Solid State Test

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Solid State Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

The effective radius of an iron atom is $$1.42\ \mathring A$$. It has FCC structure. Calculate its density (in g.cm$$^{-3}$$) if it is given that the atomic mass of $$Fe$$ is $$56$$ amu.

A
$$2.47$$

B
$$5.74$$

C
$$8.24$$

D
None of the above

Solution

$$ d = \dfrac{ zM}{N_A \times V }$$
For FCC ,
$$z=4$$
$$ a = 2\sqrt 2 r $$
$$ \implies a = 4.02 \times 10^{-8}$$ cm
$$ d = \dfrac{ 4\times 56}{6.023 \times 10^{23} \times ( 4.02 \times 10^{-8})^3 }$$
$$ d = \dfrac{ 4\times 56}{6.023 \times 64.96 \times 10^{-1} }$$
$$ d = 5.74 $$ $$ g cm^{-3} $$
Question 2
1 / -0

An element A (atomic weight = $$100$$ amu) having bcc structure has unit cell edge length of $$400$$ pm. Calculate the density (in g/cc) of A?

A
$$5.188$$

B
$$7.188$$

C
$$4.144$$

D
$$10.144$$

Solution

Density(g/mol)= $$\dfrac{Z.M}{a^3.N_A}$$ , where

Z: number of atoms per unit cell = 2 for BCC
M: Atomic Weight = 100 amu
a: edge cell length in cm = 400 x $$10^{-10}$$ cm
$$N_A$$: Avogadro Number = $$6.023\times10^{23}$$

Therefore, Density = $$\dfrac{2\times100}{400\times10^{-10}\times 6.023\times10^{23}}$$

= 5.188 g/mol

Hence, options A is correct.
Question 3
1 / -0

The interstital hole is called tetrahedral because ?

A
it is formed by four spheres

B
partly same and partly different

C
it is formed by four spheres, the centres of which form a regular tetrahedron

D
none of the above

Solution

A tetrahedral site is surrounded by four spheres and hence an interstitial hole is called tetrahedral.
Question 4
1 / -0

$$TiO_2$$ is well known example of :

A
triclinic system

B
tetragonal system

C
monoclinic system

D
none of the above

Solution

$$TiO_2$$ has tetragonal system with five plane of symmetry and five axes of symmetry.
Question 5
1 / -0

The packing factor for a simple cubic lattice is :

A
$$\displaystyle \frac{\sqrt2 \pi}{6}$$

B
$$\displaystyle \frac{\pi}{6}$$

C
$$\displaystyle \frac{\sqrt3 \pi}{8}$$

D
$$\pi$$
Question 6
1 / -0

$$KF$$ crystallizes in the $$NaCl$$ type structure. If the radius of $$K^{+}$$ ions is $$132$$ pm and that of $$F^{-}$$ ion is $$135$$ pm , what is the shortest $$K-F$$ distance?

A
$$267\ pm$$

B
$$534\ pm$$

C
$$378\ pm$$

D
$$None\ of\ the\ above$$

Solution

$$ F^- $$ ions are present in the ccp sites whereas $$ K^+ $$ ions are present in the octahedral voids.

The shortest distance d between the anion and the cation is given by
$$ d = r_+ + r_- = 132 + 135 $$
$$ d = 267 $$ pm
Question 7
1 / -0

In FCC unit cell, the percentage of unit cell not occupied by atoms is :

A
$$33\%$$

B
$$29\%$$

C
$$26\%$$

D
$$74\%$$

Solution

In FCC, $$26\%$$ of the total volume is not be recovered by atoms.
Question 8
1 / -0

Calculate the density of diamond from the fact that it has a face-centered cubic structure with two atoms per lattice point and unit cell edge length of $$3.569 \times 10 ^{-8}$$ cm.

A
3.509 g/cm$$^{3}$$

B
7.012 g/cm$$^{3}$$

C
10.12 g/cm$$^{3}$$

D
None of the above

Solution

$$\rho=\dfrac{Z\times M}{N_A\times a^3}$$

The lattice points in the FCC lattice are at the corners of the cube and face center.

Since it is given in the question that at each lattice point there are $$2$$ atoms,
So, the $$z$$ will be-

$$Z=2\times 8\times \dfrac{1}{8}+2\times 6\times \dfrac{1}{2}$$

$$Z=8$$

$$M=12\ g/mol$$ for carbon, $$N_A=6.023\times 10^{23}/mol$$

$$a=3.569\times 10^{-8},\quad a^3=(3.569\times 10^{-8})\ cm^{3}$$

$$a^3=4.546\times 10^{-23}\ cm^3$$

$$\rho=\dfrac{8\times 12}{6.023\times 10^{23}\times 4.546\times 10^{-23}}=3.506\ g/cm^3$$

$$\rho=3.5\ g/cm^3$$

Hence, the correct option is A.
Question 9
1 / -0

Which one of the following schemes of ordering closed packed sheets of equal sized spheres do not generate close packed lattice?

A
ABCABC

B
ABACABAC

C
ABBAABBA

D
ABCBCACBC

Solution

In option C, the scheme ABBAABBA of ordering closed packed sheets of equal-sized spheres do not generate close-packed lattice as BB, AA and BB represent same layers of sheets placed together.

Hence, option C is correct.
Question 10
1 / -0

The volume of a crystal in BCC structure having N atoms is :
[a = edge length]

A
$$N_Aa^3$$

B
$$0.5N_Aa^3$$

C
$$0.33N_Aa^3$$

D
$$0.25N_Aa^3$$

Solution

The body-centred cubic (bcc) has a coordination number of 8 and contains 2 atoms per unit cell.

$$V = \dfrac{N_A\times a^3}{Z}$$

$$Z = 2\because \text{BCC structure}$$

$$V = \dfrac{N_A\times a^3}{2}$$

$$V = 0.5N_Aa^3$$

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Solid State Test - 24

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Solid State Test - 24

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Question 1

$$2.47$$

$$5.74$$

$$8.24$$

None of the above

Solution

Question 2

$$5.188$$

$$7.188$$

$$4.144$$

$$10.144$$

Solution

Question 3

it is formed by four spheres

partly same and partly different

it is formed by four spheres, the centres of which form a regular tetrahedron

none of the above

Solution

Question 4

triclinic system

tetragonal system

monoclinic system

none of the above

Solution

Question 5

$$\displaystyle \frac{\sqrt2 \pi}{6}$$

$$\displaystyle \frac{\pi}{6}$$

$$\displaystyle \frac{\sqrt3 \pi}{8}$$

$$\pi$$

Question 6

$$267\ pm$$

$$534\ pm$$

$$378\ pm$$

$$None\ of\ the\ above$$

Solution

Question 7

$$33\%$$

$$29\%$$

$$26\%$$

$$74\%$$

Solution

Question 8

3.509 g/cm$$^{3}$$

7.012 g/cm$$^{3}$$

10.12 g/cm$$^{3}$$

None of the above

Solution

Question 9

ABCABC

ABACABAC

ABBAABBA

ABCBCACBC

Solution

Question 10

$$N_Aa^3$$

$$0.5N_Aa^3$$

$$0.33N_Aa^3$$

$$0.25N_Aa^3$$

Solution

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