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Homework Help: QM - Transmission coefficient for square well

  1. Aug 23, 2013 #1
    1. The problem statement, all variables and given/known data

    A steady stream of 5 eV electrons impinges on a square well of depth 10 eV. The width of the well is 7.65 * 10^-11 m. What fraction of electrons are transmitted?

    2. Relevant equations
    The following equation for the transmission coefficient, T, is given:
    [tex] T = [1 + \frac{V_0 ^2 sinh^2 κa}{4E(V_0 - E)}]^-1[/tex] (**that is meant to be ^-1 for the whole bracket - apologies, this is my first time using LaTex**)
    Where [tex] κ^2 = \frac{8mπ^2}{h^2}(V_0 - E)[/tex]

    We are also provided with a not-so-subtle hint that [tex] sinh~iθ = i~sinh~θ [/tex]

    3. The attempt at a solution

    So I have assigned the following values based on the information:

    a = 7.65 * 10^-11 m
    E = 5 eV
    V = - 10 eV
    m = 9.11 * 10^-31 kg

    It then seems like it should be very straightforward. I calculate ka and found this to be 1.52i. Then using the definition of sinh I calculate [itex] sinh^2 κa = -4.73[/itex].
    Plugging the other values in I arrive at [tex] T = 0.388 [/tex] which seemed reasonable to me, but..... The postgrad who marked my work fed back to me that the numerical answer he had was T = 0.75.
    I'd be really grateful if someone can check the calculation for me, because it's really bugging me that I can't see my error.
    Thanks in advance.
    Last edited: Aug 23, 2013
  2. jcsd
  3. Aug 23, 2013 #2


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    Staff: Mentor

    The h in the denominator for κ2 should be h2, I guess.

    I can confirm 1.52i, but if I put this in the final formula I get 0.75.
    Well, WolframAlpha does.
  4. Aug 23, 2013 #3
    Thank you, yes it should be h^2, I've now corrected that. I guess I'll go through the figures again carefully :/
  5. Aug 23, 2013 #4
    I've realised my stupid mistake. I copied down the identity incorrectly. A moments thought and I would have seen that [tex] sinh~iθ = i~sinh~θ [/tex] is nonsense :uhh: embarrassing
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