Building the Periodic Table — Question 10

Question 10

Orbital penetration breaks the same-shell hydrogenic energy degeneracy

In a supplied screened one-electron model, $E_{nl}=-13.6\,\mathrm{eV}\,Z_{\mathrm{eff,l}}^2/n^2$. For two valence orbitals with $n=3$, use $Z_{\mathrm{eff,s}}=4.20$ and $Z_{\mathrm{eff,p}}=3.50$. Assume a vacant final p spin orbital, and hold both effective charges fixed during excitation.

1. Calculate the s and p energies and identify the more tightly bound orbital.

2. Find the s-to-p excitation energy and its photon wavelength using $hc=1240\,\mathrm{eV\,nm}$.

Question 10 - Solution

Orbital penetration breaks the same-shell hydrogenic energy degeneracy

(a) Method and calculation

$$E_{3s}=-13.6(4.20)^2/9=\boxed{-26.656\,\mathrm{eV}}.$$$$E_{3p}=-13.6(3.50)^2/9=\boxed{-18.511\,\mathrm{eV}}\Rightarrow\boxed{3s\text{ more tightly bound}}.$$

(b) Interpretation and further calculation

$$\Delta E=\frac{13.6}{9}[(4.20)^2-(3.50)^2]=\boxed{8.1449\,\mathrm{eV}}.$$$$\boxed{\lambda=hc/\Delta E=152.24\,\mathrm{nm}}.$$

Physical check

Greater penetration exposes the s orbital to less screened nuclear attraction in this model. The supplied effective charges replace a self-consistent many-electron calculation; they are not universal constants for all atoms.