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Now we have a step potential

  1. Jan 4, 2006 #1
    I am taking a course in solid state electronics at the university and I was thinking about one detail in the calculation of k. In a simple situation like a step potential we say that k=sqrt(2m*abs(E-V))/hbar . Now my question is What happens if the electron approaching the barrier is having an energy E=V. According to the formula k will be zero. Does that me that the electron will not "move" in the barrier. Is (E=V) a resemblence of the infinite potential case but at a single E value???

    Thanks...
     
  2. jcsd
  3. Jan 4, 2006 #2

    HallsofIvy

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    If you are taking a course in solid state physics you should have covered this in a previous course. " Is (E=V) a resemblence of the infinite potential case but at a single E value???" I haved no idea what you mean by that but: FINITE is not the same as INFINITE! Unless your step is to an "infinite" potential, you have a finite potential and there is a probability that the electron will penetrate the potential step. If I remember correctly the standard method of solving the differential equation, to determine what that probability is, is the "WKB" approximation.
     
  4. Jan 4, 2006 #3
    Sorry if I wasn't clear. What I meant is that if at E=V our k=0 and the electron has a zero chance of penetration then this means that at this specific case the "FINITE" barrier will acts as an "INFINITE" one from the point of view of that electron having E=V...
     
  5. Jan 9, 2006 #4
    There are no stationary states, if that's your question. There are only scattering states, like with a free particle.
     
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