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Working in units of Angstroms

  1. May 22, 2015 #1
    1. The problem statement, all variables and given/known data
    j1iz5vq.jpg

    2. Relevant equations
    None.

    3. The attempt at a solution
    I understand the question, for a phonon it can excite an electron from the valance band to the conduction band between any values of k. If it were a photon it could only excite the electron at specific k values.

    Anyway so looking at the graph minimum/maximum looks (i tried differentiating Ec and Ev and setting to zero to find the exact minimum and maximum but it ended up getting complicated) like it is at k=0 for Ev and k=1 for Ec.

    So now I need to put k=1 into the equation for Ec. And this is what I'm struggling with, so I put k=1 in and get $$E_{c} = 11 - 2\AA^{-2} + \AA^{-4} = 11 - \frac{2}{(10^{-10})^{2}}+ \frac{1}{(10^{-10})^{4}}$$ which must be wrong as the angstrom terms are really small and Ec goes massive.

    Any advice on this would be appreciated.

    edit the red AA's are angstroms, dont know why its not working
     
  2. jcsd
  3. May 22, 2015 #2

    BvU

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    No, in the expression for Ec is Ec in eV and k is in ##\unicode{x212B}##.
    So when you differentiate, you do so in ##\unicode{x212B}## units. (in fact in the dimensionless units ##u = k/\unicode{x212B}## ).
    You get ##-4u+4u^3 = 0 \Leftrightarrow u(u^2-1) = 0##

    And the ##E_c## expression is scaled too, in the dimensionless quantity y = E/eV.

    Often in graphs you see axis annotations like ##E_c/{\rm eV}## and ##k * \unicode{x212B}^2## and those are the numbers (dimensionless !) at the tick marks. I like that and find it more correct than e.g. ##E (eV)
     
  4. May 22, 2015 #3
    Ahh so the Angstroms cancel I see, and thanks for the tip on the derivative!

    Thanks! I get it now but why then on the graph does it say k (##\unicode{x212B}^{-1}##) not k(##\unicode{x212B}##)
     
  5. May 22, 2015 #4

    BvU

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    Yes, (unfortunately) it's customary to write e.g Distance (km) instead of Distance/km as axis title. And the unit of ##k## is ## \unicode{x212B}^{-1} ##.
     
  6. May 23, 2015 #5
    Just one more question that's related if I may, if I had something like: $$q=\frac{\omega }{\upsilon }= \frac{36GHz}{458m/s}=786\cdot 10^{5}$$ could you then write it as $$786\cdot 10^{5}\cdot \frac{\unicode{x212B}}{\unicode{x212B}}=0.008 \unicode{x212B}^{-1}$$
     
  7. May 23, 2015 #6

    BvU

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    This looks really weird. Normally ##\omega## is not expressed in GHz (dimension 1/s) but in radians/sec. Makes a factor ##2\pi## difference, as in ##\omega = 2\pi f\;## or ##\omega = 2\pi \nu\;##. Don't know what ##q## stands for, or what its dimension is.

    Aha ! enlightened myself in the modern fashion by googling wavenumber and found ##\displaystyle k = {2\pi\over \lambda}= {2\pi\over v_p}={\omega\over v_p}##. This simply has the dimension of length-1.

    With your ##\displaystyle
    q=\frac{\omega }{\upsilon }= \frac{36GHz}{458m/s}=786\cdot 10^{5}\; {\rm \bf m^{-1}}## you can get inverse ##\unicode{x212B}## by simply multiplying by 1 :smile: !

    This 1 here taking the form ##\displaystyle 1\, {\rm m} \over 10^{10} \, {\displaystyle \unicode{x212B}} ## or ## \left ( 10^{10} \, {\displaystyle \unicode{x212B}} \right ) ^{-1} \over {\rm m}^{-1} ## , in other words, you get $$
    q=786\cdot 10^{5}\; {\rm \bf m^{-1}} = 786\cdot 10^{5} \, {\rm m^{-1}}\; 10^{-10} \, {{\displaystyle \unicode{x212B}} ^{-1} \over \displaystyle{\rm m}^{-1}} = 786\cdot 10^{-5}\,
    {\displaystyle \unicode{x212B}} ^{-1}$$

    --
     
  8. May 24, 2015 #7
    That's awesome thanks! Will just have to practice this a little more.
     
  9. May 24, 2015 #8
    Coming back to the derivative, if you set it equal to zero it makes sense however on the first part of the question below:

    H5tWDvw.jpg

    You would have for the effective mass $$m^{*}= \frac{\hbar^{2}}{\left ( \frac{\partial^2 E}{\partial k^2} \right )}$$

    If we apply what you said earlier we get $$\frac{\partial^2 E_{c}}{\partial u^2} = -4 + 12u$$ but we need $$ \frac{\partial^2 E}{\partial k^2}$$
     
  10. May 25, 2015 #9
    also in a seperate question

    bzctjgt.jpg

    if you set u=k/A then differentiate twice for the Ec term the u vanishes!

    exam on Wednesday, can anyone help further?
     
  11. May 26, 2015 #10
    I'll update this for anyone interested:

    qVZzpAV.jpg
     
  12. May 26, 2015 #11

    BvU

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    With ## {\partial \over \partial k} = {\partial \over \partial u} {du \over dk}\ ## and ##\ u = { k\over \unicode{x212B}} ## you get $$
    m^* = {\displaystyle \hbar^2 \over {\partial^2 \left ( E/eV \right ) \over \partial u^2} \left(du \over dk \right ) ^2 eV } = {\hbar^2 \over {(4-12u)} \; {\rm eV \,\unicode{x212B}^2 }} $$
    which is also dimensionally correct: ## {\rm (Js)^2 \over J m^2 } = {\rm kg}##

    --

    It's becoming a rather intermixed mess of exercises now...

    The u may vanish, but there still is a coefficient -5 10-19 ( a constant) !
    (In this exercise A3 I wonder why there appear coefficients of the order of magnitude 10-19 and ##(\unicode{x212B})^2## instead of the -2 we were used to...)

    But in your post # 10 yet other values again. A bit confusing for me, but I hope you have it clear by now. Looks like you do.

    Good luck tomorrow !

    --
     
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