Английская Википедия: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
- <math>\psi_{1s}(\zeta, \mathbf{r - R}) = \left(\frac{\zeta^3}{\pi}\right)^{1 \over 2} \, e^{-\zeta |\mathbf{r - R}|}.</math>[1]
It is a particular case of a Slater-type orbital (STO) in which the principal quantum number n is 1. The parameter <math>\zeta</math> 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 <math>e(\mathbf Z-1)</math>, where <math>\mathbf Z</math> 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
<math>\mathbf{\hat{H}}_e = - \frac{\nabla^2}{2} - \frac{\mathbf Z}{r}</math>, where <math>\mathbf Z</math> 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:
<math>\mathbf \psi_{1s} = \left (\frac{\zeta^3}{\pi} \right ) ^{0.50}e^{-\zeta r}</math>, where <math>\mathbf \zeta</math> 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 :
<math>\mathbf E_{1s} = \frac{\langle\psi_{1s}|\mathbf{\hat{H}}_e|\psi_{1s}\rangle}{\langle\psi_{1s}|\psi_{1s}\rangle}</math>, where <math>\mathbf{\langle\psi_{1s}|\psi_{1s}\rangle} = 1</math>
<math>\mathbf E_{1s} = \langle\psi_{1s}|\mathbf - \frac{\nabla^2}{2} - \frac{\mathbf Z}{r}|\psi_{1s}\rangle</math>
<math>\mathbf E_{1s} = \langle\psi_{1s}|\mathbf - \frac{\nabla^2}{2}|\psi_{1s}\rangle+\langle\psi_{1s}| - \frac{\mathbf Z}{r}|\psi_{1s}\rangle</math>
<math>\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</math>. Using the expression for Slater orbital, <math>\mathbf \psi_{1s} = \left (\frac{\zeta^3}{\pi} \right ) ^{0.50}e^{-\zeta r}</math> the integrals can be exactly solved. Thus,
<math>\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</math>
<math>\mathbf E_{1s} = \frac{\zeta^2}{2}-\zeta \mathbf Z.</math>
The optimum value for <math>\mathbf \zeta</math> is obtained by equating the differential of the energy with respect to <math>\mathbf \zeta</math> as zero.
<math> \frac{d\mathbf E_{1s}}{d\zeta}=\zeta-\mathbf Z=0</math>. Thus <math> \mathbf \zeta=\mathbf Z.</math>
Non-relativistic energy
The following energy values are thus calculated by using the expressions for energy and for the Slater exponent.
Hydrogen : H
<math> \mathbf Z=1</math> and <math> \mathbf \zeta=1</math>
<math> \mathbf E_{1s}=</math>−0.5 Eh
<math> \mathbf E_{1s}=</math>−13.60569850 eV
<math> \mathbf E_{1s}=</math>−313.75450000 kcal/mol
Gold : Au(78+)
<math> \mathbf Z=79</math> and <math> \mathbf \zeta=79</math>
<math> \mathbf E_{1s}=</math>−3120.5 Eh
<math> \mathbf E_{1s}=</math>−84913.16433850 eV
<math> \mathbf E_{1s}=</math>−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 <math> \mathbf \zeta </math>. The relativistically corrected Slater exponent <math> \mathbf \zeta_{rel} </math> is given as
<math> \mathbf \zeta_{rel}= \frac{\mathbf Z}{\sqrt {1-\mathbf Z^2/c^2}}</math>.
The relativistic energy of an electron in 1s orbital of a hydrogenic atomic systems is obtained by solving the Dirac equation.
<math>\mathbf E_{1s}^{rel} = -(c^2+\mathbf Z\zeta)+\sqrt{c^4+\mathbf Z^2\zeta^2}</math>.
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 | <math>\mathbf Z</math> | <math>\mathbf \zeta_{non rel}</math> | <math>\mathbf \zeta_{rel}</math> | <math>\mathbf E_{1s}^{non rel}</math> | <math>\mathbf E_{1s}^{rel}</math>using <math>\mathbf \zeta_{non rel}</math> | <math>\mathbf E_{1s}^{rel}</math>using <math>\mathbf \zeta_{rel}</math> |
| H | 1 | 1.00000000 | 1.00002663 | −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+) | 79 | 79.00000000 | 96.68296596 | −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
- ↑ Шаблон:Cite book
- ↑ In quantum chemistry an orbital is synonymous with "one-electron function", i.e., a function of x, y, and z.
