f
P
5+
3
– 0,6413 + 0,7673i – 0,1948 + 0,9808i – 0,6470 + 0,7624i – 0,6470 + 0,7624i
– 0,6413 + 0,7673i – 0,1948 + 0,9800i + f
O
2 –
1
0,7871 – 0,6169i – 0,1688
– 0,9856i – 0,6762 – 0,7367i – 0,6762 – 0,7367i – 0,6762 – 0,7367i – 0,1688 – 0,9856i + f
O
2 –
2
0,7518 + 0,6594i – 0,9948 – 0,1016i + 0,9966 + 0,08284i +
0,9966 + 0,08284i + 0,7518 + 0,6594i – 0,9948 – 0,1016i + f
O
2 –
3
– 0,9997 – 0,02262i + 0,8994
– 0,4371i + 0,9863 + 0,1652i + 0,9863 + 0,16512i – 0,99974 – 0,2262i + 0,8994
– 0,4371i + f
O
2 –
4
0,5949 + 0,8038i + 0,5898 + 0,8075i – 0,8677
– 0,4971i – 0,8677 – 0,4971i + 0,5949 + 0,8038i + 0,5898 + 0,8075i + f
O
2 –
5
– 0,3304 – 0,9438i + 0,8477 + 0,5305i – 0,8214 – 0,5704i – 0,8214 – 0,5704i
– 0,3304 – 0,9438i + 0,8477 + 0,5305i + f
O
2 –
6
0,08785 + 0,9961i –
0,7641 – 0,6451i – 0,7641 – 0,6451i – 0,2946 – 0,9556i + 0,08785 + 0,9961i –
0,7641 – 0,6451i
F
0210
= 3 f
Ca
2+
1
– 2,581 + 4,511i + f
Ca
2+
2
– 2,710 + 4,661i + f
Ca
2+
3
– 2,166 + 4,973i + f
Ca
2+
4
1,186 + 1,611i + f
Ca
2+
5
– 1,029 + 1,715i + f
P
5+
1
2,000 + f
P
5+
2
– 3,458 + 4,669i + f
P
5+
3
– 2,966 + 5,020i + f
O
2 –
1
– 1,579 – 4,798i + f
O
2 –
2
1,507 + 1,281i + f
O
2 –
3
1,772 – 0,7927i + f
O
2 –
4
0,6340 + 2,228i + f
O
2 –
5
– 0,6082 – 1,967i + f
O
2 –
6
– 2,411 – 0,8987i F
0210
= 316,80 – 2,581 + 4,511i + 16,80– 2,710 + 4,661i + 16,80– 2,166 +
4,973i + 16,801,186 + 1,611i + 16,80 – 1,029 + 1,715i + 9,8002,000 +
9,800 – 3,458 + 4,669i + 9,800– 2,966 + 5,020i + 8,000– 1,579 – 4,798i +
8,0001,507 + 1,281i + 8,0001,772 – 0,7927i + 8,0000,6340 + 2,228i +
8,000 – 0,6082 – 1,967i + 8,000– 2,411 – 0,8987i
F
0210
= 3 – 43,36 + 75,78i – 45,53 + 78,30i – 36,39 + 83,55i + 19,92 + 27,06i – 17,29 +
28,81i + 19,60 – 33,89 + 45,76i – 29,07 + 49,20i – 12,63 – 38,38i + 12,06 + 10,25i
+ 14,18 – 6,342i + 5,072 + 17,82i – 4,866 – 15,74i – 19,29 – 7,190i
F
0210
= 3 – 171,5 + 368,9i
F
0210
= – 514,5 + 1107i
| F
0210
| = 1221
7. 2. 1. 5. CaOH
2
Bidang 0 0 1 2 θ = 17,90
Faktor hamburan atomik =
sin θ
=
sin 8,950 1,54056 Å
= 0,1010 f
Ca
2+
= 16,80
f
O
2 –
= 8,000 f
H
+
= 0,4800 F
hkl
= f
N N
e
2 πihx
N
+ky
N
+lz
N
F
001
= f
Ca
2+
N = 1
e
2 πi0.x
Ca2+
+0.y
Ca2+
+1.z
Ca2+
+ f
O
2 –
N = 1
e
2 πi0.x
O2–
+0.y
O2–
+1.z
O2–
+ f
H
+
N = 1
e
2 πi0.x
H
+0.y
H
+1.z
H
F
001
= f
Ca
2+
e
2 πi1.0
+ f
O
2 –
e
2 πi1.0,2330
+ e
2 πi1.-0,2330
+ f
H
+
e
2 πi1.0,4180
+ e
2 πi1.-0,4180
F
001
= f
Ca
2+
1,000 + f
O
2 –
0,1066 + 0,9943i + 0,1066 – 0,9943i + f
H
+
– 0,8702 + 0,4927i
– 0,8702 – 0,4927i F
001
= f
Ca
2+
1,000 + f
O
2 –
0,2132 + f
H
+
– 1,740 F
001
= 16,801,000 + 8,0000,2132 + 0,4800 – 1,740
F
001
= 16,80 + 1,706 – 0,8352
F
001
= 17,67 |F
001
| = 17,67
7. 2. 1. 6. CaOH
2
Bidang 1 1 1 2 θ = 54,26
o
Faktor hamburan atomik =
sin θ
=
sin 27,13 1,54056 Å
= 0,2960 f
Ca
2+
= 14,00 f
O
2 –
= 5,500 f
H
+
= 0,2500 F
hkl
= f
N N
e
2 πihx
N
+ky
N
+lz
N
F
111
= f
Ca
2+
N = 1
e
2 πi1.x
Ca2+
+1.y
Ca2+
+1.z
Ca2+
+ f
O
2 –
N = 1
e
2 πi1.x
O2–
+1.y
O2–
+1.z
O2–
+ f
H
+
N = 1
e
2 πi1.x
H
+1.y
H
+1.z
H
F
111
= f
Ca
2+
e
2 πi1.0+1.0+1.0
+ f
O
2 –
e
2 πi1.0,3333+1.0,6666+1.0,2330
+ e
2 πi1.-0,3333+1.-0,6666+1.-0,2330
+ f
H
+
e
2 πi1.0,3333+1.0,6666+1.0,4180
+ e
2 πi1.-0,3333+1.-0,6666+1.-0,4180
F
111
= f
Ca
2+
1,000 + f
O
2 –
0,1072 + 0,9942i + 0,1072 – 0,9942i + f
H
+
– 0,8699 + 0,4933i
– 0,8699 – 0,4933i F
111
= f
Ca
2+
1,000 + f
O
2 –
0,2144 + f
H
+
– 1,740 F
111
= 14,001,000 + 5,5000,2144 + 0,2500 – 1,740
F
111
= 14,00 + 1,179 – 0,435
F
111
= 14,74 |F
111
| = 14,74
7. 2. 2. BCP II 7. 2. 2. 1. Menentukan Faktor Struktur
F
hkl
Karena ditinjau pada bidang yang sama, maka besarnya F
hkl
pada BCP II sama dengan besarnya F
hkl
pada BCP I.
7. 3. Menentukan R
hkl
7. 3. 1. BCP I 7. 3. 1. 1. HAp Bidang 2 1 1 2
θ = 31,68
o
V
211 –2
= 0,866.a
2
c = 0,8669,577
2
7,016
–2
= 3,220 x 10
–6
Å
–6
|F
211
|
2
= 99,96
2
= 9992 p
211
= 12 L-P
211
=
1 + cos
2
2 θ
sin
2
θ cos θ
=
1 + cos
2
31,68 sin
2
15,84 cos15,84
=
1 + 0,7242 0,074500,9620
=
1,724 0,07167
= 24,05 R
211
= V
211 –2
.
|F
211
|
2
. p
211
. L-P
211
R
211
= 3,226 x 10
–6
99921224,05 = 9,285 Å
–6
7. 3. 1. 2. HAp Bidang 0 0 4 2 θ = 52,88
o
V
004 –2
= 0,866.a
2
c = 0,8669,577
2
7,016
–2
= 3,220 x 10
–6
Å
–6
|F
004
|
2
= 249,0
2
= 6,200 x 10
4
p
004
= 2 L-P
004
=
1 + cos
2
2 θ
sin
2
θ cos θ
=
1 + cos
2
52,88 sin
2
26,44 cos26,44
=
1 + 0,3642 0,19820,8954
=
1,364 0,1775
= 7,684 R
004
= V
004 –2
.
|F
004
|
2
. p
004
. L-P
004
= 3,226 x 10
-6
6,200 x 10
4
27,684 = 3,068 Å
–6
7. 3. 1. 3. β-TCP Bidang 3 0 0 2θ = 29,54
