# In Helium, why does more tightly bound mean the electrons are further apart?

In helium, the triplets ($$S=1$$) are lower in energy (more negative) than the singlets $$S=0$$.

One reason given by my lecturer is that in a triplet the spins of the two electrons are the same, then by the Pauli exclusion principle the electron wavefunctions must no overlap too strongly otherwise the there would be two electrons with the same spin state in the same location, pushing them apart so their Coulomb energy is lower (more tightly bound).

But if the electrons are further apart, shouldn't that imply that the electron in the excited state be further away from the nucleus and hence there would be less Coulomb attraction betwen the excited electron and nucleus, decreaseing the binding enegry? (i.e. the energy of the singlet should be more negative instead)

• can you give a link for these statements? I am confused by your notation. Commented May 6, 2022 at 17:43
• @anna v these are from my lecture note, sorry for any confusion. Commented May 6, 2022 at 18:43
• The electron-nuclear and electron-electron potentials are both on the same order of magnitude so you really can't rely on qualitative argumentsm, it just so happens that in this case the $e-e$ repulsion wins out and is more important. Commented May 7, 2022 at 20:35

The energy difference is caused by the exchange interaction. For the singlet the electron interaction is $$J+K$$ en for the triplet it is $$J-K$$, where $$J = \int { \int { d^3 r_1 d^3 r_2 \frac{a^2(r_1) b^2(r_2)}{| r_1 - r_2 |}}}$$ and $$K= \int {\int {d^3 r_1 d^3 r_2 \frac{a(r_1) b(r_1) a(r_2) b(r_2)}{| r_1 - r_2 |}}}\,.$$