Re: [Wien] Super cell reg-LiMn2O4

2019-06-04 Thread delamora
I forget how to multiply...
It is "formula units" that should be taken into account and not primitive cells
With x=0.0125=1/16 would require a supercell of 8 (=2x2x2), that is 8 primitive 
cells, 16 formula units = 112 atoms

With x=0.01 = 1/100 => 100 formula units => 100*7 =700 atoms
  This could be done, but it requires a lot of computer time

So, if you do it with a 2x2x2 supercell you would have an approximation, but 
not a bad one.



LiMn1-x NdxO4
With x=0.125=1/8 would require a supercell of 8 (=2x2x2), that is 8 primitive 
cells, 16 formula units = 112 atoms

With x=0.01 = 1/100 = **1,200** => 1,400 atoms


LiMn1.99Nd0.01O4
This would require a supercell of 100 size, which would be very difficult to 
calculate due to its large size.
A larger doping would require a smaller supercell


   At present we are working on LiMn2O4 spinel Fd-3m.
Is it possible to run LiMn1.99Nd0.01O4 if so guide us in finding the order of 
super cell.

Thanks
Sundare
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Re: [Wien] Super cell reg-LiMn2O4

2019-06-04 Thread delamora
LiMn1-x NdxO4
With x=0.125=1/8 would require a supercell of 8 (=2x2x2), that is 8 primitive 
cells, 16 formula units = 112 atoms

With x=0.01 = 1/100 = **1,200** => 1,400 atoms


LiMn1.99Nd0.01O4
This would require a supercell of 100 size, which would be very difficult to 
calculate due to its large size.
A larger doping would require a smaller supercell


   At present we are working on LiMn2O4 spinel Fd-3m.
Is it possible to run LiMn1.99Nd0.01O4 if so guide us in finding the order of 
super cell.

Thanks
Sundare
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Re: [Wien] Super cell reg-LiMn2O4

2019-06-04 Thread delamora
LiMn1-x NdxO4
With x=0.125=1/8 would require a supercell of 8 (=2x2x2), that is 8 primitive 
cells, 16 formula units = 112 atoms

With x=0.01 = 1/100 = 1,200 atoms


LiMn1.99Nd0.01O4
This would require a supercell of 100 size, which would be very difficult to 
calculate due to its large size.
A larger doping would require a smaller supercell


   At present we are working on LiMn2O4 spinel Fd-3m.
Is it possible to run LiMn1.99Nd0.01O4 if so guide us in finding the order of 
super cell.

Thanks
Sundare
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Re: [Wien] Super cell reg-LiMn2O4

2019-06-04 Thread delamora
LiMn1.99Nd0.01O4
This would require a supercell of 100 size, which would be very difficult to 
calculate due to its large size.
A larger doping would require a smaller supercell


   At present we are working on LiMn2O4 spinel Fd-3m.
Is it possible to run LiMn1.99Nd0.01O4 if so guide us in finding the order of 
super cell.

Thanks
Sundare
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[Wien] Super cell reg-LiMn2O4

2019-06-04 Thread Murugan Sundareswari
Dear Professor
   At present we are working on LiMn2O4 spinel
Fd-3m.
Is it possible to run LiMn1.99Nd0.01O4 if so guide us in finding the order
of super cell.

Thanks
Sundare
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