Number of Atoms in a unit Cell
Primitive cubic unit cell has atoms only at its corner.
In all, since each cubic unit cell has 8 atoms on its corners, the total number of atoms in one unit cell is 8x1/8 = 1 atom.
Body-Centered Cubic Unit Cell
body-centred cubic (bcc)
unit cell has an atom at each of its corners and also one atom at its
body centre. Thus in a body-centered cubic
(bcc) unit cell:
(i) 8 corners 1/8 per corner atom = 8x1/8=1 atom
(ii) 1 body centre atom = 1X1= 1 atom
Total number of atoms per unit cell = 2 atoms
face-Centered Cubic Unit Cell
A face-centred cubic (fcc) unit cell contains atoms at all the
corners and at the centre of all the faces of the cube. Thus,
in a face-centred cubic (fcc) unit cell:
(i) 8 corners atoms 1/8 atom per unit cell= 8x1/8 = 1 atom
(ii) 6 face centered atoms x 1/2 atom per unit cell = 6x1/2 = 3 atoms
Total number of atoms per unit cell = 1+3=4 atoms
Edge Length – Atomic Radius Relationship
(a) Primitive Cubic (Simple Cubic)
Atoms touch each other along the edge of the cube.
Edge length = 2r
a = 2r
Packing Efficiency (Derivation)
(a) Primitive Cubic (Simple Cubic)
Number of atoms per unit cell (Z) = 1
Radius–edge relation: a = 2r
Volume occupied by atoms = 1 × (4/3)πr3
Volume of unit cell = a3 = (2r)3 = 8r3
Packing Efficiency = (Volume of atoms / Volume of unit cell) × 100
= [(4/3)πr3 / 8r3] × 100
= (π / 6) × 100 ≈ 52.4%
(b) Body-Centered Cubic (BCC)
Atoms touch each other along the body diagonal.
Body diagonal = √3 a = 4r
a = 4r/√3
(b) Body-Centered Cubic (BCC)
Number of atoms per unit cell (Z) = 2
Radius–edge relation: a = 4r/√3
Volume occupied by atoms = 2 × (4/3)πr3 = (8/3)πr3
Volume of unit cell = a3 = (4r/√3)3 = 64r3 / 3√3
Packing Efficiency = [(8/3)πr3 / (64r3/3√3)] × 100
= (π√3 / 8) × 100 ≈ 68.0%
(c) Face-Centered Cubic (FCC)
Atoms touch each other along the face diagonal.
Face diagonal = √2 a = 4r
a = 2√2 r
(c) Face-Centered Cubic (FCC)
Number of atoms per unit cell (Z) = 4
Radius–edge relation: a = 2√2 r
Volume occupied by atoms = 4 × (4/3)πr3 = (16/3)πr3
Volume of unit cell = a3 = (2√2 r)3 = 16√2 r3
Packing Efficiency = [(16/3)πr3 / 16√2 r3] × 100
= (π / 3√2) × 100 ≈ 74.0%
Density of Unit Cell
Density (ρ) of a crystalline solid is given by:
ρ = (Z × M) / (a3 × NA)
Where:
Z = Number of atoms per unit cell
M = Molar mass (g mol−1)
a = Edge length of unit cell (cm)
NA = Avogadro number (6.022 × 1023 mol−1)
Density Numericals
Question 1:
Iron crystallizes in a BCC structure. Atomic mass of iron = 56 g mol−1. Edge length of unit cell = 2.87 × 10−8 cm. Calculate the density.
Answer:
For BCC, Z = 2
ρ = (2 × 56) / ( (2.87 × 10−8)3 × 6.022 × 1023 )
ρ ≈ 7.9 g cm−3
Question 2:
Aluminium crystallizes in an FCC structure. Atomic mass = 27 g mol−1. Edge length = 4.05 × 10−8 cm. Find the density.
Answer:
For FCC, Z = 4
ρ = (4 × 27) / ( (4.05 × 10−8)3 × 6.022 × 1023 )
ρ ≈ 2.7 g cm−3
Radius → Density Mixed Numericals
Question 1:
A metal crystallizes in a BCC structure. Atomic radius = 1.25 Å. Atomic mass = 55.8 g mol−1. Calculate the density.
Answer:
a = 4r/√3 = 4(1.25 × 10−8)/√3 = 2.89 × 10−8 cm
Z = 2
ρ = (2 × 55.8) / ( (2.89 × 10−8)3 × 6.022 × 1023 )
ρ ≈ 7.7 g cm−3
Question 2:
A metal crystallizes in an FCC structure with atomic radius 1.43 Å. Atomic mass = 63.5 g mol−1. Find the density.
Answer:
a = 2√2 r = 2√2 (1.43 × 10−8) = 4.04 × 10−8 cm
Z = 4
ρ = (4 × 63.5) / ( (4.04 × 10−8)3 × 6.022 × 1023 )
ρ ≈ 8.9 g cm−3
Question 3:
A solid crystallizes in a simple cubic structure. Atomic radius = 1.60 Å. Atomic mass = 23 g mol−1. Calculate its density.
Answer:
a = 2r = 3.20 × 10−8 cm
Z = 1
ρ = (1 × 23) / ( (3.20 × 10−8)3 × 6.022 × 1023 )
ρ ≈ 1.2 g cm−3