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Ionization potentials, quantum defects

  1. Nov 23, 2007 #1
    1. The problem statement, all variables and given/known data
    The first ionization potential of potassium (Z=19) is 4.34eV, and the 4p--->4s transition occurs at approximately 768nm. Use this information to find the values of the quantum defects [tex]\delta[/tex](0) and [tex]\delta[/tex](1) for potassium, and hence to estimate the wavelength of the 6p--->4s transition.

    2. Relevant equations
    [tex]E_{nl} = -\frac{R_H}{[n-\delta(l)]^2}[/tex] will definately be used, but I can't see what else at the moment.

    3. The attempt at a solution
    I've said so far that [tex]\frac{hc}{768*10^{-9}} = R_H(\frac{1}{[4- \delta (1)]^2} - \frac{1}{[4- \delta (0)]^2})[/tex].

    I've also said that the 4.34eV ionization potential is the energy required to remove the outermost electron from its 4s state to infinity. So if I substitue in n=infinity into the equation along with n=4 and l=0 I should be able to solve the equation for [tex]\delta[/tex](0). I've done this and I get two values of [tex]\delta[/tex](0) (as I expected, because its a quadratic term on the bottom of the equation), but this seems to lead to far too much degeneracy in the answer. With 2 values for delta0 this leads to 4 values for delta1.

    Help please anyone??
  2. jcsd
  3. Nov 23, 2007 #2


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    Staff Emeritus
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    Gold Member

    Use the above equation along with:

    [tex]IE = -\frac{R_H}{[4-\delta(0)]^2}[/tex]

    That's 2 equations in 2 unknowns.
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