Solid State Question 23
Question: The number of atoms in 100 g of an fcc crystal with density, $ d=10g/cm^{3} $ and cell edge equal to 100 pm, is equal to
Options:
A) $ 1\times 10^{25} $
B) $ 2\times 10^{25} $
C) $ 3\times 10^{25} $
D) $ 4\times 10^{25} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ M=\frac{\rho \times a^{3}\times N _{A}\times {{10}^{-30}}}{4} $
$ =\frac{10\times {{(100)}^{3}}\times 6.02\times 10^{23}\times {{10}^{-30}}}{4}=15.05 $
$ \therefore $ Number of atoms in 100 g $ =\frac{6.02\times 10^{23}}{15.05}\times 100=4\times 10^{25} $
BETA
