--- title: "Effective mass can be positive or negative" subject: "Electrical Engineering Material" chapter: 1 key: "semester_3/electrical_engineering_material:effective-mass-positive-negative" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/effective-mass-positive-negative site: BCT Engineering --- # Effective mass can be positive or negative Electrical Engineering Material · Chapter 1: Theory of Metals - Asked in: 2081 Bhadra, 2+6 marks; 2071 Shrawan, 8 marks; 2068 Shrawan, 8 marks (asked 3 times) - Video: [Effective mass in semiconductors, Derivation, Difference between Mass and effective mass in Hindi](https://www.youtube.com/watch?v=Y9Akkjlo498) — Dr. Vishnu Awasthi - 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-positive-negative - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1 ## Question Derive the expression for effective mass of electron and show that it can be positive as well as negative. ## Short answer Effective mass m* = ħ²/(d²E/dk²). Near the bottom of a band, d²E/dk² > 0 gives m* > 0; near the top, d²E/dk² < 0 gives m* < 0, meaning the electron accelerates opposite to an applied force, a consequence of the band's concave downward curvature. ## Answer The **effective mass** $m^*$ is the mass an electron in a crystal appears to have when it is accelerated by an external force; it includes the effect of the periodic lattice potential. It is given by $m^* = \hbar^2/(d^2E/dk^2)$. #### Derivation An electron in a crystal is a wave packet. Its velocity is the group velocity: $$ v_g = \frac{d\omega}{dk} = \frac{1}{\hbar}\frac{dE}{dk} \qquad (E = \hbar\omega) $$ If an external force $F$ (e.g. $-eE$) acts for time $dt$, the work done raises the electron's energy: $$ dE = F\,v_g\,dt = F\frac{1}{\hbar}\frac{dE}{dk}dt $$ Since $dE = \frac{dE}{dk}dk$, this gives $$ F = \hbar\frac{dk}{dt} $$ The acceleration is $$ \begin{aligned} a &= \frac{dv_g}{dt} = \frac{1}{\hbar}\frac{d}{dt}\left(\frac{dE}{dk}\right) = \frac{1}{\hbar}\frac{d^2E}{dk^2}\frac{dk}{dt} \\ &= \frac{1}{\hbar^2}\frac{d^2E}{dk^2}\,F \end{aligned} $$ Comparing with $F = m^*a$: $$ m^* = \frac{\hbar^2}{\dfrac{d^2E}{dk^2}} $$ #### Positive and negative effective mass In a crystal the E–k curve of a band is not a simple parabola. In the tight-binding picture a band can be written as $$ E(k) = E_0 - 2\gamma\cos(ka), \qquad \gamma > 0 $$ Then $$ \frac{d^2E}{dk^2} = 2\gamma a^2\cos(ka), \qquad m^* = \frac{\hbar^2}{2\gamma a^2\cos(ka)} $$ ```text E ^ top of band: curve bends down | ___ ___ | / \ m* < 0 m*<0 / \ | \ / | \ <- inflection: / | \ m* = infinite / | \ / | \__ __/ <- bottom: m* > 0 | '--___--' +---|---------|---------|-------> k -pi/a 0 +pi/a ``` - **Near the bottom of the band** ($k \approx 0$): $\cos ka \approx 1$, the curve is concave up, $d^2E/dk^2 > 0$, so **$m^*$ is positive**. Here $E \approx E_{min} + \hbar^2k^2/2m^*$, like a free electron. - **Near the top of the band** ($k \approx \pm\pi/a$): $\cos ka \approx -1$, the curve is concave down, $d^2E/dk^2 < 0$, so **$m^*$ is negative**. An applied force makes the electron accelerate opposite to the force, because it is being Bragg-reflected by the lattice. - **At the inflection points** ($k = \pm\pi/2a$): $d^2E/dk^2 = 0$, so $m^* \to \infty$; the electron does not accelerate at all. #### Significance An electron with negative mass and negative charge near the top of the valence band behaves like a particle with **positive mass and positive charge**. This is the idea of the **hole**, used to describe conduction in the valence band of semiconductors. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the derivation, the electron is treated as a wave packet with velocity: A. v = E / p B. v_g = (1/hbar) dE/dk C. v = hbar k D. v = d^2E/dk^2 Answer: B. The group velocity v_g = d(omega)/dk = (1/hbar) dE/dk. (Key point: Group velocity) 2. Equating work done to energy gain, dE = F v_g dt, yields which relation? A. F = hbar^2 dk/dt B. F = dk/dt C. F = hbar dk/dt D. F = dE/dk Answer: C. dE = (dE/dk)dk and dE = F (1/hbar)(dE/dk) dt give F = hbar dk/dt. (Key point: Force and k) 3. Why is m* positive near the bottom of the band? A. The E-k curve is concave up, so d^2E/dk^2 > 0 B. The curve is concave down there C. The curve is flat there D. The velocity is maximum Answer: A. Near k = 0, cos ka is about 1 so d^2E/dk^2 = 2 gamma a^2 cos(ka) > 0. (Key point: Sign at band bottom) 4. What does negative m* near the top of the band imply? A. The electron stops B. The electron accelerates opposite to the applied force, leading to the hole concept C. The electron's charge is zero D. The electron becomes a free electron Answer: B. Near the top (Bragg reflection) the electron accelerates opposite to the force, equivalent to a positive-charge, positive-mass hole. 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