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Flow of electrons hit a potential hole

  1. Jan 11, 2014 #1
    1. The problem statement, all variables and given/known data
    Flow of 500 electrons per second with kinetic energy 3 eV hits a perpendicular 5 eV potential hole 0.3 nm wide. How many electrons pass per second pass the obstacle?


    2. Relevant equations



    3. The attempt at a solution

    Hmm, I checked my notes where it is written that coefficient of electrons that passes the obstacle is calculated as ##T=(1+\frac{1}{4}(\frac{k_1}{\kappa }+\frac{\kappa }{k_1})^2sinh^2(\kappa a))^{-1}##

    Where I used notation ##k_1=\sqrt{\frac{2mE}{(h)^2}}## and ##\kappa =\sqrt{\frac{2m(V-E)}{(h)^2}}##. I don't know how to write crossed h in latex, so I used (h) instead. Notation a tells how wide the hole is.

    So ##k_1=8.66 nm^{-1}## and ##\kappa =7.07 nm^{-1}## and ##sinh^2(\kappa a)=0.00137##

    Which gives me ##T=0.05377## and therefore 26 electrons should pass the obstacle. BUT the result in the book states 408 electrons as result...


    Doesn anybody know what am I doing wrong here?
     
  2. jcsd
  3. Jan 11, 2014 #2

    TSny

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    Not sure what potential "hole" means. But if it means a potential "well" of depth 5 eV, then the kinetic energy will still be positive inside the well. So, you will have an oscillatory wavefunction inside the well rather than exponential behavior. Instead of a "kappa" ##\kappa##, you'll have a ##k_2## wavevector inside the well. What happens to the Sinh function in this case?
     
  4. Jan 12, 2014 #3
    Potential well it is. In direct translation from my language it is a hole. :)

    Now here is my question. How will the kinetic energy still be positive inside the well? Before the well it is 3 eV and the well has a depth of 5 eV.

    so ##k_2 =\sqrt{\frac{2m(E-V)}{(h)^2}}=i\sqrt{\frac{2m(v-E)}{(h)^2}}=i\kappa ##

    In case you are right, which you probably are but I would like to understand why... sinh is than sin function.
     
  5. Jan 12, 2014 #4

    TSny

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    If you take the potential to be 0 outside the well, then inside the well it will be -5eV. The kinetic energy is the difference between E and V: E-V. This gives a positive value of the KE inside the well.

    Right, the sinh function becomes a sin function.
     
  6. Jan 12, 2014 #5
    So E-V=8 eV.

    Than, only if V is positive than wavevector ##k_2## will be complex, or... ?
     
  7. Jan 12, 2014 #6

    TSny

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    If V is positive and greater than E, then the wavevector will be imaginary. So, if you had a potential barrier of height 5 eV with E = 3 ev, then the kinetic energy E-V would be negative inside the barrier.
     
  8. Jan 12, 2014 #7
    How does this differ from my original post (problem)?
     
  9. Jan 12, 2014 #8

    TSny

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    The original post dealt with a potential well (I think), whereas, my last comment was for a potential barrier.
     
  10. Jan 12, 2014 #9
    For a moment a thought that's the same.

    Thank you for your help!
     
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