Solid State Test

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Solid State Test - 61

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Question 1
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The packing efficiency of the two dimensional square unit cell shown below is :

A
$$39.27\%$$

B
$$68.02\%$$

C
$$74.05\%$$

D
$$78.57\%$$

Solution

Here the diagonal is 4r, where, r is the radius of atom

We know diagonal $$= \sqrt(L^{2} +L^{2})= \sqrt{2 \times L}$$

$$4 \times r= {\sqrt2 \times L}$$

$$ L= \dfrac{ 4 \times r }{\sqrt{2}}$$

Number of spheres inside the square are $$ 1+ 4 \times \dfrac{1}{4} $$

Area of sphere $$= \pi \times r \times r $$

Packing fraction $$= \dfrac{2 \times \pi \times r \times r}{8 \times r^{2} }=\dfrac{\pi}{4}=0.7857 $$

Percentage is $$78.57\%$$
Question 2
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Lithium borohydride crystallizes in an orthorhombic system with $$4$$ molecules per unit cell. The unit cell dimensions are $$a=6.8\:\mathring{A}$$, $$b=4.4\:\mathring{A}$$ and $$c=7.2\:\mathring{A}$$. If the molar mass is
$$21.76$$ g mol$$^{-1}$$, calculate the density (in g cm$$^{-3}$$) of crystal.

A
$$1.34$$

B
$$0.87$$

C
$$0.94$$

D
$$0.67$$

Solution

The expression for density is as follows:
$$\displaystyle \text{Density}=\frac{n\times \text{Molar mass}}{V\times N_a}$$
Here,
$$n=4$$
$$M=21.76$$ g mol$$^{-1}$$
$$N_a=6.023\times10^{23}$$
$$V=a\times b\times c$$
$$\therefore V=6.8\times10^{-8}\times4.4\times10^{-8}\times7.2\times10^{-8}$$ $$=2.154\times10^{-22}$$ cm$$^3$$

$$\displaystyle\therefore \text{Density}=\frac{4\times21.76}{2.154\times10^{-22}\times6.023\times10^{23}}$$ $$=0.6709$$ g cm$$^{-3}$$
Question 3
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CsBr crystallizes in a body-centred cubic lattice of unit cell edge length $$436.6$$ pm. Atomic masses of Cs and Br are $$133$$ amu and $$80$$ amu respectively. The density (in g cm$$^{-3}$$) of CsBr is :

A
$$8.50$$

B
$$4.25$$

C
$$42.5$$

D
$$0.425$$

Solution

For bcc, z=4

The formula for density is given below:

$$\displaystyle \text{Density}=\frac{z\times M}{a^3\times N_0}$$

$$\displaystyle=\frac{2\times213}{(4.366\times10^{-8})^3\times6.02\times10^{23}}$$

$$=8.50$$ g cm$$^{-3}$$

Hence, the correct option is $$\text{A}$$
Question 4
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Directions For Questions

Packing refers to the arrangement of constituent units in such a way that the forces of attraction among the constituent particles is maximum and the constituents occupy the maximum available space. In two dimensions, there are square close packing and hexagonal close packing. In three dimensions, however, there are hexagonal close packing, cubic close packing and body-centred cubic packing.
(i) HCP : AB AB AB AB . . . arrangement
Coordination number $$= 12$$
Percentage occupied space $$= 74\%$$
(ii) CCP : ABC ABC . . . arrangement
Coordination number $$= 12$$
Percentage occupied space $$= 74\%$$
(iii) BCC : $$68\%$$ space is occupied
Coordination number $$= 8$$
Answer the following questions:

...view full instructions

The pattern of successive layers of ccp arrangement can be designated as:

A
AB AB AB

B
AB ABC AB ABC

C
ABC ABC ABC

D
AB BA AB BA

Solution

The pattern of successive layers of cubic close packing (ccp) arrangement can be designated as ABC ABC ABC.

Simple cubic arrangement can be designated as AAAAAA.

Hexagonal close packing (hcp) arrangement can be designated as ABABAB.

Hence, the correct option is $$\text{C}$$
Question 5
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Directions For Questions

The number of Schottky defects $$\left(n\right)$$ present in an ionic containing $$N$$ ions at temperature $$T$$ is given by $$n = N{ e }^{ -{ E }/{ 2kT } }$$ where $$E$$ is energy required to create $$n$$ Schottky defects and $$k$$ is Boltzmann constant. The number of Frenkel defects $$\left(n\right)$$ in an ionic crystal having $$N$$ ions is given by $$n={ \left( \dfrac { N }{ { N }_{ i } } \right) }^{ \dfrac { 1 }{ 2 } }{ e }^{ -{ E }/{ 2kT } }$$ where $$E$$ is energy required to create $$n$$ Frenkel defects and $${N}_{i}$$ is the number of interstitial sites.

...view full instructions

Absorption of photons by crystal

A
has no effect on imperfection

B
produce atomic displacement leading to imperfections

C
decreases number of defects

D
none of the above

Solution

Absorption of photons by crystal produce atomic displacement leading to imperfections. These imperfections are also called defects. These are deviations from perfectly ordered arrangement.
Question 6
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Metallic gold crystallises in fcc lattice. The edge length of the cubic unit cell is $$4.07$$ $$\mathring A$$. What is the density (in g cm$$^{-3}$$) of gold?

A
$$10.2$$

B
$$19.4$$

C
$$7.4$$

D
$$13.6$$

Solution

$$z =4 \text { (for } f C C) $$

$$M =197 \text { (for gold) } $$

$$a =4.07 \text { A( given) } $$

$$=407 \mathrm{pm}$$

$$P =\frac{2 \times M}{N a \times a^{3} \times 10^{-30}} g \mathrm{~cm}^{-3} $$

$$=\frac{4 \times 191}{6.022 \times 10^{23} \times(401)^{3} \times 10^{-30}} $$

$$=\frac{7899}{401}=19.4 \mathrm{~g} \mathrm{~cm}^{-3}$$
Question 7
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The density of solid argon is $$1.65$$ g mL$$^{-1}$$ at $$-233^\circ C$$. If the argon atom is assumed to be the sphere of radius $$1.54\times10^{-8}$$ cm, what percentage of solid argon is apparently empty space?
[Atomic mass of Ar is $$40$$ amu.]

A
$$62\%$$

B
$$76\%$$

C
$$68\%$$

D
$$52\%$$

Solution

Volume of one atom of $$\displaystyle Ar=\frac{4}{3}\pi r^3$$

Also, number of atoms in $$1.65$$ g or in $$\displaystyle1$$ mL $$=\dfrac{1.65}{40}\times6.023\times10^{23}$$

$$\therefore$$ Total volume of all atoms of $$Ar$$ in solid state $$\displaystyle=\frac{4}{3}\pi r^3\times\frac{1.65}{40}\times6.023\times10^{23}$$ $$\displaystyle=\frac{4}{3}\times\frac{22}{7}\times(1.54\times10^{-8})^3\times\frac{1.65}{40}\times6.023\times10^{23}$$ $$=0.380$$ cm$$^3$$

Volume of solid argon $$=1$$ cm$$^3$$

$$\therefore\%$$ Empty space $$\displaystyle=\frac{[1-0.380]}{1}\times100=62\%$$
Question 8
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An element (atomic mass $$= 100\ g / mol$$) having bcc structure has unit cell edge $$400\ pm$$. Then, density of the element is:

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

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

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

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

Solution

$$\rho = \dfrac {Z\times M}{N_{A} \times a^{3}} = \dfrac {2\times 100}{6.023\times 10^{23} \times (400 \times 10^{-10})^{3}}$$

$$= 5.188\ g/cm^{3}$$
Question 9
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What is not a type of crystal?

A
Ionic

B
Covalent network

C
Metallic

D
Covalent molecular

E
Amorphous

Solution

Crystals tend to have relatively sharp, well-defined melting points because all the component atoms, molecules, or ions are the same distance from the same number and type of neighbours; that is, the regularity of the crystalline lattice creates local environments that are the same.

Thus the intermolecular forces holding the solid together are uniform, and the same amount of thermal energy is needed to break every interaction simultaneously.

Amorphous has different property to crystal.

Hence the correct option is E.
Question 10
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Select the correct statement(s) about cubic structure of diamond.
1. Effective number of atoms present in a diamond cubic cell is $$8$$.
2. Each carbon is surrounded by four more carbon atoms.
3. Atomic radius of carbon is $$\displaystyle \frac{\sqrt{3}\times a}{8}$$, where a is edge length of cube.
4. Approximate packing fraction of unit cell is $$\displaystyle \frac{\sqrt{3}\times \pi }{16}$$.

A
1, 2 and 4

B
2, 3 and 4

C
2 and 3

D
All 1, 2, 3 and 4

Solution

All the four statements about cubic structure of diamond are correct.

1. Effective number of atoms present in a diamond cubic cell is $$8$$.

2. Each carbon is surrounded by four more carbon atoms. Thus each C atom is present at the centre of tetrahedron and four other C atoms are present at the corners of tetrahedron.

3. Atomic radius of carbon is $$\displaystyle \frac{\sqrt{3}\times a}{8}$$, where a is edge length of cube.

4. Approximate packing fraction of unit cell is $$\displaystyle \frac{\sqrt{3}\times \pi }{16}$$ which approximates to 34%. In other words, in the unit cell of diamond, 34% of volume is occupied and 66% of volume is empty.

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Solid State Test - 61

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Solid State Test - 61

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Question 1

$$39.27\%$$

$$68.02\%$$

$$74.05\%$$

$$78.57\%$$

Solution

Question 2

$$1.34$$

$$0.87$$

$$0.94$$

$$0.67$$

Solution

Question 3

$$8.50$$

$$4.25$$

$$42.5$$

$$0.425$$

Solution

Question 4

AB AB AB

AB ABC AB ABC

ABC ABC ABC

AB BA AB BA

Solution

Question 5

has no effect on imperfection

produce atomic displacement leading to imperfections

decreases number of defects

none of the above

Solution

Question 6

$$10.2$$

$$19.4$$

$$7.4$$

$$13.6$$

Solution

Question 7

$$62\%$$

$$76\%$$

$$68\%$$

$$52\%$$

Solution

Question 8

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

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

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

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

Solution

Question 9

Ionic

Covalent network

Metallic

Covalent molecular

Amorphous

Solution

Question 10

1, 2 and 4

2, 3 and 4

2 and 3

All 1, 2, 3 and 4

Solution

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