Question

Niobium is found to crystalline with bcc structure and found to have a density of $8.55g$ ${cm}^{-3}$. Determine the atomic radius of niobium if its atomic mass is $93u$.

• A. $1.431\times {10}^{-10}cm$
• B. $1.431\times {10}^{-8}cm$
• C. $1.431\times {10}^{-8}m$
• D. None of these

Answer 1

Answer: (B)

$1.431\times {10}^{-8}cm$

$Z=2,\rho =8.55$ $g/c{m}^3,$$M_A=93,a=?$

$\rho =\frac{Z\times M_A}{N_A\times a^3}$

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

$a^3=36.11\times {10}^{-24}$ $c{m}^3$

$\Rightarrow a=3.3\times {10}^{-8}$ $cm$

For BCC, $4r=a\sqrt{3}$

$r=\frac{3.3\times {10}^{-8}\times \sqrt{3}}{4}=1.43\times {10}^{-8}$ $cm$