For the hydrogen molecular ion treated by the LCAO method with unmodified orbitals on…

PHYSICS · Multi-part · Fermi University Physics · Atomic, Molecular & Optical Physics · Molecular structure and spectra · Molecular structure and spectra

Deriving the Spectrum, Ch The Born-Oppenheimer separation and the chemical bond

Problem

For the hydrogen molecular ion H2+\mathrm{H}_2^+ treated by the LCAO method with unmodified 1s1s orbitals on each proton, the secular problem gives the bonding-orbital energy E+(R)=E1s+e24πϵ0R+J(R)+K(R)1+S(R)E_+(R) = E_{1s} + \frac{e^2}{4\pi\epsilon_0 R} + \frac{J(R)+K(R)}{1+S(R)} where E1s=−12Eh=−13.61 eVE_{1s}=-\tfrac12E_{\rm h}=-13.61\,\mathrm{eV} is the hydrogen ground-state energy and Eh=27.211 eVE_{\rm h}=27.211\,\mathrm{eV}. With RR measured in units of a0a_0 and JJ, KK in units of EhE_{\rm h}, S(R)=e−R(1+R+R23),J(R)=−1R+e−2R(1+1R),K(R)=−e−R(1+R).\begin{aligned}S(R)&=e^{-R}\left(1+R+\frac{R^2}{3}\right),\\ J(R)&=-\frac1R+e^{-2R}\left(1+\frac1R\right),\\ K(R)&=-e^{-R}(1+R).\end{aligned} ((a)) Evaluate SS, JJ and KK at R=2.00 a0R=2.00\,a_0. ((b)) Hence find E+(R=2.00 a0)E_+(R=2.00\,a_0) in eV, and the LCAO binding energy De(R)=E1s−E+(R)D_e(R)=E_{1s}-E_+(R) at this separation. ((c)) The LCAO curve is deepest at R=2.49 a0R=2.49\,a_0, where De=1.76 eVD_e=1.76\,\mathrm{eV}; the exact curve is deepest at R=2.00 a0R=2.00\,a_0, where De=2.79 eVD_e=2.79\,\mathrm{eV}. Explain why your value of De(2.00 a0)D_e(2.00\,a_0) must lie below both.

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