{0 0 0 2} − ∥ (1 1 1) δ
[1 1 2 ̅ 0] α−Zr ∥ (1 1 ̅ 0)
2
ℎ ℎ = (4/3)
4
2
2
∆ = (∆ ℎ + ∆ ℎ + ∆ ℎ ) + 2 + 4
3 ℎ
ℎ
∆ ℎ =
∆ ℎ =
∆ ℎ =
=
=
∆ ℎ
ℎ
∗
∆ ∗
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