GATE 2026 PH – Question 32
For the energy dispersion of an electron in a one-dimensional solid $E(k)=E_0-2\gamma\cos(ka)$, the ratio of the effective mass of the electron in the solid to the free electron mass ($m_e$) at $k=0$ is $R_0$. Taking $\gamma=0.5$ eV and $a=0.5$ nm, the value of $R_0$ (rounded off to two decimal place) is _____ ($\hbar=1.054\times10^{-34}$ J s, $m_e=9.1\times10^{-31}$ kg, electron charge $=1.6\times10^{-19}$ C)
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Correct answer: 0.30 to 0.32
Explanation
Near the bottom of the band the electron behaves like a free particle with an **effective mass** defined by the curvature of $E(k)$:
$$m^*=\frac{\hbar^2}{d^2E/dk^2}.$$
**Curvature at $k=0$:** with $E=E_0-2\gamma\cos(ka)$,
$$\frac{d^2E}{dk^2}=2\gamma a^2\cos(ka)\;\Rightarrow\;\left.\frac{d^2E}{dk^2}\right|_{k=0}=2\gamma a^2 .$$
**Numbers.** $\gamma=0.5\text{ eV}=0.5\times1.6\times10^{-19}=8\times10^{-20}$ J and $a=0.5\text{ nm}=5\times10^{-10}$ m:
$$2\gamma a^2=2\times8\times10^{-20}\times(5\times10^{-10})^2=4\times10^{-38}\text{ J m}^2 .$$
$$m^*=\frac{(1.054\times10^{-34})^2}{4\times10^{-38}}=\frac{1.1109\times10^{-68}}{4\times10^{-38}}=2.777\times10^{-31}\text{ kg}.$$
**Ratio to the free electron mass:**
$$R_0=\frac{2.777\times10^{-31}}{9.1\times10^{-31}}=\mathbf{0.31}.$$