pOWER POINT PRESENTATION

1. Previous calculation:
During cleaving the slab did not contain same free surfaces .
Target:
During cleaving the slab should contain same free surfaces .

2. L10
L1
L2
L3
L4
L5
L6
L7
L8
L9
L11
L11
L12
Layers Interlayer separation (Å)
L1-L2 , L7-L8 1.439
L2-L3 , L8-L9 0.809
L3-L4, L9-L10 2.249
L4-L5 , L10-L11 2.249
L5-L6 , L11-L12 0.809
L6-L7, L12-L13 1.439
the cleavage occurs between the two parallel lattice planes,
where their lattice point density is greatest, where positive and
negative ions are neutralized, where the chemical bonding
force among the lattice points in the plate is the strongest, or
the two lattice planes have the same charges
So what are the factors ?
LuMn6
Sn6
magnetic unit (AFM)
Target:
During cleaving the slab should contain same free surfaces .
L5-L9
L1-L7
L4-L10
L3-L11

3. Stoichiometric ratio Lu:Mn:Sn=1:6:8
Unit cell height (10.434 Å)
Free surfaces are L1 and L7
L1
L7
L1
L7
L1
L7
(10.434 Å)
Next L1
Redefining unit cell for cleaving by same free surfaces(top and bottom)
Stoichiometric ratio Lu:Mn:Sn=2:6:8
Unit cell height (11.242 Å)
Free surfaces are L4 and L10
1x1x3
supercell
1x1x3
supercell
Next L4
L4
L10
L4
L10
L4

4. from Double unit after deletion of atoms
which are not in between L4 to L10
Stoichiometric ratio Lu:Mn:Sn=2:6:8
Unit cell (11.242 Å) + vacuum (6.746 Å )
Stoichiometric ratio Lu:Mn:Sn=2:6:8
Unit cell (10.434 Å) + vacuum (6.746 Å +10 Å )
L4
L10
E (AFM)= -94.1016 eV
My
(AFM)= 0.123 μB
E (FM)= -94.1996 eV
My
(FM)= 13.331 μB
Lu1
Lu2
Mn1
Mn2
Mn3
Mn4
Mn5
Mn6
Sn1
Sn2
Sn3
Sn4
Sn5
Sn6
Sn7
Sn8
C= 17.998 Å C= 27.998 Å
E (AFM)= -94.1060 eV
My
(AFM)= 0.123 μB
E (FM)= -94.2044 eV
My
(FM)= 13.331 μB
AFM-My
FM-My
C= 17.988 Å
Lu1
Lu2
Mn1
Mn2
Mn3
Mn4
Mn5
Mn6
Sn1
Sn2
Sn3
Sn4
Sn5
Sn6
Sn7
Sn8
AFM-My
FM-My

5. E (AFM)= -94.1016 eV
My
(AFM)= 0.123 μB
E (FM)= -94.1996 eV
My
(FM)= 13.331 μB
E (AFM)= -88.1531 eV
My
(AFM)= 0.012 μB
E (FM)= -88.2568 eV
My
(FM)= 13.028 μB
FM-My
(Bulk)
Cleavage energy = {E(slab)- E(bulk)}/A = -3.13 J/m2
Stoichiometric ratio Lu:Mn:Sn=2:6:8
Unit cell height (11.242 Å)
Free surfaces are L4 and L10
FM-My
(Slab)
Lu1
Lu2
Mn1
Mn2
Mn3
Mn4
Mn5
Mn6
Sn1
Sn2
Sn3
Sn4
Sn5
Sn6
Sn7
Sn8
BULK SLAB

7. In Double unit
Bulk (AB)
Lu:Mn:Sn=2:12:12
L10
L1
L2
L3
L4
L5
L6
L7
L8
L9
L11
L12
E (AFM)= -166.608 eV
E(FM) = -166.508 eV
In Double unit
Slab A (L4 to L10)
Lu:Mn:Sn=2:6:8
In Double unit
Slab B (L1 to L3
& L11 to L12)
Lu:Mn:Sn=0:6:4
E (AFM)= -94.1016 eV
E(FM)= -94.1996 eV
L11
L3
E(FM)= -42.0375 eV
C= 17.998 Å
MODEL-2

8. Slab- L4 to L8
Lu:Mn:Sn=2:6:8
My
(AFM)= 0.123 μB
My
(FM)= 13.331 μB
Lu1
Lu2
Mn1
Mn2
Mn3
Mn4
Mn5
Mn6
Sn1
Sn2
Sn3
Sn4
Sn5
Sn6
Sn7
Sn8
C= 17.998 Å
AFM-My
FM-My
C= 17.998 Å
L4
L8
Slab- L1 to L3 &
L11 to L12
Lu:Mn:Sn=0:6:4
Atom My
( μB)
Mn1 4.407
Mn2 4.020
Mn3 0.648
Mn4 0.542
Mn5 0.668
Mn6 0.635
Sn1 -0.123
Sn2 -0.121
Sn3 0.006
Sn4 0.001
My
(FM)= 10.323 μB
17.998 Å
MODEL-2

9. Cleavage enrgy from model 2
Cleavage Energy= {E(slab-A)+E(slab-B)-E(bulk)}/2A
Bulk AB : E (AFM)= -166.608 eV
Slab A: E(FM)= -94.1996 eV
Slab B: E(FM)= -42.0375 eV
A= (5.507*10-10
)2
m2
Cleavage Energy= 8.002 J/m2

10. C= 17.988 Å
E (AFM) = -88.7325 eV
My
(AFM)= 0.000 μB
E(FM) = -88.8244 eV
My
(FM)= 14.359 μB
AFM
L1
L7 Lu1
Mn1
Mn2
Mn3
Mn4
Mn5
Mn6
Sn1
Sn2
Sn3
Sn4
Sn5
Sn6
Sn7
Sn8
AFM-My
C= 27.988 Å
Lu1
Mn1
Mn2
Mn3
Mn4
Mn5
Mn6
Sn1
Sn2
Sn3
Sn4
Sn5
Sn6
Sn7
Sn8
E(AFM)= -88.7356 eV
My
(AFM)= 0.000
E(FM)= -88.8278 eV
My
(FM) = 14.359 μB
AFM-My
from Double unit after deletion of atoms
which are not in between L1 to L7
Stoichiometric ratio Lu:Mn:Sn=1:6:8
Unit cell (10.434 Å) + vacuum (7.554 Å )
Stoichiometric ratio Lu:Mn:Sn=1:6:8
Unit cell (11.242 Å) + vacuum (7.754 Å +10 Å )
FM-My
FM-My

11. In single unit
Lu:Mn:Sn=1:6:8 unit cell
L7
L1
10.434 Å
E(AFM)= -5.5896 eV
E(FM)= -5.4730 eV
single unit+ 10 Å vacuum
Lu:Mn:Sn=1:6:8
E (AFM) = -88.7325 eV
E(FM) = -88.8278 eV
20.434 Å
L7
L1
unit cell
(AFM)
unit cell
(FM)
Lu1
Mn1
Mn2
Mn3
Mn4
Mn5
Mn6
Sn1
Sn2
Sn3
Sn4
Sn5
Sn6
Sn7
Sn8
unit cell (AFM)+
10 Å vacuum

12. Position
ion
AFM
My
[-0.0003 μB]
AFM with 10 A vac
My
=-0.01[ μB ]
L1 Sn9 -0.191 -0.071
L1 Sn10 -0..191 -0.071
L2 Sn1 -0.182 -0.060
L3 Mn1 2.416 2.446
L3 Mn3 2.416 2.445
L3 Mn5 2.420 2.446
L4 Lu1 0.000 0.000
L4 Sn5 0.00 0.00
L4 Sn7 0.00 0.00
L5 Mn8 -2.420 -2.449
L5 Mn10 -2.416 -2.449
L5 Mn12 -2.416 -2.450
L6 Sn4 0.182 0.059
L7 Sn11 0.191 0.071
L7 Sn12 0.191 0.071
L8 Sn3 0.182 0.355
L9 Mn7 -2.416 -3.109
L9 Mn9 -2.416 -3.109
L9 Mn11 -2.420 -3.109
L10 Lu2 0.000 0.001
L10 Sn6 0.000 0.000
L10 Sn8 0.000 0.000
L11 Mn2 2.416 3.110
L11 Mn4 2.416 3.110
L11 Mn6 2.420 3.110
L12 Sn2 -0.182 -0.354
L10
L1
L2
L3
L4
L5
L6
L7
L8
L9
L12
3 types of Sn layers:
L1,L7-spacer
L2,L6.L8,L12-adjacent to Mn
L4,L10-Lu layer
AFM bulk: E= -166.608 eV My
= -0.0003 μB
FM bulk E=-166.508 eV My
= 25.584 μB
Energy and Magnetic moment calculation of LuMn6
Sn6
bulk
L11
When 10 Å vac is inserted
between L12 and next L1
E(AFM-slab)= -159.2950 eV
Cleavage energy=
{E(slab)-E(bulk)}/A
A= (5.507*10-10
)2
m2
Cleavage energy= 3.85 J/m2

14. Lu1
Lu2
Mn1
Mn2
Mn3
Mn4
Mn5
Mn6
Sn1
Sn2
Sn3
Sn4
Sn5
Sn6
Sn7
Sn8
Bulk
AFM-My
Slab
AFM-My