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Applications for hydrogen-like atomic systems

A hydrogen-like atom or a hydrogenic atom is an atom with one electron. Except for the hydrogen atom itself (which is neutral) these atoms carry positive charge e ( Z − 1 ) {\displaystyle e(\mathbf {Z} -1)} , where Z {\displaystyle \mathbf {Z} } is the atomic number of the atom. Because hydrogen-like atoms are two-particle systems with an interaction depending only on the distance between the two particles, their (non-relativistic) Schrödinger equation can be exactly solved in analytic form. The solutions are one-electron functions and are referred to as hydrogen-like atomic orbitals.² The electronic Hamiltonian (in atomic units) of a Hydrogenic system is given by H ^ e = − ∇ 2 2 − Z r {\displaystyle \mathbf {\hat {H}} _{e}=-{\frac {\nabla ^{2}}{2}}-{\frac {\mathbf {Z} }{r}}} , where Z {\displaystyle \mathbf {Z} } is the nuclear charge of the hydrogenic atomic system. The 1s electron of a hydrogenic systems can be accurately described by the corresponding Slater orbital: ψ 1 s = ( ζ 3 π ) 0.50 e − ζ r {\displaystyle \mathbf {\psi } _{1s}=\left({\frac {\zeta ^{3}}{\pi }}\right)^{0.50}e^{-\zeta r}} , where ζ {\displaystyle \mathbf {\zeta } } is the Slater exponent. This state, the ground state, is the only state that can be described by a Slater orbital. Slater orbitals have no radial nodes, while the excited states of the hydrogen atom have radial nodes.

Exact energy of a hydrogen-like atom

The energy of a hydrogenic system can be exactly calculated analytically as follows : E 1 s = ⟨ ψ 1 s | H ^ e | ψ 1 s ⟩ ⟨ ψ 1 s | ψ 1 s ⟩ {\displaystyle \mathbf {E} _{1s}={\frac {\langle \psi _{1s}|\mathbf {\hat {H}} _{e}|\psi _{1s}\rangle }{\langle \psi _{1s}|\psi _{1s}\rangle }}} , where ⟨ ψ 1 s | ψ 1 s ⟩ = 1 {\displaystyle \mathbf {\langle \psi _{1s}|\psi _{1s}\rangle } =1} E 1 s = ⟨ ψ 1 s | − ∇ 2 2 − Z r | ψ 1 s ⟩ {\displaystyle \mathbf {E} _{1s}=\langle \psi _{1s}|\mathbf {-} {\frac {\nabla ^{2}}{2}}-{\frac {\mathbf {Z} }{r}}|\psi _{1s}\rangle } E 1 s = ⟨ ψ 1 s | − ∇ 2 2 | ψ 1 s ⟩ + ⟨ ψ 1 s | − Z r | ψ 1 s ⟩ {\displaystyle \mathbf {E} _{1s}=\langle \psi _{1s}|\mathbf {-} {\frac {\nabla ^{2}}{2}}|\psi _{1s}\rangle +\langle \psi _{1s}|-{\frac {\mathbf {Z} }{r}}|\psi _{1s}\rangle } E 1 s = ⟨ ψ 1 s | − 1 2 r 2 ∂ ∂ r ( r 2 ∂ ∂ r ) | ψ 1 s ⟩ + ⟨ ψ 1 s | − Z r | ψ 1 s ⟩ {\displaystyle \mathbf {E} _{1s}=\langle \psi _{1s}|\mathbf {-} {\frac {1}{2r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial }{\partial r}}\right)|\psi _{1s}\rangle +\langle \psi _{1s}|-{\frac {\mathbf {Z} }{r}}|\psi _{1s}\rangle } . Using the expression for Slater orbital, ψ 1 s = ( ζ 3 π ) 0.50 e − ζ r {\displaystyle \mathbf {\psi } _{1s}=\left({\frac {\zeta ^{3}}{\pi }}\right)^{0.50}e^{-\zeta r}} the integrals can be exactly solved. Thus, E 1 s = ⟨ ( ζ 3 π ) 0.50 e − ζ r | − ( ζ 3 π ) 0.50 e − ζ r [ − 2 r ζ + r 2 ζ 2 2 r 2 ] ⟩ + ⟨ ψ 1 s | − Z r | ψ 1 s ⟩ {\displaystyle \mathbf {E} _{1s}=\left\langle \left({\frac {\zeta ^{3}}{\pi }}\right)^{0.50}e^{-\zeta r}\right|\left.-\left({\frac {\zeta ^{3}}{\pi }}\right)^{0.50}e^{-\zeta r}\left[{\frac {-2r\zeta +r^{2}\zeta ^{2}}{2r^{2}}}\right]\right\rangle +\langle \psi _{1s}|-{\frac {\mathbf {Z} }{r}}|\psi _{1s}\rangle } E 1 s = ζ 2 2 − ζ Z . {\displaystyle \mathbf {E} _{1s}={\frac {\zeta ^{2}}{2}}-\zeta \mathbf {Z} .}

The optimum value for ζ {\displaystyle \mathbf {\zeta } } is obtained by equating the differential of the energy with respect to ζ {\displaystyle \mathbf {\zeta } } as zero. d E 1 s d ζ = ζ − Z = 0 {\displaystyle {\frac {d\mathbf {E} _{1s}}{d\zeta }}=\zeta -\mathbf {Z} =0} . Thus ζ = Z . {\displaystyle \mathbf {\zeta } =\mathbf {Z} .}

Non-relativistic energy

The following energy values are thus calculated by using the expressions for energy and for the Slater exponent.

Hydrogen : H Z = 1 {\displaystyle \mathbf {Z} =1} and ζ = 1 {\displaystyle \mathbf {\zeta } =1} E 1 s = {\displaystyle \mathbf {E} _{1s}=} −0.5 Eh E 1 s = {\displaystyle \mathbf {E} _{1s}=} −13.60569850 eV E 1 s = {\displaystyle \mathbf {E} _{1s}=} −313.75450000 kcal/mol

Gold : Au(78+) Z = 79 {\displaystyle \mathbf {Z} =79} and ζ = 79 {\displaystyle \mathbf {\zeta } =79} E 1 s = {\displaystyle \mathbf {E} _{1s}=} −3120.5 Eh E 1 s = {\displaystyle \mathbf {E} _{1s}=} −84913.16433850 eV E 1 s = {\displaystyle \mathbf {E} _{1s}=} −1958141.8345 kcal/mol.

Relativistic energy of Hydrogenic atomic systems

Hydrogenic atomic systems are suitable models to demonstrate the relativistic effects in atomic systems in a simple way. The energy expectation value can calculated by using the Slater orbitals with or without considering the relativistic correction for the Slater exponent ζ {\displaystyle \mathbf {\zeta } } . The relativistically corrected Slater exponent ζ r e l {\displaystyle \mathbf {\zeta } _{rel}} is given as ζ r e l = Z 1 − Z 2 / c 2 {\displaystyle \mathbf {\zeta } _{rel}={\frac {\mathbf {Z} }{\sqrt {1-\mathbf {Z} ^{2}/c^{2}}}}} . The relativistic energy of an electron in 1s orbital of a hydrogenic atomic systems is obtained by solving the Dirac equation. E 1 s r e l = − ( c 2 + Z ζ ) + c 4 + Z 2 ζ 2 {\displaystyle \mathbf {E} _{1s}^{rel}=-(c^{2}+\mathbf {Z} \zeta )+{\sqrt {c^{4}+\mathbf {Z} ^{2}\zeta ^{2}}}} . Following table illustrates the relativistic corrections in energy and it can be seen how the relativistic correction scales with the atomic number of the system.

References

1. Attila Szabo & Neil S. Ostlund (1996). Modern Quantum Chemistry - Introduction to Advanced Electronic Structure Theory. Dover Publications Inc. pp. 153. ISBN 0-486-69186-1. 0-486-69186-1 ↩

2. In quantum chemistry an orbital is synonymous with "one-electron function", i.e., a function of x, y, and z. ↩