1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Flow of electrons hit a potential hole