---
title: "Copper FCC: atoms, packing, concentration and density"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:copper-fcc-atoms-density"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/copper-fcc-atoms-density
site: BCT Engineering
---

# Copper FCC: atoms, packing, concentration and density

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2075 Asoj, 8 marks
- Video: [Unit Cell Chemistry   Simple Cubic, Body Centered Cubic, Face Centered Cubic Crystal Lattice Structu](https://www.youtube.com/watch?v=HCWwRh5CXYU) — The Organic Chemistry Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/copper-fcc-atoms-density
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

Draw a neat diagram of face centered cubic (FCC) unit cell crystal structure for copper and find (i) Number of atoms per unit cell (ii) Packing density (iii) Atomic concentration if radius of copper atom is 0.128 nm (iv) Density of crystal given that atomic mass of Cu is 63.55 g mol⁻¹

## Short answer

For FCC copper: 4 atoms per unit cell; packing density 0.74; atomic concentration 8.43×10^28 m⁻³ using a = 2√2 R = 0.362 nm; crystal density 8.90×10^3 kg m⁻³.

## Answer

Copper crystallises in the **face centred cubic (FCC)** structure: atoms at the 8 corners and at the centres of the 6 faces, touching along the face diagonal.

```text
         o------------o
        /|    x      /|
       / |          / |     o = corner atom
      o------------o  |     x = face-centre atom
      |  |  x      |x |
      |x |    x    |  |
      |  o---------|--o
      | /    x     | /
      |/           |/
      o------------o

  One face:   o-------o
              | \   / |    face diagonal
              |   x   |    = R + 2R + R = 4R
              | /   \ |
              o-------o
```

**Given:** $R = 0.128$ nm, $M = 63.55$ g/mol, $N_A = 6.022\times10^{23}$ mol⁻¹.

#### (i) Number of atoms per unit cell
- Corner atoms: each shared by 8 cells, $8\times\frac{1}{8} = 1$
- Face atoms: each shared by 2 cells, $6\times\frac{1}{2} = 3$

$$
N = 1 + 3 = 4
$$

#### Lattice constant
Along the face diagonal $\sqrt{2}\,a = 4R$:

$$
\begin{aligned}
a &= 2\sqrt{2}\,R = 2\sqrt{2}\times0.128 = 0.3620\ \text{nm} = 3.620\times10^{-10}\ \text{m} \\
a^3 &= 4.745\times10^{-29}\ \text{m}^3
\end{aligned}
$$

#### (ii) Packing density

$$
\text{APF} = \frac{4\times\frac{4}{3}\pi R^3}{(2\sqrt{2}R)^3} = \frac{\pi}{3\sqrt{2}} = 0.74
$$

#### (iii) Atomic concentration

$$
n_{at} = \frac{N}{a^3} = \frac{4}{4.745\times10^{-29}} = 8.43\times10^{28}\ \text{atoms m}^{-3}
$$

#### (iv) Density

$$
\begin{aligned}
\rho &= \frac{N M}{N_A a^3} = \frac{4\times63.55\times10^{-3}}{6.022\times10^{23}\times4.745\times10^{-29}} \\
&= 8.90\times10^{3}\ \text{kg m}^{-3} = 8.90\ \text{g cm}^{-3}
\end{aligned}
$$

**Answer:** (i) 4 atoms per cell; (ii) packing density 0.74; (iii) $8.43\times10^{28}$ m⁻³; (iv) $\rho = 8.90\times10^3$ kg m⁻³ (close to the measured 8.96 g cm⁻³).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The number of atoms per FCC unit cell is:
   A. 2
   B. 4
   C. 6
   D. 8
   Answer: B. 8 x 1/8 + 6 x 1/2 = 4. (Key point: Atoms per cell)

2. With R = 0.128 nm, lattice constant a of Cu is:
   A. 0.256 nm
   B. 0.362 nm
   C. 0.512 nm
   D. 0.181 nm
   Answer: B. a = 2 sqrt(2) R = 0.362 nm. (Key point: Lattice constant)

3. Atomic concentration of Cu is calculated as:
   A. N a^3
   B. N/a^3
   C. 1/a^3
   D. N/(4/3 pi R^3)
   Answer: B. n = N/a^3 = 4/(4.745x10^-29) = 8.43 x 10^28 m^-3. (Key point: Atomic concentration)

4. Density formula and result for Cu:
   A. rho = N M/(N_A a^3) = 8.90 x 10^3 kg/m^3
   B. rho = M/(N_A a^3) = 2.2 x 10^3 kg/m^3
   C. rho = N M a^3/N_A = 8.90 kg/m^3
   D. rho = N_A a^3/(N M) = 1 kg/m^3
   Answer: A. Mass of 4 atoms divided by cell volume gives about 8900 kg/m^3. (Key point: Density)

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