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How do we know that the solutions of the time-independent Schrodinger equation for multielectron atoms are the same with the solutions for the hydrogen atom if we add a effective potential due to the shielding of the charge of the nucleus from inner shell electrons?

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    $\begingroup$ We know the opposite: they are not the same. There is a vague resemblance, hardly more. $\endgroup$ Commented Dec 9, 2022 at 20:10
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    $\begingroup$ But if they are not why do we still use them to predict behaviour of chemical elements? $\endgroup$
    – Volpina
    Commented Dec 9, 2022 at 20:11
  • $\begingroup$ Because of the said resemblance, that's why. $\endgroup$ Commented Dec 9, 2022 at 21:01
  • $\begingroup$ @IvanNeretin, Would you mind expanding your comment into a useful answer? Who is "we" and what is the meaning of vague resemblance? I know at least one theoretical physical chemist who considers all this orbital teaching business for higher elements as complete non-sense. $\endgroup$
    – ACR
    Commented Dec 10, 2022 at 4:09
  • $\begingroup$ Does they have a better approach that is somewhat understandable? $\endgroup$
    – jimchmst
    Commented Dec 10, 2022 at 4:16

1 Answer 1

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The electronic Hamiltonian for a hydrogen atom can be written as

$$\hat{H}_e = -\frac{Ze^2}{r_{Ne}} - \frac{\hbar^2}{2m_e} \nabla^2$$

The relative nuclear charge for an H-atom is $Z=1$ (the atomic number). When approximating the single electron Hamiltonian using an effective nuclear potential and otherwise ignoring electron-electron repulsion, which is missing in the above Hamiltonian, $Z$ is set to $Z_{eff}$, the difference of the nuclear charge and the screening constant.

The effect of an increase in $Z$ is to contract the size of an atom. The Hamiltonian could be written as

$$\hat{H}_e = \frac{e^2}{r^*_{Ne}} - \frac{\hbar^2}{2m^*_e} \nabla^{*2}$$

where $r^*_{Ne} = r_{Ne}/Z$ is an effective distance from the nucleus, $m^*_e = m_e Z^2$ is an effective electron mass and $\nabla^* = \sum_i \vec{e}_i \frac{\partial}{\partial x_i^*}$. The change in the dimensions of $r$ therefore does not alter the mathematical form of the eigenfunctions of the Hamiltonian above, it only rescales the mass and distance dimensions. The solutions will be the single electron hydrogenic wavefunctions but with the effective radius $r^*_{Ne} = r_{Ne}/Z$ in place of $r$ and the momentum of the electrons altered by the change in effective mass. An effective Bohr radius can be computed under these circumstances as (setting $\mu=Z^2 m_e$) $$a_\circ ^* = \frac{4 \pi \epsilon _\circ \hbar^2}{Z^2 m_e e^2} = \frac{a_\circ}{Z^2}$$ illustrating the contraction in the extent of the wavefunction due to the increase in $Z$.

Note that this is an oversimplification of the true situation, but the approximation suggests for instance that Slater-type orbitals might be useful as a first step to describing the properties of atoms. For molecules it is likewise inaccurate but can be helpful to grasp how behavior scales with nuclear charge.

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    $\begingroup$ Do we know that the additional contribution of electron-electron interactions in bigger atoms does not make a big contribution to the orbital shapes? $\endgroup$
    – matt_black
    Commented Dec 10, 2022 at 17:19
  • $\begingroup$ And not just the electrostatic repulsion which is stated in this answer maybe the Pauli repulsion plays a role as well? $\endgroup$
    – Volpina
    Commented Dec 10, 2022 at 21:15
  • $\begingroup$ @matt_black An effective nuclear charge attempts to account for a large chunk of the e-e repulsion, but does not account for correlation. Screening effects can predict electronic properties (although semi-empirically) with a fair degree of accuracy for moderately sized atoms. $\endgroup$
    – Buck Thorn
    Commented Dec 11, 2022 at 8:28
  • $\begingroup$ So "How do we know that the solutions of the time-independent Schrodinger equation for multielectron atoms are ..." - the answer I give is that the math for the simplified version of the Hamiltonian for a single electron that sees an effective nuclear charge is the same. But that doesn't mean that the result is physically very accurate, especially for molecules. However you can look at Slater's papers and see what nice predictions (such as IEs) are possible for atoms using that model. $\endgroup$
    – Buck Thorn
    Commented Dec 11, 2022 at 8:32
  • $\begingroup$ Also if you properly include the electron-electron interactions the whole orbital picture no longer applies - an orbital is a single electron wavefunction and in the strictest sense only applies to hydrogenic atoms. Their use in many electron systems is a (very useful) approximation. $\endgroup$
    – Ian Bush
    Commented Dec 11, 2022 at 9:10

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