--- title: "Negative effective mass at top of valence band" subject: "Electrical Engineering Material" chapter: 1 key: "semester_3/electrical_engineering_material:negative-mass-valence-band-top" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/negative-mass-valence-band-top site: BCT Engineering --- # Negative effective mass at top of valence band Electrical Engineering Material · Chapter 1: Theory of Metals - Asked in: 2072 Chaitra, 2+4 marks - 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/negative-mass-valence-band-top - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1 ## Question What is effective mass? The electron at the top of valence band is said to have negative effective mass. Explain with the help of E-k diagram. ## Short answer Effective mass m* = ħ²/(d²E/dk²) accounts for lattice forces. At the top of the valence band the E–k curvature is negative, so m* is negative; the electron accelerates opposite to the applied field. This behavior is treated as a hole with positive charge and positive mass. ## Answer **Effective mass** $m^*$ is the apparent mass of an electron in a crystal when it is acted on by an external force, including the effect of the lattice: $$ m^* = \frac{\hbar^2}{d^2E/dk^2} $$ It depends on the curvature of the E–k curve. #### Negative effective mass at the top of the valence band ```text E ^ Conduction band | \ / | \ m* > 0 / | `-.__ __.-' | ''--....--'' bottom of CB | (concave up) | - - - - - - Eg - - - - - - - - - | .--~~~~--. top of VB | _.-' m* < 0 '-._ (concave down) | / \ | / Valence band \ +---------------|------------------> k 0 ``` - At the **top of the valence band** the E–k curve is an inverted parabola: $E \approx E_v - \dfrac{\hbar^2k^2}{2|m^*|}$. - So $\dfrac{d^2E}{dk^2} = -\dfrac{\hbar^2}{|m^*|} < 0$, and therefore $m^* = \dfrac{\hbar^2}{d^2E/dk^2}$ is **negative**. #### Physical meaning An electron near the top of the band has a wave vector close to the Bragg reflection condition. When an electric field pushes it, the lattice reflects it so strongly that the electron accelerates **opposite** to the direction expected for a normal negative particle. An electron with negative charge ($-e$) and negative mass ($-|m^*|$) accelerates in the same direction as a particle with **positive charge and positive mass**: $$ a = \frac{(-e)E}{-|m^*|} = \frac{(+e)E}{+|m^*|} $$ So instead of tracking the many electrons with negative mass at the top of the valence band, we track the few **empty states** there and treat each one as a **hole** with charge $+e$ and positive effective mass $m_h^*$. This is why holes in p-type semiconductors move in the direction of the field and carry current like positive particles. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Near the top of the valence band, d^2E/dk^2 is: A. Positive B. Zero C. Infinite D. Negative Answer: D. The E-k curve is an inverted parabola, so m* is negative. (Key point: Curvature) 2. The electron near the band top accelerates opposite to the force because: A. It is repelled by the field B. The lattice reflects it (Bragg condition) more strongly than the applied force C. Its charge is positive D. It has zero velocity Answer: B. Strong lattice reflection gives an effective negative mass. (Key point: Why opposite) 3. Treating the empty state as a hole gives it: A. Negative charge and negative mass B. Positive charge and negative mass C. Positive charge and positive effective mass D. No charge Answer: C. (-e)/(-|m*|) = (+e)/(+|m*|). 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