Band Theory of Solids Questions and Answers

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.

Answer: D

The correct order is \(m_{a}^{*} \gt m_{b}^{*} \gt m_{c}^{*}\).

For free-electron-like bands,

\[ \begin{aligned} E &= \frac{\hbar^{2}k^{2}}{2m^{*}} \\ \frac{1}{m^{*}} &= \frac{1}{\hbar^{2}}\frac{d^{2}E}{dk^{2}} \end{aligned} \]

so the effective mass is inversely proportional to the curvature of \(E(k)\). The \(m_{c}^{*}\) band is the narrowest, most sharply curved parabola, so its effective mass is the smallest; the \(m_{a}^{*}\) band is the flattest, so its mass is the largest; \(m_{b}^{*}\) lies between.

Equal masses would need identical parabolas, and the other two orders put a sharply curved band above a flatter one, which reverses the relation.

Answer: C

The effective mass is infinite in the middle of the energy band.

The effective mass is set by the curvature of the band, \( \frac{1}{m^{*}} = \frac{1}{\hbar^{2}}\frac{d^{2}E}{dk^{2}} \). Near the bottom of a band \( E(k) \) curves upward and \( m^{*} \) is positive; near the top it curves downward and \( m^{*} \) is negative. In between there is an inflection point where the curvature is zero, so \( m^{*} \) becomes infinite: an applied force there produces no acceleration.

So it is finite (positive) at the bottom, finite (negative) at the top, and 'never' is wrong because every band has such an inflection point.

Answer: D

The band width is 12β.

Each cosine ranges from -1 to +1, so their sum ranges from -3 to +3:

\[ \begin{aligned} E_{min} &= E_{0} - 2\beta(3) = E_{0} - 6\beta \quad (\vec{k} = 0) \\ E_{max} &= E_{0} + 6\beta \quad (k_{x}, k_{y}, k_{z} = \pm\pi/a) \\ \Delta E &= 12\beta \end{aligned} \]

(The paper names the lattice constant α in the text but writes a in the formula.) This is the tight-binding band of a simple cubic lattice, whose width is 2zβ with z = 6 nearest neighbours. 4β is the width of a one-dimensional chain, 8β that of a square lattice, and 2β is a single hopping energy.

Answer: A

The correct statement is that the slope of E versus k is proportional to the group velocity.

The velocity of an electron wave packet in a band is the group velocity

\[ v_{g} = \frac{1}{\hbar}\frac{dE}{dk} \]

so the slope of the E-k curve is \(\hbar v_{g}\).

At the zone boundary \(|k| = \pi/a\) Bragg reflection produces standing waves and the slope is zero, not maximum. The curve is parabolic only near the band edges, where the effective mass approximation holds, not across the whole zone. The slope is also zero at k = 0, so it is not non-zero everywhere in the interval.

Answer: C

With no barrier, there are no forbidden energy regions.

In the Kronig-Penney model, the band gaps come from Bragg reflection by the periodic potential barriers. If the barrier strength is zero, the electron is free: \(E = \frac{\hbar^{2}k^{2}}{2m}\), a continuous parabola, and every energy is allowed.

So there is no periodic dependence of energy on \(k\) (beyond the trivial zone folding), energy is a continuous function of \(k\), and all energies are allowed, which rules out the other three statements.