Finding Potassium's ionisation potential using Sharp, Principal etc. Series

[EmmaLemming]

Homework Statement

Make a rough estimate of the ionisation potential of potassium (19) given the following infomation relating to its optical spectrum:

wavelength of the first line of the Sharp Series: 1248nm
wavelength of the first line of the Principal Series: 767nm
wavelength of the first line of the Diffuse Series: 1173nm
wavelength of the first line of the Fundamental series: 1516nm

speed of light = 2.998 x 10^8 m/s
Planck Constant = 6.626 x 10^(-34)
Rydberg constant R∞ = 1.097 x 10^7
electronic charge = -1.602 x 10^(-19)

Homework Equations

I = (Rhc)/((n-δ)^2)

The Attempt at a Solution

I have no idea how to go about this at all. I have tried but got nothing..

I was told that δ was a ratio of some sort.. but I don't know what of..

A method of how to go about this would be great and an explanation of why would be handy too.

 

[NateD]

The equation for the ionization energy is given by I = Rhc/((n-δ)^2), where n is the principal quantum number of the outermost electron in the atom, δ is a small correction term that takes into account relativistic and quantum effects, R is the Rydberg constant and h is Planck's constant. The value of δ is typically 0.25 for atoms with Z > 10, and can be neglected for atoms with lower atomic numbers. Now, since potassium has an atomic number of 19, we can assume that δ is approximately 0.25. To find the ionization energy of potassium, we must first calculate the principal quantum number of the outermost electron. This can be done by using the following formula:n = 1 + (R∞/λ),where λ is the wavelength of the line observed in the spectrum. We can choose any of the wavelengths given to us in the problem statement, so let's choose the wavelength of the first line of the Sharp Series, which is 1248nm. Thus, our equation becomes:n = 1 + (1.097 x 10^7/1248 x 10^(-9)) = 9.489.Now, we can plug this value of n as well as the values of R and h into the equation for the ionization energy:I = (Rhc)/((n-0.25)^2)= (6.626 x 10^(-34) x 2.998 x 10^8)/((9.489 - 0.25)^2)= 4.11 x 10^(-18) Joules.Therefore, the estimate for the ionization energy of potassium is 4.11 x 10^(-18) Joules.