Solid State Question 279
Question: $ CsBr $ has bcc structure with edge length 4.3. The shortest interionic distance in between $ C{{s}^{+}} $ and $ B{{r}^{-}} $ is
Options:
A) 3.72
B) 1.86
C) 7.44
D) 4.3
Show Answer
Answer:
Correct Answer: A
Solution:
[a] For bcc structure, atomic radius, $ r=\frac{\sqrt{3}}{4}a $
$ =\frac{\sqrt{3}}{4}\times 4.3=1.86 $ Since, r = half the distance between two nearest neighbouring atoms.
$ \therefore $ Shortest interionic distance $ =2\times 1.86=3.72 $
BETA
