[Pw_forum] work function of Metal transition metal oxide.
giovanni.cantele at spin.cnr.it
Tue May 19 10:26:04 CEST 2015
you are calculating the electronic properties of the interface between ZnO and Au(111), using an hexagonal lattice with lattice parameter 6.1413449010 au ~ 3.25 A, that is the gold film is strained so as to match the ZnO lattice parameter. The work function of gold calculated from first principles is about 5.1-5.2 eV, but with that strain it reduces to about 4.98 eV (see Table 2 of J. Phys.: Condens. Matter 27 (2015) 015006), which is exactly what you get. So, reverting back to your question (why if I remove some O’s in the ZnO slab nothing changes?), it could be that the Fermi level is pinned by the metal slab, so even if you do induce changes within the ZnO, the Au film does not change its properties and you always “measure” the same work function (namely, distance between the pinned Fermi level and the potential in the vacuum at the Au side).
So, I suggest you to get help from available literature and textbooks on Schottky (metal-semiconductor) barriers and Fermi level pinning mechanisms, to understand the relevant properties of your system.
You might check if this answer makes sense by removing all the Zn and O atoms from your unit cell, and recalculating the work function. Likely, you’ll not see any change!
Hope this helps,
> On 18 May 2015, at 13:16, Bipul Rakshit <bipulrr at gmail.com> wrote:
> Dear Giovanni,
> I am using 8 Double layer of ZnO and 6layers of Au. and U value of 12 eV on Zn and 6.5eV on Oxygen. I am getting the workfunction values of 4.95 eV without oxygen vacancy
> and 4.96 with oxygen vacancy.
> The corresponding graphs I attached with it. and I also attached the input file where I include the dipole correction.
> I remove the oxygen from 3rd and 5th layer from Au-ZnO interface. While keeping the 7th and 8th layer fixed at bulk position.
> On Mon, May 18, 2015 at 1:46 PM, Giovanni Cantele <giovanni.cantele at spin.cnr.it <mailto:giovanni.cantele at spin.cnr.it>> wrote:
> What is the value of the work function you obtain in the two cases? Could you please post a plot of the electrostatic potential in the two cases, with relevant energies (Fermi level) highlighted?
> > On 17 May 2015, at 11:50, Bipul Rakshit <bipulrr at gmail.com <mailto:bipulrr at gmail.com>> wrote:
> > Dear users,
> > I am calculation metal (Au) - ZnO work-function. Au(111) is over the ZnO(0001) surface. Since ZnO (0001) surface is polar, so i applied dipole correction.
> > I am using 6Layers of Au and 8 double layer of ZnO. So is the workfunction is just the difference between fermi energy Ef and the constant potential in vacuum. Which i got after running pp.x and then average.x Or something else.
> > I am asking this because when after removing two oxygen's from ZnO's 3rd and 5th layer, there is no change in the workfunction. But I am expecting a change due to excess electrons of ZnO.
> > --
> > Dr. Bipul Rakshit
> > Research Associate,
> > Institute of Physics (IOP),
> > Bhubaneswar- 751 005
> > Orissa
> > India
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> Giovanni Cantele, PhD
> c/o Dipartimento di Fisica
> Universita' di Napoli "Federico II"
> Complesso Universitario M. S. Angelo - Ed. 6
> Via Cintia, I-80126, Napoli, Italy
> e-mail: giovanni.cantele at spin.cnr.it <mailto:giovanni.cantele at spin.cnr.it>
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> Dr. Bipul Rakshit
> Research Associate,
> Institute of Physics (IOP),
> Bhubaneswar- 751 005
> Pw_forum mailing list
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Giovanni Cantele, PhD
c/o Dipartimento di Fisica
Universita' di Napoli "Federico II"
Complesso Universitario M. S. Angelo - Ed. 6
Via Cintia, I-80126, Napoli, Italy
e-mail: giovanni.cantele at spin.cnr.it
Phone: +39 081 676910
Skype contact: giocan74
Web page: http://people.na.infn.it/~cantele
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