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Ground State Energy of an Electron

  1. Oct 1, 2008 #1
    This is what I've tried to work out and I'm not getting -13.7 eV. What am I doing wrong?


    E 2 Π m e^4 / (4 Π ε0 )^2 h^2 6.90E-19 J=4.31eV

    m 9.11 x 10-31 kg 9.11E-31
    e 1.60 x 10-19 C 1.60E-19
    ε0 8.85 x 10-12 C2/Nm2 8.85E-12
    h 6.63 x 10-34 J S 6.63E-34

    1 joule = 6.24150974 × 10^18 electron volts
     
  2. jcsd
  3. Oct 1, 2008 #2
    Sorry but I have absolutely no idea what that equation is supposed to be, or where you got it from. Try TeXing it perhaps? How did you arrive at it?
     
  4. Oct 2, 2008 #3
    Need more information, have no idea what you're saying.
     
  5. Oct 3, 2008 #4
    The equation is based on the Bohr atom for energy levels. The version I wrote above is for n=1. The development in the text I'm using:

    AUTHOR Mortimer, Robert G.
    TITLE Physical chemistry / Robert G. Mortimer.
    PUB INFO San Diego, Calif. : Academic Press, c2000.
    pgs. 511-520 roughly

    uses En = 2 Π m e4 / n2 (4 Π ε0 )2 h2

    When I plug in the constants, n=1, the value is off from 13.7 eV, after conversion from Joules, by a factor of 3.14, as if Pi doesn't belong in the denominator. I'm thinking that it is already included in the permitivity constant ε0.
     
  6. Oct 3, 2008 #5
    I presume you mean [tex]E_n = \frac{2 \pi m_e e^4}{n^2 (4 \pi \epsilon_{0}) h^2}[/tex]

    Incidentally, my quantum mech book gives the equation for the energy according to the Bohr model as [tex]\frac{m_e Z^2 e^4}{(4 \pi \epsilon_0)^2 2 \hbar^2}\frac{1}{n^2}[/tex]

    so you're missing a factor of [tex]2 \pi[/tex] up top and you're missing a 2 from down below.... in other words, you're missing a factor of pi. Which is what you say you're missing. =)
     
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