We know that the density of the crystal material is given by
q ¼
Mass
Volume
¼
M Â n
N Â a 3
¼
4 Â 58:5
6:023 Â 10 26 Â 5:64 Â 10 À10
ð
Þ
3
¼ 2165 kg=m
3
Example 2 The density of NaCl is 2:18 Â 10
3
kg=m
3 , and the atomic weights of
sodium and chlorine are 23 and 35.5, respectively. Sodium chloride has fcc
structure, calculate its interatomic separation.
Solution:
Given:
Molecular
weight
of
NaCl = 23 + 35.5 = 58.5,
q ¼ 2:18 Â 10
3 kg/m
3 , structure is fcc, so that n = 4, a = ?, d ¼
a
2
À Á ¼ ?
We know that the density of the crystal material is given by
q ¼
Mass
Volume
¼
n  M
N Â a
3
where a is the side of the unit cell.
) a
3
¼
M Â n
N Â q
or
a ¼
M Â n
N Â q
1=3
¼
4 Â 58:5
6:023 Â 10 26 Â 2:18 Â 10 3
1=3
¼ 5:63 Â 10
À10 m
Therefore, the interatomic distance
d ¼
a
2
¼
5:63 Â 10
À10
2
¼ 2:81 Â 10
À10 m
Example 3 The density of potassium chloride is 1:98 Â 10
3 kg=m
3 , and its
molecular weight is 74.55. Calculate its interatomic separation.
Solution: Given: Since KCl has NaCl structure, so that n = 4,
q ¼ 1:98 Â 10
3 kg=m
3 , M = 74.55, a = ?, d ¼
a
2
À Á ¼ ?
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9 Determination of Crystal Structure Parameters