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Ground State and Energy

  1. Nov 28, 2008 #1
    1. The problem statement, all variables and given/known data
    Calculate the energy of the electron in the ground state of triple ionized beryllium, which has an atomic number Z = 4.

    2. Relevant equations

    -Z2 (13.6)

    3. The attempt at a solution

    Z=4 as given and it is tripled so our N=3. We throw our numbers in there and we should have -(16)(13.6). Take that product and divide it by 9. I got an answer of -24.18, but I am incorrect.

    Unfortunately, I do not have my textbook with me right now so I might have the equation all wrong. I would really appreciate it if someone could give me a hint or a good website to learn from.

    THANK YOU!!!
  2. jcsd
  3. Nov 28, 2008 #2
    [tex]\frac{m_e Z^2 e^4}{(4 \pi \epsilon_0)^2 2 \hbar^2}\frac{1}{n^2}[/tex]
  4. Nov 28, 2008 #3
    Z=4 is correct, however the electron is in the ground state, so n=1.
    The triple ionized means that the normally neutrally charged beryllium atom now has a +3e charge, so the situation is idealized to a +4e core (Z=4) with just one -1e electron in it's orbit, obviously being in the ground state (n=1).
    This is because the equation you gave doesn't hold for high Z, high N and disregards electron-electron interaction, since the Bohr model (the equation is an extrapolated result of the Bohr model of the hydrogen atom) doesn't properly account for these kind of interactions.
    I'm not completing the calculation because I don't feel like it, but I'm fairly positive that you'll arrive at the correct answer.
  5. Nov 28, 2008 #4
    Sjorris, I would just like to THANK YOU!!! I succesfully solved my problem because of your help.

    I really appreciate your help.

    THANKS!!! again

    Happy Holidays
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