Question

Niobium crystallizes in a body-centred Cubic structure having density $8.55gcm^{- 3}$ (atomic mass of Niobium $93g/mol\left.,$ then calculate the radius of the niobium atom? $\left(N_{A} = 6 \times \left(10\right)^{23} , \sqrt[3]{36} = 3 . 3\right)$

• A. $3.3\times 10^{- 8}cm$
• B. $1.16\times 10^{- 8}cm$
• C. $1.43\times 10^{- 8}cm$
• D. $1.65\times 10^{- 8}cm$

Answer 1

Answer: (C)

$d=\frac{Z M}{a^{3} N_{A}}$
Z for $BCC=2$
$d=8.55gcm^{- 3}$
$N_{A}=6\times 10^{23}$
$M=93$
$a^{3}=\frac{2 \times 93}{8 . 55 \times 6 \times 10^{23}}$
$=36.25\times 10^{- 24}cm^{3}$
$a=3.3\times 10^{- 8}cm$
In BCC structure
$r=\frac{\sqrt{3} a}{4}$
$=\frac{1 . 732 \times 3 . 3 \times 10^{- 8}}{4}$
$=1.43\times 10^{- 8}cm$ .