---
title: "Fermi energy from electron concentration"
subject: "Electrical Engineering Material"
chapter: 2
key: "semester_3/electrical_engineering_material:fermi-energy-electron-concentration"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/fermi-energy-electron-concentration
site: BCT Engineering
---

# Fermi energy from electron concentration

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

- Asked in: 2076 Asoj, 4 marks
- Video: [Fermi level Energy (Fermi Level) Fermi Energy (PHYSICS) BE / Btech 1st year engineering](https://www.youtube.com/watch?v=XCgCPWCN4qA) — Gautam Varde
- 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-electron-concentration
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-2

## Question

Calculate the Fermi energy level at absolute zero for the copper having electron concentration of 8.43×10²⁸ m⁻³.

## Short answer

Using E_{F0} = (h²/8m_e)(3n/π)^{2/3} with n = 8.43×10²⁸ m⁻³ gives 1.12×10⁻¹⁸ J = 7.01 eV.

## Answer

For free electrons in a metal at 0 K, all states up to $E_{F0}$ are filled, giving

$$
E_{F0} = \frac{h^2}{8 m_e}\left(\frac{3n}{\pi}\right)^{2/3}
$$

**Given:** $n = 8.43 \times 10^{28}\ \text{m}^{-3}$, $h = 6.626 \times 10^{-34}\ \text{J s}$, $m_e = 9.109 \times 10^{-31}\ \text{kg}$.

#### Calculation

$$
\begin{aligned}
\left(\frac{3n}{\pi}\right)^{2/3} &= \left(\frac{3 \times 8.43 \times 10^{28}}{\pi}\right)^{2/3} = (8.050 \times 10^{28})^{2/3} = 1.864 \times 10^{19}\ \text{m}^{-2} \\
\frac{h^2}{8m_e} &= \frac{(6.626 \times 10^{-34})^2}{8 \times 9.109 \times 10^{-31}} = 6.025 \times 10^{-38}\ \text{J m}^2 \\
E_{F0} &= 6.025 \times 10^{-38} \times 1.864 \times 10^{19} = 1.123 \times 10^{-18}\ \text{J}
\end{aligned}
$$

In electron-volts:

$$
E_{F0} = \frac{1.123 \times 10^{-18}}{1.602 \times 10^{-19}} = 7.01\ \text{eV}
$$

This agrees with the measured value for copper (about 7 eV). The Fermi temperature $T_F = E_{F0}/k \approx 8.1 \times 10^{4}$ K, far above any working temperature, so $E_F$ stays almost the same at room temperature.

**Answer:** $E_{F0} \approx 1.12 \times 10^{-18}\ \text{J} = 7.01\ \text{eV}$.

## Check yourself

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

1. At 0 K, E_F0 is found by
   A. Setting kT equal to E_F0
   B. Filling all states up to E_F0 with n electrons
   C. Equating E_F0 to work function
   D. Using the electron kinetic energy at room temperature
   Answer: B. n = integral of g(E) dE from 0 to E_F0 gives the formula. (Key point: Basis of formula)

2. E_F0 is proportional to
   A. n^(2/3)
   B. n^(3/2)
   C. n
   D. 1/n
   Answer: A. E_F0 = (h^2/8m)(3n/π)^(2/3). (Key point: Formula)

3. For n = 8.43e28 m^-3 the Fermi energy is about
   A. 70.1 eV
   B. 0.701 eV
   C. 1.12 eV
   D. 7.01 eV
   Answer: D. E_F0 = 1.123e-18 J = 7.01 eV. (Key point: Result)

4. Why does E_F stay almost the same at room temperature?
   A. Electrons do not move at room temperature
   B. n decreases with T to compensate
   C. Fermi temperature (~8e4 K) is far above working temperature
   D. Fermi energy is defined at room temperature only
   Answer: C. kT is much less than E_F, so thermal effect is small. (Key point: Temperature independence)

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