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Energy Levels of Quantum Dots

  1. May 8, 2015 #1
    Say we have a Quantum Dot with $$n=10^{26}m^{-3}$$ and radius $$R = 3nm$$ then this will give us of the order of 13 electrons. My question is how do you relate the number of electrons to the quantum numbers n and l in order to use the spherical Bessel function values?

    In class for 13 electrons we were given $$n,l=0,2 (^{1}D)$$ which he then inserted into the formula as $$E_{0,2}= \frac{\hbar^{2} \beta_{0,2}^{2}}{2m^{*}R^{2}}$$ which is the spacing between energy levels.


    Just not sure how he got the n=0 l=2 and the shell 1D, any help would really be appreciated.
     
  2. jcsd
  3. May 9, 2015 #2

    mfb

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    That looks like an example.
    At ground state, the electrons will occupy the lowest 13 states (note that you have two spin orientations). You can find allowed values of n and l in your lecture notes, book or similar.
     
  4. May 9, 2015 #3
    Thanks for the reply, yes I think you are right, if 13 electrons filled the first s p d orbits then 2 in s, 6 in p and 5 in d, so he has said it's the d orbit we want. No idea how he got the n and l values with that info, pretty sure he wants us (somehow) to do it though.

    thanks for the reply!

    edit think it's something to do with that for 1s2 n=0 l=0, for 1p6 n=0 l=1 for 1d10 n=0 l=2 etc etc
     
  5. May 24, 2015 #4
    I'll update this for anyone else who has this problem in future.

    You would be given

    IUQpHGB.jpg

    You have electronic structure of SPDSPF.....

    So for example if you have 1s2 1p6 1d10 (which is 18 electrons), you start with the LOWEST number for beta from the chart for 1s2, then the next lowest for 1p6, then the next lowest for 1d10. So for 1s2 1p6 1d10 you would be at n=0 l=2. BUT be careful if you had a longer configuration (i.e. more electrons) the next beta value would be 39.5 and so l=0 n=1. NOT l=3 n=0. This is where I was getting mixed up.
     
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