Solid State Question 11
Question: If calcium crystallizes in bcc arrangement and the radius of Ca atom is 96 pm, then the volume of unit cell of Ca is
Options:
A) $ 10.9\times {{10}^{-36}}m^{3} $
B) $ 10.9\times {{10}^{-30}}m^{3} $
C) $ 21.8\times {{10}^{-30}}m^{3} $
D) $ 21.8\times {{10}^{-36}}m^{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] For bcc lattice, $ \sqrt{3}a=4R\Rightarrow a=\frac{4\times 96}{\sqrt{3}}pm=221.7pm $ (where R is the radius of Ca atom) Colume of unit cell $ =a^{3}={{(211.7\times {{10}^{-12}})}^{3}}m^{3} $
$ =10.9\times {{10}^{-30}}m^{3} $
BETA
