Niobium crystallizes in a bcc structure. If i | vedclass

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(A) For a $bcc$ structure, the number of atoms per unit cell $Z = 2$. The density formula is given by $ho = \frac{Z 	imes M}{N_A 	imes a^3}$. Subs

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$143.1$

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(A) For a $bcc$ structure, the number of atoms per unit cell $Z = 2$. The density formula is given by $ho = \frac{Z 	imes M}{N_A 	imes a^3}$. Substituting the values: $8.55 = \frac{2 	imes 93}{6.022 	imes 10^{23} 	imes a^3}$. $a^3 = \frac{186}{8.55 	imes 6.022 	imes 10^{23}} = 3.614 	imes 10^{-23} \, cm^3$. $a = (36.14 	imes 10^{-24})^{1/3} = 3.306 	imes 10^{-8} \, cm = 330.6 \, pm$. For $bcc$, the relationship between radius $r$ and edge length $a$ is $r = \frac{\sqrt{3}}{4} a$. $r = \frac{1.732}{4} 	imes 330.6 \, pm = 0.433 	imes 330.6 \, pm = 143.15 \, pm$.

Niobium crystallizes in a $bcc$ structure. If its density is $8.55 \, g/cm^3$, calculate the atomic radius of niobium. [Atomic mass of $Nb = 93 \, u$] (in $pm$)

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