Effective nuclear charge for outer electron of Lithium

[frozenguy]

Homework Statement

C. The ionization energy of lithium is 5.39 eV. Assume that r = 4a0 for the outer electron.
a) Use this fact to calculate the effective nuclear charge Zeff for the outer electron in lithium. (Hint: This is easy.)
b) Give one reason that supports the assumption that r = 4a0, and one reason that challenges it.

Homework Equations

r=\frac{n^{2}a_{0}}{Z}

r=4a_{0}

E=\frac{13.606Z_{eff}}{n^{2}}

The Attempt at a Solution

So from the first and second equation, I get n=\sqrt{12}??

With that I find Z_{eff}=2.16

Am I doing this right? n=\sqrt{12}=3.46 Just doesn't feel right for an n value

Thanks for your help.

 

[Redbelly98]

For (a), perhaps you just need to calculate the energy to move an electron that is 4a₀ away from the nuclear charge to infinity. Then n doesn't even come into the picture.

 

[frozenguy]

For (a), perhaps you just need to calculate the energy to move an electron that is 4a₀ away from the nuclear charge to infinity. Then n doesn't even come into the picture.

Hi, thanks for the response.
Just so I can make sure I have this right in my head:
The problem states that the ionization energy is 5.39eV. That means it takes 5.39eV to remove an electron from the INNER shell and send it to infinity. So they want to know how much energy it will take to remove an outer electron, which happens to be located at r=4a₀, and move it to infinity.

I assume it will be less energy. But isn't that because n is larger? I don't know how not to use n in this case because isn't r=4a₀ related to n?