13
$\begingroup$

TL; DR

I am asking for

  • an analytic potential curve function for the triplet state of $\ce{H2}$ (preferred) with experimental parameters, or
  • experimental data (enough points to realise a smooth-looking curve).

Also see solid red curve pictured below.


To characterise the energy of a two-atom system, Morse potential is applied as a second approximation after the harmonic oscillator model. This is an analytic function of the form $$E=D_\mathrm{e}\left\{1-\exp[-a(r-r_\mathrm{c})]\right\}$$

where the parameters $D_e$, $a$ and $r_c$ depend on the atoms in play. Surely enough, better results are achievable. For examples on simple molecules, I found the paper by Hulburt and Hirschfelder, 'Potential Energy Functions for Diatomic Molecules', J. Phys. Chem., ($1940$) to be rather useful. Have a look at equations $(14)$ and ($15$) specifically (available more easily here). The one I used was

$$E_{\mathrm{P}}=\mathrm{D_e}[(1-\mathrm{e}^{-z})^2+\mathrm{c}x^3\mathrm{e}^{-2z}(1+\mathrm{b}z)],$$

where

$$z=\frac{\omega_\mathrm{e}}{{2}\sqrt{\mathrm{B_eD_e}}}\left(\frac{x-\mathrm{r_0}}{\mathrm{r_0}}\right)$$

This is for the case where spins are antiparallel.

If atoms of parallel spin approach one another, the qualitative picture is quite different. A smaller distance will equal higher energy regardless of separation, i.e., no stable equilibrium state exists. Unfortunately, finding a nice function for the antibonding situation proved less fruitful. An example of what I'm after is the solid red curve found in this Wolfram demonstration.

Neutral triplet state of a dihydrogen failing to form, Wolfram demonstrations, http://demonstrations.wolfram.com/ValenceBondTheoryOfTheHydrogenMolecule/

$\endgroup$
1
  • $\begingroup$ I did contact a representative of Wolfram Research Europe, but (for some reason) failed to receive an answer. Furthermore, I was initally reluctant to post this question due to the fact that it is essentially a data question... but here it is anyway. If it's off-topic, closing will not offend. $\endgroup$ Jul 31, 2017 at 17:34

2 Answers 2

7
$\begingroup$

I believe that you wont be able to find a closed analytic expression for $\mathrm{H_2}$, since no closed analytic expresion is possible for a many-body problem (more than 2 bodies). That said you can calculate as many points on the triplet curve as you like by using a QM program able of doing unrestricted Hartree-Fock (UHF). Here you should be aware that UHF does not give very good potential curves, because the Hartree-Fock approximation lacks the electron correleation energy. $\mathrm{H_2}$ is a quite small system, so you will be able to make a potential curve using unrestricted MP2 or unrectriced CCSD. I do believe that PySCF is capable of doing such calculations. Regarding the experimental data, I am not aware of any databases for such kind of data. I would suggest finding a paper where a curve is presented and then use engauge digitizer to extract the points you need.

$\endgroup$
1
  • $\begingroup$ Thank you for the insight! Apologies for the delay, I was swamped. Didn't even think of the engauge digitiser approach. Will give it a go in the coming days, and depending on the outcome (probably) will accept the answer :). $\endgroup$ Aug 13, 2017 at 23:45
1
$\begingroup$

Potentiology is a field that began in the early days of quantum mechanics:

  • 1924: John Lennard-Jones publishes the Lennard-Jones pontial
  • 1929: Philip M. Morse publishes the Morse potenital.

The most recent reference you gave was a potential from:

  • 1940: by Hugh Hulburt and Joseph Hirschfelder

You seem to be hoping that in the last 80 years, someone might have come up with something better.
The answer is the 2009 Morse/Long-range, which is the most flexible potential energy function known. It has the theoretically correct long-range behaviour built into it. You can easily do the anti-bonding potential, and also double-well potentials, potentials with a shelf, potentials with a shoulder, potentials with a barrier, etc.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.