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Calculating binding energies of ground state electrons

  1. Oct 10, 2009 #1
    1. The problem statement, all variables and given/known data

    What is the calculated binding energy of the electron in the ground state of (a) deuterium, (b) [tex]He^{+}[/tex] and (c) [tex]Be^{+++}[/tex]?

    2. Relevant equations

    For the hydrogen atom, [tex]E_{n}[/tex] = - [tex]E_{o}[/tex] / [tex]n^{2}[/tex]

    [tex]E_{o}[/tex] = [tex]me^{4}[/tex] / 2[tex]hbar[/tex]^2(4[tex]\pi[/tex]Eo)^2

    3. The attempt at a solution

    Not sure how to do this. Can I apply the above equations which are meant for a hydrogen atom to these other atoms? I dont believe so but I'm unsure how to change them.

    I know how to correct the Rydberg constant for the different atoms, but it doesnt even appear in any of these formulas so I don't think that would change anything.

    Not sure where to start. Help?
     
  2. jcsd
  3. Oct 12, 2009 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    Since you aren't dealing with Hydrogen, shouldn't you be using the equation for the approximate energy levels for any Hydrogenic atom with [itex]Z[/itex] protons instead of the equation for Hydrogen's energy levels?
     
  4. Oct 12, 2009 #3
    Yes, that would make sense, although I can't find that equation anywhere.
     
  5. Oct 12, 2009 #4
    Ok, I finally found it online. [tex]E_{n}[/tex] = ([tex]-Z^{2}[/tex][tex]R_{e}[/tex]) / [tex]n^{2}[/tex]
    Where [tex]R_{e}[/tex] = 0.5([tex]m_{e}[/tex][tex]c^{2}[/tex])[tex]\alpha^{2}[/tex]
    Where [tex]\alpha[/tex] = 1/137
    Therefore [tex]R_{e}[/tex] = 0.5 [(9.11x10^-31 kg)(3.00x10^8 m/s)^2](1/137)^2
    =2.184x10^-18 J = 13.65 ev

    So for deuterium,
    [tex]E_{n}[/tex] = [[tex](-1)^{2}[/tex](13.65ev)] / [tex]1^{2}[/tex]
    [tex]E_{n}[/tex] = 13.65ev

    For He, the difference is that Z = 2, and thus E = 4(-13.65ev) = -54.6ev
    and for Be Z = 4, and thus E = 16 (-13.65ev) = -218.4 ev
     
    Last edited: Oct 12, 2009
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