Niobium crystallizes in body - centered cubic structure . If density is 8.55 g `cm^(-3)` , calculate atomic radius of niobium using its atomic mass 93 U .

Explanation

Radius of unit cell in bcc structure = `(sqrt3)/(4) a`
First we have to calculate edge length of unit cell 'a'
Given Atomic mass of Niobium = 93 g/mole
No. of particles in bcc type unit cell (Z) = 2
mass of unit cell = `(ZM)/(N_(A)) = ( 2 xx 93)/(6.023 xx 10^(23)) = 30.89 xx 10^(23)` gms
Given density (d) = 8.55 gm/`cm^(3)`
Volume of unit cell `(a^(3)) = ("mass")/("density") = (30.89 xx 10^(23))/(8.55)`
= `36.16 xx 10^(-24) cm^(3)`
Edge length of unit cell (a) = `(36.13 xx 10^(-24) )^(1//3)`
`= 3.31 xx 10^(-8)` cm
radius of unit cell (r) = `(sqrt3)/(4) a`
`= (sqrt3 xx 3.31 xx 10^(-8))/(4)`
`= 1.43 xx 10^(-8) cm = 143` pm .