1s Slater-type function

A normalized 1s Slater-type function is a function which is used in the descriptions of atoms and in a broader way in the description of atoms in molecules. It is particularly important as the accurate quantum theory description of the smallest free atom, hydrogen. It has the form

\psi_{1s}(\zeta, \mathbf{r - R}) = \left(\frac{\zeta^3}{\pi}\right)^{1 \over 2} \, e^{-\zeta |\mathbf{r - R}|}.[1]

It is a particular case of a Slater-type orbital (STO) in which the principal quantum number n is 1. The parameter \zeta is called the Slater orbital exponent. Related sets of functions can be used to construct STO-nG basis sets which are used in quantum chemistry.

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(\mathbf Z-1), where \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
\mathbf{\hat{H}}_e = -  \frac{\nabla^2}{2} - \frac{\mathbf Z}{r}, where \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:
\mathbf \psi_{1s} = \left (\frac{\zeta^3}{\pi} \right ) ^{0.50}e^{-\zeta r}, where \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 :
\mathbf E_{1s} = \frac{<\psi_{1s}|\mathbf{\hat{H}}_e|\psi_{1s}>}{<\psi_{1s}|\psi_{1s}>}, where \mathbf{<\psi_{1s}|\psi_{1s}>} = 1
\mathbf E_{1s} = <\psi_{1s}|\mathbf -  \frac{\nabla^2}{2} - \frac{\mathbf Z}{r}|\psi_{1s}>
\mathbf E_{1s} = <\psi_{1s}|\mathbf -  \frac{\nabla^2}{2}|\psi_{1s}>+<\psi_{1s}| - \frac{\mathbf Z}{r}|\psi_{1s}>
\mathbf E_{1s} = <\psi_{1s}|\mathbf -  \frac{1}{2r^2}\frac{\partial}{\partial r}\left (r^2 \frac{\partial}{\partial r}\right )|\psi_{1s}>+<\psi_{1s}| - \frac{\mathbf Z}{r}|\psi_{1s}>. Using the expression for Slater orbital, \mathbf \psi_{1s} = \left (\frac{\zeta^3}{\pi} \right ) ^{0.50}e^{-\zeta r} the integrals can be exactly solved. Thus,
\mathbf E_{1s} = <\left (\frac{\zeta^3}{\pi} \right ) ^{0.50}e^{-\zeta r}|-\left (\frac{\zeta^3}{\pi} \right )^{0.50}e^{-\zeta r}\left[\frac{-2r\zeta+r^2\zeta^2}{2r^2}\right]>+<\psi_{1s}| - \frac{\mathbf Z}{r}|\psi_{1s}> \mathbf E_{1s} = \frac{\zeta^2}{2}-\zeta \mathbf Z.

The optimum value for \mathbf \zeta is obtained by equating the differential of the energy with respect to \mathbf \zeta as zero.
 \frac{d\mathbf E_{1s}}{d\zeta}=\zeta-\mathbf Z=0. Thus  \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
 \mathbf Z=1 and  \mathbf \zeta=1
 \mathbf E_{1s}=−0.5 Eh
 \mathbf E_{1s}=−13.60569850 eV
 \mathbf E_{1s}=−313.75450000 kcal/mol

Gold : Au(78+)
 \mathbf Z=79 and  \mathbf \zeta=79
 \mathbf E_{1s}=−3120.5 Eh
 \mathbf E_{1s}=−84913.16433850 eV
 \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  \mathbf \zeta . The relativistically corrected Slater exponent  \mathbf \zeta_{rel} is given as
 \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.
\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.

Atomic system \mathbf Z\mathbf \zeta_{non rel}\mathbf \zeta_{rel}\mathbf E_{1s}^{non rel}\mathbf E_{1s}^{rel}using \mathbf \zeta_{non rel}\mathbf E_{1s}^{rel}using \mathbf \zeta_{rel}
H11.000000001.00002663−0.50000000 Eh−0.50000666 Eh−0.50000666 Eh
−13.60569850 eV−13.60587963 eV−13.60587964 eV
−313.75450000 kcal/mol−313.75867685 kcal/mol−313.75867708 kcal/mol
Au(78+)7979.0000000096.68296596−3120.50000000 Eh−3343.96438929 Eh−3434.58676969 Eh
−84913.16433850 eV −90993.94255075 eV −93459.90412098 eV
−1958141.83450000 kcal/mol−2098367.74995699 kcal/mol−2155234.10926142 kcal/mol

References

  1. Attila Szabo & Neil S. Ostlund (1996). Modern Quantum Chemistry - Introduction to Advanced Electronic Structure Theory. Dover Publications Inc. p. 153. ISBN 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.
This article is issued from Wikipedia - version of the Wednesday, December 30, 2015. The text is available under the Creative Commons Attribution/Share Alike but additional terms may apply for the media files.