<html><body><div style="color:#000; background-color:#fff; font-family:times new roman, new york, times, serif;font-size:12pt"><div><span>Dear Abolore,</span></div><div><span>Read this article it contains all the atomic positions for the compound. You need to know that for this compound Z=8.<br></span></div><div><span> http://www.springerlink.com/content/p27870432mt31374/fulltext.pdf</span></div><div> </div><div>Adetunji Bamidele Ibrahim(PhD Student)<br>Department of physics,University of Agriculture,<br>Abeokuta, Ogun State,Nigeria.</div><div><br></div> <div style="font-family: times new roman, new york, times, serif; font-size: 12pt;"> <div style="font-family: times new roman, new york, times, serif; font-size: 12pt;"> <div dir="ltr"> <font face="Arial" size="2"> <hr size="1"> <b><span style="font-weight:bold;">From:</span></b> Abolore Musari <abmus007@gmail.com><br> <b><span style="font-weight: bold;">To:</span></b> PWSCF Forum
<pw_forum@pwscf.org> <br> <b><span style="font-weight: bold;">Sent:</span></b> Monday, August 6, 2012 4:42 PM<br> <b><span style="font-weight: bold;">Subject:</span></b> Re: [Pw_forum] equilibrium lattice constant of magnetite<br> </font> </div> <br>
<meta http-equiv="x-dns-prefetch-control" content="off"><div id="yiv275358972">Thanks so much Jia<br>But Sir the atomic positions in the input is for 3 atoms, should l leave the atomic position and change the number of atom or my atomic position is wrong??<br>thanks<br>Musari A. A<br>UNAAB Nigeria<br>
<br><div class="yiv275358972gmail_quote">On Mon, Aug 6, 2012 at 3:34 PM, jia chen <span dir="ltr"><<a rel="nofollow" ymailto="mailto:jiachen@princeton.edu" target="_blank" href="mailto:jiachen@princeton.edu">jiachen@princeton.edu</a>></span> wrote:<br><blockquote class="yiv275358972gmail_quote" style="margin:0pt 0pt 0pt 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex;">
Dear Abolore Musari,<div><br></div><div>I guess the number of atoms should be 14, instead of 3.</div><div><br></div><div>Regards</div><div>Jia<br><br><div class="yiv275358972gmail_quote"><div><div class="yiv275358972h5">On Mon, Aug 6, 2012 at 10:24 AM, Abolore Musari <span dir="ltr"><<a rel="nofollow" ymailto="mailto:abmus007@gmail.com" target="_blank" href="mailto:abmus007@gmail.com">abmus007@gmail.com</a>></span> wrote:<br>
</div></div><blockquote class="yiv275358972gmail_quote" style="margin:0pt 0pt 0pt 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex;"><div><div class="yiv275358972h5">Dear QE users<br>Sir, I am try to find the lattice constant of magnetite (Fe3O4) which is a spinel ferrimagnetic material. The expt lattice constant was 15.87 and I have been trying to find the equilibrium lattice constant (celldm(1)) between 15.80 - 16.00 and all i get is a straight line graph instead of the usual exponential curve graph. I have tried all l could most esp with starting_magnetization but the results are the same. My input is pasted below pls help me to get the appropriate graph.<br>
The kind of the graph I got is attached.<br>Thanks<br><br>&control<br> calculation = 'scf',<br> restart_mode ='from_scratch',<br> prefix = 'Fe3O4',<br> pseudo_dir = '/home/abolore/Programs/Pseudos/',<br>
outdir = './tmp/'<br> /<br> &system<br> ibrav = 2, <br> celldm(1)= $15.80-16.00,<br> nat = 3, <br> ntyp = 3,<br> ecutwfc = 40,<br> ecutrho = 400,<br> starting_magnetization(1) = 0.7,<br>
starting_magnetization(2) = -0.5,<br> starting_magnetization(3) = 0.0,<br> nspin = 2,<br> occupations='smearing',<br> smearing='gaussian',<br> degauss = 0.05,<br> /<br> &electrons<br>
diagonalization = 'david',<br> mixing_mode = 'plain',<br> mixing_beta = 0.7<br> /<br>ATOMIC_SPECIES<br> Fe1 55.845 Fe.pz-nd-rrkjus.UPF<br> Fe2 55.845 Fe.pz-nd-rrkjus.UPF<br> O 16.000 O.pz-rrkjus.UPF<br>
<br>ATOMIC_POSITIONS<br>Fe1 0.125 0.125 0.125 <br>Fe2 0.500 0.500 0.500<br>O 0.2548 0.2548 0.2548<br> <br>K_POINTS (automatic)<br> 8 8 8 1 1 1<br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br>
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<br></blockquote></div><span class="yiv275358972HOEnZb"><font color="#888888"><br><br clear="all"><div><br></div>-- <br>Jia Chen<br><br><br>
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