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
V
300 –2
= 0,866.a
2
c = 0,86610,38
2
37,19
–2
= 8,305 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
29,54 sin
2
14,77 cos14,77
=
1 + 0,7569 0,064990,9670
=
1,757 0,06284
= 27,96 R
300
= V
300 –2
.
|F
300
|
2
. p
300
. L-P
300
= 8,305 x 10
–8
4,678 x 10
4
627,96 = 0,6518 Å
–6
7. 3. 1. 4. β-TCP Bidang 0 2 10 2θ = 30,98
V
0210 –2
= 0,866.a
2
c = 0,86610,38
2
37,19
–2
= 8,305 x 10
–8
Å
–6
|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
30,98 sin
2
15,56 cos15,56
=
1 + 0,7350 0,071960,9634
=
1,735 0,06933
= 25,02 R
0210
= V
0210 –2
.
|F
0210
|
2
. p
0210
. L-P
0210
= 8,305 x 10
–8
1,491 x 10
6
625,02 = 18,59 Å
–6
7. 3. 1. 5. CaOH
2
Bidang 0 0 1 2 θ = 17,90
V
001 –2
= 0,866.a
2
c = 0,8663,893
2
5,507
–2
= 1,914 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
17,90 sin
2
8,950 cos8,950
=
1 + 0,9055 0,024200,9878
=
1,906 0,023900
= 79,75 R
001
= V
001 –2
.
|F
001
|
2
. p
001
. L-P
001
= 1,914 x 10
–4
312,2279,75 = 9,532 Å
–6
7. 3. 1. 6. CaOH
2
Bidang 1 1 1 2 θ = 54,26
V
111 –2
= 0,866.a
2
c = 0,8663,893
2
5,507
–2
= 1,914 x 10
–4
Å
–6
|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
54,26 sin
2
27,13 cos27,13
=
1 + 0,3412 0,20790,8900
=
1,341 0,1850
= 7,249 R
111
= V
111 –2
.
|F
111
|
2
. p
111
. L-P
111
= 1,914 x 10
–4
217,367,249 = 1,809 Å
–6
7. 3. 2. BCP II 7. 3. 2. 1. HAp Bidang 2 1 1 2
θ = 31,87
V
211 –2
= 0,866.a
2
c = 0,8669,436
2
6,906
–2
= 3,527 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
31,87 sin
2
15,94 cos15,94
=
1 + 0,7212 0,075420,9615
=
1,712 0,07252
= 23,61 R
211
= V
211 –2
.
|F
211
|
2
. p
211
. L-P
211
= 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
