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

# Copper FCC: packing, density and Fermi energy

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2081 Baisakh, 2+2+2+2 marks
- Video: [Atomic Packing Factor Calculation for SC, BCC, and FCC Materials](https://www.youtube.com/watch?v=PQGpHvw9O2g) — Dr. Joshua Paul Steimel
- 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-fermi-energy
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

Copper has FCC structure unit cell, atomic mass = 63.55 gm/mol and radius (R) = 0.13nm. Calculate (i) Packing density (ii) Density of copper (iii) Atomic concentration (iv) Fermi energy

## Short answer

FCC Cu: a = 2√2 R = 0.3677 nm. Packing density = 0.74. Density ρ = (4×63.55×10⁻³)/(N_A a³) = 8.49×10³ kg/m³. Atomic concentration n = 4/a³ = 8.05×10²⁸ m⁻³. Fermi energy E_F = (h²/8m_e)(3n/π)^(2/3) = 1.09×10⁻¹⁸ J (6.81 eV).

## Answer

Copper has an FCC unit cell: 4 atoms per cell, atoms touching along the face diagonal, so $\sqrt{2}a = 4R$.

**Given:** $R = 0.13$ nm, $M = 63.55$ g/mol $= 63.55\times10^{-3}$ kg/mol, $N_A = 6.022\times10^{23}$ mol⁻¹, $h = 6.626\times10^{-34}$ J s, $m_e = 9.1\times10^{-31}$ kg. Copper is monovalent (one free electron per atom).

Lattice constant:

$$
a = 2\sqrt{2}\,R = 2\sqrt{2}\times0.13 = 0.3677\ \text{nm} = 3.677\times10^{-10}\ \text{m}
$$

$$
a^3 = 4.971\times10^{-29}\ \text{m}^3
$$

#### (i) Packing density

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

#### (ii) Density of copper

$$
\begin{aligned}
\rho &= \frac{N M}{N_A a^3} = \frac{4\times63.55\times10^{-3}}{6.022\times10^{23}\times4.971\times10^{-29}} \\
&= 8.49\times10^{3}\ \text{kg m}^{-3} = 8.49\ \text{g cm}^{-3}
\end{aligned}
$$

#### (iii) Atomic concentration

$$
n_{at} = \frac{N}{a^3} = \frac{4}{4.971\times10^{-29}} = 8.05\times10^{28}\ \text{m}^{-3}
$$

#### (iv) Fermi energy
With one conduction electron per atom, electron concentration $n = n_{at} = 8.05\times10^{28}$ m⁻³. At 0 K,

$$
\begin{aligned}
E_F &= \frac{h^2}{8m_e}\left(\frac{3n}{\pi}\right)^{2/3} \\
&= \frac{(6.626\times10^{-34})^2}{8\times9.1\times10^{-31}}\left(\frac{3\times8.05\times10^{28}}{\pi}\right)^{2/3} \\
&= 6.03\times10^{-38}\times1.807\times10^{19} = 1.09\times10^{-18}\ \text{J} \\
&= \frac{1.09\times10^{-18}}{1.6\times10^{-19}} = 6.81\ \text{eV}
\end{aligned}
$$

**Answer:** (i) APF $= 0.74$; (ii) $\rho = 8.49\times10^3$ kg m⁻³ (8.49 g cm⁻³); (iii) $n_{at} = 8.05\times10^{28}$ m⁻³; (iv) $E_F = 1.09\times10^{-18}$ J $= 6.81$ eV.

(With the more accurate $R = 0.128$ nm these become 8.90 g cm⁻³, $8.43\times10^{28}$ m⁻³ and 7.0 eV, close to measured values.)

## Check yourself

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

1. For FCC copper with radius R, how is the lattice constant found?
   A. a = 4R/sqrt(3)
   B. a = 2R
   C. a = 2 sqrt(2) R from sqrt(2) a = 4R
   D. a = R/sqrt(2)
   Answer: C. Atoms touch along the face diagonal, so sqrt(2) a = 4R. (Key point: Lattice constant)

2. Which formula gives the density of copper?
   A. rho = M / (N_A a^3)
   B. rho = N M / (N_A a^3) with N = 4
   C. rho = N_A M / (4 a^3)
   D. rho = 4 N_A / (M a^3)
   Answer: B. Mass of 4 atoms per cell divided by cell volume: rho = 4M/(N_A a^3) = 8.49e3 kg/m^3. (Key point: Density formula)

3. How is the free electron concentration found for Cu?
   A. n = 4 per atom times the atomic concentration
   B. n equals the atomic concentration 4/a^3 since Cu is monovalent
   C. n = 2 times the atomic concentration
   D. n is the density in kg/m^3
   Answer: B. One conduction electron per atom gives n = 4/a^3 = 8.05e28 m^-3. (Key point: Electron concentration)

4. Which formula gives the Fermi energy at 0 K and about what value for Cu?
   A. E_F = (h^2/8m)(3n/pi)^(2/3), about 6.8 eV
   B. E_F = (h^2/8m)(3n/pi)^(3/2), about 6.8 eV
   C. E_F = (h^2/8m)(3n/pi)^(2/3), about 68 eV
   D. E_F = kT, about 0.025 eV
   Answer: A. E_F = (h^2/8m)(3n/pi)^(2/3) = 1.09e-18 J = 6.81 eV. (Key point: Fermi energy)

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