Übungen zur Kristallographie/Kristallchemie Nr. 1 Musterlösung

= 3.4 · 10 6 = 3.4 · 10 3 = 0.34 · 10 9 = 340 · 10 6 = 34 · 10 15
m = ρ · V = 3.529 3 · 45.21 · 10 −24 3 = 1.595 · 10 −22
V = m
ρ =
N = V
V
0.2
3.529 3 = 0.05667 3 = 5.667 · 10 22 3
= 5.667·1022 3
45.21 3
= 0.125 · 10 22
N = n · N = m
M · N
0.2
= 12 · 6 · 1023 −1 = 1 · 1022
Z = N
N
= 1·1022
0.125·10 22 = 8
|a| = a 2 1 + a2 2 + a2 3 = √ 1 2 + 1 2 + 0 2 = √ 2
| b| = b 2 1 + b2 2 + b2 3 = √ 1 2 + 0 2 + 2 2 = √ 5
a · b = a1b1 + a2b2 + a3b3 = 1 · 1 + 1 · 0 + 0 · 2 = 1
φ = arccos a· b
|a|·| = √ 1√
= 0.3162 = 71.565
b| 2· 5 ◦
|a × b| = |a| · | b| · sin φ = √ 2 · √ 5 · sin(71.565 ◦ ) = 3

t = − = (0, 1, −2) t = − = (0, −1, 2)
t
⎛
|t| = (t1t2t3) ⎝
a 2 ab cos(γ) ac cos(β)
ba cos(γ) b 2 bc cos(α)
ca cos(β) cb cos(α) c 2
⎞ ⎛
⎠ ⎝
t 2 1 a 2 + t 2 2 b2 + t 2 3 c2 + 2t1t2ab cos(γ) + 2t1t3ac cos(β) + 2t2t3bc cos(α) = 0 2 2 2 + 1 2 3 2 + (−2) 2 4 2 = √ 73
= 8.544
t1
t2
t3
⎞
⎠ =