Question

Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm−3 calculate atomic radius of niobium using its atomic mass 93 u.

Answer 1

Answer:  1.428 ×$10^{-8}$

Given,

Density = 8.55 g/cm^3

Atomic mass (z) =  93 amu

$\rho$ = $\frac{z*m}{a^{3}*N_{a}}$

8.55 =  $\frac{2*93}{a^{3})6.023*10^{23}}$

$a^{3}$ = $\frac{2*93}{8.55*6.023*10^{23}}$

a = ∛6.311 × $10^{-23}$

a = 3.30 ×$10^{-8}$

thus ,

in body centred cubic ,

radius = $\frac{a\sqrt{3}}{4}$

          = $\frac{3.30*10^{-8}*\sqrt{3}}{4}$

         = $1.4289 *10^{-8} \\\\= 1.428 *10^{-8}$