The document provides solutions to two tutorial problems on atomic structure: 1) It derives expressions for the atomic packing factor (APF) of face-centered cubic (FCC) and body-centered cubic (BCC) solids, showing the APF is 74% for FCC and 68% for BCC. 2) It estimates the density of face-centered cubic copper to be 8.887x103 g/m3 based on the number of copper atoms in the FCC unit cell, copper's atomic weight, and Avogadro's number.

1. Danyuo Yiporo
Dept. MSE
African Univ. of Science & Tech.
Abuja
Tutorials on Atomic structure

2. (a) Using possible sketches derive the expressions for the atomic packing factor of a face
centered cubic solid and body centered cubic solid and show that their atomic packing factors
are respectively, 74 % and 68 %.
Solution
• FCC
• By computing the unit cell size, there are 4R along the diagonal axis of an FCC crystal
• 𝑎2
+ 𝑎2
= (4𝑅)2
• 2𝑎2
= 16𝑅2
• 𝑎 =
16𝑅2
2
= 𝑅 8 = 2𝑅 2 (1mk)
• 𝑉𝑐 = 𝑎3
= 2𝑅 2
3
= 16𝑅3
2 (1mk)
• 𝐴𝑡𝑜𝑚𝑖𝑐 𝑃𝑎𝑐𝑘𝑖𝑛𝑔 𝐹𝑎𝑐𝑡𝑜𝑟, 𝐴𝑃𝐹 =
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑖𝑛 𝑎 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙 (𝑉𝑠)
𝑇𝑜𝑡𝑎𝑙 𝑐𝑒𝑙𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 (𝑉𝑐)
• 𝑇ℎ𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑝ℎ𝑒𝑟𝑒 𝑉𝑠 =
4𝜋𝑅3
3
𝑎𝑛𝑑 𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 4 𝑎𝑡𝑜𝑚𝑠 𝑜𝑛 𝑒𝑎𝑐ℎ 𝑓𝑎𝑐𝑒 𝑜𝑓 𝐹𝐶𝐶
• 𝑇ℎ𝑖𝑠 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑉𝑠 = 4𝑥
4𝜋𝑅3
3
= 𝑉𝑠 =
16𝜋𝑅3
3
• 𝑇ℎ𝑖𝑠 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝐴𝑃𝐹 =
𝑉𝑠
𝑉𝑐
=
16𝜋𝑅3
3
16𝑅3 2
=
𝜋
3 2
= 0.74 = 74% (1 mk)

3. • BCC
• Unit cell size:
• 𝒂 𝟐
+ 𝒂 𝟐
= 𝟐𝒂 𝟐
• This implies 𝑎2
+ 2𝑎2
= (4𝑅)2
• 3𝑎2
= (4𝑅)2
• 𝑇ℎ𝑖𝑠 𝑔𝑖𝑣𝑒𝑠 𝑎2
=
(4𝑅)2
3
• 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑎 =
4𝑅
3
(1mk)
• 𝐴𝑡𝑜𝑚𝑖𝑐 𝑃𝑎𝑐𝑘𝑖𝑛𝑔 𝐹𝑎𝑐𝑡𝑜𝑟, 𝐴𝑃𝐹 =
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑖𝑛 𝑎 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙 (𝑉𝑠)
𝑇𝑜𝑡𝑎𝑙 𝑐𝑒𝑙𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 (𝑉𝑐)
• 𝑉𝑐 = 𝑎3
=
4𝑅
3
3
=
64𝑅3
3 3
• There are 2 atoms on the corner faces of BCC, which makes 𝑉𝑠 = 2𝑥
4𝜋𝑅3
3
= 𝑉𝑠 =
8𝜋𝑅3
3
(1 mk)
• This gives 𝐴𝑃𝐹 = 𝐴𝑃𝐹 =
𝑉𝑠
𝑉𝑐
=
8𝜋𝑅3
3
64𝑅3
3 3
=
8𝜋𝑅3
3
3 3
64𝑅3 =
𝜋 3
8
= 0.680 (1 mk)
• Therefore APF = 68%.

4. (a) Estimate the density of face centered cubic copper from the knowledge of
the number of atoms in its unit cell. You may assume that the atomic weight
of copper is 63.5 g/mol and the Avogadro’s number is 6.022 x 1023
atoms/mol. (4 mks)
• Solution
• Copper atom has a radius of 0.128 nm = 0.128𝑥10−9 𝑚
• Atomic weight, A = 63.5 g/mol
• Avogadro’s number, NA = 6.022 x 1023 atoms/mol
• There are 4 atoms per each face of an FCC, n = 4
• 𝑉𝑐 = 16𝑅3
2 (1 mk)
• Density, ρ, of copper =
n A
VcNA
=
4
atoms
unit cell
63.5 g/mol
16x(0.128x10−9 m)3 2 6.022 x 1023 atoms/mol
=
0.254
2.858x10−5
• Therefore, the density, ρ, of copper = 𝟖. 𝟖𝟖𝟕𝒙𝟏𝟎 𝟑
𝒈/𝒎 𝟑
(3 mks)