# Ionization potentials, quantum defects

1. Nov 23, 2007

### Brewer

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

2. Relevant equations
$$E_{nl} = -\frac{R_H}{[n-\delta(l)]^2}$$ 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 $$\frac{hc}{768*10^{-9}} = R_H(\frac{1}{[4- \delta (1)]^2} - \frac{1}{[4- \delta (0)]^2})$$.

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.

2. Nov 23, 2007

### Gokul43201

Staff Emeritus
Use the above equation along with:

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

That's 2 equations in 2 unknowns.