Übungen zur Kristallographie/Kristallchemie Nr. 9 Musterlösung

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ρ0 = m
V0 ⇔ V0 = m
ρ0
VF,29 = V0 · 0.65
VP,29 = VF,29 · 0.94 = V0 · 0.65 · 0.94
V m
ρ
m m·0.65·0.94
= ⇒ ρP,29 =
ρP,29 ρ0
ρ0
2.16
0.65·0.94 = 0.65·0.94 = 3.535 3
ρP,29 = M·Z
V·N ⇒ V = a 3 = M
ρP,29·N = 58.4
3.535·6.022·10 23 = 2.743 · 10 −23 3
⇒ a = 3.016
V ∗ = a ∗ b ∗ c ∗ 1 − cos 2 (α ∗ ) − cos 2 (β ∗ ) − cos 2 (γ ∗ ) + 2 cos(α ∗ ) cos(β ∗ ) cos(γ ∗ )
V ∗ = 0.149 · 0.1327 · 0.1832 1 − cos 2 (86.78) − cos 2 (78.44) − cos 2 (73.66) + 2 cos(86.78) cos(78.44) cos(73.66)
V ∗ = 3.405 · 10 −3 −3 V = 1
V ∗ = 1
3.405·10 −3 = 293.6 3
a = b∗ ·c ∗ ·sin(α ∗ )
V ∗
c = a∗ ·b ∗ ·sin(γ ∗ )
V ∗
= 0.1327·0.1832·sin(86.78)
3.405·10−3 = 7.128 b = a∗ ·c ∗ ·sin(β ∗ )
V ∗ = 0.149·0.1832·sin(78.44)
3.405·10−3 = 7.854
= 0.149·0.1327·sin(73.66)
3.405·10 −3
cos(α) = cos(β∗ ) cos(γ ∗ )−cos(α ∗ )
sin(β ∗ ) sin(γ ∗ )
cos(β) = cos(α∗ ) cos(γ ∗ )−cos(β ∗ )
sin(α ∗ ) sin(γ ∗ )
cos(γ) = cos(α∗ ) cos(β ∗ )−cos(γ ∗ )
sin(α ∗ ) sin(β ∗ )
= cos(78.44) cos(73.66)−cos(86.78)
= cos(86.78) cos(73.66)−cos(78.44)
= cos(86.78) cos(78.44)−cos(73.66)
= 5.572
sin(78.44) sin(73.66)
sin(86.78) sin(73.66)
sin(86.78) sin(78.44)
= −3.368 · 10 −3 ⇒ α = 90.19 ◦
= −0.1934 ⇒ β = 101.1 ◦
= −0.2764 ⇒ γ = 106.04 ◦

θ
1
d2 =
hkl
h 2 = h2a∗2 + k2b∗2 + l2c∗2 + 2hka∗b∗ cos(γ∗ ) + 2hla∗c∗ cos(β∗ ) + 2klb∗c∗ cos(α∗ )
sin 2 (θ) = λ2
4d 2 hkl
⇒ sin 2 (θ) = λ2
4
h 2 a ∗2 + k 2 b ∗2 + l 2 c ∗2 + 2hka ∗ b ∗ cos(γ ∗ ) + 2hla ∗ c ∗ cos(β ∗ ) + 2klb ∗ c ∗ cos(α ∗ )
sin 2 (θ) = 1.54182
4 [22 · 0.1492 + 12 · 0.13272 + 12 · 0.18322 + 2 · 2 · 1 · 0.149 · 0.1327 · cos(73.66) + 2 · 2 · 1 · 0.149 · 0.1832 ·
cos(78.44) + 2 · 1 · 1 · 0.1327 · 0.1832 · cos(86.78)]
sin 2 (θ) = 0.111 ⇒ sin(θ) = 0.3332 ⇒ θ = 19.46 ◦
Ihkl ∝ |Fhkl| 2
I020 = 7500 ⇒ |F020| = 86.60
I110 = 35000 ⇒ |F110| = 187.08
0, y, 1/4 0, −y, −3/4 1/2, 1/2 + y, 1/4 1/2, 1/2 − y, 3/4
Fhkl =
N
k=1 f exp 2πi xk
h = N k=1 f
cos 2π xk =0
h + i sin 2π xk
h
= 2f cos 2π 0 · h + y · k + 1
4 · l + cos 2π 1
2 · h + 1
2 + y · k + 1
4 · l
= 2f cos π 2yk + l
1
2 + cos π h + 2k 2 + y + l
2
ya = 0.1
F110 = 2 · 80 cos [π (2 · 0.1 · 1)] + cos π 1 + 2 · 1 1
2 + 0.1 = 258.89
F020 = 2 · 70 cos [π (2 · 0.1 · 2)] + cos π 2 · 2 1
2 + 0.1 = 86.52
yb = 0.15
F110 = 2 · 80 cos [π (2 · 0.15 · 1)] + cos π 1 + 2 · 1 1
2 + 0.15 = 188.09
F020 = 2 · 70 cos [π (2 · 0.15 · 2)] + cos π 2 · 2 1
2 + 0.15 = −86.52
yb = 0.15

h 2 + k 2 + l 2 hkl
2ϑ/ ◦ sin(ϑ) sin 2 (ϑ)
sin 2 (ϑ)
0.0451 h2 + k2 + l2
hkl
a
F h, k, l
h + k = 2n, h + l = 2n, k + l = 2n
a 6.291(2)