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Electron energy increases with N?

  1. Apr 24, 2010 #1
    I'm stuck on a pretty basic concept. I'm thinking in terms of the Bohr model, which I know is rather archaic now, but I think the outputs of the model for basic understanding still hold? In any event, to my understanding, at least in hydrogen, N = 1 corresponds to the electron having zero eV, and N = 2 to 10.2 eV, so an electron that absorbs a photon of 10.2 eV will change to N = 2 at a greater radius and now have 10.2 eV in terms of its energy, which I assume is essentially kinetic. So an electron's kinetic energy increases with radius or N. Yet all the while its velocity decreases with radius or N. And yet its kinetic energy is supposed to be a function of its velocity, right, even in QM, or no? The only other place for the energy to come I would think is potential energy due to its presence in the electrostatic field, but that decreases with radius too, right by V = kQ/r? That's the contradiction that I have in my head, although I know it's a matter of me missing something essential here. Any guidance would be appreciated and thanks for your patience.
     
  2. jcsd
  3. Apr 24, 2010 #2
    No, n=1 corresponds to an energy of -13.6 eV.
    Zero eV would correspond to the nuclei and the electron infinitely separated. :smile:
     
  4. Apr 24, 2010 #3

    Matterwave

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    This is the regular convention we use. Although we could use another convention, this one is more useful.

    In regards to your question, the potential energy increases as distance from the nucleus. Just think, the farther the electron is, the more work can be done by it falling farther. This is analogous to raising a stone farther from the Earth's surface. The higher you raise the stone, the more potential energy it has.

    Your confusion arises because you used the wrong formula. Using our regular convention, the potential energy is V=-kq^2/r^2 notice the negative sign! As r increases, the potential gets less negative which means it's increasing!
     
  5. Apr 24, 2010 #4
    Appreciate your help, and thanks for the clarification. Btw, your formula above looks like Coulomb's force equation, but it can't be, right?
     
  6. Apr 24, 2010 #5

    jtbell

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    Yes, that should be r, not r^2. Sometimes I mix up the formulas for Coulomb force and potential energy, myself!
     
  7. Apr 25, 2010 #6

    Matterwave

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    Yea, sorry. =P
     
  8. Apr 25, 2010 #7
    Np, so now using U = -kq^2/r I get -4.34 x 10^-18J or -27 eV for the ground state N=1 and -1.09 x 10-18J or -6.8eV for N=2. But that doesn't seem right. When I use the formula U = 13.6 eV - (13.6 / N^2) that generates 0 eV for N=1 and -10.2eV for N=2. Even if it was just a conventional of making U = 0 for the ground state, the U difference b/w N = 1 and N = 2 should be the same in these two approaches but it isn't, so I must still doing something wrong :(

    Also, if U = -kq^2/r, than V = -kq/r right, or no?

    Thanks again for your patience and help.
     
    Last edited: Apr 25, 2010
  9. Apr 25, 2010 #8

    jtbell

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    The electron has kinetic energy as well as potential energy.
     
  10. Apr 25, 2010 #9

    Matterwave

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    Also, the energies are U=-13.6eV/n^2

    There is no 13.6- in front of that. Thus for n=1, U=-13.6eV as n goes higher U becomes less negative (higher), and eventually reaches 0 for n=infinity.
     
  11. Apr 26, 2010 #10
    Thanks, all adds up now :)
     
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