How to calculate the first and second ionization energies of helium?

Apr 9, 2015

I saw this question and I was interested in the answer as well. I teach high school (IB) chemistry and physics and have never encountered a national curriculum that requires this. Are you a college level student?

I did a little research which confirmed my thoughts that this is an incredibly complex problem. If you were presented with a simplified equation by your instructor, please pass this on and we will try to help you.

The paper I found that was most closely related to your question is located at http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3671562/

As you can see, the math is quite complex.

You should know that the 2nd ionization energy for He will be much higher than the first. This is because
1) the second electron is now being pulled away from a positive ion
2)helium's two electrons are both located in the 1s orbital and when both present repel each other which makes removing the first one easier than the second.

Good luck

Apr 9, 2015

Bohr's equation cannot be applied to multi-electron atoms like helium because it simply cannot account for experimental data.

It is valid for hydrogen-like atoms, however, this being the reason for why the second ionization energy of helium is calculated correctly.

In helium's case, Bohr's equation predicts the same value for both the first, and the second ionization energies: 5276 kJ/mol, with only the second ionization energy being correct.

For multi-electron atoms, you must replace $Z$ with ${Z}_{\text{eff}}$, the effective nuclear charge.

So, the equation becomes

${E}_{n} = - {R}_{H} \cdot {Z}_{\text{eff}}^{2} / {n}^{2}$, where

$- {R}_{H} = 2.178 \cdot {10}^{- 18} \text{J}$

Here's where it gets interesting. I won't go into details about it, but, according to calculations, helium's second 1s electron feels an effective nuclear charge equal to 1.7 - more on that here:

However, when you plug 1.7 into the above equation you get a first ionization energy approximately equal to 3791 kJ/mol, which is closer to the experimental value of 2372 kJ/mol, but not quite the same.

If you work backwards and plug the experimental 2372 kJ/mol first ionization energy into the equation and solve for ${Z}_{\text{eff}}$, you'd get a value of 1.34, lower than the 1.7 value the calculations would predict.

Like David said in one of the answers already posted, the calculations for the first ionization energy are very complex (very, very messy), so I don't know if you'd ever be asked to actually go through them and get the elusive 2372 kJ/mol value.