Discussion
The energy dispersion for electrons in one dimensional lattice with lattice parameter \(a\) is given by \(E(k) = E_{0} - \frac{1}{2}W\cos ka\), where \(W\) and \(E_{0}\) are constants. The effective mass of the electron near the bottom of the band is
Answer: A
The effective mass at the band bottom is \(\frac{2\hbar^{2}}{Wa^{2}}\).
The effective mass is set by the curvature of the band: \(m^{*} = \hbar^{2}/(d^{2}E/dk^{2})\). The band bottom is at \(k = 0\), where \(E = E_{0} - W/2\) is least.
\[ \begin{aligned} \frac{d^{2}E}{dk^{2}} &= \tfrac{1}{2}Wa^{2}\cos ka \to \tfrac{1}{2}Wa^{2} \\ m^{*} &= \frac{\hbar^{2}}{\tfrac{1}{2}Wa^{2}} = \frac{2\hbar^{2}}{Wa^{2}} \end{aligned} \]
Equivalently, near \(k = 0\), \(\cos ka \approx 1 - k^{2}a^{2}/2\), so \(E \approx E_{0} - W/2 + Wa^{2}k^{2}/4\), and matching to \(\hbar^{2}k^{2}/2m^{*}\) gives the same result. \(\hbar^{2}/Wa^{2}\) drops the factor \(\tfrac{1}{2}\) in the dispersion, and the smaller options misplace it further.
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