Module 6 : PHYSICS OF SEMICONDUCTOR DEVICES
Lecture 31 : Electron in a Potential Well
 
$\displaystyle \sin ka =0\ \ \ {\rm so\ that}\ \ k = \frac{n\pi}{a}$
  where $ n$ is an integer $ 1,2,3,\ldots$. ( $ n=0$ is not possible because it would make the wavefunction inside the well vanish; negative values of $ n$ does not give new solution as it amounts to multiplying the whole wavefunction by $ -1$.) Thus the electron energy in the potential well are
 
$\displaystyle E = \frac{\hbar^2k^2}{2m} = \frac{n^2\pi^2\hbar^2}{2ma^2}$
  The energy levels corresponding wave functions are shown in the figure.
 
\includegraphics{fig5.07.eps}
 
See the animation
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