[Wien] Representative number for O2 energy

Jeff Spirko spirko at lehigh.edu
Fri Mar 11 18:15:57 CET 2005


You are effecitvely using the difference in E(O) at two RMTs to
correct the E(O2) from one RMT to the other.  You might also use the
energy of a stretched O2 molecule to make this correction.  That is,

  E(O2, min, RMT=1.6) = E(O2, min, RMT=1.1) +
     E(O2, stretch, RMT=1.6)-E(O2, stretch, RMT=1.1)

If the correction is in line with what you are already using, it
might be okay.

Best Regards,
-Jeff Spirko

On Fri, Mar 11, 2005 at 10:46:27AM -0600, L. D. Marks wrote:
> I would like to get a representative energy (i.e. not the correct one, but
> one which is consistent with other calculations) for the O2 (molecule)
> energy for an RMT of 1.6 and RKMAX of 7, GGA. (These values are what I am
> using in some other calculations of bulk oxides, and the energy of O2
> has a significant effect on the theoretical heats of formation.) Obviously
> I cannot use these for the O2 molecule, the RMT has to be about 1.1. What
> I have done is:
> 	a) Calculate O2 with RMT=1.1 RKMAX=5, ~30 a.u. cell: E_a
> 	b) Calculate O  with RMT=1.1 RKMAX=5, ~30 a.u. cell: E_b
> 	c) Calculate O  with RMT=1.6 RKMAX=7, ~30 a.u. cell: E_c
> Then estimate the effective O2 energy as
> 	E_a - 2*E_b + 2*E_c
> I would be interested to know if anyone has a better idea/method, apart
> from using 2*E_c + an experimental number for the bond energy (or
> meta-GGA).
> -----------------------------------------------
> Laurence Marks
> Department of Materials Science and Engineering
> MSE Rm 2036 Cook Hall
> 2220 N Campus Drive
> Northwestern University
> Evanston, IL 60201, USA
> Tel: (847) 491-3996 Fax: (847) 491-7820
> email: L - marks @ northwestern . edu
> http://www.numis.northwestern.edu
> -----------------------------------------------
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Jeff Spirko   spirko at lehigh.edu   spirko at yahoo.com   WD3V   |=>


The study of non-linear physics is like the study of non-elephant biology.

All theoretical chemistry is really physics;
and all theoretical chemists know it. -- Richard P. Feynman 

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