Question

Niobium forms bcc structure.
The density of niobium is 8.55 g/cm³ and length of unit cell edge is 330.6 pm. how many atoms and unit cell are present in 0.5 g of niobium?​

Answer 1

The number of atoms in 0.5 gm of niobium is $3.25$×$10^{21}$ and the number of unit cells in $0.5 gm$ of niobium is$1.62$×$10^{21}$

Explanation:

Number of atoms in $x gm$ of niobium=$\frac{xn}{pa^{3} }$

• where $x=0.5gm ,n=2,p=8.55gm.cm3$
• $a=330.6pm=3.306$×$10^{-8}$
• Number of atoms in $0.5 gm$ of niobium
• =$\frac{(0.5(2))}{(8.55gcm3)(3.306*10^{-8)} }$  
• Number of atoms =$3.25$×$10^{21}$

ii)Number of unit cells in $x gm$=$\frac{x}{pa^{3} }$

• number of unit cells in 0.5gm of niobium = $\frac{(0.5)}{(8.55gcm3)(3.306*10^{-8)} }$
• Number of unit cells =$1.62$×$10^{21}$
• Unit cell is the smallest part of the crystal lattice which shows the three-dimensional pattern of crystal.
• It can be the same unit cell which is repeated over and over in three dimensions.