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

Density `=(zxxmol. wt. )/(VxxAv. no . )`
`a=5xx10^(-8)cm, rho =2g//cm^(3),mol. wt.=75`
`:. 2=(zxx75)/((5xx10^(-8))^(3)xx6xx10^(23))`
`implies :. Z=2`
`i.e.,` the metal crystallizes in `b.c.c.` thus
Radius `=(sqrt(3))/(4)xxa=(sqrt(3))/(4)xx5=2.165Å`
`=2.165xx10^(-8)=216.5p m`