Solid State Test

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Solid State Test - 25

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Weekly Quiz Competition

Question 1
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If a is the lattice parameter, the volume of an fcc crystal having $$N$$ atoms is :

A
$$\displaystyle \frac{Na^3}{4}$$

B
$$\displaystyle \frac{Na^3}{2}$$

C
$$\displaystyle Na^3$$

D
$$\displaystyle 2Na^2$$
Question 2
1 / -0

The physical dimension of unit cells in a crystal lattice:

A
lattice parameter

B
packing factor

C
atomic radius

D
number of constituent particles

Solution

Lattice parameter which refers to the physical dimension of unit cells in a crystal lattice. Lattices in three dimensions generally have three lattice constants, referred to as a, b, and c.
Question 3
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If A is the molecular mass of an element, a is the edge length and $$N$$ is the Avogadro's number, then the density for simple cubic lattice is :

A
$$\displaystyle\frac{2A}{Na^3}$$

B
$$\displaystyle\frac{4A}{Na^3}$$

C
$$\displaystyle\frac{A}{Na^3}$$

D
$$\displaystyle\frac{A}{Na^2}$$

Solution

$$\text{Density}$$ $$=$$ $$\dfrac{\text{Number of molecules present in lattice (Z)} \times \text{Mass of each molecule (m)}}{\text{Volume of lattice (V)}}$$

In simple lattice, each lattice has one molecule and mass of each molecule is $$\dfrac{\text{Molar mass (A)}}{\text{Avogadro's number (N)}}$$

For simple cubic lattice, $$Z = 1$$

Hence, $$\text{density} = \dfrac{Z\times A}{N\times a^{3}}=\dfrac{A}{N\times a^3}$$, where $$a$$ is the edge length.
Question 4
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The lattices of Na and Al are bcc and fcc respectively. Presuming them to be closed packed, their packing fractions are, respectively :

A
$$0.68$$ and $$0.34$$

B
$$0.68$$ and $$0.74$$

C
$$0.34$$ and $$0.34$$

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

Solution

Packing fraction is known as the fraction of volume occupied by the constituent particles in crystal structure.
GIVEN
Lattices of $$Na$$ is Body centred cubic $$BCC$$
Lattices of $$Al$$ is Face centred cubic $$FCC$$

SOLUTION

$$Packing\hspace{0.1cm}fraction$$ =$$\dfrac{N_{atom}\hspace{0.1cm}V_{atom}}{V_{unit\hspace{0.05cm}cell}}$$

Packing fraction of BCC ($$Na$$)

Number of atoms ($$N_{atom}$$) = $$2$$
Volume of atoms ($$V_{atom}$$) = $$\dfrac{4}{3}$$$$\pi$$$$r^{2}$$
Volume of unit cell ($$V_{unit\hspace{0.05cm}cell}$$) = $$($$$$\dfrac{4r}{\sqrt3}$$$$)^{3}$$

Particle fraction $$_{BCC}$$ =$$\dfrac{N_{atom}\hspace{0.1cm}V_{atom}}{V_{unit\hspace{0.05cm}cell}}$$

= $$\dfrac{2\times\dfrac{4}{3}\pi r^{2}}{(\dfrac{4r}{\sqrt3})^{3}}$$

= $$\dfrac{\pi \sqrt{3}}{8}$$

$$\mathbf Particle fraction $$$$_{BCC}$$$$\cong$$ $$\mathbf 0.6802$$

Packing fraction of FCC ($$Al$$)

Number of atoms ($$N_{atom}$$) = $$4$$
Volume of atoms ($$V_{atom}$$) = $$\dfrac{4}{3}$$$$\pi$$$$r^{30}$$
Volume of unit cell ($$V_{unit\hspace{0.05cm}cell}$$) = $$($$$$\dfrac{4r}{\sqrt2}$$$$)^{3}$$

Particle fraction $$_{FCC}$$ = $$\dfrac{4\times\dfrac{4}{3}\pi r^{3}}{(\dfrac{4r}{\sqrt2})^{3}}$$

=$$\dfrac{\pi \sqrt{2}}{12}$$

Particle fraction $$_{FCC}$$ = $$0.74$$

The particle fractions for $$Na$$ and $$Al$$ are $$0.68$$ and $$0.74$$.

The correct option : B
Question 5
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The density of solid argon (atomic mass $$=40$$ amu) is $$1.68$$ g/ml at $$40$$ K. If the argon atom is assumed to be a sphere of radius $${1.50 \times 10^{-8}}$$ cm, then the percentage of solid Ar which is space will be:
[use $$N_A=6\times 10^{23}$$]

A
$$35.64$$

B
$$64.36$$

C
$$73$$

D
None of the above

Solution

As we know, density $$=1.68 = \dfrac{nM}{VN_A}=\dfrac{n\times 40}{V\times6\times 10^{23}}$$
$$\implies V = 39.53 \times 10^{-24}\ n$$
The radius of argon is $$1.5 \times 10^{-8}$$ cm and so, the empty space is $$\left (1-\dfrac{4}{3} \right ) \times n\times \pi r^3 \times \dfrac{100}{V} = 64.3\%$$
Question 6
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An atomic solid crystallizes in a body centre cubic lattice and the inner surface of the atoms at the adjacent corner are separated by $$60.3$$ pm. If the atomic weight of A is $$48$$, then the density of the solid is nearly:

A
$$2.7$$ g/cc

B
$$5.07$$ g/cc

C
$$3.5$$ g/cc

D
$$1.75$$ g/cc

Solution

According to the given diagram,

The separation between the inner surface of the atoms at adjacent corners is given by-
$$a - 2r =60.3$$

For BCC,

$$\sqrt3 a = 4r$$

So, density $$= \dfrac{nM}{VN_A}$$$$ = \dfrac{2 \times 48 \times 10^{30}}{(463.8)^3 \times 6.023 \times 10^{23}} = 1.7$$ $$g/cc$$
Question 7
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The packing efficiency of a simple cubic crystal with an interstitial atom exactly fitting at the body center is:

A
$$0.48$$

B
$$0.52$$

C
$$0.73$$

D
$$0.91$$

Solution

For a simple cubic, $$ a=2r $$
Therefore, the radius of the interstitial atom exactly fitting at the centre of the cube is $$R= \dfrac{1}{2} \times (a\sqrt 3 -2r) $$

$$R= \dfrac{1}{2} \times (a\sqrt 3 -a) $$

$$R= 0.366a $$

Packing fraction $$ = \dfrac{\text{volume occupied by spheres}}{a^3} $$

Packing fraction $$ = \dfrac{ (\dfrac{1}{8} \times 8 \times \dfrac{4 \pi r^3}{3}) + ( 1 \times \dfrac{4 \pi R^3}{3} ) }{a^3} $$

Packing fraction$$ = \dfrac{4\pi ({(\dfrac{a}{2})}^3 + (0.366a)^3)}{3a^3} $$

Packing fraction $$= 0.7285$$.
Question 8
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Sodium (atomic mass $$=23$$ amu) crystallizes in a bcc arrangement with the interfacial separation between the atoms at the edge $$53.6$$ pm. The density (in g cm$$^{-3}$$) of sodium crystal is:

A
$$2.07$$

B
$$2.46$$

C
$$1.19$$

D
none of the above

Solution

According to given condition,
$$ {a-2R} = 53.6$$ pm
$$ a(1 - {2(\sqrt 3 )\over 4}) = 53.6 $$ pm
$$\implies a=400$$ pm
So, density $$= \dfrac{nM}{VN_A}$$$$ = \dfrac{2 \times 23 }{(400)^3 \times 6.023 \times 10^{23}\times 10^{-24}} = 1.19$$ g/cc
Question 9
1 / -0

If A is the molecular mass of an element, a is the edge length and $$N$$ is the Avogadro's number, then the density for face-centered cubic lattice is :

A
$$\displaystyle\frac{4A}{Na^3}$$

B
$$\displaystyle\frac{2A}{Na^3}$$

C
$$\displaystyle\frac{A}{Na^3}$$

D
$$1$$

Solution

$$\text{Density}$$ $$=$$ $$\dfrac{\text{Number of molecules present in lattice (Z)} \times \text{Mass of each molecule (m)}}{\text{Volume of lattice (V)}}$$

In face-centered cubic lattice, each lattice has four molecules and mass of each molecule is $$\dfrac{\text{Molar mass (A)}}{\text{Avogadro's number (N)}}$$

Hence, $$\text{Density} = \dfrac{Z\times A}{N\times a^{3}}=\dfrac{4\times A}{N\times a^3}$$, where $$a$$ is the edge length.
Question 10
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Directions For Questions

Packing fraction of a unit cell is defined as the fraction of the total volume of the unit cell occupied by the atom(s).$$\displaystyle \text{Packing fraction (P.F)} = \dfrac{\text{Volume of the atom (s) present in a unit cell}}{\text{Volume of unit cell}}$$ $$=\dfrac {Z \times \dfrac{4}{3} \pi r^{3}}{a^{3}}$$ Percentage of empty space $$=100-\text{P.F}\times 100$$ where $$Z$$ is the effective number of atoms in a cube, $$r$$ is the radius of an atom and $$a$$ is the edge length of the cube.

...view full instructions

Packing fraction in face-centered cubic unit cell is :

A
$$0.7406$$

B
$$0.6802$$

C
$$0.5236$$

D
none of the above

Solution

Packing efficiency (P.E.) $$=\dfrac {n\times\dfrac 43 \pi r^3}{V} \times 100$$
For a FCC lattice, $$n=4$$
Therefore, P.E. $$=\dfrac{4\times \dfrac 43 \times {(\dfrac {a}{2\sqrt 2}}^3)}{a^3} =\dfrac {\pi}{3\sqrt2} \times 100=74\%$$
Hence. in FCC, $$74\%$$ of the total volume is occupied by atoms.

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Solid State Test - 25

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Solid State Test - 25

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Question 1

$$\displaystyle \frac{Na^3}{4}$$

$$\displaystyle \frac{Na^3}{2}$$

$$\displaystyle Na^3$$

$$\displaystyle 2Na^2$$

Question 2

lattice parameter

packing factor

atomic radius

number of constituent particles

Solution

Question 3

$$\displaystyle\frac{2A}{Na^3}$$

$$\displaystyle\frac{4A}{Na^3}$$

$$\displaystyle\frac{A}{Na^3}$$

$$\displaystyle\frac{A}{Na^2}$$

Solution

Question 4

$$0.68$$ and $$0.34$$

$$0.68$$ and $$0.74$$

$$0.34$$ and $$0.34$$

$$0.52$$ and $$0.52$$

Solution

Question 5

$$35.64$$

$$64.36$$

$$73$$

None of the above

Solution

Question 6

$$2.7$$ g/cc

$$5.07$$ g/cc

$$3.5$$ g/cc

$$1.75$$ g/cc

Solution

Question 7

$$0.48$$

$$0.52$$

$$0.73$$

$$0.91$$

Solution

Question 8

$$2.07$$

$$2.46$$

$$1.19$$

none of the above

Solution

Question 9

$$\displaystyle\frac{4A}{Na^3}$$

$$\displaystyle\frac{2A}{Na^3}$$

$$\displaystyle\frac{A}{Na^3}$$

$$1$$

Solution

Question 10

$$0.7406$$

$$0.6802$$

$$0.5236$$

none of the above

Solution

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