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Band Theory of Solids Questions and Answers

1.

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

Exam Year(s)

GATE Physics - 2018,

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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2.

For nonrelativistic electrons in a solid, different energy dispersion relations (with effective masses \(m_{a}^{*}\), \(m_{b}^{*}\), and \(m_{c}^{*}\)) are schematically shown in the plots. Which one of the following options is CORRECT?

Three parabolas E against k with minimum at k = 0: the dotted mc curve is the narrowest, the dashed mb curve is wider, and the solid ma curve is the widest

Exam Year(s)

GATE Physics - 2023,

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.

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3.

In an allowed band of semiconductor the effective mass m* of the electron is infinite :

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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.

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