Energy Levels in a Crystalline Solid — Question 3

Question 3

The tight-binding band carries fastest electrons away from its extrema

Use $E(k)=E_0-2t\cos(ka)$ and group velocity $v=(1/\hbar)dE/dk$. Take $t=1.00\,\mathrm{eV}$, $a=.300\,\mathrm{nm}$, and $\hbar=6.582\times10^{-16}\,\mathrm{eV\,s}$.

1. Derive v and its maximum magnitude.

2. Find velocities at $k=0$, $k=\pi/(2a)$ and $k=\pi/a$.

Question 3 - Solution

The tight-binding band carries fastest electrons away from its extrema

(a) Method and calculation

$$\boxed{v(k)=\frac{2ta}{\hbar}\sin(ka)}.$$$$\boxed{v_{\max}=2ta/\hbar=9.116\times10^5\,\mathrm{m/s}}.$$

(b) Interpretation and further calculation

$$\sin0=\sin\pi=0\Rightarrow\boxed{v(0)=v(\pi/a)=0}.$$$$\sin(\pi/2)=1\Rightarrow\boxed{v(\pi/2a)=+v_{\max}}.$$

Physical check

An electron at the top of this band has zero group velocity despite high energy. Crystal transport depends on the dispersion slope rather than energy alone.