In order to solve this, let us put
h þ 2k
ð
Þ
3
þ
l
2
!
¼ n; where n is an integer including zero
Therefore the structure factor becomes
F hkl
ð Þ ¼ ð1 þ exp2pinÞ f Zn þ f S :exp2pi:
3l
8
!
The exponents contain fractional values and the expression therefore will remain
complex. To overcome this difficulty, let us simplify this as in case of hcp discussed
above, we obtain
F hkl
ð Þ
j
j
2 ¼ 4cos
2
pn: f
2
Zn þ f
2
S þ 2f Zn :f S cos2pi:
3l
8
&
'
¼ 0 when h þ 2k
ð
Þis a multiple of 3 and l is odd
Let us further analyze with different values (h + 2 k) and l.
The results obtained from considering all possible h, k and l values are summarized as follows:
h þ 2k
ð
Þ
l
FðhklÞ
j
j
2
3n
2p + 1 (as 1, 3, 5, 7 …)
0
3n
8p (as 8, 16, 24…)
4 f Zn þ f S
½
2
3n
4(2p + 1) (as 4, 12, 20, 28, …)
4 f Zn À f S
½
2
3n
2(2p + 1) (as 2, 6, 10, 14, …)
4 f
2
Zn þ f
2
S
Â
Ã
3n Æ1
8 pÆ1 (as 1, 7, 9, 15, 17…)
3 f
2
Zn þ f
2
S À
ffiffi ffi
2
p
f Zn :f S
Â
Ã
3n Æ1
4(2p + 1)Æ1 (as 3, 5, 11, 13, 19, 21 …)
3 f
2
Zn þ f
2
S þ
ffiffi ffi
2
p
f Zn :f S
Â
Ã
3n Æ1
8 p
f Zn þ f S
½
2
3n Æ1
4(2p + 1)
f Zn À f S
½
2
3n Æ1
2(2p + 1)
f
2
Zn þ f
2
S
Â
Ã
where p is any integer including zero.
Example 16 Graphite shows hexagonal structure with four atoms per unit cell.
Atoms are positioned at the following locations:
0; 0; 0
ð
Þ; 1=3; 2=3; 0
ð
Þ ; 0; 0; 1=2
ð
Þ ; 2=3; 1=3; 1=2
ð
Þ
Show that the structure factor is given by
8.2 Determination of Phase Angle, Amplitude …
323
h þ 2k
ð
Þ
3
þ
l
2
!
¼ n; where n is an integer including zero
Therefore the structure factor becomes
F hkl
ð Þ ¼ ð1 þ exp2pinÞ f Zn þ f S :exp2pi:
3l
8
!
The exponents contain fractional values and the expression therefore will remain
complex. To overcome this difficulty, let us simplify this as in case of hcp discussed
above, we obtain
F hkl
ð Þ
j
j
2 ¼ 4cos
2
pn: f
2
Zn þ f
2
S þ 2f Zn :f S cos2pi:
3l
8
&
'
¼ 0 when h þ 2k
ð
Þis a multiple of 3 and l is odd
Let us further analyze with different values (h + 2 k) and l.
The results obtained from considering all possible h, k and l values are summarized as follows:
h þ 2k
ð
Þ
l
FðhklÞ
j
j
2
3n
2p + 1 (as 1, 3, 5, 7 …)
0
3n
8p (as 8, 16, 24…)
4 f Zn þ f S
½
2
3n
4(2p + 1) (as 4, 12, 20, 28, …)
4 f Zn À f S
½
2
3n
2(2p + 1) (as 2, 6, 10, 14, …)
4 f
2
Zn þ f
2
S
Â
Ã
3n Æ1
8 pÆ1 (as 1, 7, 9, 15, 17…)
3 f
2
Zn þ f
2
S À
ffiffi ffi
2
p
f Zn :f S
Â
Ã
3n Æ1
4(2p + 1)Æ1 (as 3, 5, 11, 13, 19, 21 …)
3 f
2
Zn þ f
2
S þ
ffiffi ffi
2
p
f Zn :f S
Â
Ã
3n Æ1
8 p
f Zn þ f S
½
2
3n Æ1
4(2p + 1)
f Zn À f S
½
2
3n Æ1
2(2p + 1)
f
2
Zn þ f
2
S
Â
Ã
where p is any integer including zero.
Example 16 Graphite shows hexagonal structure with four atoms per unit cell.
Atoms are positioned at the following locations:
0; 0; 0
ð
Þ; 1=3; 2=3; 0
ð
Þ ; 0; 0; 1=2
ð
Þ ; 2=3; 1=3; 1=2
ð
Þ
Show that the structure factor is given by
8.2 Determination of Phase Angle, Amplitude …
323