Module 6 : PHYSICS OF SEMICONDUCTOR DEVICES
Lecture 32 : Covalent Bond
  Example 3
Calculate the effective mass for a simple cubic lattice whose band structure is given by
 
$\displaystyle E(k) = E_0(\cos k_xa + \cos k_ya + \cos k_za)$
  at the point (0,0,0) of the k-space.
  Solution
  Because of symmetry, we have $ m^\ast_{ij}$ is the same for all $ i$ and $ j$ for $ i\ne j$. Thus $ m^\ast_{xy}=m^\ast_{yz}= m^\ast_{zx}$
 
$\displaystyle \frac{1}{m^\ast_{xy}}$ $\displaystyle =$ $\displaystyle \frac{1}{\hbar^2} \frac{\partial^2E}{\partial k_x \partial k_y}$
$\displaystyle =$ $\displaystyle \frac{1}{\hbar^2}E_0(-a\sin(k_xa)-a\sin(k_ya))$
$\displaystyle =$ $\displaystyle 0\ \ {\rm for}\ \ (0,0,0)$
  On the other hand $ m^\ast_{xx}=m^\ast_{yy}=m^\ast_{zz}$ is given by
 
$\displaystyle \frac{1}{m^\ast_{xx}}$ $\displaystyle =$ $\displaystyle \frac{1}{\hbar^2} \frac{\partial^2E}{\partial k_x^2}$
$\displaystyle =$ $\displaystyle \frac{1}{\hbar^2}E_0(-a\cos(k_xa))$
$\displaystyle =$ $\displaystyle -\frac{E_0a^2}{\hbar^2}\ \ {\rm for}\ \ (0,0,0)$
 

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