Solid State Question 229
Question: The number of atoms in $ 100g $ of an $ fcc $ crystal with density $ d=10g/cm^{3} $ and cell edge equal to $ 100pm, $ is equal to
[CBSE PMT 1994; KCET 2002]
Options:
A) $ 4\times 10^{25} $
B) $ 3\times 10^{25} $
C) $ 2\times 10^{25} $
D) $ 1\times 10^{25} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ M=\frac{\rho \times a^{3}\times N_0\times {{10}^{-30}}}{z} $
$ =\frac{10\times {{(100)}^{3}}\times (6.02\times 10^{23})\times {{10}^{-30}}}{4}=15.05 $ No. of atoms in 100 g $ =\frac{6.02\times 10^{23}}{15.05}\times 100 $
$ =4\times 10^{25} $ .
BETA
