Question

5. A typical paper clip weighs 0.59 g and consists of BCC iron. Calculate (a) the number of
unit cells and (b) the number of iron atoms in the paper clip. (The lattice parameter for
BCC iron is 2.866 x 10^-8 cm and The density is 7.87 g/cm^3)​

Answer 1

Answer:

(a) $3.185*10^{21} cells$

(b) $6.37*10^{21} atoms$

Explanation:

(a)

Volume, V of unit cell

$V=(2.866*10^{-8})^{3}=2.354*10^{-23}$

Number of unit cells, N

$N=\frac {W_{mat}}{V\rho_{mat}}$ Where $W_{mat}$ is weight of material and $\rho_{mat}$ is density of material

$N=\frac{0.59}{7.87*(2.354*10^{-23}}=3.185*10^{21} cells$

(b)

Number of atoms in paper clip

This is a product of number of unit cells and number of atoms per cell

Since iron has 2 atoms per cell

Number of atoms of iron=$3.185*10^{21} cells*2 atoms/cell=6.37*10^{21} atoms$