How to calculate the potential energy of an electron in the #"nth"# shell (also take example #"n=3"# shell) of a copper atom using the Hartree Fock Theory? Also calculate its kinetic energy

Question

Wyatt Adkins · Accepted Answer

You can't calculate the kinetic energy of an electron that hasn't left the atom. So we could only calculate the orbital potential energy.

DISCLAIMER: COMPLICATED ANSWER!

Since copper is not that heavy, we don't have to worry too much about scalar relativistic effects or spin-orbit coupling. In fact, its ground electronic state is a #""^2 S_"1/2"#, with no low-lying excited states to include in a more multi-referenced level of theory like MCSCF or CCSD(T).

The ground-state electron configuration would be:

#[Ar] 3d^10 4s^1#

INPUT FILE SETUP

At the Hartree-Fock level of theory (a single-reference method), we assume a single configuration with doubly-occupied core and #3d# valence orbitals, and a singly-occupied #4s# valence orbital. We don't expect good agreement with experiment with this level of theory...
MolPro, Psi4, and other computational software could do this, but I have MolPro with me right now. A basic input for copper atom would look like this:

[
Here,

#400# megawords are used for the memory (32 bits per word on a 32-bit machine, etc).

Only the Abelian point groups are used, and by default for atoms, #bb(D_(2h))# is used.

copper is using an augmented correlation-consistent polarized valence triple zeta basis set, with basic atomic geometry specified.

The atom has #29# electrons, the unpaired electron is of #A_g# symmetry (symmetry 1), and has twice the total spin, #2 xx S#, of #1#, as specified by the wave function card:

#"wf, 29, 1, 1"#

VERIFYING THE ORBITAL OCCUPATION

The orbitals in the basis set are being printed in order by energy within each irreducible representation of the #D_(2h)# point group, whose irreducible representations are listed as #A_g#, #B_(3u)#, #B_(2u)#, #B_(1g)#, #B_(1u)#, #B_(2g)#, #B_(3g)#, #A_u#.

So, orbital #6.1# is the sixth orbital (starting from the lowest energy) of #A_g# symmetry, orbital #2.5# is the second orbital of #B_(1u)# symmetry, etc.

We expect the orbital occupation is ordered as follows:

#ul(A_g" "B_(3u)" "B_(2u)" "B_(1g)" "B_(1u)" "B_(2g)" "B_(3g)" "A_u)#
#1s" "2p_x" "2p_y" "3d_(xy)" "2p_z" "3d_(xz)" "3d_(yz)" "-#
#2s" "3p_x" "3p_y" "" "" "color(white)(.)3p_z#
#3s" "#
#3d_z^2" "#
#3d_(x^2 - y^2)#
#color(red)(4s)" "#
where red indicates the singly-occupied orbital with one spin-up electron.
MolPro believes it to be:

[

The orbital occupation looks correct on the first go, but in practice, you should specify the occ and closed cards anyway. It should be an occupied configuration of:

#(A_g, B_(3u), B_(2u), B_(1g), B_(1u), B_(2g), B_(3g), A_u) = (6, 2, 2, 1, 2, 1, 1, 0)#, specified by

#"occ, 6, 2, 2, 1, 2, 1, 1, 0"# within the brackets for the HF command.

and a closed-shell (doubly-occupied) configuration of:

#(A_g, B_(3u), B_(2u), B_(1g), B_(1u), B_(2g), B_(3g), A_u) = (5, 2, 2, 1, 2, 1, 1, 0)#, specified by

#"closed, 5, 2, 2, 1, 2, 1, 1, 0"# within the brackets for the HF command
indicating one spin-up electron in the #4s# orbital (since it is the sixth orbital in symmetry 1).

ORBITAL ENERGIES (BE CRITICAL!)

The orbital output was:

[
The default energies listed are in Hartrees, so you'll have to convert to #"eV"#s if you want them in #"eV"#s. There are #"27.2114 eV"#s per Hartree.

Apparently, the HOMO, the #4s# (orbital #6.1#), was (poorly) calculated to have an energy of #-"6.49 eV"#, which is too high in energy by about #"2 eV"#...

The #3d#, orbital #5.1# (orbital five of symmetry #A_g#), lists an energy of #bb(-"13.37 eV")#, which is quite close to the actual #-"13.47 eV"# here.

The #3p# orbitals are the last orbitals in symmetry 2, 3, and 5, so they are orbitals #2.2, 2.3, 2.5#. These have calculated energies of #-"90.48 eV"#, which is quite low and certainly core-like. I am not at all sure if these #3p# energies are correct, but who really cares about the core orbitals in a light atom like this (especially since the #3d# subshell is full)?

For a basic calculation at a low level of theory, this isn't that bad...

Cameron Andrews · Answer

undefined The potential energy of an electron in the nth shell of an atom can be calculated using the Hartree-Fock Theory formula:

$$E_{\text{potential}} = -\frac{{Z \cdot e^2}}{{2n^2}}$$

Where:
- $E_{\text{potential}}$ = potential energy
- $Z$ = atomic number of the atom
- $e$ = elementary charge
- $n$ = principal quantum number (shell number)

For the copper atom ($Z = 29$) and $n = 3$, the potential energy can be calculated as follows:

$$E_{\text{potential}} = -\frac{{29 \cdot (1.602 \times 10^{-19})^2}}{{2 \cdot 3^2}}$$

For kinetic energy, you can use the formula:

$$E_{\text{kinetic}} = -\frac{{E_{\text{potential}}}}{{2}}$$

Substitute the value of $E_{\text{potential}}$ into the kinetic energy formula to find the kinetic energy.