Übungen zur Kristallographie/Kristallchemie Nr. 9 Musterlösung
Übungen zur Kristallographie/Kristallchemie Nr. 9 Musterlösung Übungen zur Kristallographie/Kristallchemie Nr. 9 Musterlösung
• • • • • 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 ◦
- Seite 2 und 3: θ 1 d2 = hkl h 2 = h2a∗2 + k
• <br />
<br />
• <br />
• <br />
<br />
• <br />
<br />
• <br />
<br />
ρ0 = m<br />
V0 ⇔ V0 = m<br />
ρ0<br />
VF,29 = V0 · 0.65<br />
VP,29 = VF,29 · 0.94 = V0 · 0.65 · 0.94<br />
V m<br />
ρ <br />
m m·0.65·0.94<br />
= ⇒ ρP,29 =<br />
ρP,29 ρ0<br />
ρ0<br />
2.16<br />
0.65·0.94 = 0.65·0.94 = 3.535 3 <br />
ρP,29 = M·Z<br />
V·N ⇒ V = a 3 = M<br />
ρP,29·N = 58.4<br />
3.535·6.022·10 23 = 2.743 · 10 −23 3<br />
⇒ a = 3.016 <br />
<br />
<br />
V ∗ = a ∗ b ∗ c ∗ 1 − cos 2 (α ∗ ) − cos 2 (β ∗ ) − cos 2 (γ ∗ ) + 2 cos(α ∗ ) cos(β ∗ ) cos(γ ∗ )<br />
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)<br />
V ∗ = 3.405 · 10 −3 −3 V = 1<br />
V ∗ = 1<br />
3.405·10 −3 = 293.6 3 <br />
a = b∗ ·c ∗ ·sin(α ∗ )<br />
V ∗<br />
c = a∗ ·b ∗ ·sin(γ ∗ )<br />
V ∗<br />
= 0.1327·0.1832·sin(86.78)<br />
3.405·10−3 = 7.128 b = a∗ ·c ∗ ·sin(β ∗ )<br />
V ∗ = 0.149·0.1832·sin(78.44)<br />
3.405·10−3 = 7.854 <br />
= 0.149·0.1327·sin(73.66)<br />
3.405·10 −3<br />
cos(α) = cos(β∗ ) cos(γ ∗ )−cos(α ∗ )<br />
sin(β ∗ ) sin(γ ∗ )<br />
cos(β) = cos(α∗ ) cos(γ ∗ )−cos(β ∗ )<br />
sin(α ∗ ) sin(γ ∗ )<br />
cos(γ) = cos(α∗ ) cos(β ∗ )−cos(γ ∗ )<br />
sin(α ∗ ) sin(β ∗ )<br />
= cos(78.44) cos(73.66)−cos(86.78)<br />
= cos(86.78) cos(73.66)−cos(78.44)<br />
= cos(86.78) cos(78.44)−cos(73.66)<br />
= 5.572 <br />
sin(78.44) sin(73.66)<br />
sin(86.78) sin(73.66)<br />
sin(86.78) sin(78.44)<br />
<br />
= −3.368 · 10 −3 ⇒ α = 90.19 ◦ <br />
= −0.1934 ⇒ β = 101.1 ◦ <br />
= −0.2764 ⇒ γ = 106.04 ◦
θ <br />
1<br />
d2 =<br />
hkl<br />
h 2 = h2a∗2 + k2b∗2 + l2c∗2 + 2hka∗b∗ cos(γ∗ ) + 2hla∗c∗ cos(β∗ ) + 2klb∗c∗ cos(α∗ )<br />
sin 2 (θ) = λ2<br />
4d 2 hkl<br />
⇒ sin 2 (θ) = λ2<br />
4<br />
h 2 a ∗2 + k 2 b ∗2 + l 2 c ∗2 + 2hka ∗ b ∗ cos(γ ∗ ) + 2hla ∗ c ∗ cos(β ∗ ) + 2klb ∗ c ∗ cos(α ∗ ) <br />
sin 2 (θ) = 1.54182<br />
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 ·<br />
cos(78.44) + 2 · 1 · 1 · 0.1327 · 0.1832 · cos(86.78)]<br />
sin 2 (θ) = 0.111 ⇒ sin(θ) = 0.3332 ⇒ θ = 19.46 ◦ <br />
<br />
Ihkl ∝ |Fhkl| 2<br />
<br />
I020 = 7500 ⇒ |F020| = 86.60<br />
I110 = 35000 ⇒ |F110| = 187.08<br />
0, y, 1/4 0, −y, −3/4 1/2, 1/2 + y, 1/4 1/2, 1/2 − y, 3/4 <br />
<br />
<br />
<br />
Fhkl = <br />
N<br />
k=1 f exp 2πi xk <br />
h = N k=1 f<br />
<br />
cos 2π xk =0<br />
<br />
h + i sin 2π xk <br />
h<br />
<br />
= 2f cos 2π 0 · h + y · k + 1<br />
4 · l + cos 2π 1<br />
2 · h + 1<br />
2 + y · k + 1<br />
4 · l<br />
= 2f cos π 2yk + l<br />
<br />
1<br />
2 + cos π h + 2k 2 + y + l<br />
<br />
2<br />
ya = 0.1<br />
F110 = 2 · 80 cos [π (2 · 0.1 · 1)] + cos π 1 + 2 · 1 1<br />
2 + 0.1 = 258.89 <br />
F020 = 2 · 70 cos [π (2 · 0.1 · 2)] + cos π 2 · 2 1<br />
2 + 0.1 = 86.52 <br />
yb = 0.15<br />
F110 = 2 · 80 cos [π (2 · 0.15 · 1)] + cos π 1 + 2 · 1 1<br />
2 + 0.15 = 188.09 <br />
F020 = 2 · 70 cos [π (2 · 0.15 · 2)] + cos π 2 · 2 1<br />
2 + 0.15 = −86.52 <br />
yb = 0.15
h 2 + k 2 + l 2 hkl <br />
2ϑ/ ◦ sin(ϑ) sin 2 (ϑ)<br />
sin 2 (ϑ)<br />
0.0451 h2 + k2 + l2 <br />
hkl<br />
<br />
a<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
F h, k, l <br />
h + k = 2n, h + l = 2n, k + l = 2n <br />
a 6.291(2)