The density of lead is `11.35 g cm^(-3)` and crystallise with foc unit cell. Estimate the radius of lead atom.
(At. mass of lead `= 207 g mol^(-1) and N_A 6.02 xx 10^(23) mol^(-1)` )

Explanation

`d=(ZxxM)/(a^3xxN_A)implies a^3 = (ZxxM)/(d xx N_A)` ...(i)
For foc unit cell, Z=4
Given, `M=207 g mol^(-1), N_A = 6.02 x 10^(23) mol^(-1)`
`d=11.35 g cm^(-3)`
Substituting these values in equation (i), we get
`a^3 = (4xx207 g mol^(-1))/(11.35g cm^(-3) xx6.02xx10^(23) mol^(-1))`
`a^3 = (4xx 207 xx10)/(11.35 xx 6.02 xx10^(24))cm^3`
`implies a=((8280)/(11.35xx6.02))^(1//3)xx10^(-8) cm`
`:. a = 4.949 xx 10^(-8) cm implies a=494.9` pm
For fcc, `r= a/(2sqrt2)`
`:. r = (494.9)/(2sqrt2) " pm " =(494.9sqrt2)/4 " pm " = (494.9 xx 1.414)/4` pm
`:. r = 174.95` pm