Atomic Energy of P

Message

Oscar_rp · #1 Post by **Oscar_rp** » Mon Oct 06, 2008 2:58 pm

Hello,

I want to calculated the Free atom energy of P (Phosphorus atom)
in its ground state [Ne] 3s2 3p3 with
PBE_PAW in a big box.

VASP gives me an energy about -1.888409 eV,

and I wonder if this atomic energy is correct or I have done a mistake in the inputs

My input files are
parameter for this Run:
LPETIM=F write-flag & timer
ISTART = 0 job : 0-new 1-cont 2-samecut

Electronic Relaxation 1
PREC = normal
ENCUT = 550.00 eV
NELM = 99
EDIFF = 1E-06
INIWAV = 1 random initial wavefunctions
MAXMIX = 88

Ionic Relaxation
EDIFFG = -0.02 stopping-criterion for IOM
NSW = 99 number of steps for IOM
NBLOCK = 1; KBLOCK = 40 inner block; outer block
IBRION = 2 ionic relax: 0-MD 1-quasi-New 2-CG
ISIF = 2
IWAVPR = 1 prediction: 0-non 1-charg 2-wave 3-comb
ISYM = 1 0-nonsym 1-usesym
SYMPREC = 0.0001
LCORR = F Harris-correction to forces
ISMEAR = -2
LREAL = F
ISPIN = 2
ALGO = Fast
LDIAG = F

POSCAR
P
1.00000000000000
10.0079999999999991 0.0000000000000000 0.0000000000000000
0.0000000000000000 8.6669999999999998 0.0000000000000000
0.0000000000000000 0.0000000000000000 47.0780000000000030
1
Direct
0.0000000000000000 0.0000000000000000 0.0000000000000000

0.00000000E+00 0.00000000E+00 0.00000000E+00

Many thanks
Oscar

physicalattraction · #2 Post by **physicalattraction** » Mon Oct 06, 2008 3:22 pm

Hello Oscar,

I do not know whether you have calculated this correctly, but I have one question for you: why do you choose an ionic relaxation procedure when you only have one isolated atom in a large vacuum?

Erwin

Oscar_rp · #3 Post by **Oscar_rp** » Mon Oct 06, 2008 5:51 pm

Hello Erwin,

That's true, it is not necessary, and it only makes one iteration.
The large vacuum is because I need to compare to other system which has such large box

Thanks
Oscar

#4 Post by **admin** » Tue Oct 07, 2008 2:31 pm

your result looks very reasonable.