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Quantum Electron Energy Level for Hydrogenic atom

  1. Jul 4, 2013 #1
    1. Consider doubly ionized Lithium (Li++), which has one electron orbiting a +3 charge nucleus. Assuming the electron is in the ground state (n=1), what is the maximum wavelength of light, λ, that would completely ionize the Li++? (free the electron from the nucleus),
    all variables and given/known data




    2. Energy of a Hydrogen atom is [itex]\frac{-13.606eV}{n^{2}}[/itex] where 'n' is the energy level. The general formula is:
    E = [itex]\frac{-1}{4\pi r^{2}}[/itex][itex]\frac{me^{4}}{2\bar{h}n^{2}}[/itex]
    for atoms with one electron.

    For light:
    E = [itex]\frac{1240eV-nm}{λ}[/itex]




    3. I assumed that Li++ would be like a hydrogen atom with 3 times the mass. Since E is proportional to m, the energy of Li should be -13.606eV*(3)/n^2 = 40.818eV. (n=1) I plugged this into the equation λ = 1240eV-nm/E, and got λ = 30.38nm.
    The correct answer is 10.1nm... so I am off by a factor of three. So does this mean that the doubly ionized lithium actually has 9 times the mass?

    Thank you!
     
    Last edited: Jul 4, 2013
  2. jcsd
  3. Jul 4, 2013 #2
    Mass is not the only difference between H and Li.
     
  4. Jul 4, 2013 #3

    TSny

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    In the energy formula, does m stand for the mass of the atom or the mass of the electron?
     
  5. Jul 4, 2013 #4
    Hi guys,
    I figured it out. As it turns out m is mass of the electron, and e^4 is actually the charge of the nucleus squared times the charge of the electron squared. Which happens to be e^4 in an H atom. In this case it was e^2*(3e)^2 That's where the factor of 9 comes from! Thanks
     
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