---
title: "Effective mass of a free electron"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:free-electron-effective-mass"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/free-electron-effective-mass
site: BCT Engineering
---

# Effective mass of a free electron

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2078 Kartik, 4 marks; 2075 Asoj, 1+3 marks; 2071 Chaitra, 1+3 marks (asked 3 times)
- Video: [Effective mass of Electron (Solid state Physics) PHYSICS (in हिन्दी)](https://www.youtube.com/watch?v=__nOVPc7FIU) — 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/free-electron-effective-mass
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

What is effective mass? Show that the effective mass is same as mass of electron in vacuum.

## Short answer

Effective mass m* = ℏ²/(d²E/dk²). For a free electron, E = ℏ²k²/(2m_e), so d²E/dk² = ℏ²/m_e, giving m* = m_e (9.1×10⁻³¹ kg). The effective mass equals the vacuum electron mass because no periodic potential is present.

## Answer

**Effective mass** $m^*$ is the apparent mass of an electron in a crystal when it is accelerated by an external force; it includes the effect of the lattice's internal forces. It is defined as

$$
m^* = \frac{\hbar^2}{d^2E/dk^2}
$$

#### Free electron
For an electron moving freely in vacuum (or a "free" electron with $V = 0$), all its energy is kinetic:

$$
E = \frac{1}{2}m_ev^2 = \frac{p^2}{2m_e}
$$

By de Broglie, $p = \hbar k$, so

$$
E = \frac{\hbar^2k^2}{2m_e}
$$

This is a parabola in the E–k diagram. Differentiating twice:

$$
\frac{dE}{dk} = \frac{\hbar^2k}{m_e}, \qquad \frac{d^2E}{dk^2} = \frac{\hbar^2}{m_e}
$$

Substituting in the definition:

$$
m^* = \frac{\hbar^2}{\hbar^2/m_e} = m_e
$$

Hence, for a free electron the effective mass equals the ordinary mass of the electron in vacuum, $m_e = 9.1\times10^{-31}$ kg. This is expected: with no lattice there are no internal forces, so the only force is the external one and Newton's law $F = m_ea$ holds directly. In a real crystal the E–k curve is not exactly parabolic, so $m^*$ differs from $m_e$ (e.g. $1.08\,m_e$ for electrons in Si).

## Check yourself

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

1. What is the E-k relation of a free electron?
   A. E = hbar k / 2m
   B. E = hbar^2 k^2 / (2 m_e), a parabola
   C. E = hbar^2 k / m_e
   D. E = constant
   Answer: B. With p = hbar k, E = p^2/2m_e = hbar^2 k^2 / 2m_e. (Key point: Free electron E-k)

2. What is d^2E/dk^2 for the free electron?
   A. hbar^2 k / m_e
   B. m_e / hbar^2
   C. hbar^2 / m_e
   D. zero
   Answer: C. dE/dk = hbar^2 k / m_e and differentiating again gives hbar^2 / m_e. (Key point: Second derivative)

3. Substituting into m* = hbar^2 / (d^2E/dk^2) gives:
   A. m* = 2 m_e
   B. m* = m_e / 2
   C. m* = hbar^2 m_e
   D. m* = m_e
   Answer: D. m* = hbar^2 / (hbar^2/m_e) = m_e, as there are no internal lattice forces. (Key point: Result)

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