Solid State Test

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Solid State Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

The side length of diamond unit cell is (in pm):

A
$$154$$

B
$$1422.63$$

C
$$711.32$$

D
$$355.66$$

Solution

Given, $$C-C$$ bond length $$=154$$ $$pm$$
Let, side length be $$a$$
$$\cfrac{a\sqrt{3}}{4}=154$$
$$\Rightarrow a=355.66$$ $$pm$$
Question 2
1 / -0

If $$fcc$$ unit cell is converted in to End centred unit cell, then its packing efficiency will decrease by:

A
$$25\%$$

B
$$50\%$$

C
$$40\%$$

D
$$60\%$$

Solution

$$Z=4$$ for FCC unit cell
In end centered unit cell, $$8$$ atoms at body corners, $$2$$ atoms at any $$2$$ of the face centres.
$$Z_{eff}$$ for end centred unit cell $$=\cfrac{1}{8}\times 8+2\times \cfrac{1}{2}=2$$
As $$packaging\quad fraction\propto Z_{effective}$$
$$\%$$ decrease in packaging fraction $$=\cfrac{4-2}{4}\times 100=50\%$$
Question 3
1 / -0

The coefficient of linear expansion of crystal in x-direction is $$6\times {10}^{-5}/^{o}C$$ and that in y and z- directions is $$4\times {10}^{-5}/^{o}C$$. The coefficient of cubical expansion of the crystal is?

A
$$18\times {10}^{-5}/^{o}C$$

B
$$16\times {10}^{-5}/^{o}C$$

C
$$14\times {10}^{-5}/^{o}C$$

D
$$32\times {10}^{-5}/^{o}C$$

Solution

$$\gamma =\alpha _x+\alpha_y+\alpha_z$$
$$\implies \gamma=6\times 10^{-5}+4\times 10^{-5}+4\times 10^{-5}$$
$$\implies \gamma= 14\times 10^{-5}/^oC$$
Question 4
1 / -0

If metallic radius of metal having $$fcc$$ structure is $$127\ pm$$, then the length of edge of its unit cell will be.

A
$$288\ pm$$

B
$$359\ pm$$

C
$$576\ pm$$

D
$$432\ pm$$

Solution

In FCC unit cell
$$4r=a\sqrt{2}$$ (along the face diagonal)
$$a=2\sqrt{2}r$$
$$a=2\sqrt{2}\times 127=359$$ $$pm$$
Question 5
1 / -0

Number of unit cells in $$10\ g\ NaCl$$ is:

A
$$\dfrac {1.5}{58.5}\times 10^{24}$$

B
$$\dfrac {2.5}{58.5}\times 10^{23}$$

C
$$\dfrac {5.6}{58.5}\times 10^{20}$$

D
$$\dfrac {5.6}{58.5}\times 10^{21}$$

Solution

Moles of $$NaCl=\cfrac{10}{58.5}$$

Number of molecules of $$NaCl=\cfrac{10}{58.5}\times 6.023\times 10^{23}$$
In one unit cell $$4$$ molecules of $$NaCl$$ are present.

Number of unit cells $$=\cfrac{10\times 6.023\times 10^{23}}{58.5}\times \cfrac{1}{4}=\cfrac{1.5}{58.5}\times 10^{24}$$
Question 6
1 / -0

$$NaBr$$ has same crystal structure as $$NaCl$$. If density is $$3.203g\ cm^{-3}$$. Calculate the unit cell side length. [$$Na=23, Br=79.9$$]

A
$$5.976A^{o}$$

B
$$6.976A^{o}$$

C
$$7.976A^{o}$$

D
$$4.976A^{o}$$

Solution

$$\rho=\cfrac{Z\times M_A}{N_A\times a^3}$$
$$Z=4$$, $$\rho=3.203$$ $$g/cm^3$$,
$$M_A=102.9$$ $$amu$$
$$\implies 3.203=\cfrac{4\times 102.9}{6.023\times 10^{23}\times a^3}$$
$$\implies a^3=\cfrac{4\times 102.9}{6.023\times 3.203}\times 10^{-23}$$
$$\implies a^3=213.356\times 10^{-24}$$
$$\implies a=5.976\mathring A$$
Question 7
1 / -0

Which parameters are correct for unit cell of $$Na_{2}SO_{4}, 10H_{2}O$$ crystal?

A
$$a\neq b\neq c$$ and $$\alpha =\gamma =90^{o}, \beta \neq 120^{o}$$

B
$$a= b= c$$ and $$\alpha =\beta =\gamma = 90^{o}$$

C
$$a= b\neq c$$ and $$\alpha =\beta =\gamma = 90^{o}$$

D
$$a\neq b\neq c$$ and $$\alpha \neq \beta \neq \gamma =90^{o}$$

Solution

$$Na,SO_4,H_2O$$ belongs to monoclinic sulphade as for molecule.
Axial length:$$a\neq b\neq c$$
Axial angle: $$\alpha=\gamma=90^o,\, \beta\neq 120^o$$
Question 8
1 / -0

How many unit cells are present in a cube-shaped ideal crystal of $$NaCI$$ of mass $$1.00\ g$$?

A
$$5.14\times { 10 }^{ 21 }$$ unit cell

B
$$1.28\times { 10 }^{ 21 }$$ unit cell

C
$$1.71\times { 10 }^{ 21 }$$ unit cell

D
$$2.57\times { 10 }^{ 21 }$$ unit cell

Solution

Overall $$4$$ units $$NaCl$$ are present in one cell
$$n$$ of moles = $$\dfrac{1}{(23 + 35.5)}$$
$$= \dfrac{1}{58.5}$$
$$n$$ of molecular = $$6.022 \times 10^{23} \times \dfrac{1}{58.5}$$
$$\therefore n$$ of unit cells =$$\dfrac{molecules}{4}$$
$$= \dfrac{6.022 \times 10^{23}}{4 \times 58.5}$$
$$= \dfrac{602.2 \times 10^{21}}{234}$$
$$= 2.57 \times 10^{21}$$
Question 9
1 / -0

If the edge length of a KCl cell is 488 pm, what is the length of KCl bond if it crystallizes in the fcc structure?

A
122 pm

B
244 pm

C
488 pm

D
974 pm

Solution

In FCC structure of NaH, Na atom is present at the corner and H is present at the edge center.
Hence,
$$Bond\ length=\dfrac{Edge\ length}{2}=\dfrac{488}{2}=244\ pm$$
Question 10
1 / -0

The crystal structure of pure silicon is like:

A
Diamond

B
Graphite

C
Silver

D
Zinc

Solution

Silicon crystallizes in the same pattern as $$diamond$$, in a structure which Ashcroft and Mermin call "two interpenetrating face-centered cubic" primitive lattices.

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Solid State Test - 46

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Solid State Test - 46

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SHARING IS CARING

Question 1

$$154$$

$$1422.63$$

$$711.32$$

$$355.66$$

Solution

Question 2

$$25\%$$

$$50\%$$

$$40\%$$

$$60\%$$

Solution

Question 3

$$18\times {10}^{-5}/^{o}C$$

$$16\times {10}^{-5}/^{o}C$$

$$14\times {10}^{-5}/^{o}C$$

$$32\times {10}^{-5}/^{o}C$$

Solution

Question 4

$$288\ pm$$

$$359\ pm$$

$$576\ pm$$

$$432\ pm$$

Solution

Question 5

$$\dfrac {1.5}{58.5}\times 10^{24}$$

$$\dfrac {2.5}{58.5}\times 10^{23}$$

$$\dfrac {5.6}{58.5}\times 10^{20}$$

$$\dfrac {5.6}{58.5}\times 10^{21}$$

Solution

Question 6

$$5.976A^{o}$$

$$6.976A^{o}$$

$$7.976A^{o}$$

$$4.976A^{o}$$

Solution

Question 7

$$a\neq b\neq c$$ and $$\alpha =\gamma =90^{o}, \beta \neq 120^{o}$$

$$a= b= c$$ and $$\alpha =\beta =\gamma = 90^{o}$$

$$a= b\neq c$$ and $$\alpha =\beta =\gamma = 90^{o}$$

$$a\neq b\neq c$$ and $$\alpha \neq \beta \neq \gamma =90^{o}$$

Solution

Question 8

$$5.14\times { 10 }^{ 21 }$$ unit cell

$$1.28\times { 10 }^{ 21 }$$ unit cell

$$1.71\times { 10 }^{ 21 }$$ unit cell

$$2.57\times { 10 }^{ 21 }$$ unit cell

Solution

Question 9

122 pm

244 pm

488 pm

974 pm

Solution

Question 10

Diamond

Graphite

Silver

Zinc

Solution

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