Fall 2008, problem 19: Estimate the energy of full ionization of Lithium (LiàLi3+), given that the energy of first ionization of Helium (HeàHe1+) = 24.5 eV
Use the idea of “screening”: Ignore the electron-electron repulsion and assume the electron about to be removed doesn’t experience the full Coulomb attraction from the nucleus because the other electrons partially “screen” the nuclear charge.
Remember unscreened energy levels: En un= - Z2 (13.6 eV)/n2.
Posit screened energy levels: En= - (Z-s)2 (13.6 eV)/n2, where s is a constant.
Lithium: Z=3
Li à Li1+: Outer electron, in n=2 state, fully screened by two n=1 electrons (ie assume they are always between it and the nucleus). Take s=2.
E0à1+=(energy with electron removed) – (energy with electron present)
= 0 eV - - (3-2)2 *(13.6 eV)/22 = 13.6 eV/4 = 3.4 eV (cf experimentally 5.4 eV)
Li1+ à Li2+: Deduce appropriate s from Helium data. For Helium, as given:
E0à1+=E1+ - E0 = - (2-0)2 * 13.6 eV/12 - - 2 (2-s)2*13.6 eV/12 = 24.5 eV
= -54.4 eV + (2-s)2 27.2 eV = 24.5 eV (2-s)2 = 2.9 s = 0.3
(in the first line, the extra factor of 2 in the term for E0 comes because there are two electrons in this energy level—each screens the other one by the same amount).
Now use s=0.3 to calculate E1+à2+ for Lithium:
E1+à2+= ELi2+ - ELi+ = - 32 *13.6 eV/12 - - 2 (3-0.3)2 *13.6 eV/12
= 76.4 eV (cf experimentally 75.6 eV)
Finally, calculate the energy needed to remove the final electron:
E2+à3+=ELi3+ - ELi2+= 0 - - 32 13.6 eV = 122.4 eV (cf experimentally 122.4 eV)
Thus the energy needed to completely ionize neutral Lithium is estimated to be:
E0à3+= 3.4eV + 76.4 eV + 122.4 eV = 202.2 eV (cf experimentally 203.5 eV)
Thus agreement to ~1% between our estimate and experiment!
Classic Text: Quantum Mechanics of One- and Two-electron Atoms, Bethe & Salpeter. See section 32.
Variational calculation gives s=5/16, corresponding to wavefunction u ~ e-(Z-s)r/a0.