Niobium crystallises in body-centred cubic structure. If density is `8.55g*cm^(-3)`, calculate atomic radius of nionium using its atomic mass 93 u.

Explanation

We know, `rho=(ZxxM)/(Nxxa^(3))`
Given: `rho=8.55g*cm^(-3),M=93g*mol^(-1)`
For a body-centred cubic unit cell, Z = 2
`therefore" "a^(3)=(ZxxM)/(rhoxxN)=(2xx93)/(8.55xx6.022xx10^(23))`
`therefore" "a=3.305xx10^(-8)cm`
In case of a body-centred cubic unit cell, the relation between the radius ( r) of the atom and the edge length (a) of the unit cell is `r=sqrt3/4a`
`therefore" "` Radius of a niobium atom = `sqrt3/4xx3.305xx10^(-8)cm`
`=1.43xx10^(-8)cm`