The edge length of unit cell of metal having molecular weight `75g//mol` is `5Å` which crystallises in simple cubic lattice. If the density is 2g/cc then the radius of metal atom in pm is `x xx 10^(2)` then 'x' is `(N_(A) = 6 xx 10^(23))` 

Answer

2

Explanation

`d=(Zm)/(a^(3)N),2=(1xx75)/(a^(3)xx6xx10^(23))`
In simple cube `2r=a,r=(a)/(2)`, original density = 1
`(d_(1))/(d_(2))=((a_(2))^(3))/(5xx10^(-8)xx5xx10^(-8)xx5xx10^(-8)),(1)/(2)=(a_(2)^(3))/(5xx10^(-8)xx5xx10^(-8)xx5xx10^(-8))`
On simplification we weight r = 2pm