The density of lead is `11.35 g cm^(-3)` and the metal crystallizes with fcc 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

Apply the relation
`d=(zM)/(a^(3)N_(A))`
Given that `z=4" (fcc structure), M = 207 g mol"^(-1), d="11.35 g cm"^(-3), N_(A)=6.022xx10^(23)`
Substituting the values, we have
`"11.35 g cm"^(-3)=(4xx"207 g mol"^(-1))/(a^(3)xx6.022xx10^(23))`
`"or "a^(3)=(4xx"207 g mol"^(-1))/("11.35 g cm"^(-3)xx6.022xx10^(23))=12.1142xx10^(-23)cm^(3)`
`"or "a=4.948xx10^(-8)cm`
In fcc structure
`r=(a)/(2sqrt2) or r=(4.948xx10^(-8))/(2xx1.4142) cm or r=1.7494 xx10^(-8)cm`.