= 3,527 x 10
–6
99921223,61 = 9,985 Å
–6
7. 3. 2. 2. HAp Bidang 0 0 4 2 θ = 53,21
V
004 –2
= 0,866.a
2
c = 0,8669,436
2
6,906
–2
= 3,527 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
53,21 sin
2
26,60 cos26,60
=
1 + 0,3587 0,20050,8942
=
1,359 0,1793
= 7,579 R
004
= V
004 –2
.
|F
004
|
2
. p
004
. L-P
004
= 3,527 x 10
–6
6,200 x 10
4
27,579 = 3,315 Å
–6
7. 3. 2. 3.
β-TCP Bidang 3 0 0 2θ = 29,62
V
300 –2
= 0,866.a
2
c = 0,86610,44
2
37,45
–2
= 8,003 x 10
–8
Å
–6
|F
300
|
2
= 216,3
2
= 4,678 x 10
4
p
300
= 6 L-P
300
=
1 + cos
2
2 θ
sin
2
θ cos θ
=
1 + cos
2
29,62 sin
2
14,81 cos14,81
=
1 + 0,7557 0,065340,9668
=
1,756 0,06317
= 27,80 R
300
= V
300 –2
.
|F
300
|
2
. p
300
. L-P
300
= 8,003 x 10
–8
4,678 x 10
4
627,80 = 0,6245 Å
–6
7. 3. 2. 4. β-TCP Bidang 0 2 10 2θ = 31,13
V
0210 –2
= 0,866.a
2
c = 0,86610,44
2
37,45
–2
= 8,003 x 10
–8
|F
0210
|
2
= 1221
2
= 1,491 x 10
6
p
0210
= 6 L-P
0210
=
1 + cos
2
2 θ
sin
2
θ cos θ
=
1 + cos
2
31,13 sin
2
15,56 cos15,56
=
1 + 0,7327 0,071960,9634
=
1,733 0,06933
= 25,00 R
0210
= V
0210 –2
.
|F
0210
|
2
. p
0210
. L-P
0210
= 8,003 x 10
–8
1,491 x 10
6
625,00 = 17,90 Å
–6
7. 3. 2. 5. CaOH
2
Bidang 0 0 1 2 θ = 18,07
V
001 –2
= 0,866.a
2
c = 0,8663,590
2
4,911
–2
= 3,328 x 10
–4
Å
–6
|F
001
|
2
= 17,67
2
= 312,2 p
001
= 2
L-P
001
=
1 + cos
2
2 θ
sin
2
θ cos θ
=
1 + cos
2
18,07 sin
2
9,035 cos9,035
=
1 + 0,9038 0,024660,9856
=
1,904 0,02430
= 78,35 R
001
= V
001 –2
.
|F
001
|
2
. p
001
. L-P
001
as = 3,328 x 10
–4
312,2278,35 = 16,28 Å
–6
7. 3. 2. 6. CaOH
2
Bidang 1 1 1 2 θ =54,24
V
111 –2
= 0,866.a
2
c = 0,8663,590
2
4,911
–2
= 3,328 x 10
–4
|F
111
|
2
= 14,74
2
= 217,3 p
111
= 6 L-P
111
=
1 + cos
2
2 θ
sin
2
θ cos θ
=
1 + cos
2
54,24 sin
2
27,12 cos27,12
=
1 + 0,3415 0,20780,8900
=
1,342 0,1849
= 7,258 R
111
= V
111 –2
.
|F
111
|
2
. p
111
. L-P
111
= 3,328 x 10
–4
217,367,258 = 3,149 Å
–6
7. 4. Menentukan Komposisi Fasa di dalam BCP
I
hkl HAp
I
hkl -TCP
=
R
hkl HAp .
C
hkl HAp
R
hkl -TCP .
C
hkl -TCP
I
hkl HAp
I
hkl CaOH2
=
R
hkl HAp .
C
hkl HAp
R
hkl CaOH2 .
C
hkl CaOH2
C
hkl HAp
+ C
hkl -TCP
+ C
hkl CaOH
2
= 1 2
BM
HAp
V
HAp
= 2
502,3 528,8
= 3,800 21
BM
-TCP
V
-TCP
= 21
310,2 3521
= 1,850 1
BM
CaOH2
V
CaOH2
= 1
74,09 54,88
= 1,350 Misalkan W = fraksi berat, maka fraksi berat untuk masing-masing fasa, yaitu:
W
HAp
= C
HAp
2 BM
HAp
V
HAp
C
HAp
2 BM
HAp
V
HAp
+ C
-TCP
21 BM
-TCP
V
-TCP
+ C
CaOH
2
1 BM
CaOH
2
V
CaOH
2
W
-TCP
= C
-TCP
21 BM
-TCP
V
-TCP
C
HAp
2 BM
HAp
V
HAp
+ C
-TCP
21 BM
-TCP
V
-TCP
+ C
CaOH
2
1 BM
CaOH
2
V
CaOH
2
W
CaOH
2
= C
CaOH
2
1 BM
CaOH
2
V
CaOH
2
C
HAp
2 BM
HAp
V
HAp
+ C
-TCP
21 BM
-TCP
V
-TCP
+ C
CaOH
2
1 BM
CaOH
2
V
CaOH
2
7. 4. 1. BCP I Tabel 46 Perhitungan fraksi berat HAp, -TCP, dan CaOH
2
pada BCP I Fasa
Bidang I
hkl
R
hkl
C
hkl
W HAp
2 1 1 114,0
9,285 0,1670
29,44 -TCP
3 0 0 38,00
0,6518 0,7930
68,06 CaOH
2
0 0 1 28,00
9,532 0,04000
2,500 HAp
2 1 1 114,0
9,258 0,1391
26,15 -TCP
3 0 0 38,00
0,6518 0,6605
60,46 CaOH
2
1 1 1 32,00
1,809 0,2004
13,39 HAp
2 1 1 114,0
9,258 0,4879
67,59 -TCP
0 2 10 185,0
18,59 0,3954
26,67 CaOH
2
0 0 1 28,00
9,532 0,1167
5,744 HAp
2 1 1 114,0
9,258 0,3076
52,45 -TCP
0 2 10 185,0
18,59 0,2493
20,70 CaOH
2
1 1 1 32,00
1,809 0,4431
26,85 HAp
0 0 4 40,00
3,074 0,1755
30,70 -TCP
3 0 0 38,00
0,6518 0,7849
66,84 CaOH
2
0 0 1 28,00
9,532 0,03960
2,460 HAp
0 0 4 40,00
3,074 0,1464
27,33 -TCP
3 0 0 38,00
0,6518 0,6549
59,50 CaOH
2
1 1 1 32,00
1,809 0,1987
13,17 HAp
0 0 4 40,00
3,074 0,5029
68,89 -TCP
0 2 10 185,0
18,59 0,3838
25,60 CaOH
2
0 0 1 28,00
9,532 0,1133
5,510 HAp
0 0 4 40,00
3,074 0,3205
53,95 -TCP
0 2 10 185,0
18,59 0,2446
20,05 CaOH
2
1 1 1 32,00
1,809 0,4539
26,00 Persentase fraksi berat rata-rata sebesar:
W
HAp
= 44,56 W
-TCP
= 43,48 W
CaOH
2
= 11,96
7. 4. 2. BCP II Tabel 47 Perhitungan fraksi berat HAp
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