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Gap energy and changing diameters

  1. Jan 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi,

    I'm trying to show how the band gap energy changes with the radius as Part 2 of this questions asks us to find: http://screencast.com/t/N7R42aS4l [Broken]

    The full solution is here: http://screencast.com/t/y1K3hqOus4H [Broken]

    2. Relevant equations

    See solution link above.

    3. The attempt at a solution

    This question should relatively straightforward and should just be a simple case of inputting the values into the equation given in the link above. I think we also need to pick a radius at set intervals (ie. 1,2,3,4.. nm) for use in the equation. The problem is that I always get 1.42 for my Eg(R) value. I tested my equation and this seems to be due to the value I get being so low that when I add the first term of Eg to it I always end up with that as my final answer.

    Any thoughts on what I could be doing wrong?

    Thanks
    David
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 6, 2013 #2

    TSny

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    You'll need to watch your units carefully. For example, can you show what you get for just the last term ##1.8e_o^2/(\epsilon R)## for ##R = 1## nm? This looks like a Coulomb energy term. In SI units the ##e_o^2## part would have an additional factor of Coulomb's constant ##k = 1/4\pi \epsilon_o##
     
  4. Jan 6, 2013 #3
    Hi TSny,

    For that last term I get [itex]\frac{1.8 * (1.602*10^-19)^2}{(12.9 * 1*10^-9)}[/itex]

    = 3.58 *10-30 c2/nm

    I'm not sure what you mean about an additional factor for Coulomb's constant?

    Thanks!
     
  5. Jan 6, 2013 #4

    TSny

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    The Coulomb energy of two point charges ##e_o## is ##U = k \;e_o^2/R## where ##k ## is Coulomb's constant. Note that you are getting units of C2/nm which is not an energy unit. However, if you multiply by ##k##, you will get energy since ##k## has units of J m /C2 where J is Joules. You'll also need to convert over to eV.
     
  6. Jan 6, 2013 #5
    Hi,

    Thanks for that I think I understand the logic now, I re-did the equation and now have: http://screencast.com/t/yByLIrwdi9 [Broken]

    The problem is that no matter what I change the radius to all the answers are always 1.41...J

    Sorry for all the trouble!
    Thanks
     
    Last edited by a moderator: May 6, 2017
  7. Jan 6, 2013 #6

    TSny

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    Almost there. For the last term where you included ##k = 1/(4 \pi \epsilon_o)##, you need to use ##\epsilon_o = 8.85 \times 10^{-12}## SI units. For the second term, you just need to convert from J to eV. You should then be able to express the result as ##E_G(R) = 1.42 + a/R^2-b/R## where ##a## and ##b## are numerical factors (roughly of the order of 1), E is in eV, and R is in nm.
     
  8. Jan 6, 2013 #7
    Thanks! I think we might have it :)

    My answer using a radius of 4 gives 1.8ev which from the graph in the answer looks about right?

    A few quick questions:

    How did you know how to convert the value of Eo to 8.85*10^-12?

    If I had to draw a graph like this in an exam would I keep picking radii in increments of 1 until the graph levels out? How would I know what value to start with in this case?

    My current working: http://screencast.com/t/oOKJqB6rmEdX [Broken]
     
    Last edited by a moderator: May 6, 2017
  9. Jan 6, 2013 #8

    TSny

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    I think you have it. See if you can multiply out all your numbers and get

    ##E_G(R) = 1.42 + a/R^2-b/R##

    where ##a## abd ##b## are fixed numbers of roughly the order of 1. Then you can easily evaluate the expression for different values of R (in nm).

    Generally, you would probably want to choose values of R on the order of 1 nm. So, maybe R = 0.5, 1, 2, 3, 4, 5 nm.

    For ##\epsilon_0## see the value for permitivity of free space.
     
  10. Jan 6, 2013 #9
    Thanks for all your help Tsny, I was going crazy trying to figure this out =)
     
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