Estimating Distances Between Electrons in He Atom

[strangequark]

b]1. Homework Statement [/b]
Given the observed spectrum of helium, estimate the distance between two electrons in a helium atom (a) in the ground state and (b) in the first excited state. Neglect the exchange energy.

Homework Equations

[tex]E_{1}=-78eV[/tex]
[tex]E_{2}=-58eV[/tex]

Given in my textbook,
"Hartree theory predicts that the radius of the n=1 shell is smaller than that of the n=1 shell of hydrogen by approximately a factor of [tex]\frac{1}{Z-2}[/tex]"

[tex]r~\frac{r_{hydrogen}}{Z-2}[/tex]

The Attempt at a Solution

I'm confused by the text above. Doesn't [tex]Z_{helium}=2[/tex], and hence [tex]Z-2=0[/tex]?

Another source I found says that,

[tex]E_{n}=\frac{n^{2}(-13.6eV)}{(Z-2)^{2}}[/tex]

and,

[tex]r~\frac{a_{0}}{Z-2}[/tex]

I can solve the first eq. for Z-2, but again this doesn't really make sense to me because I get Z-2=2.39 (for ground state) which would imply that Z=4.39? Also using this, I don't get the answers that are in the key.

Please point me in the right direction... thanks.

 

[Jhero]

Thank you for your question. The text you referenced from your textbook is correct. The Z in the equation refers to the atomic number of the element, which in this case is 2 for helium. Therefore, the calculation for the distance between two electrons in a helium atom would be as follows:

(a) In the ground state:
Using the equation r~\frac{r_{hydrogen}}{Z-2}, we can calculate the distance between two electrons in a helium atom as:
r~\frac{a_{0}}{Z-2} = \frac{a_{0}}{2-2} = \frac{a_{0}}{0} = undefined

This result makes sense, as the ground state of helium is a singlet state where the two electrons are in the same orbital, thus the distance between them is undefined.

(b) In the first excited state:
Using the equation E_{n}=\frac{n^{2}(-13.6eV)}{(Z-2)^{2}}, we can calculate the energy of the electron in the first excited state as:
E_{2}=\frac{2^{2}(-13.6eV)}{(2-2)^{2}} = -13.6eV

Now we can use this energy value in the equation E_{n}=\frac{n^{2}(-13.6eV)}{(Z-2)^{2}} to solve for the distance between two electrons in the first excited state:

-13.6eV = \frac{2^{2}(-13.6eV)}{(Z-2)^{2}}
(Z-2)^{2} = 2^{2}
Z-2 = 2
Z = 4

Therefore, the distance between two electrons in the first excited state of helium is:
r~\frac{a_{0}}{Z-2} = \frac{a_{0}}{4-2} = \frac{a_{0}}{2}

I hope this helps clarify the confusion you had. Let me know if you have any further questions. Keep up the good work in your studies!