QB_PH_SEC7_002: Band Theory of Solids & Semiconductors

The energy-wavevector dispersion relation for an electron in the conduction band of a 1D crystal is given by $E(k) = E_0 - 2t \cos(ka)$, where $t = 1.0\text{ eV}$ and $a = 3.0\text{ Å}$. What is the effective mass of the electron $m^*/m_e$ at the bottom of the band ($k = 0$)? Take $\hbar = 1.054 \times 10^{-34}\text{ J}\cdot\text{s}$ and $m_e = 9.11 \times 10^{-31}\text{ kg}$. (Round off your answer to two decimal places)

Correct Answer

### Step-by-Step Derivation & Calculation: 1. **Dispersion Derivatives**: $E(k) = E_0 - 2t \cos(ka)$ $\frac{dE}{dk} = 2t a \sin(ka)$ $\frac{d^2 E}{dk^2} = 2t a^2 \cos(ka)$ 2. **Evaluation at Band Bottom ($k = 0$)**: $\left.\frac{d^2 E}{dk^2}\right|_{k=0} = 2t a^2$ 3. **Effective Mass Definition**: $m^* = \frac{\hbar^2}{d^2 E / dk^2} = \frac{\hbar^2}{2 t a^2}$ 4. **Numerical Evaluation**: - $t = 1.0\text{ eV} = 1.602 \times 10^{-19}\text{ J}$ - $a = 3.0\text{ Å} = 3.0 \times 10^{-10}\text{ m} \implies a^2 = 9.0 \times 10^{-20}\text{ m}^2$ - $\hbar = 1.05457 \times 10^{-34}\text{ J}\cdot\text{s} \implies \hbar^2 = 1.112 \times 10^{-68}\text{ J}^2\text{s}^2$ - Denominator $= 2 \times (1.602 \times 10^{-19}) \times (9.0 \times 10^{-20}) = 2.8836 \times 10^{-38}\text{ J}\cdot\text{m}^2$ $m^* = \frac{1.112 \times 10^{-68}}{2.8836 \times 10^{-38}} = 3.856 \times 10^{-31}\text{ kg}$ 5. **Mass Ratio**: $\frac{m^*}{m_e} = \frac{3.856 \times 10^{-31}}{9.11 \times 10^{-31}} \approx 0.42$ 6. **Official Range**: $[0.40, 0.45]$.