Solid State Question 301
Question: $ CsBr $ crystallises in a body centered cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs=133 and that of Br=80 amu and Avogadro number being $ 6.02\times 10^{23}mo{{l}^{-1}} $ the density of $ CsBr $ is
Options:
A) $ 0.425g/cm^{3} $
B) $ 8.5g/cm^{3} $
C) $ 4.25g/cm^{3} $
D) $ 82.5g/cm^{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] For body centred cubic lattice Z = 2 Atomic mass of unit cell $ =133+80=213 $ a.m.u Volume of cell $ ={{( 436.6\times {{10}^{-10}} )}^{3}}cm^{3} $ Density, $ \rho =\frac{ZM}{a^{3}N _{A}} $
$ =\frac{2\times 213}{{{( 436.6\times {{10}^{-10}} )}^{3}}\times 6.02\times 10^{23}} $
$ =8.50g/cm^{3} $
BETA
