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

# Copper FCC: packing density and atomic concentration

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2068 Baisakh, 4 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-packing-concentration
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

Copper has FCC (Face-centered cubic) structure. Find the packing density and atomic concentration for copper if radius of copper atom is 0.128nm.

## Short answer

FCC has 4 atoms per cell, atoms touch on face diagonal: a = 2√2 R. For Cu with R=0.128nm, packing density = π/(3√2) ≈ 0.74 (74%). Atomic concentration = 4/a³ = 8.43×10²⁸ m⁻³ (8.43×10²² cm⁻³).

## Answer

Copper has an FCC structure: 4 atoms per unit cell ($8\times\frac{1}{8} + 6\times\frac{1}{2}$), with atoms touching along the face diagonal, so $\sqrt{2}\,a = 4R$.

**Given:** $R = 0.128$ nm.

#### Lattice constant

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

#### Packing density

$$
\begin{aligned}
\text{APF} &= \frac{4\times\frac{4}{3}\pi R^3}{a^3} = \frac{4\times\frac{4}{3}\pi\times(0.128\times10^{-9})^3}{4.745\times10^{-29}} \\
&= \frac{3.513\times10^{-29}}{4.745\times10^{-29}} = 0.74
\end{aligned}
$$

(In general, for FCC, $\text{APF} = \pi/(3\sqrt{2}) = 0.74$.)

#### Atomic concentration

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

**Answer:** packing density $= 0.74$ (74 %); atomic concentration $= 8.43\times10^{28}$ m⁻³ $= 8.43\times10^{22}$ cm⁻³.

## Check yourself

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

1. How many atoms are there per FCC unit cell?
   A. 2
   B. 4
   C. 6
   D. 8
   Answer: B. 8 x 1/8 + 6 x 1/2 = 4. (Key point: FCC atoms per cell)

2. Atoms touch along the face diagonal in FCC. What is a in terms of R?
   A. a = 4R/sqrt(3)
   B. a = 2R
   C. a = 2 sqrt(2) R
   D. a = 4R
   Answer: C. sqrt(2)a = 4R gives a = 2 sqrt(2) R = 0.362 nm for R = 0.128 nm. (Key point: Lattice constant)

3. What is the packing density of FCC copper?
   A. 0.52
   B. 0.68
   C. 0.74
   D. 0.91
   Answer: C. APF = 4(4/3)pi R^3 / a^3 = pi/(3 sqrt(2)) = 0.74. (Key point: Packing density)

4. How is the atomic concentration of copper found?
   A. n = 1/a^3
   B. n = 4/a^3
   C. n = 4/a
   D. n = 2/a^3
   Answer: B. Four atoms per cell of volume a^3 gives n = 4/a^3 = 8.43e28 per m^3. (Key point: Atomic concentration)

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