Ionization potentials, quantum defects

[Brewer]

Homework Statement

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 \delta(0) and \delta(1) for potassium, and hence to estimate the wavelength of the 6p--->4s transition.

Homework Equations

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

The Attempt at a Solution

I've said so far that \frac{hc}{768*10^{-9}} = R_H(\frac{1}{[4- \delta (1)]^2} - \frac{1}{[4- \delta (0)]^2}).

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 \delta(0). I've done this and I get two values of \delta(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??

 

[Gokul43201]

Use the above equation along with:

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

That's 2 equations in 2 unknowns.