---
title: "Concept of effective mass in a crystal"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:effective-mass-concept-crystal"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/effective-mass-concept-crystal
site: BCT Engineering
---

# Concept of effective mass in a crystal

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2082 Kartik (new course), 5 marks; 2075 Chaitra, 6 marks; 2070 Asar, 6 marks; 2068 Baisakh, 6 marks (asked 4 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/effective-mass-concept-crystal
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

Explain the concept of effective mass in crystal with necessary mathematical expression.

## Short answer

The effective mass m* = ħ²/(d²E/dk²) describes an electron's response to external forces, incorporating the lattice's periodic potential. It depends on the curvature of the E-k band: positive at band bottom, negative at top, and infinite at inflection points, enabling a free-particle-like description.

## Answer

The **effective mass** $m^*$ of an electron in a crystal is the mass it appears to have when it responds to an **external** force, after the effect of all the internal forces from the lattice ions and other electrons has been included in it. It lets us treat a crystal electron as if it were a free particle with mass $m^*$.

#### Why it is needed
Inside a crystal an electron feels the external force $F_{ext}$ (for example $-eE$ from an applied field) and also the internal periodic force $F_{int}$ from the lattice. Newton's law gives

$$
F_{ext} + F_{int} = m_e a
$$

$F_{int}$ is very complicated and changes from point to point. We therefore write

$$
F_{ext} = m^* a
$$

and put all the effect of the lattice into $m^*$.

#### Expression for $m^*$
The electron is a wave packet that moves with the group velocity

$$
v_g = \frac{d\omega}{dk} = \frac{1}{\hbar}\frac{dE}{dk}
$$

Its acceleration is

$$
a = \frac{dv_g}{dt} = \frac{1}{\hbar}\frac{d^2E}{dk^2}\frac{dk}{dt}
$$

In time $dt$ the external force does work $dE = F_{ext}v_g\,dt$, so $\frac{dE}{dk}dk = F_{ext}\frac{1}{\hbar}\frac{dE}{dk}dt$, giving

$$
F_{ext} = \hbar\frac{dk}{dt}
$$

Substituting $dk/dt = F_{ext}/\hbar$:

$$
a = \frac{1}{\hbar^2}\frac{d^2E}{dk^2}F_{ext} \quad\Rightarrow\quad m^* = \frac{\hbar^2}{d^2E/dk^2}
$$

#### Meaning
- $m^*$ depends on the **curvature** of the E–k curve. A sharply curved band gives a small $m^*$ (light, fast electron); a flat band gives a large $m^*$.
- For a free electron $E = \hbar^2k^2/2m_e$, so $d^2E/dk^2 = \hbar^2/m_e$ and $m^* = m_e$.
- Near the bottom of a band $m^*$ is positive; near the top it is negative (this is why holes are used); at the inflection point it is infinite.

#### Typical values

| Material | $m_e^*/m_e$ | $m_h^*/m_e$ |
|---|---|---|
| Si | 1.08 | 0.56 |
| Ge | 0.55 | 0.37 |
| GaAs | 0.067 | 0.48 |

The small $m^*$ of electrons in GaAs is why GaAs devices are faster than Si devices.

## Check yourself

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

1. What does the effective mass m* of a crystal electron represent?
   A. The actual rest mass of the electron changes in a crystal
   B. The mass of the lattice ions
   C. The mass it appears to have under an external force, with lattice effects included
   D. The mass of an electron-hole pair
   Answer: C. m* absorbs all internal lattice forces so that F_ext = m* a. (Key point: Meaning of m*)

2. Which formula gives m*?
   A. m* = hbar^2 / (d^2E/dk^2)
   B. m* = hbar^2 x (d^2E/dk^2)
   C. m* = hbar / (dE/dk)
   D. m* = (d^2E/dk^2) / hbar^2
   Answer: A. From a = (1/hbar^2)(d^2E/dk^2)F, m* = hbar^2 / (d^2E/dk^2). (Key point: Formula)

3. How does m* depend on the E-k curve?
   A. Sharper curvature gives larger m*
   B. Sharper curvature gives smaller m* (lighter electron)
   C. m* is independent of curvature
   D. A flat band gives zero m*
   Answer: B. m* is inversely proportional to curvature, so a sharply curved band gives a light electron and a flat band a heavy one. (Key point: Curvature)

4. What is the sign of m* near the top of a band?
   A. Positive
   B. Zero
   C. Infinite
   D. Negative
   Answer: D. At the band top the curve is concave down, d^2E/dk^2 < 0, so m* is negative, which leads to the hole concept. (Key point: Sign of m*)

## Related in Electrical Engineering Material

- [Negative and infinite effective mass](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Anegative-infinite-effective-mass) — Electrical Engineering Material, chapter 1
- [Effective mass can be positive or negative](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aeffective-mass-positive-negative) — Electrical Engineering Material, chapter 1, asked 4×
- [Infinite effective mass and its E-k explanation](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Ainfinite-effective-mass-ek) — Electrical Engineering Material, chapter 1, asked 4×
- [Effective mass of a free electron](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Afree-electron-effective-mass) — Electrical Engineering Material, chapter 1, asked 3×
- [Effective mass of electron in gold](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Agold-effective-mass-justify) — Electrical Engineering Material, chapter 1
- [Negative effective mass at top of valence band](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Anegative-mass-valence-band-top) — Electrical Engineering Material, chapter 1

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