Dear Herbert,

If the unit cell for the ion is sufficiently large, Siesta prints an estimate 
of this interaction energy as Emadel (rather than printing the warning).  I 
have seen values like 2 eV for a 10x10x10 Angstrom cell.

For the slab+ion you will always get the warning.

Your simple approach of using identical unit cells for all calculations may 
work (after substracting the Emadel from the ion calculation) in case the 
charge remains localized on the ion.  However, if it delocalizes onto the slab 
or induces shielding charges there is a problem. And to my knowledge no 
solution.

Regards, Ulrich.



----- Original Message -----
From: Herbert Fruchtl
[mailto:[email protected]]
To: [email protected]
Sent: Thu, 12
Apr 2012 11:44:31 +0200
Subject: [SIESTA-L] Adsorption energy of an ion


> Folks,
> 
> The manual says not to do it unless you know what you're doing, and since I 
> don't, I thought I'd ask...
> 
> I want to calculate the adsorption energy of an ion (let's say OH- ) on a
> metal 
> surface (let's call it M, denoting a neutral slab of a few layers). The 
> straight-forward calculation would be
> 
> dE = E(MOH-) - (E(M) + E(OH-))
> 
> Ideally all in the same size of unit cell. There is the problem of BSSE, but
> I 
> understand that and know more or less how to handle it. My question concerns
> the 
> two charged calculations (MOH- and OH-) and the effect the background charge
> 
> that will compensate for the excess electron has on the energy. Is this
> change 
> in the Hamiltonian big enough to make the calculated dE meaningless? I have
> seen 
> cases where dE (after correction for BSSE, including relaxation energy) came
> out 
> positive, which seemed chemically unlikely.
> 
> Thanks in advance,
> 
>    Herbert
> -- 
> Herbert Fruchtl
> Senior Scientific Computing Officer
> School of Chemistry, School of Mathematics and Statistics
> University of St Andrews
> --
> The University of St Andrews is a charity registered in Scotland:
> No SC013532
> 

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