Question

Silver metal crystallizes with a face-centred cubic lattice. The length of the unit cell is found to be $3.0\times {10}^{-8}cm$. Calculate atomic radius and density of silver.

(Atomic mass of $Ag=108u,N_A=6.02\times {10}^{23}mo{l}^{-1}$)

Answer 1

Given,

Atomic mass of silver $=108u$ "or" $=108g/mol$

Atomic cell length, $a=3.0\times {10}^{-8}cm$

Atomic radius in FCC, $r=\frac{a\sqrt{2}}{4}=\frac{3.0\times {10}^{-8}\sqrt{2}}{4}=1.06\times {10}^{-8}cm$

Each silver crystal unit cell contains $4$ atom.

Density, $d=\frac{4\times atomicmassofsilver}{N_A\times a^3}=\frac{4\times 108}{6.02\times {10}^{23}\times {(3.0\times {10}^{-8})}^3}=26.58g.{cm}^{-3}$

Hence, atomic radius is $1.06\times {10}^{-8}cm$ and density is $26.58g.{cm}^{-3}$.