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

Density = `8.55 g cm^(-3)`
Let length of the edge = a cm
Number of atoms per unit cell, Z = 2 (bcc)
Atomic mass, `M = 93 g mol^(-1)`
Density , `rho = (Z xx M)/(a^3 xx N_A)`
`8.55 g cm^(-3) = (2 xx (93 g mol^(-1)))/(a^3 xx (6.022 xx 10^(23) mol^(-1)))`
`:. " " a^3 = (2 xx (93 g mol^(-1)))/((8.55 g cm^(-3)) xx (6.022 xx 10^(23) mol^(-1)))`
`= 36.12 xx 10^(-24) cm^(3)`
Edge length, `a = (36.12 xx 10^(-24))^(1//3) = 3.306 xx 10^(-8) cm`
`= 3.306 xx 10^(-10) m`
Now, radius in body centred cubic, `r = (sqrt3)/(4) a`
`= (sqrt(3) xx 3.306 xx 10^(-10)m)/(4) = 1.431 xx 10^(-10) m`
`= 0.143 nm`.