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Ground state energy for exciton in Si

  1. Oct 24, 2016 #1
    1. The problem statement, all variables and given/known data
    Estimate the ground state energy (eV) for an exciton in Si.
    εSi = 12
    ε = 1.0359×10−10

    Effective masses
    me* = 0.26me
    mh* = 0.36me
    effective mass = 0.15me

    Values of h
    6.626×10−34 J⋅s
    4.136×10−15 eV⋅s
    Values of ħ Units
    1.055×10−34 J⋅s
    6.582×10−16 eV⋅s

    2. Relevant equations
    E1 = me4/(8*h2ε02)
    E = m*e4/(8*h2ε2)

    3. The attempt at a solution
    0.15(9.11e-31)e^4/[8(6.626e-34)^2(1.0359e-10)^2] =
    0.15(9.11e-31)54.59815/[8(6.626e-34)^2(1.0359e-10)^2]
    [link to wolfram for this calculation]
    = 1.98×1056 J
    This seems way too big!
    It's even bigger in eV: 1.24×1075


    I would really appreciate it if someone could help point out where I went wrong. It seems like most want to beat around the bush...or maybe everyone that's helped me here is just in a rush. That epsilon I'm not 100% sure about either.
     
  2. jcsd
  3. Oct 25, 2016 #2

    kuruman

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    0.15(9.11e-31)e^4/[8(6.626e-34)^2(1.0359e-10)^2]
    What do you understand the term highlighted in red to be?
    Wolfram understands to be (Exp[1.0])4. The same understanding applies to all the e's in your expression.
     
  4. Oct 25, 2016 #3
    Yes. I don't see any problem with that. The link to Wolfram clearly shows that without the caret operator (^) is needed for an exponential. 10e10 = 100 billion.
     
  5. Oct 25, 2016 #4

    kuruman

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    What does "e4" mean in your posted equation and how do you implement it?
     
  6. Oct 25, 2016 #5
    That's a typo for e4.
     
  7. Oct 25, 2016 #6

    kuruman

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    And does "e" stand for?
     
  8. Oct 25, 2016 #7
    I believe it's the constant: 2.71828
     
  9. Oct 25, 2016 #8

    kuruman

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    Which constant? What is its value?
     
  10. Oct 25, 2016 #9
    Okay, do you actually know the answer to this minor tangent?! If so, why are you beating around the bush?

    I understand probing for thinking and such is the main idea of Physics Forums, but this is a mere tangent about an equation.
     
  11. Oct 25, 2016 #10
    1.6×10−19
    the charge of an electron?
    That's what someone else said from a different site.
     
  12. Oct 25, 2016 #11

    kuruman

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    I don't think so. It's the charge of the electron.
     
  13. Oct 25, 2016 #12
    0.15(9.11e-31)1.6e-19/[8(6.626e-34)^2(1.0359e-10)^2] =
    5.8×1035 J
    or 36×1045 eV

    I don't know what the ballpark number should be (so I have no idea if this is near correct).
     
  14. Oct 25, 2016 #13

    kuruman

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    Maybe so, but it looks like you did not understand what the symbols in the equation stood for and therefore you did not understand the equation and couldn't use it correctly. I wouldn't call that a "mere tangent".
     
  15. Oct 25, 2016 #14

    kuruman

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    Where is it that you raise the charge of the electron to the fourth power in this expression?
     
  16. Oct 25, 2016 #15
    Ya, I got it. Ate up about 8 hours though.
    14.8mJ
    It is a mere equation, but I don't have time to explain out why. It should not take an hour to point out that the equation is wrong, and what it actually should be! I mean, a guy from another site is the one who actually helped.
     
  17. Oct 25, 2016 #16

    kuruman

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    You are welcome.
     
  18. Oct 25, 2016 #17
     
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