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^(-3)`, then find the radius of metal atom `(N_(A) = 6 xx 10^(23))`. Give the answer in pm.

Explanation

` rho = (ZM)/(N_(A) a^(3))`
`rArr " " z = (rho N_(A) a^(3))/(M) = (2 xx 6 xx 10^(23) xx (5 xx 10^(-8) cm )^(3))/( 75) = 2 ` (bcc)
`rArr " "` In bcc, 4 r = `sqrt(3)` a
`rArr" " r = (sqrt(3))/(4) a = (sqrt(3))/(4) xx 5 xx 10^(-10) ` m
`= 2.17 xx 10^(-10) m = 217 ` pm