---
title: "Fermi level of copper from conductivity"
subject: "Electrical Engineering Material"
chapter: 2
key: "semester_3/electrical_engineering_material:fermi-level-from-conductivity"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/fermi-level-from-conductivity
site: BCT Engineering
---

# Fermi level of copper from conductivity

Electrical Engineering Material · Chapter 2: Free Electron Theory of Conduction in Metal

- Asked in: 2074 Chaitra, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/fermi-level-from-conductivity
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-2

## Question

The conductivity and drift mobility of copper conductor is 63.5×10⁶ S/m and 43 cm²/V/s. Calculate Fermi level for copper conductor.

## Short answer

Find n = σ/(eμ_d) = 63.5×10⁶ / (1.602×10⁻¹⁹ × 4.3×10⁻³) = 9.22×10²⁸ m⁻³. Then E_{F0} = (h²/8m_e)(3n/π)^{2/3} ≈ 7.44 eV.

## Answer

The conductivity gives the free-electron concentration, and the concentration gives the Fermi energy:

$$
\sigma = e n \mu_d \;\Rightarrow\; n = \frac{\sigma}{e\mu_d}, \qquad E_{F0} = \frac{h^2}{8m_e}\left(\frac{3n}{\pi}\right)^{2/3}
$$

**Given:** $\sigma = 63.5 \times 10^{6}\ \text{S/m}$, $\mu_d = 43\ \text{cm}^2\text{V}^{-1}\text{s}^{-1} = 43 \times 10^{-4}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$.

#### Electron concentration

$$
n = \frac{63.5 \times 10^{6}}{1.602 \times 10^{-19} \times 43 \times 10^{-4}} = 9.22 \times 10^{28}\ \text{m}^{-3}
$$

#### Fermi level

$$
\begin{aligned}
E_{F0} &= \frac{(6.626 \times 10^{-34})^2}{8 \times 9.109 \times 10^{-31}} \left(\frac{3 \times 9.22 \times 10^{28}}{\pi}\right)^{2/3} \\
&= 6.025 \times 10^{-38} \times 1.979 \times 10^{19} \\
&= 1.19 \times 10^{-18}\ \text{J} = \frac{1.19 \times 10^{-18}}{1.602 \times 10^{-19}} = 7.44\ \text{eV}
\end{aligned}
$$

The Fermi level lies 7.44 eV above the bottom of the conduction band (close to the accepted 7.0 eV for copper; the difference comes from the slightly high conductivity value given).

**Answer:** $n = 9.22 \times 10^{28}\ \text{m}^{-3}$, $E_F \approx 1.19 \times 10^{-18}\ \text{J} = 7.44\ \text{eV}$.

## Check yourself

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

1. Electron concentration is obtained from
   A. n = σ e μ_d
   B. n = σ/(e μ_d)
   C. n = e μ_d/σ
   D. n = σ μ_d/e
   Answer: B. Rearranging σ = e n μ_d. (Key point: Finding n)

2. 43 cm^2/V/s in SI is
   A. 43e-4 m^2/V/s
   B. 43e-2 m^2/V/s
   C. 43e4 m^2/V/s
   D. 43 m^2/V/s
   Answer: A. 1 cm^2 = 1e-4 m^2. (Key point: Unit conversion)

3. Once n is known, Fermi energy is
   A. E_F0 = n e^2/m
   B. E_F0 = kT ln(n)
   C. E_F0 = h^2 n/8m
   D. E_F0 = (h^2/8m)(3n/π)^(2/3)
   Answer: D. Same free-electron relation. (Key point: Fermi level formula)

4. The Fermi level is about
   A. 1.19 eV
   B. 74.4 eV
   C. 7.44 eV
   D. 0.74 eV
   Answer: C. n = 9.22e28 m^-3, E_F = 1.19e-18 J = 7.44 eV. (Key point: Result)

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