3 Spin-Polarized Plasmonics: Fresh View …
59
TEM of [100] crystal structure of some cobalt nanoparticles have lattice spacing
2.102 Å and inter-planar angle 83.6° typical for hcp. Figure 3.4b shows high resolution TEM image of [111] crystal structure of other cobalt nanoparticles has lattice
spacing 1.95 Å and inter-planar angle 25.63°, typical for fcc.
For calculation expressions explaining the basic crystallography of cobalt was
used [53]. Following by definition, if two planes have indices (hkl) the distance
perpendicularly between them is shown by the inverse of the magnitude of the lattice
vector.
d hkl =
1
ha ∗ + kb
∗
+ l c ∗
(3.1)
Then:
ha
∗
+ kb
∗
+ l c
∗
2 =
ha
∗
+ kb
∗
+ l c
∗
·
ha
∗
+ kb
∗
+ l c
∗
= h
2 a
∗2
+ k
2 b
∗2
+ l
2 c
∗2
+ 2kl b
∗
· c
∗
+ 2lhc
∗
· a
∗
+ 2hka
∗
· b
∗
= h
2 a
∗2
+ k
2 b
∗2
+ l
2 c
∗2
+ 2klb
∗ c
∗ cos α
∗
+ 2lhc
∗ a
∗ cos β
∗
+ 2hka
∗ b
∗ cos γ
∗
here, a
∗ ,b
∗ , and c
∗ are lattice vectors, α
∗ is the angle between lattice vectors b
∗ and
c
∗ , β
∗ is the angle between c
∗ and a
∗ . Knowing that cobalt is found primarily in two
phases, hexagonal close packed and face-centered cubic further expansion of this
formula was used only for those two overall structures. The lattice constants used
for cobalt are found in a variety of literatures. For hcp lattice constants of a = b =
0.2507 nm, c = 0.4069 nm. For fcc lattice constants of a = b = c = 0.35446 nm.
First calculations were done to families of standard planes for both fcc and hcp
structures. Hexagonal first:
α
∗
= β
∗
= 90
◦
, γ
∗
= 60
◦ and a
∗
= b
∗
, d
2
hkl =
1
(h 2 + k 2 + hk)a ∗2 + l 2 c ∗2 (3.2)
cos φ = d hkl d h k l
hh
+ kk
+
1
2
(hk
+ kh
)
a
∗2
+ ll
c
∗2
In which:
a
∗
=
2
a
√
3
, c
∗
=
1
c
Next for cubic:
α
∗
= β
∗
= γ
∗
= 90
◦ and a
∗
= b
∗
= c
∗
, d
2
hkl =
1
(h 2 + k 2 + l 2 )a ∗2
(3.3)
59
TEM of [100] crystal structure of some cobalt nanoparticles have lattice spacing
2.102 Å and inter-planar angle 83.6° typical for hcp. Figure 3.4b shows high resolution TEM image of [111] crystal structure of other cobalt nanoparticles has lattice
spacing 1.95 Å and inter-planar angle 25.63°, typical for fcc.
For calculation expressions explaining the basic crystallography of cobalt was
used [53]. Following by definition, if two planes have indices (hkl) the distance
perpendicularly between them is shown by the inverse of the magnitude of the lattice
vector.
d hkl =
1
ha ∗ + kb
∗
+ l c ∗
(3.1)
Then:
ha
∗
+ kb
∗
+ l c
∗
2 =
ha
∗
+ kb
∗
+ l c
∗
·
ha
∗
+ kb
∗
+ l c
∗
= h
2 a
∗2
+ k
2 b
∗2
+ l
2 c
∗2
+ 2kl b
∗
· c
∗
+ 2lhc
∗
· a
∗
+ 2hka
∗
· b
∗
= h
2 a
∗2
+ k
2 b
∗2
+ l
2 c
∗2
+ 2klb
∗ c
∗ cos α
∗
+ 2lhc
∗ a
∗ cos β
∗
+ 2hka
∗ b
∗ cos γ
∗
here, a
∗ ,b
∗ , and c
∗ are lattice vectors, α
∗ is the angle between lattice vectors b
∗ and
c
∗ , β
∗ is the angle between c
∗ and a
∗ . Knowing that cobalt is found primarily in two
phases, hexagonal close packed and face-centered cubic further expansion of this
formula was used only for those two overall structures. The lattice constants used
for cobalt are found in a variety of literatures. For hcp lattice constants of a = b =
0.2507 nm, c = 0.4069 nm. For fcc lattice constants of a = b = c = 0.35446 nm.
First calculations were done to families of standard planes for both fcc and hcp
structures. Hexagonal first:
α
∗
= β
∗
= 90
◦
, γ
∗
= 60
◦ and a
∗
= b
∗
, d
2
hkl =
1
(h 2 + k 2 + hk)a ∗2 + l 2 c ∗2 (3.2)
cos φ = d hkl d h k l
hh
+ kk
+
1
2
(hk
+ kh
)
a
∗2
+ ll
c
∗2
In which:
a
∗
=
2
a
√
3
, c
∗
=
1
c
Next for cubic:
α
∗
= β
∗
= γ
∗
= 90
◦ and a
∗
= b
∗
= c
∗
, d
2
hkl =
1
(h 2 + k 2 + l 2 )a ∗2
(3.3)