From equations (i) and (ii), we have
sin
2
h ¼
k
2
4a 2 h
2
þ k
2 + l
2
À
Á
or sin
2
h / h
2
þ k
2 + l
2
À
Á
ðiiiÞ
Now, for
h
2
þ k
2
þ l
2
À
Á ¼ 3; sin
2
h / 0:45 and h ¼ 42
h
2
þ k
2
þ l
2
À
Á ¼ 4; sin
2
h / 0:60 and h ¼ 50:6
This gives us
d 200 ¼
k
2 sin h
¼
1:54
2 Â sin 50:6
¼ 0:996 ˚
A
A similar exercise for third reflection (220) will show that d 220 is greater than 1
because sin h ¼
nk
2d
À Á
1; which is not a possible reflection. Thus the possible
reflections are (111) and (200) and the corresponding interplanar spacing’s are
1:15 ˚
A and 0:996 ˚
A; respectively.
Example 10 A powder pattern was obtained for an fcc crystal by using X-ray of
wavelength 1:79 ˚
A: The lattice parameter was found to be 3:52 ˚
A: Determine the
lowest and highest reflections possible.
Solution Given: Crystal is fcc, k = 1.79 Å, a = 3.52 Å, lowest reflection = ?,
highest reflection = ?
We know that the ratios of h
2
þ k
2 + l
2
À
Á
for allowed reflections in fcc are:
3 : 4 : 8 : 11 : 12 : 16 : 19 : 20
They correspond to (111), (200), (220), (113), (222), (400), etc. planes,
respectively.
For h
2
þ k
2
þ l
2
À
Á ¼ 3; we have
d 111 ¼
a
ffiffi ffi
3
p and sin h 111 ¼
k
2d 111
¼
ffiffi ffi
3
p  1:79
2 Â 3:25
¼ 0:48
Similarly, for h
2
þ k
2 + l
2
À
Á ¼ 4; we have
d 200 ¼
a
2
and sin h 200 ¼
k
2d 200
¼
k
a
¼
1:79
3:25
¼ 0:51
350
9 Determination of Crystal Structure Parameters