# Estimate ionization energy

• John Greger
What is Zeff for the outer electron?In summary, the excitation energy for He-like Carbon in the 1s3s^1S state is 2851180 cm^-1 and for the 1s4s^1S state it is 2988246 cm^-1. To estimate the ionization energy, we need to use the equation Eio = T + Eexcitation, where T = R/(n-δ)^2 and δ is the quantum defect. In this case, the effective nuclear charge is +1, as the inner electrons screen Z-1 nuclear charges. Using the correct value for T, we can solve for δ and then plug it into the equation to obtain the ionization energy for

## Homework Statement

In He-like Carbon, C V, the excitation energy of 1s3s ##^1S## is 2851180 ##cm^-1## and for 1s4s ##^1S## it is 2988246 ##cm^-1##. Estimate the ionization energy. Compare with the value in the NIST database!

## Homework Equations

##E_{io}= T + E_{excitation}, ## ##T = ##\frac{R}{(n- \delta)^2} ## where delta is the quantum defect.

## The Attempt at a Solution

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I started with setting up the equations $$E_{io} = E_{exc}(3s) + \frac{R}{(3- \delta)^2}$$
$$E_{io} = E_{exc}(4s) + \frac{R}{(4-\delta)^2}$$

I subtracted the two equations so I could get rid of ##E_{io}## and just solve for delta, assuming that delta is the same for 3s and 4s.

Numerically in my pocket calculator I got that delta = ca 2.199.

I now plugged that into the equations for T, hence got a value for ##E_{io}##.

But that was wrong..

Apparently the quantum defect and ionization energy was very wrong.

I also assumed that the ionization energy was same fro 3s and 4s.

The answer is " With ##\delta## = 0.03 3s gives 3162180 cm-1 and 4s 3162300 cm-1"

My question is, what did I do wrong and how should I alternatively set up equations to solve for delta? If the ionization energy is not the same for 3s and 4s I can think of any other way.

I think the difference is due to the rounding of δ. I get δ = 0.03096, and Eio = 3162395 cm-1 for both 3s and 4s.
Your equations are incorrect. They refer to a neutral Rydberg atom, where the inner electrons screen Z-1 nuclear charges, so the effective nuclear charge felt by the outer electron is +1 (the quantum defect corrects for imperfect screening). What is the effective nuclear charge here?

• BvU
mjc123 said:
I think the difference is due to the rounding of δ. I get δ = 0.03096, and Eio = 3162395 cm-1 for both 3s and 4s.
Your equations are incorrect. They refer to a neutral Rydberg atom, where the inner electrons screen Z-1 nuclear charges, so the effective nuclear charge felt by the outer electron is +1 (the quantum defect corrects for imperfect screening). What is the effective nuclear charge here?

Okey. So the idea is to get the same Eionzation energy for both 3s and 4s since it element-specific?

Sorry, I was a little sloppy in my expression for T, but $$Z-N_{inner}$$
= 6-5 = 1 so I didn't write it out. I use ##R=R_{\infty}## , don't you? But even with the "right" rydberg constant it should not differ that much. Do you use the same method as I used above because I can't get that delta.

Is Ninner 5? Do you know what "C V" is? Does "He-like" give you a clue? Or the electron configuration 1s3s?

mjc123 said:
Is Ninner 5? Do you know what "C V" is? Does "He-like" give you a clue? Or the electron configuration 1s3s?
Actually I no not know what C V mean, what does it mean? When I know that I can figure out number of "free" electrons. But was my method otherwise right?

(you get the NIST answer thrown in ...)

• John Greger
Roman numerals are used in atomic spectroscopy to indicate successive degrees of ionisation, beginning with the neutral atom. So "C I" is the C atom, "C II" is C+, "C III" is C2+ etc. "He-like" means it has the same number of electrons as helium, which is another clue.
Your method is right, except for the wrong effective nuclear charge.

• John Greger
mjc123 said:
Roman numerals are used in atomic spectroscopy to indicate successive degrees of ionisation, beginning with the neutral atom. So "C I" is the C atom, "C II" is C+, "C III" is C2+ etc. "He-like" means it has the same number of electrons as helium, which is another clue.
Your method is right, except for the wrong effective nuclear charge.

Thank you. Then we have Z= 6 and the shelled electrons, ##N_{inner}## = 0. Since we only have two electrons, which fills the first s-shell. so now ##T=\frac{R*6^2}{(n-\delta)^2}##

But that still makes my results terrible wrong. It's impossible to obtain that delta. And even if I plug in the delta the ionization energy gets wrong. ##E_{exc}(3s) + T=\frac{R*6^2}{(3-0.03)^2} \neq 3162180 cm^{-1} ##as the solutions-manual suggests .

I tried with a similar exercise but here we had F VII (##F^{+6}##). Here the Z_N(inner) = 9-1 = 8. But same here, the results get awfully wrong. What am I doing wrong that mess up my results?

How much charge does this 3s electron 'see' ?

No. In the Rydberg states (one electron in a high energy orbital) there is one inner electron. That's what the configuration 1s3s tells you. So Zeff = 5. Likewise F6+ has 3 electrons - 2 inner and one outer.