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

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Solution

Given :- Density $ρ = 8.55 g c m^{- 3}$

No. of atoms in the unit cell of BCC lattice $= 2$

Avagadro's no. $N_{a} = 6.022 \times 10^{23}$

$ρ = \frac{Z M}{a^{3} N_{A}} \Rightarrow ρ a^{3} N_{A} = Z M \Rightarrow a^{3} = \frac{Z M}{ρ N_{A}}$

$a^{3} = \frac{93 \times 2}{8.55 \times 6.022 \times 10^{23}} = 3.61 \times 10^{- 23} c m^{3}$

$∴ a = 3.3 \times 10^{- 8} c m$ ( by taking cube root)
For BCC structure, $r = \frac{a \sqrt{3}}{4}$

$r = \frac{3.3 \times 10^{- 8} \sqrt{3}}{4} = \frac{3.3 \times 10^{- 8} \times 1.732}{4} = 1.43 \times 10^{- 8} c m$

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Niobium is found to crystallize with bcc structure and found to have the density of 8.55 g cm−3 Determine the atomic radius of niobium if its atomic mass is 93u.

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