Solid State Test

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Solid State Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

If an element crystallizes as a simple cube, what is the volume of an element provided its density is 1.5 g/cm³ and atomic mass of the element is 63?

A
69.7 x 10^-23 cm³

B
6.97 x 10^-23 cm³

C
63 x 10^-23 cm³

D
5.97 x 10^-23 cm³

Solution

Simple cube has 1 atom

Therefore volume of cube i.e. a³ = 6.97 x 10^-23 cm³
Question 2
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Mass of an atom in an unit cell is calculated by which of the following formula if M is the molar mass:

A

B

C

D

Solution

(Molar mass = M) (1 molar of atom = M 6.023×10²³)( atom = M/Na)
Question 3
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In the formula to calculate the density of a unit cell what is z ?

A
Volume of unit cell.

B
Number of atoms in a unit cell.

C
Mass of atoms in a unit cell.

D
Molar mass of unit cell.

Solution

The correct answer is Option B.
Mass of unit cell = number of atoms in unit cell × mass of each atom = z × m
Where, z = number of atoms in unit cell,
m = Mass of each atom
Mass of an atom can be given with the help of Avogadro number and molar mass as:
Where, M = molar mass
N_A = Avogadro’s number
Volume of unit cell, V = a³
⇒ Density of unit cell =
⇒ Density of unit cell =
Question 4
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Volume of a body centred cubic unit cell is

A

B

C

D
r³

Solution

In a body centered cubic unit cell ,√3a = 4r a= 4r/√3 where a = side of the cube volume = a³ = (4r/√3)³.
Question 5
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Density of a unit cell is equal to mass of unit cell divided by

A
volume of 4 unit cells.

B
volume of unit cell.

C
volume of one molecule.

D
volume of an atom.

Solution

Volume of unit cell, V = a3
=> Density of unit cell = mass of unit cell /volume of unit cell
Question 6
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Length a in unit cell is along

A
x axis

B
y axis

C
z axis

D
no axis

Solution

The correct answer is option A
Unit cell is essentially considered a cube, so a cube has 3 axes. X,Y,Z so X represents length , Y represents breadth, Z represents height.
Question 7
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The number of atoms (z) in simple cubic unit cell is

A
one

B
two

C
three

D
four

Solution

The face-centered cubic (fcc) has a coordination number of 12 and contains 4 atoms per unit cell. The body-centered cubic (bcc) has a coordination number of 8 and contains 2 atoms per unit cell. The simple cubic has a coordination number of 6 and contains 1 atom per unit cell.
Question 8
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X-ray diffraction studies show that copper crystallizes in an fcc unit cell with cell edge of 3.608 x 10^-8 cm. In a separate experiment, copper is determined to have a density of 8.92 g/cm³, calculate the atomic mass of copper.

A
63.1u

B
73.1u

C
53.1u

D
43.1u

Solution

In case of fcc lattice,number of atoms per unit cell,x=4 atoms
∴ Atomic mass of copper =63.1u
Question 9
1 / -0

An element with bcc geometry has atomic mass 50 and edge length 290 pm. The density of unit cell will be

A
8.16 g/cm³

B
6.81 g/cm³

C
68.1 g/cm³

D
1.86 g/cm³

Solution

Length of the edge , a = 290 pm =290 x 10^-10 cm
Volume of unit cell = ( 290 x 10^-10 cm )³ = 24.39 x 10^-24 cm³

Since it is bcc arrangement,
Number of atoms in the unit cell, Z = 2
Atomic mass of the element = 50

Mass of the atom = atomic mass/ Avogadro number = M/N_o = 50/6.02 x 10²³

Mass of the unit cell = Z x M/N_o = 2 x 50/6.02 x 10²³ = 100/6.23 x 10²³
Therefore , density = mass of unit cell / volume of unit cell
= 100/6.023 x 10²³ x 24.39 x 10^-24 = 6.81 g cm^-3
Question 10
1 / -0

The number of octahedral void(s) per atom present in a cubic close-packed structure is

A
2

B
4

C
1

D
3

Solution

The total number of octahedral void(s) per atom present in a cubic close packed structure is 4. Besides the body centre, there is one octahedral void at the centre of each of the 12 edges. It is surrounded by six atoms, four belonging to the same unit cell (2 on the corners and 2 on face centres) and two belonging to two adjacent unit cells. Since each edge of the cube is shared between four adjacent unit cells, so is the octahedral void located on it. Only 1/4 th of each void belongs to a particular unit cell. Thus, in cubic close packed structure, octahedral void at the body-centre of the cube is 1.
12 octahedral voids located at each edge and shared between four unit cells=12× 1/4=3
Total number of octahedral voids =4
We know that in ccp structure, each unit cell has 4 atoms. Thus, the number of octahedral voids is equal to 4/4=1.