The density of lead is `11.35 g cm^(-3)` and the metal crystallizes with fee unit cell. Estimate the radius of lead atom. (At. Mass of lead `= 207 g mol^(-1) and NA = 6.02xx10^23 mol^(-1))`

Explanation

Let length of edge = a cm
Density = `11.35 g cm^(-3)`
No. of atoms per unit cell in fcc lattice = 4
Atomic mass, `M = 207 g mol^(-1)`
Density , `rho = (Z xx M)/(a^3 xx N_A)`
`11.35 g cm^(-3) = (4 xx (207 g mol^(-1)))/(a^3 xx (6.022 xx 10^(23) mol^(-1)))`
`a^3 = (4 xx (207 g mol^(-1)))/((11.35 g cm^(-3)) xx (6.022 xx 10^(23)mol^(-1)))`
`= 121.14 xx 10^(-24) cm^(3)`
Edge length, `a= (121.14)^(1//3) xx 10^(-8) = 4.948 xx 10^(-8) cm`
or `= 4.948 xx 10^(-10) m`
or `494.8 xx 10^(-12)m = 494.8 p m`
Now, radius in fcc = `a/(2sqrt(2)) = (494.8 p m)/(2 xx 1.414) = 174.96 p m`.