfrac{\sqrt3a}{4}\end{bmatrix}^3=\cfrac{\sqrt3}{8}\pi{a^3}$$

The volume of unit cell $$=a^3$$

$$\therefore$$ Packing factor $$\displaystyle=\cfrac{\sqrt3}{8}\cfrac{\pi a^3}{a^3}=0.68$$

$$\text{Simple Cubic}:$$

Simple cubic has radius $$=\displaystyle r=\cfrac{a}{2}$$
The unit cell of sc has one atom.

$$\therefore$$ Volume of atom $$=\displaystyle\cfrac{4}{3}\pi\begin{pmatrix}\cfrac{a}{2}\end{pmatrix}^3=\cfrac{4\pi a^3}{24}$$

The volume of unit cell $$=a^3$$

$$\therefore$$ Packing factor $$=\displaystyle\cfrac{4}{24}\cfrac{\pi a^3}{a^3}=0.524$$

Hence, the correct option is $$\text{A}$$
Question 5
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The efficiency of packing is $$68\%$$ in:

A
hcp structure

B
ccp structure

C
fcc structure

D
bcc structure

Solution

$$\begin{array}{l}\text{ FCC or CCP packing efficiency }=74\%\\\text{ HCP Packing Efficiency }=74\%\\\text{ BCC Packing Efficiency }=68\%\end{array}$$
Question 6
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The density of solid argon is $$1.65 g$$ per cc at $$-233$$. If the argon atom is assumed to be a sphere of radius $$1.54\times { 10 }^{ -8 } cm$$, what percent of solid argon is apparently empty space? $$\left( Ar=40 \right) $$

A
$$16.5$$%

B
$$38$$%

C
$$50$$%

D
$$62$$%

Solution

Volume of one molecule $$=\dfrac { 4 }{ 3 } \pi { r }^{ 3 }=\dfrac { 4 }{ 3 } \pi { \left( 1.54\times { 10 }^{ -8 } \right) }^{ 3 }{ cm }^{ 3 }$$
$$=1.53\times { 10 }^{ -23 }{ cm }^{ 3 }$$
Volume of molecules in $$1.65 g$$ $$Ar$$ $$=\dfrac { 1.65 }{ 40 } \times { N }_{ 0 }\times 1.53\times { 10 }^{ -23 }=0.380{ cm }^{ 3 }$$
Volume of solid containing $$1.65 g$$ $$Ar$$ $$=1{ cm }^{ 3 }$$
$$\therefore$$ Empty space $$ =1 - 0.380=0.620$$
$$\therefore$$ Percent of empty space $$ =62$$%
Question 7
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Certain crystals produce electric signals on application of pressure. This phenomenon is called:

A
Ferroelectricity

B
Ferrielectricity

C
Pyroelectricity

D
Piezoelectricity

Solution

When a polar crystal is subjected to a mechanical stress, electricity is produced a case of piezoelectricity. Reversely, if an electric field is applied, then a mechanical stress is developed. Piezoelectric crystal acts as a mechanical-electrical transductor.
Question 8
1 / -0

Shortest distance between two tetrahedral voids is:

A
$$a\sqrt {3}$$

B
$$\dfrac {a\sqrt {3}}{2}$$

C
$$\dfrac {a}{2}$$

D
$$\dfrac {a\sqrt {3}}{4}$$

Solution

If $$a$$ is the edge length of the cube, then the shortest distance between two tetrahedral voids is $$\dfrac a2$$.
Question 9
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Which of the following compounds represents an inverse $$2 : 3$$ spinel structure?

A
$$Fe^{III} [Fe^{II} Fe^{III}]O_{4}$$

B
$$PbO_{2}$$

C
$$Al_{2}O_{3}$$

D
$$Mn_{3}O_{4}$$

Solution

Inverse spinel is generally denoted as $$A^{+2} {B_2}^{+3}O_4$$

$$A^{+2}$$ $$\cfrac {1}{8}$$ tetrahedral voids

$$A^{+3}$$ $$\cfrac {1}{8}$$ tetrahedral voids + $$\cfrac {1}{4}$$ octahedral voids

$$O^{2-}$$ at FCC lattice points.

In $$Fe^{+3}[Fe^{+3}Fe^{+2}]O_4$$ or $${Fe}_3O_4$$

$$Fe^{+3}$$ present at $$\cfrac {1}{8}$$ tetrahedral void and $$\cfrac {1}{4}$$ octahedral void $${Z_{Fe}}^{+3}=2$$

$$Fe^{+2}$$ present at $$\cfrac {1}{8}$$ tetrahedral void $${Z_{Fe}}^{+2}=8\times \cfrac {1}{8}=1$$

$$O^{2-}$$ at FCC lattice points $$Z_{eff}=4$$

Formula: $${Fe}^{+2} {Fe}_2^{+3}O_4\Rightarrow {Fe}_{3}O_4$$

Hence, the correct option is $$\text{A}$$
Question 10
1 / -0

At room temperature, sodium crystallises in body-centred cubic lattice with $$a = 4.24\overset {\circ}{A}$$.
Calculate the theoretical density of sodium. (Atomic mass of $$Na = 23.0$$)

A
$$1.002 \ g \ cm^{-3}$$

B
$$2.002 \ g \ cm^{-3}$$

C
$$3.002 \ g \ cm^{-3}$$

D
$$None$$

Solution

A body-centred cubic unit cell contains $$8$$ atoms at the $$8$$ corners and $$1$$ in the centre.
Hence,
Total number of atoms in a unit cell $$= 8\times \dfrac {1}{8} + 1 = 2$$

Volume of unit cell $$= a^{3} = (4.24 \times 10^{-8})^{3} cm^{3}$$

So, $$Density = \dfrac {Z\times M}{N_{0}\times V} $$

$$= \dfrac {2\times 23}{(6.023\times 10^{23})(4.24\times 10^{-8})^{3}}$$

$$= 1.002\ g\ cm^{-3}$$.

Option A is correct.

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Solid State Test - 62

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Solid State Test - 62

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SHARING IS CARING

Question 1

0.236

0.67

0.45

0.82

Solution

Question 2

$$2.144\ g/cm^{3}$$

$$7.289\ g/cm^{3}$$

$$5.188\ g/cm^{3}$$

$$10.376\ g/cm^{3}$$

Solution

Question 3

cubic

hexagonal

orthohombic

rhombohedral

Solution

Question 4

$$0.68$$ and $$0.524$$

$$0.524$$ and $$0.68$$

$$0.48$$ and $$0.52$$

$$0.52$$ and $$0.48$$

Solution

Question 5

hcp structure

ccp structure

fcc structure

bcc structure

Solution

Question 6

$$16.5$$%

$$38$$%

$$50$$%

$$62$$%

Solution

Question 7

Ferroelectricity

Ferrielectricity

Pyroelectricity

Piezoelectricity

Solution

Question 8

$$a\sqrt {3}$$

$$\dfrac {a\sqrt {3}}{2}$$

$$\dfrac {a}{2}$$

$$\dfrac {a\sqrt {3}}{4}$$

Solution

Question 9

$$Fe^{III} [Fe^{II} Fe^{III}]O_{4}$$

$$PbO_{2}$$

$$Al_{2}O_{3}$$

$$Mn_{3}O_{4}$$

Solution

Question 10

$$1.002 \ g \ cm^{-3}$$

$$2.002 \ g \ cm^{-3}$$

$$3.002 \ g \ cm^{-3}$$

$$None$$

Solution

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