Niobium crystallises in a body-centred cubic | vedclass

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(B) Given: Density $(\rho) = 8.55 \ g \ cm^{-3}$ Atomic mass $(M) = 93 \ g \ mol^{-1}$ For $bcc$ structure, number of atoms per unit cell $(Z) = 2$. A

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$143$

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(B) Given: Density $(ho) = 8.55 \ g \ cm^{-3}$ Atomic mass $(M) = 93 \ g \ mol^{-1}$ For $bcc$ structure, number of atoms per unit cell $(Z) = 2$. Avogadro's number $(N_A) = 6.022 	imes 10^{23} \ mol^{-1}$. Using the formula: $ho = \frac{Z 	imes M}{a^3 	imes N_A}$ $8.55 = \frac{2 	imes 93}{a^3 	imes 6.022 	imes 10^{23}}$ $a^3 = \frac{186}{8.55 	imes 6.022 	imes 10^{23}} = 3.614 	imes 10^{-23} \ cm^3$ $a = (3.614 	imes 10^{-23})^{1/3} \ cm = 3.306 	imes 10^{-8} \ cm = 330.6 \ pm$ For $bcc$ structure, the relation between radius $(r)$ and edge length $(a)$ is: $r = \frac{\sqrt{3}}{4} a$ $r = \frac{1.732}{4} 	imes 330.6 \ pm = 0.433 	imes 330.6 \ pm \approx 143 \ pm$.

Niobium crystallises in a body-centred cubic $(bcc)$ structure. If the density is $8.55 \ g \ cm^{-3}$, then the atomic radius of niobium is (atomic mass of niobium $= 93 \ u$) (in $pm$)

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