 Solution

Write the one-electron Coulomb energies as $E_n=-C/n^2$, $C>0$. The Pauli exclusion principle allows $2n^2$ electrons in shell $n$, including spin. The ground configuration therefore has two electrons in $n=1$ and seven in $n=2$, giving $E_g=-C(2+7/4)=-15C/4$. The least energy-increasing promotion is from $n=2$ to $n=3$:Rearranging the partly occupied degenerate $n=2$ shell does not produce a higher energy. Both total energies are negative, so this ratio is below one even though $E_e>E_g$.