---
title: "Fermi energy of copper at 0 K and 15°C"
subject: "Electrical Engineering Material"
chapter: 2
key: "semester_3/electrical_engineering_material:fermi-energy-copper-15c"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/fermi-energy-copper-15c
site: BCT Engineering
---

# Fermi energy of copper at 0 K and 15°C

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

- Asked in: 2078 Bhadra, 4 marks
- Video: [Density of states and Derivation of Fermi Energy at 0 K](https://www.youtube.com/watch?v=M7Z6B9MJOb4) — Ravi C.B
- 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-energy-copper-15c
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-2

## Question

Calculate the Fermi energy level in copper at 0K, if its density is 8.96 gm.cm⁻³ and atomic weight is 63.5 gm.mol⁻¹. What will be its new Fermi energy at 15°C?

## Short answer

Electron concentration n = ρN_A/M = 8.50×10²⁸ m⁻³. At 0 K, E_{F0} = (h²/8m_e)(3n/π)^{2/3} ≈ 7.05 eV. At 15°C the temperature correction is ~10⁻⁵, so E_F ≈ 7.05 eV.

## Answer

**Given:** $\rho = 8.96\ \text{g/cm}^3 = 8960\ \text{kg/m}^3$, $M = 63.5\ \text{g/mol}$; copper gives one free electron per atom.

#### Fermi energy at 0 K

Electron concentration:

$$
n = \frac{\rho N_A}{M} = \frac{8960 \times 6.022 \times 10^{23}}{0.0635} = 8.50 \times 10^{28}\ \text{m}^{-3}
$$

Free-electron Fermi energy:

$$
\begin{aligned}
E_{F0} &= \frac{h^2}{8m_e}\left(\frac{3n}{\pi}\right)^{2/3} \\
&= \frac{(6.626 \times 10^{-34})^2}{8 \times 9.109 \times 10^{-31}} \times \left(\frac{3 \times 8.50 \times 10^{28}}{\pi}\right)^{2/3} \\
&= 6.025 \times 10^{-38} \times 1.874 \times 10^{19} = 1.129 \times 10^{-18}\ \text{J} = 7.05\ \text{eV}
\end{aligned}
$$

#### Fermi energy at 15°C

The temperature dependence of the Fermi energy in a metal is

$$
E_F(T) = E_{F0}\left[1 - \frac{\pi^2}{12}\left(\frac{kT}{E_{F0}}\right)^2\right]
$$

At $T = 15 + 273.15 = 288.15\ \text{K}$: $kT = \frac{1.381 \times 10^{-23} \times 288.15}{1.602 \times 10^{-19}} = 0.02484\ \text{eV}$.

$$
\begin{aligned}
\frac{kT}{E_{F0}} &= \frac{0.02484}{7.0487} = 3.524 \times 10^{-3} \\
\frac{\pi^2}{12}\left(\frac{kT}{E_{F0}}\right)^2 &= 0.8225 \times 1.242 \times 10^{-5} = 1.02 \times 10^{-5} \\
E_F(288\ \text{K}) &= 7.0487 \times (1 - 1.02 \times 10^{-5}) = 7.0486\ \text{eV}
\end{aligned}
$$

The decrease is only about $7.2 \times 10^{-5}$ eV (0.001%), so the Fermi energy of a metal is practically independent of temperature.

**Answer:** $E_{F0} = 7.05\ \text{eV}$; at 15°C, $E_F \approx 7.05\ \text{eV}$ (lower by only $\approx 7.2 \times 10^{-5}$ eV).

## Check yourself

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

1. How is n for copper found?
   A. n = M/(ρ N_A)
   B. n = ρ N_A/M, one free electron per atom
   C. n = 2 ρ N_A/M
   D. n = ρ/M
   Answer: B. Copper is monovalent, giving 8.50e28 m^-3. (Key point: Electron concentration)

2. The 0 K Fermi energy is
   A. E_F0 = (h^2/8m)(3n/π)^(2/3)
   B. E_F0 = (h^2/8m)(3n/π)^(3/2)
   C. E_F0 = (h^2/8m) n
   D. E_F0 = kT ln n
   Answer: A. Filling states up to E_F0 gives the 2/3 power of n. (Key point: Fermi energy formula)

3. E_F(T) is given by
   A. E_F0 [1 + kT/E_F0]
   B. E_F0 exp(-kT)
   C. E_F0 - kT
   D. E_F0 [1 - (π^2/12)(kT/E_F0)^2]
   Answer: D. The Sommerfeld expansion gives a second-order small decrease. (Key point: Temperature correction)

4. At 15°C the Fermi energy of copper is
   A. About 0.025 eV higher
   B. About 1 eV lower
   C. Practically unchanged, 7.05 eV (drop ~7e-5 eV)
   D. Zero
   Answer: C. kT is much smaller than E_F, so the correction is about 1e-5 of E_F0. (Key point: Temperature dependence)

## Related in Electrical Engineering Material

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- [Fermi energy of copper at 0 K](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Afermi-energy-copper-0k) — Electrical Engineering Material, chapter 2, asked 2×
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