The density of lead is 11.35 g `cm^(-3)` and the metal crystallises with fcc unit cell. Estimate radius of lead atom.

Explanation

Step-I. Calculation of edge length (a)
We know that `rho=(ZxxM)/(a^(3)xxN_(0)xx10^(-30))ora^(3)=(ZxxM)/(rhoxxN_(0)xx10^(-30))`
According to available data,
No. of atoms in fcc unit cell (Z) = 4
Density of unit cell `(rho)=11.35"g cm"^(-3)`
Atomic mass (M) of lead = 207 g `mol^(-1)`
Avogadro's No. `(N_(0))=6.022xx10^(23)"mol"^(-1)`
`a^(3)=(4xx("207 g mol"^(-1)))/((11.35"g cm")^(-3)xx(6.022xx10^(23)"mol"^(-1))xx(10^(-30)"cm"^(3)))`
`=121xx10^(6)"(pm)"^(3)`
`a=(121xx10^(6)"pm"^(3))^(1//3)=4.95xx10^(2)"pm"="495 pm"`
Step-II. Calculation of radius (r) of unit cell
For a face centred unit cell,
`r=(sqrt2)/(2)a=(1.414xx("495 pm"))/(2)=349.9"pm"`