The edge length of unit cell of a metal having molecular weight `75 g mol^(-1)` is `5 Å` which crystallizes in cubic lattice. If the density is `2 g cc^(-1)`, then find the radius of metal atom `(N_(A) = 6 xx 10^(23))`. Give the answer in pm.

Explanation

Let 'z' be the number lattice points per unit cell.
`:.` Density of the metal whose `M=75 "gmol"^(-1)` and `a=5Å` is given by
`d=(ZM)/(N_(A)(a)^(3)),2=(Zxx75)/(6xx10^(23)(5xx10^(-8))^(3))`
On solving z=2
`:.` The metal has unit cell in which `4r=sqrt(3)a`
`r = (1.732xx5xx100)/(4) `pm = 216.5